College Algebra [11 ed.] 9781111990909, 1111990905

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Properties of Real Numbers Associative properties: of addition 1a 1 b2 1 c 5 a 1 1b 1 c2 of multiplication 1ab2 c 5 a 1bc2 Commutative properties: of addition a 1 b 5 b 1 a of multiplication ab 5 ba

Properties of Inequalities If a, b, and c are real numbers:

Distributive Property: a 1b 1 c2 5 ab 1 ac

If a , b, then a 1 c , b 1 c and a 2 c , b 2 c.

Properties of Radicals If all radicals are real numbers and there are no divisions by 0, then "ab 5 "a "b n

n

The midpoint of the line segment joining 1x1, y12 and 1x2, y22 is the point M with coordinates x 1 x2 y 1 1 y 2 M5a 1 , b 2 2

Midpoint Formula

If a , b and c . 0, then ac , bc and

a b , . c c

If a , b and c , 0, then ac . bc and

a b . . c c

n

Trichotomy Property: a , b, a 5 b, or a . b

a "a 5 n Åb "b n

n

Transitive Property: If a , b and b , c, then a , c.

#"a 5 #"a 5 "a m

n

mn

n m

Slope

The slope of the nonvertical line passing through points P 1x1, y12 and Q 1x2, y22 is

Properties of Equality If a 5 b and c is a number, then a1c5b1c

and

ac 5 bc

and

a2c5b2c a b 1c 2 02 5 c c

Quadratic Formula x5

2b 6 "b2 2 4ac 2a

1a 2 02

Rules of Exponents If there are no divisions by 0,

1xm2 n 5 xmn

xmxn 5 xm1n 1xy2 n 5 xnyn

x 5 1 1x 2 02 0

x n xn a b 5 n y y x

xm 5 xm2n xn

2n

1 5 n x

y n x 2n a b 5a b y x

Factoring Formulas Factoring the difference of two squares: x2 2 y2 5 1x 1 y2 1x 2 y2 Factoring trinomial squares: x2 1 2xy 1 y2 5 1x 1 y2 2 x2 2 2xy 1 y2 5 1x 2 y2 2

Factoring the sum and difference of two cubes: x3 1 y3 5 1x 1 y2 1x2 2 xy 1 y22 x3 2 y3 5 1x 2 y2 1x2 1 xy 1 y22

m5

change in y y2 2 y1 5 change in x x2 2 x1

1x2 2 x12

Equations of Lines Point-slope form: An equation of the line passing through P 1x1, y12 and with slope m is y 2 y1 5 m 1x 2 x12

Slope-intercept form: An equation of the line with slope m and y-intercept 10, b2 is y 5 mx 1 b Standard form of an equation of a line: Ax 1 By 5 C General form of an equation of a line: Ax 1 By 1 C 5 0

Vertical and Horizontal Lines

Equation of a vertical line through 1a, b2 : x5a

Equation of a horizontal line through 1a, b2 : y5b Slopes of horizontal and vertical lines: • The slope of a horizontal line (a line with an equation of the form y 5 b) is 0. • The slope of a vertical line (a line with an equation of the form x 5 a) is not defined.

The distance d between points 1x1, y12 and 1x2, y22 is given by d 5 " 1x2 2 x12 2 1 1y2 2 y12 2

The Distance Formula

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Definition of Absolute Value

Graphs of Common Functions

x when x $ 0 0x0 5 b 2x when x , 0

y

y

• If k $ 0, then 0 x 0 5 k is equivalent to x 5 k or x 5 2k. • If a and b are algebraic expressions, 0 a 0 5 0 b 0 is equivalent to a 5 b or a 5 2b.

Absolute Value Equations

f(x) = x

x

• If k . 0, then 0 x 0 , k is equivalent to 2k , x , k. • If k . 0, then 0 x 0 . k is equivalent to x . k or x , 2k.

Absolute Value Inequalities

f (x) = x 2

y

y

These two properties hold for # and $ also.

f(x) = |x|

Circles

The standard form of an equation of a circle with center 1h, k2 and radius r: 1x 2 h2 2 1 1y 2 k2 2 5 r2 The standard form of an equation of a circle with center 10, 02 and radius r: x2 1 y2 5 r2

x f (x) = x 3

x

y

Quadratic Functions

y

f(x) = x x

A quadratic function is a second-degree polynomial function in one variable of the form

3

f(x) = x

f 1x2 5 ax2 1 bx 1 c or y 5 ax2 1 bx 1 c,

where a, b, and c are real numbers and a 2 0.

x

Greatest-Integer Function

The graph of a quadratic function of the form f 1x2 5 ax2 1 bx 1 c

x

1a 2 02

y

is a parabola with vertex at a2

b b2 ,c2 b 2a 4a

x y = [[x]]

• If a . 0, the parabola opens upward. • If a , 0, the parabola opens downward. The standard form of an equation of a quadratic function is y 5 f 1x2 5 a 1x 2 h2 2 1 k

1a 2 02

The vertex is at 1h, k2 . • The parabola opens upward when a . 0 and downward when a , 0. • The axis of symmetry of the parabola is the vertical line graph of the equation x 5 h.

Tests for Symmetry Test for x-axis symmetry: To test for x-axis symmetry, replace y with 2y. If the resulting equation is equivalent to the original one, the graph is symmetric about the x-axis. Test for y-axis symmetry: To test for y-axis symmetry, replace x with 2x. If the resulting equation is equivalent to the original one, the graph is symmetric about the y-axis.

Interest Formulas Compound interest formula: If P dollars are deposited in an account earning interest at an annual rate r, compounded n times each year, the amount A in the account after t years is given by r nt A 5 Pa1 1 b n Continuous compound interest formula: If P dollars are deposited in an account earning interest at an annual rate r, compounded continuously, the amount A after t years is given by the formula A 5 Pert

Test for origin symmetry: To test for symmetry about the origin, replace x with 2x and y with 2y. If the resulting equation is equivalent to the original one, the graph is symmetric about the origin.

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College Algebra Eleventh Edition R. David Gustafson Rock Valley College

Jeffrey D. Hughes Hinds Community College

Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

This is an electronic version of the print textbook. Due to electronic rights restrictions, some third party content may be suppressed. Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. The publisher reserves the right to remove content from this title at any time if subsequent rights restrictions require it. For valuable information on pricing, previous editions, changes to current editions, and alternate formats, please visit www.cengage.com/highered to search by ISBN#, author, title, or keyword for materials in your areas of interest.

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College Algebra, Eleventh Edition R. David Gustafson, Jeffrey D. Hughes

© 2013, 2010 Brooks/Cole, Cengage Learning

Editorial Assistant: Sabrina Black

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Printed in the United States of America 1 2 3 4 5 6 7 15 14 13 12 11

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Contents Chapter 0

© Istockphoto.com/Lucas Rucchin

0.1 0.2 0.3 0.4 0.5 0.6

1

Sets of Real Numbers 2 Integer Exponents and Scientific Notation Rational Exponents and Radicals 27 Polynomials 40 Factoring Polynomials 53 Rational Expressions 63

15

Chapter Review 74 Chapter Test 82

Chapter 1

Equations and Inequalities

© Istockphoto.com/Arthur Kwiatkowski

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

85

Linear Equations and Rational Equations 86 Applications of Linear Equations 95 Quadratic Equations 105 Applications of Quadratic Equations 119 Complex Numbers 128 Polynomial and Radical Equations 140 Inequalities 149 Absolute Value 165 Chapter Review 174 Chapter Test 185 Cumulative Review Exercises

Chapter 2

© Stock Connection Blue/Alamy

A Review of Basic Algebra

186

The Rectangular Coordinate System and Graphs of Equations 2.1 2.2 2.3 2.4 2.5

189

The Rectangular Coordinate System 190 The Slope of a Nonvertical Line 206 Writing Equations of Lines 217 Graphs of Equations 234 Proportion and Variation 255 Chapter Review 263 Chapter Test 275

iii Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

iv

Contents

Chapter 3

Functions 3.1 3.2 3.3

Benis Arapovic/Shutterstock.com

3.4 3.5 3.6 3.7

Kiselev Andrey Valerevich/Shutterstock.com

326

410

Exponential and Logarithmic Functions 4.1 4.2 4.3 4.4 4.5 4.6

413

Exponential Functions and Their Graphs 414 Applications of Exponential Functions 429 Logarithmic Functions and Their Graphs 437 Applications of Logarithmic Functions 451 Properties of Logarithms 457 Exponential and Logarithmic Equations 470 Chapter Review 484 Chapter Test 497

Chapter 5

Solving Polynomial Equations 5.1 5.2 5.3

AVAVA/Shutterstock.com

5.4

499

The Remainder and Factor Theorems; Synthetic Division Descartes’ Rule of Signs and Bounds on Roots 510 Roots of Polynomial Equations 520 Approximating Irrational Roots of Polynomial Equations Chapter Review 537 Chapter Test 546 Cumulative Review Exercises

Chapter 6

6.2 6.3 6.4 6.5 6.6

500

530

547

Linear Systems 6.1

© Istockphoto.com/nullplus

Functions and Function Notation 278 Quadratic Functions 294 Polynomial and Other Functions 309 Transformations of the Graphs of Functions Rational Functions 344 Operations on Functions 366 Inverse Functions 381 Chapter Review 393 Chapter Test 409 Cumulative Review Exercises

Chapter 4

277

Systems of Linear Equations 550 Gaussian Elimination and Matrix Methods Matrix Algebra 577 Matrix Inversion 591 Determinants 600 Partial Fractions 613

549 564

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Contents

6.7 6.8

Graphs of Inequalities Linear Programming

v

621 632

Chapter Review 641 Chapter Test 650

Chapter 7

Conic Sections and Quadratic Systems 7.1

Nomad_Soul/Shutterstock.com

7.2 7.3 7.4

The Circle and the Parabola 654 The Ellipse 671 The Hyperbola 687 Solving Nonlinear Systems of Equations Chapter Review 710 Chapter Test 719 Cumulative Review Exercises

© Istockphoto.com/Steve Cole

Chapter 8

653

702

720

Sequences, Series, and Probability 8.1 8.2 8.3 8.4 8.5 8.6 8.7

The Binomial Theorem 724 Sequences, Series, and Summation Notation Arithmetic Sequences and Series 740 Geometric Sequences and Series 746 Mathematical Induction 755 Permutations and Combinations 761 Probability 772

723 731

Chapter Review 778 Chapter Test 786

Appendix I Appendix II

A Proof of the Binomial Theorem 787 Tables 789 Table A Powers and Roots 789 Table B Base-10 Logarithms 790 Table C Base-e Logarithms 791 Answers to Selected Exercises A1 (in Student Edition only) Index I1

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Preface To the Instructor It is with great delight that we present the eleventh edition of College Algebra. This edition maintains the same philosophy of the highly successful previous editions but is enhanced to meet the current expectations of today’s students and instructors. Our goal is to increase students’ problem-solving skills while at the same time preparing them for success in trigonometry, calculus, statistics, or other disciplines of study. The textbook has been revised for greater clarity of design and instruction. Our goal is to write a textbook that • • • • •

presents solid mathematics written in a way that is easy to understand for students with a broad range of abilities and backgrounds; emphasizes the important concept of a function; uses real-life applications to motivate learning and problem solving; improves critical-thinking abilities in all students; develops algebra skills needed for future success in mathematics courses.

We believe that we have accomplished this goal through a successful blending of content and pedagogy. We present a thorough coverage of classic college algebra topics, incorporated into a contemporary framework of tested teaching strategies. Features are written to appeal to both students and instructors. In keeping with the spirit of the NCTM and AMATYC standards, this book emphasizes conceptual understanding, problem solving, and the appropriate use of technology.

New Features •

Strategy Boxes To enable students to build on their mathematical reasoning and approach problems with confidence, Strategy boxes offer problem-solving techniques and steps at appropriate points in the material. Strategy for Solving Quadratic Inequalities

•

Caution

Method 1: Constructing a Table and Testing Numbers • Solve the quadratic equation and use the roots of the equation to establish intervals on a number line. • Construct a table. To do so, write down each interval, select a number to test from each interval, test the selected value to determine if it satisfies the inequality, and then write the result. • Use the results from the table and write the solution of the quadratic inequality. Method 2: Constructing a Sign Graph • Solve the quadratic equation and use the roots of the equation to establish intervals on the number line.

Caution Boxes To alert students to common errors and misunderstandings, and reinforce correct mathematics, Caution boxes appear throughout the text. Be sure to write the Quadratic Formula correctly. Do not write the Quadratic Formula as . x 5 2b 6

"b2 2 4ac 2a

vi

0

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Preface

•

vii

Now Try Exercises To provide students an additional opportunity to assess their understanding of the concept related to each worked example, a reference to an exercise follows all Examples and Self Check problems. These references also show students a correspondence between the examples in the book and the exercises sets. EXAMPLE 3

Finding the Domain of a Function Find the domain of the function defined by the equation y 5

SOLUTION

1 . x2 2 5x 2 6

We can factor the denominator to see what values of x will give 0’s in the denominator. These values are not in the domain. x2 2 5x 2 6 5 0 1x 2 62 1x 1 12 5 0 x 2 6 5 0 or x 1 1 5 0 x56

x 5 21

The domain is 12`, 212 c 121, 62 c 16, ` 2 . Self Check 3

Find the domain of the function defined by the equation y 5

2 . x2 2 16

Now Try Exercise 41.

•

Titled Examples To clearly identify the topic and purpose of each example, descriptive titles have been added to example identifiers. 1x2

1x2 Continued and Updated Features 2 2 We have kept and updated the1xpedagogical features that made the previous editions 122 122 x x of the book so successful.

•

•

Student-Friendly Writing 12Style To alleviate student anxiety about reading a 2 mathematics textbook, the exposition is clear, concise, and reader-friendly. The writing level is informal yet accurate. Students and instructors alike should find the reading both interesting and inviting. Careers and Mathematics Chapter Openers To encourage students to explore careers that use mathematics and make a connection between math and real life, each chapter opens with Careers and Mathematics. New, exciting careers are featured in this edition. These snapshots include information on how the professionals use math in their work and who employs them. Most information is taken from the Occupational Outlook Handbook. A web address is provided for students to learn more about the career.

Kiselev Andrey Valerevich/Shutterstock.com

CAREERS AND MATHEMATICS: Epidemiologists Epidemiologists investigate and describe the determinants and distribution of disease, disability, and other health outcomes. They also develop means for prevention and control. Applied epidemiologists typically work for state health agencies and are responsible for responding to disease outbreaks and determining the cause and method of containment. Research epidemiologists work in laboratories studying ways to prevent future outbreaks. This career can be quite rewarding, both mentally and financially. Epidemiologists spend a lot of time saving lives and finding solutions for better health.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Preface

•

Section Openers To pique interest and motivate students to read the material, each section begins with a contemporary photo and a real-life application that will appeal to students of varied interests.

•

Numbered Objectives To keep students focused, numbered learning objectives are given at the beginning of each section and appear as subheadings in the section.

4.3 Logarithmic Functions and Their Graphs In this section, we will learn to 1. Evaluate logarithms. 2. Evaluate common logarithms. 3. Evaluate natural logarithms. 4. Graph logarithmic functions. 5. Use transformations to graph logarithmic functions. Guests aboard the Royal Caribbean’s cruise ship Freedom of the Seas can now “hang ten” while out to sea. The flowrider surf simulator allows riders to body board surf against a wave-like water flow of 34,000 gallons per minute.

•

•

•

•

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Example Structure To help students gain a deeper understanding of how to solve each problem, solutions begin with a stated approach. The examples are engaging, and step-by-step solutions with annotations are provided. Application Examples To answer the student question, When will I ever use this math? applications from a wide range of disciplines demonstrate how mathematics is used to solve real problems. These applications motivate the student and help students become better problem solvers. An eye-catching photo accompanies many of these modern examples. Self Checks To actively reinforce student understanding of concepts and example solutions, each example is followed immediately by a Self Check exercise. The answers for students are offered at the end of each section. To assist instructors, the answers to Self Checks appear next to the problem in the AIE, printed in blue. Comments To provide additional insights into specific content, Comment boxes appear throughout the textbook. These noteworthy statements provide clarification on a specific step or concept in an example. Some offer a tip for studying the material. Accents on Technology and Calculators To encourage students to become intelligent users of technology and grasp concepts graphically, Accents on Technology appear throughout the textbook. These illustrate and guide the use of a TI-84 graphing calculator for specific problems, and many are new. Although graphing calculators are incorporated into the book, their use is not required. All graphing topics are fully discussed in traditional ways.

ACCENT ON TECHNOLOGY

Exponential Regression The table below shows the cooling temperatures of a hot cup of coffee after it is made. Time in minutes

Temperature °F

0

179.5

5

168.7

8

158.1

11

149.2

15

141.7

18

134.6

22

125.4

25

123.5

30

116.3

34

113.2

graphx/Shutterstock.com

viii

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

ix

Preface

•

Getting Ready Exercises To test student understanding of concepts and proper use of mathematical vocabulary, each problem set begins with Getting Ready exercises. Students should be able to answer these fill-in-the blank questions before moving on to the Practice exercises.

•

Comprehensive Exercise Sets To improve mathematical skills and cement understanding, the exercise sets progress from routine to more challenging. The mathematics in each exercise set is sound, but not so rigorous that it will confuse students. All exercise sets include Getting Ready, Practice, Application, Discovery and Writing, and Review problems. The book contains more than 4,000 exercises, many of which are new. Interesting and Contemporary Application Exercises To continue the emphasis on solving problems through realistic applications, new Application exercises have been added and others updated. All Application problems have titles. Chapter Reviews and Chapter Tests To give students the best opportunity for study and exam preparation, each chapter closes with a Chapter Review and Chapter Test. Chapter Reviews are comprehensive and consist of three parts: definitions and concepts, examples, and review exercises. Chapter Tests cover all the important topics and yet are brief enough to emulate a real-time test, so students can practice not only the math but also their test-taking aptitude. As an additional quick reference, endpapers offer the important formulas and graphs developed in the book.

•

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CHAPTER REVIEW SECTION 4.1 Definitions and Concepts

Use properties of exponents to simplify.

Graph: f 1x2 5 6x. We will find several points 1x, y2 that satisfy the equation, plot the points, and join them with a smooth curve, as shown in the figure.

1b . 0, b 2 12

f 1x2 5 6

y 6

(1, 6)

4

f(x) = 6 x

(–1, 1–6) 2 –2

x 21

(0, 1) 2

x

f 1x2 1 6

0

1

1

6

Graph the function defined by each equation. 3. f 1x2 5 3x

1 x 4. f 1x2 5 a b 3 y

y

x

1x, f 1x2 2

1 a21, b 6 10, 12 11, 62

x x

5.

1

1

p2 bx

•

2. Q2"5R"2

1. 5"2 ? 5"2

Examples

An exponential function with base b is defined by the equation y 5 f 1x2 5 bx

EXERCISES

Exponential Functions and Their Graphs

q2

7x p 0

q 1

Cumulative Reviews To reinforce student learning and improve student retention of concepts, Cumulative Review exercises appear after every two chapters. These b comprehensive reviews revisit all the essential topics covered in prior chapters. 2

Content Changes Each section in the text has been edited to fine-tune the presentation of topics for better flow of concepts and for clarity. There are many new exercises and applications. Some changes made to specific chapters include: Chapter 0: A Review of Basic Algebra • Section 0.1—Four new exercises have been added. Students are asked to determine whether the decimal form of a given fraction terminates or repeats. • Section 0.2—The Accents on Technology in this section now show TI-84 Plus A t graphing calculator screen shots. Two contemporary applications were added. / • Section 0.3—Four new exercises were added to provide students additional A practice rationalizing the denominator. 0 h • Section 0.4—Four new exercises were added. Two exercises ask students to multiply binomials containing radicals. Two exercises are long division problems and help improve proficiency of this skill. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

Preface

• •

Section 0.5—Two new exercises, factoring the difference of two squares, were added. Examples 11 and 12 were reversed for a better flow of the content of the section. Section 0.6—A new Strategy box has been added to clearly exhibit two methods for simplifying complex fractions.

Chapter 1: Equations and Inequalities • Section 1.1—The title of the section has been changed to “Linear Equations and Rational Equations.” The title better reflects the content in the section. Students are now asked to identify restrictions on the values of variables for only linear and rational equations; radical equations have been deleted. • Section 1.2—Seven Self Checks have been added to this section. These are contemporary applications. Two new Application exercises have been added. • Section 1.3—The discriminant material has been rewritten for greater clarity. Eight additional Square Root Property exercises and two new Quadratic Formula exercises have been added. These help strengthen students’ ability to solve equations and simplify radicals. • Section 1.4—Six Self Checks have been added to this section. These are contemporary applications. Two new Application exercises were also added. • Section 1.5—Example 11, “Factoring the Sum of Two Squares,” is new. An Accent on Technology showing operations of complex numbers has been added. The exercise set has been extensively revised and 42 new exercises have been added. There is now a greater emphasis on solving quadratic equations with complex solutions, using the methods previously taught. • Section 1.6—Two Strategy boxes were added in this section, one for solving polynomial equations and one for solving radical equations. • Section 1.7—A Strategy box for solving quadratic inequalities has been added to the section. Two methods for solving the inequalities are displayed in the Strategy box. • Section 1.8—The forms of the absolute value inequalities have been rewritten to clearly include less than or equal to and greater than or equal to. Two new absolute value equation exercises have been added and two new Discovery and Writing exercises have been added. • End of Chapter—Four new Square Root Property exercises were added to the Chapter Review. Two new exercises, a compound inequality and a quadratic inequality, were added to the Chapter Test. Chapter 2: The Rectangular Coordinate System and Graphs of Equations • Section 2.1—The Accent on Technology “Finding Intercepts Using Zoom and Trace” has been streamlined for efficiency. The distance and midpoint exercises have been revised. • Section 2.2—Four new exercises, corresponding to Example 5, have been added. • Section 2.3—A new Accent on Technology, “Linear Regression,” has been added. Two new exercises have also been added. • Section 2.4—Four contemporary applications involving circles have been added. Two Discovery and Writing exercises have been added. • Section 2.5—Two new inverse variation exercises have been added. Chapter 3: Functions • Section 3.1—In this section, a new Accent on Technology, “Evaluating a Function,” was added. Four new domain exercises, involving square roots or cube roots, were added. Two new function exercises were added. The functions have two x terms. Also, two cube-root graphing exercises were added. Two contemporary applications were also added. • Section 3.2—A new Accent on Technology, “Finding the Maximum Point or Minimum Point (Vertex) of a Parabola,” was added. • Section 3.3—The definitions of even and odd functions are formally defined. The definitions of increasing and decreasing are formally defined. Six exercises Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Preface

• • •

xi

were added to identify the graph of a function as being even, odd, or neither. A new Accent on Technology, “Piecewise-Defined Functions,” now appears in the section. Section 3.4—The title of the section has been changed to “Transformations of the Graphs of Functions.” An Accent on Technology box has been added to illustrate translations. Section 3.5—There is one new Accent on Technology in this section. Self Checks 5–10 are new. The applications have been revised in this section. Section 3.6—There are now Accents on Technology demonstrating combinations of functions and the domain of composite functions. Sixteen new exercises were added. These include eight exercises asking students to identify function values given a graph and two contemporary applications.

Chapter 4: Exponential and Logarithmic Functions • Section 4.1—Example 1, “Approximating Exponential Expressions,” is new. The Compound Interest Formula now uses n for the number of years instead of k. A new Accent on Technology, “Graphing Exponential Functions,” has been added. • Section 4.2—A new Accent on Technology, “Exponential Regression,” has been added. • Section 4.3—Two new examples have been added to cover converting from exponential form to logarithmic form and converting from logarithmic form to exponential form. TI-84 Plus graphing calculator screens are now shown extensively throughout this section. Six new graphing exercises have been added. • Section 4.4—The section opener has been revised and includes Haiti and Japan. • Section 4.5—Three new Accent on Technology boxes were added. These illustrate the Product, Quotient, and Power Properties of Logarithms and using the Change-of-Base Formula to graph a logarithmic function. • Section 4.6—The Accents on Technology are new in this section. They are “Verifying Solutions of an Exponential Equation,” “Finding Approximate Solutions to an Exponential Equation,” and “Finding Approximate Solutions to a Logarithmic Equation.” Example 9, “Solving Logarithmic Equations,” is new. A summary of strategies used to solve exponential and logarithmic equations is now given. Eight new exercises have been added to help students master solving exponential equations using like bases. • End of Chapter—One new exercise has been added to the Chapter Review. Chapter 5: Solving Polynomial Equations • Section 5.1—Four new exercises were added. These require synthetic division using complex numbers. • Section 5.2—A Strategy box for finding a polynomial equation when given a partial list of roots has been added. • Section 5.3—A new Accent on Technology, “Using the Table Feature on a Graphing Calculator to Find Roots of a Polynomial Equation,” has been added. The exercises have been reordered based on the degree of the polynomial equation. Two new polynomial equation exercises have been added. These are fifth-degree equations and the leading coefficient is not one. Two contemporary applications have also been added. • Section 5.4—Two new exercises have been added. • End of Chapter—Three new exercises have been added to the Chapter Review. These exercises require the Rational Roots Theorem. In the Chapter Test, two exercises were changed from logarithms to natural logarithms. Chapter 6: Linear Systems • Section 6.1—A new Accent on Technology, “Solving a System of Linear Equations,” has been added. Self Check 9 was added. There are four new exercises that have been added. These require graphing and have either no solution or an infinite number of solutions. Two contemporary applications have been added. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

xii

Preface

• • • • • •

Section 6.2—A new Accent on Technology, “Reduce a Matrix to Row Reduced Echelon Form,” has been added. Section 6.3—Self Check 6 has been added. Section 6.4—Self Check 6 has been added. Section 6.5—Self Check 7 has been added. Section 6.6—Property boxes have been added to display the various types of partial-fraction decomposition problems that surface and the techniques used to solve the problem Section 6.8—Self Checks 3, 4, and 5 have been added.

Chapter 7: Conic Sections and Quadratic Systems • Section 7.1—Six new exercises have been added. Four ask the student to identify the conic as a circle or a parabola. There are two new exercises asking students to use a graphing calculator to graph a parabola. • Section 7.2—There are nine new exercises. Six ask students to identify the conic as a circle, parabola, or ellipse. Two ask students to use a graphing calculator and graph an ellipse. One contemporary application has been added. • Section 7.3—Ten new exercises have been added. Eight exercises ask students to identify the conic as a circle, parabola, ellipse, or hyperbola. Two ask students to graph a hyperbola using a graphing calculator. Chapter 8: Sequences, Series, and Probability • Section 8.1—A new Accent on Technology, “Factorials,” has been added. • Section 8.2—A new Accent on Technology, “Sequences, Series, and Summation,” has been added. • Section 8.3—The section title has been revised. It is now titled “Arithmetic Sequences and Series.” • Section 8.4—Self Checks 6, 7, and 8 have been added. These are contemporary applications. • Section 8.6—Two Accents on Technology have been added. One involves permutations and the other involves combinations. Three contemporary applications and two new exercises have been added. Chapter 9: The Mathematics of Finance *This chapter is now available online at cengagebrain.com. Students will also have access to Answers to Selected Exercises for this chapter, and instructors will have access to the Annotated Instructor’s Edition of this chapter.

Organization and Coverage This text can be used in a variety of ways. To maintain optimum flexibility, many chapters are sufficiently independent to allow you to pick and choose topics that are relevant to your students. After teaching Chapters 0–3 in order, you can teach Chapters 4–9 in any order.

Ancillaries for the Instructor Enhanced WebAssign® (ISBN-10: 0-538-73810-3; ISBN-13: 978-0-538-73810-1) Exclusively from Cengage Learning, Enhanced WebAssign offers an extensive online program for College Algebra to encourage the practice that’s so critical for concept mastery. The meticulously crafted pedagogy and exercises in this text become even more effective in Enhanced WebAssign, supplemented by multimedia tutorial support and immediate feedback as students complete their assignments. Algorithmic problems allow you to assign unique versions to each student. The Practice Another Version feature (activated at your discretion) allows students to attempt the questions with new sets of values until they feel confident enough to work the original problem. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Preface

xiii

Students benefit from a new YouBook with highlighting and search features; Personal Study Plans (based on diagnostic quizzing) that identify chapter topics they still need to master; and links to video solutions, interactive tutorials, and even live online help. PowerLecture with ExamView® (ISBN-10: 1-133-10345-6; ISBN-13: 978-1-133-10345-5) This CD-ROM provides the instructor with dynamic media tools for teaching. Create, deliver, and customize tests (both print and online) in minutes with ExamView Computerized Testing Featuring Algorithmic Equations. Easily build solution sets for homework or exams using Solution Builder’s online solutions manual. Microsoft® PowerPoint® lecture slides and figures from the book are also included on this CD-ROM. Solution Builder (www.cengage.com/solutionbuilder) This online instructor database offers complete worked solutions to all exercises in the text, allowing you to create customized, secure solutions printouts (in PDF format) matched exactly to the problems you assign in class. Complete Solutions Manual (ISBN-10: 1-133-10342-1; ISBN-13: 978-1-133-10342-4) This manual contains solutions to all exercises from the text, including Chapter Review Exercises, Chapter Tests, and Cumulative Review Exercises. Test Bank (ISBN-10: 1-133-10343-X; ISBN-13: 978-1-133-10343-1) The test bank includes six tests per chapter as well as three final exams. The tests are made up of a combination of multiple-choice, free-response, true/false, and fillin-the-blank questions.

Ancillaries for the Student Enhanced WebAssign® (ISBN-10: 0-538-73810-3; ISBN-13: 978-0-538-73810-1) Exclusively from Cengage Learning, Enhanced WebAssign offers an extensive online program for College Algebra to encourage the practice that’s so critical for concept mastery. You’ll receive multimedia tutorial support as you complete your assignments. You’ll also benefit from a new Premium eBook with highlighting and search features; Personal Study Plans (based on diagnostic quizzing) that identify chapter topics you still need to master; and links to video solutions, interactive tutorials, and even live online help. Student Solutions Manual (ISBN-10: 1-133-10347-2; ISBN-13: 978-1-133-10347-9) Go beyond the answers—see what it takes to get there and improve your grade! This manual provides worked-out, step-by-step solutions to the odd-numbered problems in the text. This gives you the information you need to truly understand how these problems are solved. Text-Specific DVDs (ISBN 10: 1-133-10344-8; ISBN-13: 978-1-133-10344-8) These text-specific instructional videos provide students with visual reinforcement of concepts and explanations given in easy-to-understand terms with detailed examples and sample problems. A flexible format offers versatility for quickly accessing topics or catering lectures to self-paced, online, or hybrid courses. Closed captioning is provided for the hearing impaired.

To get access, visit CengageBrain.com

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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

xiv

Preface

To the Student Congratulations! You now own a state-of-art textbook that has been written especially for you. We have written a book that you can easily read and understand and that you will find interesting as well. The book includes carefully written sections that will appeal to you because of our use of popular culture and contemporary examples. The book also contains real-life applications that help you see how mathematics is currently used in our world today. Our goal is for you to use this textbook as we designed it—as a study tool—so that you will be successful in learning mathematics and reaching your career goals. So what are you waiting for? Grab your favorite pencil and some paper and dive in! You can begin reading, working exercises, and mastering skills today. We wish you the very best on this exciting mathematics journey you are about to take.

Acknowledgments We are grateful to the following people who reviewed previous editions of the text or the current manuscript in its various stages. All of them had valuable suggestions that have been incorporated into this book. Catherine Aguilar-Morgan, New Mexico State University–Alamogordo Ebrahim Ahmadizadeh, Northampton Community College Ricardo Alfaro, University of Michigan– Flint Richard Andrews, University of Wisconsin James Arnold, University of Wisconsin Ronald Atkinson, Tennessee State University Wilson Banks, Illinois State University Chad Bemis, Riverside Community College Anjan Biswas, Tennessee State College Jerry Bloomberg, Essex Community College Elaine Bouldin, Middle Tennessee State University Dale Boye, Schoolcraft College Eddy Joe Brackin, University of North Alabama Susan Williams Brown, Gadsden State Community College Jana Bryant, Manatee Community College Lee R. Clancy, Golden West College Krista Blevins Cohlmia, Odessa College Dayna Coker, Southwestern Oklahoma State University–Sayre campus Jan Collins, Embry Riddle College Cecilia Cooper, William & Harper College John S. Cross, University of Northern Iowa Charles D. Cunningham, Jr., James Madison University

M. Hilary Davies, University of Alaska– Anchorage Elias Deeba, University of Houston– Downtown Grace DeVelbiss, Sinclair Community College Lena Dexter, Faulkner State Junior College Emily Dickinson, University of Arkansas Mickey P. Dunlap, University of Tennessee, Martin Gerard G. East, Southwestern Oklahoma State University Eric Ellis, Essex Community College Eunice F. Everett, Seminole Community College Dale Ewen, Parkland College Harold Farmer, Wallace Community College–Hanceville Ronald J. Fischer, Evergreen Valley College Mary Jane Gates, University of Arkansas at Little Rock Lee R. Gibson, University of Louisville Marvin Goodman, Monmouth College Edna Greenwood, Tarrant County College Jerry Gustafson, Beloit College Jerome Hahn, Bradley University Douglas Hall, Michigan State University Robert Hall, University of Wisconsin David Hansen, Monterey Peninsula College Shari Harris, John Wood Community College Kevin Hastings, University of Delaware

William Hinrichs, Rock Valley College Arthur M. Hobbs, Texas A&M University Jack E. Hofer, California Polytechnic State University Ingrid Holzner, University of Wisconsin Wayne Humphrey, Cisco College Warren Jaech, Tacoma Community College Joy St. John Johnson, Alabama A&M University Nancy Johnson, Broward Community College Patricia H. Jones, Methodist College William B. Jones, University of Colorado Barbara Juister, Elgin Community College David Kinsey, University of Southern Indiana Helen Kriegsman, Pittsburg State University Marjorie O. Labhart, University of Southern Indiana Betty J. Larson, South Dakota State University Paul Lauritsen, Brown College Jaclyn LeFebvre, Illinois Central College Susan Loveland, University of Alaska– Anchorage James Mark, Eastern Arizona College Marcel Maupin, Oklahoma State University, Oklahoma City Robert O. McCoy, University of Alaska– Anchorage Judy McKinney, California Polytechnic Institute at Pomona

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Preface

Sandra McLaurin, University of North Carolina Marcus McWaters, University of Southern Florida Donna Menard, University of Massachusetts, Dartmouth James W. Mettler, Pennsylvania State University Eldon L. Miller, University of Mississippi Stuart E. Mills, Louisiana State University, Shreveport Mila Mogilevskaya, Wichita State University Gilbert W. Nelson, North Dakota State Marie Neuberth, Catonsville City College Charles Odion, Houston Community College C. Altay Özgener, State College of Florida Anthony Peressini, University of Illinois David L. Phillips, University of Southern Colorado William H. Price, Middle Tennessee State University Ronald Putthoff, University of Southern Mississippi

Brooke P. Quinlan, Hillsborough Community College Leela Rakesh, Carnegie Mellon University Janet P. Ray, Seattle Central Community College Robert K. Rhea, J. Sargeant Reynolds Community College Barbara Riggs, Tennessee Technological University Minnie Riley, Hinds Community College Renee Roames, Purdue University Paul Schaefer, SUNY, Geneseo Vincent P. Schielack, Jr., Texas A&M University Robert Sharpton, Miami Dade Community College L. Thomas Shiflett, Southwest Missouri State University Richard Slinkman, Bemidji State University Merreline Smith, California Polytechnic Institute at Pomona John Snyder, Sinclair Community College Sandra L. Spain, Thomas Nelson Community College

xv

Warren Strickland, Del Mar College Paul K. Swets, Angelo State College Ray Tebbetts, San Antonio College Faye Thames, Lamar State University Douglas Tharp, University of Houston– Downtown Carolyn A. Wailes, University of Alabama, Birmingham Carol M. Walker, Hinds Community College William Waller, University of Houston– Downtown Richard H. Weil, Brown College Carroll G. Wells, Western Kentucky University William H. White, University of South Carolina at Spartanburg Clifton Whyburn, University of Houston Charles R. Williams, Midwestern State University Harry Wolff, University of Wisconsin Roger Zarnowski, Angelo State University Albert Zechmann, University of Nebraska

We wish to thank the staff at Cengage Learning, especially Gary Whalen and Jennifer Risden for their support in the production process. Cynthia Ashton, Sabrina Black, and Lynh Pham have worked tirelessly to pull together all the pieces behind the scenes. Leslie Lahr, our development editor, provided constant encouragement and valuable input. We appreciate you so much for being by our side during this entire journey. We also thank Lori Heckelman for her fine artwork, and special thanks goes to both Rhoda Bontrager and Craig Beffa at Graphic World for providing excellent publishing services. We appreciate the important work of accuracy reviewing done by John Samons. R. David Gustafson Jeffrey D. Hughes

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

About the Authors R. David Gustafson is professor emeritus of mathematics at Rock Valley College in Illinois and coauthor of several best-selling math texts, including Gustafson/Frisk’s Beginning Algebra, Intermediate Algebra, Beginning and Intermediate Algebra: A Combined Approach, College Algebra, and the Tussy/Gustafson developmental mathematics series. His numerous professional honors include Rock Valley Teacher of the Year and Rockford’s Outstanding Educator of the Year. He earned a Master of Arts from Rockford College in Illinois, as well as a Master of Science from Northern Illinois University.

Jeff Hughes is a mathematics instructor at Hinds Community College in Mississippi and has degrees from Mississippi College and the University of Mississippi. He has worked at Hinds CC for 23 years and his favorite courses to teach are College Algebra, the Calculus Sequence, and Differential Equations. He sees teaching as a privilege and enjoys the challenge of finding new ways to present topics while engaging students. He has been the recipient of Hinds CC’s Outstanding Academic Instructor Award twice and is active in AMATYC. He currently serves as president of LaMsMATYC, the Mississippi-Louisiana affiliate of AMATYC. He has worked part-time as Minister of Students at his church for over 15 years and enjoys traveling. He taught English in China for seven summers with ELIC, the English Language Institute/China.

About the Cover The wonderful cover art is by Van Evan Fuller. Van is not only a brilliant artist but also a gifted writer. He lives in Baton Rouge, Louisiana, and is a graduate of LSU. The cover art was initially created as a traditional, though digital, abstract. The artist then processed it through deforming software into which he programmed a series of variables. After creating about a dozen such compositions, all based on the same abstract painting, the final design was created using four of these. In the art, numerous variations of mathematical curves such as circles, parabolas, ellipses, and hyperbolas can be seen intertwined in an amazing, colorful array. In the words of the artist, “Every beautiful thing is somehow rooted in mathematics, and the expression of mathematics is always beautiful.” The authors were struck by this beauty and the way Van was able to capture it. You can see more of Van’s creative designs at his online gallery at www.vanevanfuller.com.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

0

A Review of Basic Algebra

CAREERS AND MATHEMATICS:

Pharmacist

© Istockphoto.com/Lucas Rucchin

Pharmacists distribute prescription drugs to individuals. They also advise patients, physicians, and other healthcare workers on the selection, dosages, interactions, and side effects of medications. They also monitor patients to ensure that they are using their medications safely and effectively. Some pharmacists specialize in oncology, nuclear pharmacy, geriatric pharmacy, or psychiatric pharmacy.

Education and Mathematics Required •

•

0.1

Sets of Real Numbers

0.2

Integer Exponents and Scientific Notation

0.3

Rational Exponents and Radicals

0.4

Polynomials

0.5

Factoring Polynomials

0.6

Rational Expressions Chapter Review Chapter Test

Pharmacists are required to possess a Pharm.D. degree from an accredited college or school of pharmacy. This degree generally takes four years to complete. To be admitted to a Pharm.D. program, at least two years of college must be completed, which includes courses in the natural sciences, mathematics, humanities, and the social sciences. A series of examinations must also be passed to obtain a license to practice pharmacy. College Algebra, Trigonometry, Statistics, and Calculus I are courses required for admission to a Pharm.D. program.

How Pharmacists Use Math and Who Employs Them •

•

Pharmacists use math throughout their work to calculate dosages of various drugs. These dosages are based on weight and whether the medication is given in pill form, by infusion, or intravenously. Most pharmacists work in a community setting, such as a retail drugstore, or in a healthcare facility, such as a hospital.

Career Outlook and Salary • •

Employment of pharmacists is expected to grow by 17 percent between 2008 and 2018, which is faster than the average for all occupations. Median annual wages of wage and salary pharmacists is approximately $106,410.

For more information see: www.bls.gov/oco

In this chapter, we review many concepts and skills learned in previous algebra courses. Be sure to master this material now, because it is the basis for the rest of this course.

1 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

2

Chapter 0

A Review of Basic Algebra

0.1 Sets of Real Numbers In this section, we will learn to 1. Identify sets of real numbers. 2. Identify properties of real numbers. 3. Graph subsets of real numbers on the number line. 4. Graph intervals on the number line. 5. Define absolute value. 6. Find distances on the number line.

5 3 7 1 9 5 6 9 8 6 8 6 4 8 3 7 2 6 2 8 4 1 9 8 7

Sudoku, a game that involves number placement, is very popular. The objective is to fill a 9 by 9 grid so that each column, each row and each of the 3 by 3 blocks contains the numbers from 1 to 9. A partially completed Sudoku grid is shown in the margin. To solve Sudoku puzzles, logic and the set of numbers, 5 1, 2, 3, 4, 5, 6, 7, 8, 9 6 are used. Sets of numbers are important in mathematics, and we begin our study of algebra with this topic. A set is a collection of objects, such as a set of dishes or a set of golf clubs. The set of vowels in the English language can be denoted as 5 a, e, i, o, u 6 , where the braces 5 6 are read as “the set of.” If every member of one set B is also a member of another set A, we say that B is a subset of A. We can denote this by writing B ( A, where the symbol ( is read as “is a subset of.” (See Figure 0.1 below.) If set B equals set A, we can write B 8 A. If A and B are two sets, we can form a new set consisting of all members that are in set A or set B or both. This set is called the union of A and B. We can denote this set by writing, A c B where the symbol c is read as “union.” (See Figure 0.1 below.) We can also form the set consisting of all members that are in both set A and set B. This set is called the intersection of A and B. We can denote this set by writing A d B, where the symbol d is read as “intersection.” (See Figure 0.1 below.)

3 1 6 5 9

SUDOKU

A B

A

BA

B

A

AB

B

AB

FIGURE 0-1

EXAMPLE 1

Understanding Subsets and Finding the Union and Intersection of Two Sets

Let A 5 5 a, e, i 6 , B 5 5 c, d, e 6 , and V 5 5 a, e, i, o, u 6 . a. Is A ( V ?

SOLUTION

b. Find A c B.

c. Find A d B.

a. Since each member of set A is also a member of set V, A ( V .

b. The union of set A and set B contains the members of set A, set B, or both. Thus, A c B 5 5 a, c, d, e, i 6 .

c. The intersection of set A and set B contains the members that are in both set A and set B. Thus, A d B 5 5 e 6 .

Self Check 1

a. Is B ( V ? c. Find A d V

b. Find B c V

Now Try Exercise 33. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.1

Sets of Real Numbers

3

1. Identify Sets of Real Numbers There are several sets of numbers that we use in everyday life.

Basic Sets of Numbers

Natural numbers The numbers that we use for counting: 5 1, 2, 3, 4, 5, 6, % 6

Whole numbers The set of natural numbers including 0: 5 0, 1, 2, 3, 4, 5, 6, % 6 Integers The set of whole numbers and their negatives: 5 % , 25, 24, 23, 22, 21, 0, 1, 2, 3, 4, 5, % 6

In the definitions above, each group of three dots (called an ellipsis) indicates that the numbers continue forever in the indicated direction. Two important subsets of the natural numbers are the prime and composite numbers. A prime number is a natural number greater than 1 that is divisible only by itself and 1. A composite number is a natural number greater than 1 that is not prime. • •

The set of prime numbers: 5 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, % 6 The set of composite numbers: 5 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, % 6

Two important subsets of the set of integers are the even and odd integers. The even integers are the integers that are exactly divisible by 2. The odd integers are the integers that are not exactly divisible by 2. • •

The set of even integers: 5 % , 210, 28, 26, 24, 22, 0, 2, 4, 6, 8, 10, % 6 The set of odd integers: 5 % , 29, 27, 25, 23, 21, 1, 3, 5, 7, 9, % 6

So far, we have listed numbers inside braces to specify sets. This method is called the roster method. When we give a rule to determine which numbers are in a set, we are using set-builder notation. To use set-builder notation to denote the set of prime numbers, we write 5 x  0  x is a prime number 6 ▲ ▲

▲

Caution Remember that the denominator of a fraction can never be 0.

Rational Numbers

variable such that

Read as “the set of all numbers x such that x is a prime number.” Recall that when a letter stands for a number, it is called a variable.

rule that determines membership in the set

The fractions of arithmetic are called rational numbers.

Rational numbers are fractions that have an integer numerator and a nonzero integer denominator. Using set-builder notation, the rational numbers are e

a 0 a is an integer and b is a nonzero integer f b

Rational numbers can be written as fractions or decimals. Some examples of rational numbers are 5 55 , 1

3 5 0.75, 4

2

1 5 20.333 % , 3

2

5 5 20.454545 % 11

The = sign indicates that two quantities are equal.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

4

Chapter 0

A Review of Basic Algebra

These examples suggest that the decimal forms of all rational numbers are either terminating decimals or repeating decimals.

EXAMPLE 2

Determining whether the Decimal Form of a Fraction Terminates or Repeats Determine whether the decimal form of each fraction terminates or repeats: a.

SOLUTION

7 16

65 99

b.

In each case, we perform a long division and write the quotient as a decimal. 7 7 a. To change 16 to a decimal, we perform a long division to get 16 5 0.4375. Since 7 0.4375 terminates, we can write 16 as a terminating decimal. 65 b. To change 65 99 to a decimal, we perform a long division to get 99 5 0.656565 % . 65 Since 0.656565 ... repeats, we can write 99 as a repeating decimal.

We can write repeating decimals in compact form by using an overbar. For example, 0.656565 % 5 0.65. Self Check 2

Determine whether the decimal form of each fraction terminates or repeats: a.

38 99

b.

7 8

Now Try Exercise 35. Some numbers have decimal forms that neither terminate nor repeat. These nonterminating, nonrepeating decimals are called irrational numbers. Three examples of irrational numbers are "2 5 1.414213562 %

1.010010001000010 %

and

p 5 3.141592654 %

The union of the set of rational numbers (the terminating and repeating decimals) and the set of irrational numbers (the nonterminating, nonrepeating decimals) is the set of real numbers (the set of all decimals).

Real Numbers

A real number is any number that is rational or irrational. Using set-builder notation, the set of real numbers is 5 x 0 x is a rational or an irrational number 6

EXAMPLE 3

Classifying Real Numbers

In the set 5 23, 22, 0, 12, 1, "5, 2, 4, 5, 6 6 , list all a. even integers

SOLUTION

b. prime numbers

c. rational numbers

We will check to see whether each number is a member of the set of even integers, the set of prime numbers, and the set of rational numbers. a. even integers: –2, 0, 2, 4, 6 b. prime numbers: 2, 5 c. rational numbers: –3, –2, 0, 12, 1, 2, 4, 5, 6

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Section 0.1

Self Check 3

In the set in Example 3, list all c. irrational numbers.

Sets of Real Numbers

5

a. odd integers b. composite numbers

Now Try Exercise 43. Figure 0-2 shows how the previous sets of numbers are related. Real numbers 8 11 −3, – – , −√2, 0, –– , π, 9.9 5 16

Rational numbers 2 3 −6, –1.25, 0, – , 5 – , 80 3 4

Noninteger rational numbers 111 3.17 2 –––, – 13 ––, – 0.1, –, 7 5 53

Irrational numbers −√5, π, √21, 2√101

Integers −4, −1, 0, 21

Negative integers −47, −17, −5, −1

Whole numbers 0, 1, 4, 8, 10, 53, 101

Zero 0

Natural numbers (positive integers) 1, 12, 38, 990

FIGURE 0-2

2. Identify Properties of Real Numbers When we work with real numbers, we will use the following properties.

Properties of Real Numbers

If a, b, and c are real numbers, The Commutative Properties for Addition and Multiplication a1b5b1a

ab 5 ba

The Associative Properties for Addition and Multiplication 1a 1 b2 1 c 5 a 1 1b 1 c2

1ab2 c 5 a 1bc2

The Distributive Property of Multiplication over Addition or Subtraction a 1b 1 c2 5 ab 1 ac

or

a 1b 2 c2 5 ab 2 ac

The Double Negative Rule 2 12a2 5 a

Caution

When the Associative Property is used, the order of the real numbers does not change. The real numbers that occur within the parentheses change.

The Distributive Property also applies when more than two terms are within parentheses. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

6

Chapter 0

A Review of Basic Algebra

EXAMPLE 4

Identifying Properties of Real Numbers Determine which property of real numbers justifies each statement.

b. 3 1x 1 y 1 22 5 3x 1 3y 1 3 ? 2

a. 19 1 22 1 3 5 9 1 12 1 32 SOLUTION

We will compare the form of each statement to the forms listed in the properties of real numbers box. a. This form matches the Associative Property of Addition. b. This form matches the Distributive Property.

Self Check 4

Determine which property of real numbers justifies each statement: a. mn 5 nm b. 1xy2 z 5 x 1yz2

c. p 1 q 5 q 1 p

Now Try Exercise 17.

3. Graph Subsets of Real Numbers on the Number Line We can graph subsets of real numbers on the number line. The number line shown in Figure 0-3 continues forever in both directions. The positive numbers are represented by the points to the right of 0, and the negative numbers are represented by the points to the left of 0. Negative numbers

Comment

–5

–4

–3

–2

Positive numbers

–1

0

1

2

3

4

5

FIGURE 0-3

Zero is neither positive nor negative.

Figure 0-4(a) shows the graph of the natural numbers from 1 to 5. The point associated with each number is called the graph of the number, and the number is called the coordinate of its point. Figure 0-4(b) shows the graph of the prime numbers that are less than 10. Figure 0-4(c) shows the graph of the integers from 24 to 3. Figure 0-4(d) shows the graph of the real numbers 273, 234, 0.3, and "2. –1

Comment

"2 can be shown as the diagonal of a

0

1

2

3

4

5

6

0

1

–5 – 4 –3 –2 –1

0

1

2

3

4

–3

(c) 1

4

5

6

7

8

9 10

(b) –7/3

1

3

(a)

square with sides of length 1.

2

2

–2

–3/4 –1

– 0.3 0

2 1

2

(d) FIGURE 0-4

1

The graphs in Figure 0-4 suggest that there is a one-to-one correspondence between the set of real numbers and the points on a number line. This means that to each real number there corresponds exactly one point on the number line, and to each point on the number line there corresponds exactly one real-number coordinate.

1

EXAMPLE 5

Graphing a Set of Numbers on a Number Line

Graph the set U23, 243, 0, "5V.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.1

SOLUTION

Sets of Real Numbers

We will mark (plot) each number on the number line. To the nearest tenth, "5 5 2.2. 5

– 4/3 –3

Self Check 5

7

–2

–1

0

1

2

3

Graph the set U22, 34, "3V. (Hint: To the nearest tenth, "3 5 1.7.) Now Try Exercise 51.

4. Graph Intervals on the Number Line To show that two quantities are not equal, we can use an inequality symbol. Symbol

Read as

Examples

2

“is not equal to”

528

and

0.25 2 13

,

“is less than”

12 , 20

and

0.17 , 1.1

.

“is greater than”

15 . 9

and

1 2

#

“is less than or equal to”

25 # 25

and

1.7 # 2.3

$

“is greater than or equal to”

19 $ 19

and

15.2 $ 13.7


1 x is greater than 1.

(a)

(b) FIGURE 0-7

Caution

The symbol ` (infinity) is not a real number. It is used to indicate that the graph in Figure 0-7(b) extends infinitely far to the right.

A compound inequality such as 22 , x , 4 can be written as two separate inequalities: x . 22

and

x,4

This expression represents the intersection of two intervals. In interval notation, this expression can be written as 122, ` 2 d 12`, 42

Read the symbol d as “intersection.”

Since the graph of 22 , x , 4 will include all points whose coordinates satisfy both x . 22 and x , 4 at the same time, its graph will include all points that are larger than 22 but less than 4. This is the interval 122, 42 , whose graph is shown in Figure 0-7(a).

EXAMPLE 6

Writing an Inequality in Interval Notation and Graphing the Inequality Write the inequality 23 , x , 5 in interval notation and graph it.

SOLUTION

This is the interval 123, 52 . Its graph includes all real numbers between 23 and 5, as shown in Figure 0-8.

(

−3

) 5

FIGURE 0-8

Self Check 6

Write the inequality 22 , x # 5 in interval notation and graph it. Now Try Exercise 63.

If an interval extends forever in one direction, it is called an unbounded interval.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.1

9

Sets of Real Numbers

Unbounded Intervals 1a, ` 2

Interval 12, ` 2

3 a, ` 2 3 2, ` 2

(

)

−3

2

(−3, 2) −3 < x < 2 (a)

[

)

−2

3

[−2, 3) −2 ≤ x < 3 (b) FIGURE 0-9

[

]

−2

4

[−2, 4] −2 ≤ x ≤ 4 FIGURE 0-10

12`, a2 12`, 22

12`, a 4 12`, 2 4

12`, ` 2

Inequality

Graph

(

x.a

a

(

x.2 –2 –1

0

1

2

3

4

5

6

0

1

2

3

4

5

6

– 4 –3 –2 –1

0

1

3

4

0

1

3

4

[

x$a

a

x$2

[ –2 –1

(

x,a

a

)

x,2

2

[

x#a

a

]

x#2 –4 –3 –2 –1

2

2` , x , `

A bounded interval with no endpoints is called an open interval. Figure 0-9(a) shows the open interval between 23 and 2. A bounded interval with one endpoint is called a half-open interval. Figure 0-9(b) shows the half-open interval between 22 and 3, including 22. Intervals that include two endpoints are called closed intervals. Figure 0-10 shows the graph of a closed interval from 22 to 4. Open Intervals, Half-Open Intervals, and Closed Intervals Interval 1a, b2

Inequality

Graph

Open 122, 32 3 a, b2

a,x,b

(

)

a

b

(

22 , x , 3

)

–4 –3 –2 –1

0

1

2

0

1

2

0

1

2

0

1

2

3

4

5

4

5

4

5

4

5

Half-Open 3 22, 32 1a, b 4

a#x,b

[

)

a

b

[

22 # x , 3

)

–4 –3 –2 –1

3

Half-Open 122, 3 4

3 a, b 4

a,x#b

(

]

a

b

(

22 , x # 3

]

–4 –3 –2 –1

3

Closed

3 22, 3 4

a#x#b 22 # x # 3

[

]

a

b

[ –4 –3 –2 –1

] 3

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

10

Chapter 0

A Review of Basic Algebra

EXAMPLE 7

Writing an Inequality in Interval Notation and Graphing the Inequality Write the inequality 3 # x in interval notation and graph it.

SOLUTION

The inequality 3 # x can be written in the form x $ 3. This is the interval 3 3, ` 2 . Its graph includes all real numbers greater than or equal to 3, as shown in Figure 0-11.

[ 3 FIGURE 0-11

Self Check 7

Write the inequality 5 . x in interval notation and graph it. Now Try Exercise 65.

EXAMPLE 8

Writing an Inequality in Interval Notation and Graphing the Inequality Write the inequality 5 $ x $ 21 in interval notation and graph it.

SOLUTION

The inequality 5 $ x $ 21 can be written in the form 21 # x # 5

This is the interval 3 21, 5 4 . Its graph includes all real numbers from 21 to 5. The graph is shown in Figure 0-12.

[

]

−1

5

FIGURE 0-12

Self Check 8

Write the inequality 0 # x # 3 in interval notation and graph it. Now Try Exercise 69. The expression x , 22 or x $ 3

Read as “x is less than 22 or x is greater than or equal to 3.”

represents the union of two intervals. In interval notation, it is written as 12`, 222 c 3 3, ` 2

Read the symbol c as “union.”

Its graph is shown in Figure 0-13.

)

[

−2

3

FIGURE 0-13

5. Define Absolute Value

The absolute value of a real number x (denoted as 0 x 0 ) is the distance on a number line between 0 and the point with a coordinate of x. For example, points with coordinates of 4 and 24 both lie four units from 0, as shown in Figure 0-14. Therefore, it follows that 0 24 0 5 0 4 0 5 4

4 units

–5 –4 –3 –2 –1

4 units

0

1

2

3

4

5

FIGURE 0-14 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.1

Sets of Real Numbers

11

In general, for any real number x, 0 2x 0 5 0 x 0

We can define absolute value algebraically as follows.

Absolute Value

If x is a real number, then 0x0 5 x

when x $ 0

0 x 0 5 2x

Caution Remember that x is not always positive and 2x is not always negative.

EXAMPLE 9

when x , 0

This definition indicates that when x is positive or 0, then x is its own absolute value. However, when x is negative, then 2x (which is positive) is its absolute value. Thus, 0 x 0 is always nonnegative. 0x0 $ 0

for all real numbers x

Using the Definition of Absolute Value Write each number without using absolute value symbols: a. 0 3 0

SOLUTION

Self Check 9

b. 0 24 0

c. 0 0 0

d. 2 0 28 0

In each case, we will use the definition of absolute value. a. 0 3 0 5 3

b. 0 24 0 5 4

c. 0 0 0 5 0

d. 2 0 28 0 5 2 182 5 28

Write each number without using absolute value symbols: a. 0 210 0 b. 0 12 0 c. 2 0 6 0 Now Try Exercise 85. In Example 10, we must determine whether the number inside the absolute value is positive or negative.

EXAMPLE 10

Simplifying an Expression with Absolute Value Symbols Write each number without using absolute value symbols: a. 0 p 2 1 0

SOLUTION

b. 0 2 2 p 0

c. 0 2 2 x 0 if x $ 5

a. Since p < 3.1416, p 2 1 is positive, and p 2 1 is its own absolute value. 0p 2 10 5 p 2 1

b. Since 2 2 p is negative, its absolute value is 2 12 2 p2 .

0 2 2 p 0 5 2 12 2 p2 5 22 2 1 2 p2 5 22 1 p 5 p 2 2

c. Since x $ 5, the expression 2 2 x is negative, and its absolute value is 2 12 2 x2 . 0 2 2 x 0 5 2 12 2 x2 5 22 1 x 5 x 2 2 provided x $ 5

Self Check 10

Write each number without using absolute value symbols. 1Hint: "5 < 2.236.2 a. P 2 2 "5 P

b. 0 2 2 x 0 if x # 1

Now Try Exercise 89. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

12

Chapter 0

A Review of Basic Algebra

6. Find Distances on the Number Line On the number line shown in Figure 0-15, the distance between the points with coordinates of 1 and 4 is 4 2 1, or 3 units. However, if the subtraction were done in the other order, the result would be 1 2 4, or 23 units. To guarantee that the distance between two points is always positive, we can use absolute value symbols. Thus, the distance d between two points with coordinates of 1 and 4 is d 5 04 2 10 5 01 2 40 5 3

d = |4 − 1| = 3

–1

0

1

2

3

4

5

FIGURE 0-15

In general, we have the following definition for the distance between two points on the number line.

Distance between Two Points

If a and b are the coordinates of two points on the number line, the distance between the points is given by the formula d 5 0b 2 a0

EXAMPLE 11

Finding the Distance between Two Points on a Number Line Find the distance on a number line between points with coordinates of a. 3 and 5 b. 22 and 3 c. 25 and 21

SOLUTION

We will use the formula for finding the distance between two points. a. d 5 0 5 2 3 0 5 0 2 0 5 2

b. d 5 0 3 2 1222 0 5 0 3 1 2 0 5 0 5 0 5 5

c. d 5 0 21 2 1252 0 5 0 21 1 5 0 5 0 4 0 5 4 Self Check 11

Find the distance on a number line between points with coordinates of a. 4 and 10 b. 22 and 27 Now Try Exercise 99.

Self Check Answers

1. a. no b. {a, c, d, e, i, o, u} c. {a, e, i} 2. a. repeats b. terminates 3. a. 23, 1, 5 b. 4, 6 c. "5 4. a. Commutative Property of Multiplication b. Associative Property of Multiplication 3 –2 3/4 c. Commutative Property of Addition 5. 6. 122, 5 4 8. 3 0, 3 4

(

]

−2

[

10. a. "5 2 2

0

5

]

7. 12`, 52 9. a. 10

–3

–2

–1

0

1

2

) 5

b. 12

c. 26

3

b. 2 2 x

11. a. 6

b. 5

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Section 0.1

Sets of Real Numbers

13

Exercises 0.1 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

24. The between two distinct points on a number line is always positive. Let N 5 the set of natural numbers

Fill in the blanks. 1. A is a collection of objects. 2. If every member of one set B is also a member of a second set A, then B is called a of A. 3. If A and B are two sets, the set that contains all members that are in sets A and B or both is called the of A and B. 4. If A and B are two sets, the set that contains all members that are in both sets is called the of A and B. 5. A real number is any number that can be expressed as a . 6. A is a letter that is used to represent a number. 7. The smallest prime number is . 8. All integers that are exactly divisible by 2 are called integers. 9. Natural numbers greater than 1 that are not prime are called numbers. 2 8 10. Fractions such as 3 , 2 , and 279 are called numbers. 11. Irrational numbers are that don’t terminate and don’t repeat. 12. The symbol is read as “is less than or equal to.” 13. On a number line, the numbers are to the left of 0. 14. The only integer that is neither positive nor negative is . 15. The Associative Property of Addition states that 1x 1 y2 1 z 5 . 16. The Commutative Property of Multiplication states that xy 5 . 17. Use the Distributive Property to complete the state. ment: 5 1m 1 22 5 1 2 1 18. The statement m 1 n p 5 p m 1 n2 illustrates the Property of . 19. The graph of an line.

is a portion of a number

W 5 the set of whole numbers Z 5 the set of integers Q 5 the set of rational numbers R 5 the set of real numbers Determine whether each statement is true or false. Read the symbol ( as “is a subset of.” 26. Q ( R 25. N ( W 27. Q ( N 28. Z ( Q 29. W ( Z 30. R ( Z Let A 5 5 a, b, c, d, e 6 , B 5 1d, e, f, g 6 , and C 5 5 a, c, e, f 6 . Find each set.

Practice

31. A c B

32. A d B

33. A d C

34. B c C

Determine whether the decimal form of each fraction terminates or repeats. 9 3 35. 36. 16 8 3 5 37. 38. 11 12 Consider the following set: 5 25, 24, 223, 0, 1, "2, 2, 2.75, 6, 7 6 .

39. Which numbers are natural numbers? 40. Which numbers are whole numbers? 41. Which numbers are integers? 42. Which numbers are rational numbers? 43. Which numbers are irrational numbers? 44. 45. 46. 47. 48.

Which numbers are prime numbers? Which numbers are composite numbers? Which numbers are even integers? Which numbers are odd integers? Which numbers are negative numbers?

20. The graph of an open interval has

endpoints. 21. The graph of a closed interval has endpoints. 22. The graph of a interval has one endpoint. 23. Except for 0, the absolute value of every number is .

Graph each subset of the real numbers on a number line. 49. The natural numbers between 1 and 5

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14

Chapter 0

A Review of Basic Algebra

50. The composite numbers less than 10 51. The prime numbers between 10 and 20

Write each inequality as the union of two intervals and graph the result. 77. x , 22 or x . 2

52. The integers from –2 to 4

78. x # 25 or x . 0

53. The integers between –5 and 0

79. x # 21 or x $ 3

54. The even integers between –9 and –1

80. x , 23 or x $ 2

55. The odd integers between –6 and 4 56. 20.7, 1.75, and

Write each expression without using absolute value symbols. 82. 0 217 0 81. 0 13 0

378

Write each inequality in interval notation and graph the interval. 57. x . 2

58. x , 4

59. 0 , x , 5

60. 22 , x , 3

61. x . 24

62. x , 3

63. 22 # x , 2

64. 24 , x # 1

65. x # 5

66. x $ 21

67. 25 , x # 0

68. 23 # x , 4

83. 0 0 0 85. 2 0 28 0

84. 2 0 63 0 86. 0 225 0

87. 89. 91. 93.

88. 90. 92. 94.

2 0 32 0 0p 2 50 0p 2 p0 0 x 1 1 0 and x $ 2

95. 0 x 2 4 0 and x , 0

2 0 26 0 08 2 p0 0 2p 0 0 x 1 1 0 and x # 2 2

96. 0 x 2 7 0 and x . 10

Find the distance between each pair of points on the number line. 97. 3 and 8 98. 25 and 12 99. 28 and 23 100. 6 and 220

Applications 101. What subset of the real numbers would you use to describe the populations of several cities?

69. 22 # x # 3 71. 6 $ x $ 2

70. 24 # x # 4 72. 3 $ x $ 22

102. What subset of the real numbers would you use to describe the subdivisions of an inch on a ruler? 103. What subset of the real numbers would you use to report temperatures in several cities?

Write each pair of inequalities as the intersection of two intervals and graph the result. 73. x . 25 and x , 4

104. What subset of the real numbers would you use to describe the financial condition of a business?

74. x $ 23 and x , 6

Discovery and Writing

75. x $ 28 and x # 23 76. x . 1 and x # 7

105. Explain why 2x could be positive. 106. Explain why every integer is a rational number. 107. Is the statement 0 ab 0 5 0 a 0 ? 0 b 0 always true? Explain. 0a0 a 1b 2 02 always true? 108. Is the statement ` ` 5 0b0 b Explain.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.2

109. Is the statement 0 a 1 b 0 5 0 a 0 1 0 b 0 always true? Explain. 111. Explain why it is incorrect to write a , b . c if a , b and b . c.

Integer Exponents and Scientific Notation

15

110. Under what conditions will the statement given in Exercise 109 be true? 112. Explain why 0 b 2 a 0 5 0 a 2 b 0 .

0.2 Integer Exponents and Scientific Notation In this section, we will learn to 1. Define natural-number exponents. 2. Apply the rules of exponents. 3. Apply the rules for order of operations to evaluate expressions. 4. Express numbers in scientific notation. 5. Use scientific notation to simplify computations.

bioraven/Shutterstock.com

The number of cells in the human body is approximated to be one hundred trillion or 100,000,000,000,000. One hundred trillion is 1102 1102 1102 # # # 1102 , where ten occurs fourteen times. Fourteen factors of ten can be written as 1014. In this section, we will use integer exponents to represent repeated multiplication of numbers.

1. Define Natural-Number Exponents When two or more quantities are multiplied together, each quantity is called a factor of the product. The exponential expression x4 indicates that x is to be used as a factor four times. 4 factors of x       

x4 5 x ? x ? x ? x In general, the following is true. Natural-Number Exponents

For any natural number n, n factors of x           

xn 5 x ? x ? x ? c ? x

In the exponential expression xn, x is called the base, and n is called the exponent or the power to which the base is raised. The expression xn is called a power of x. From the definition, we see that a natural-number exponent indicates how many times the base of an exponential expression is to be used as a factor in a product. If an exponent is 1, the 1 is usually not written: x1 5 x

EXAMPLE 1

Using the Definition of Natural-Number Exponents Write each expression without using exponents: a. 42

b. 1242 2

c. 253

d. 1252 3

e. 3x4

f. 13x2 4

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

16

Chapter 0

A Review of Basic Algebra

SOLUTION

In each case, we apply the definition of natural-number exponents. a. 42 5 4 ? 4 5 16

Read 42 as “four squared.”

b. 1242 2 5 1242 1242 5 16

Read 1242 2 as “negative four squared.”

c. 253 5 25 152 152 5 2125

Read 253 as “the negative of five cubed.”

d. 1252 3 5 1252 1252 1252 5 2125 e. 3x4 5 3 ? x ? x ? x ? x

Read 1252 3 as “negative five cubed.”

Read 3x4 as “3 times x to the fourth power.”

f. 13x2 4 5 13x2 13x2 13x2 13x2 5 81 ? x ? x ? x ? x Self Check 1

Read 13x2 4 as “3x to the fourth power.”

Write each expression without using exponents: a. 73 b. 1232 2 c. 5a3 d. 15a2 4 Now Try Exercise 19.

Comment

Note the distinction between axn and 1ax2 n:

n factors of x 1 2x 5 2 x ? x ? x ? c ? x2

n factors of 2x 12x2 n 5 12x2 12x2 12x2 ? c ? 12x2

            

n factors of ax 1ax2 n 5 1ax2 1ax2 1ax2 ? c ? 1ax2

        

n factors of x axn 5 a ? x ? x ? x ? c ? x

Also note the distinction between 2xn and 12x2 n:

              

        

n

ACCENT ON TECHNOLOGy

Using a Calculator to Find Powers We can use a graphing calculator to find powers of numbers. For example, consider 2.353. • •

Input 2.35 and press the Input 3 and press ENTER .

key.

Comment To find powers on a scientific calculator use the yx key.

FIGURE 0-16

We see that 2.353 5 12.977875 as the figure shows above.

2. Apply the Rules of Exponents We begin the review of the rules of exponents by considering the product xmxn. Since xm indicates that x is to be used as a factor m times, and since xn indicates that x is to be used as a factor n times, there are m 1 n factors of x in the product xmxn.                       

m 1 n factors of x

m factors of x

n factors of x

          

          

m n

x x 5 x ? x ? x ? c ? x ? x ? x ? x ? c ? x 5 xm1n Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.2

Integer Exponents and Scientific Notation

17

This suggests that to multiply exponential expressions with the same base, we keep the base and add the exponents.

Product Rule for Exponents

If m and n are natural numbers, then xmxn 5 xm1n

Comment

mn factors of x               

The Product Rule applies to exponential expressions with the same base. A product of two powers with different bases, such as x4y3, cannot be simplified.

To find another property of exponents, we consider the exponential expression 1xm2 n. In this expression, the exponent n indicates that xm is to be used as a factor n times. This implies that x is to be used as a factor mn times. 1xm2 n 5 1xm2 1xm2 1xm2 ? c ? 1xm2 5 xmn               

n factors of xm

This suggests that to raise an exponential expression to a power, we keep the base and multiply the exponents. To raise a product to a power, we raise each factor to that power. 1xy2 5 1xy2 1xy2 1xy2 ? c ? 1xy2 5 1x ? x ? x ? c ? x2 1y ? y ? y ? c ? y2 5 xnyn n factors of y

          

          

n factors of x

            

n factors of xy

n

To raise a fraction to a power, we raise both the numerator and the denominator to that power. If y 2 0, then n factors of

x y

            

x n x x x x a b 5 a ba ba b ? c ? a b y y y y y n factors of x

      

xxx ? c ? x yyy ? c ? y

      

5

n factors of y

5

xn yn

The previous three results are called the Power Rules of Exponents.

Power Rules of Exponents

EXAMPLE 2

If m and n are natural numbers, then 1xm2 n 5 xmn

1xy2 n 5 xnyn

x n xn a b 5 n y y

1y 2 02

Using Exponent Rules to Simplify Expressions with Natural-Number Exponents Simplify: a. x5x7 e. a

x 5 b y2

f. a

b. x2y3x5y

5x2y 2 b z3

c. 1x42 9

d. 1x2x52 3

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

18

Chapter 0

A Review of Basic Algebra

SOLUTION

In each case, we will apply the appropriate rule of exponents. a. x5x7 5 x517 5 x12 c. 1x42 9 5 x4?9 5 x36 e. a f. a

Self Check 2

x 5 x5 x5 5 5 b 1y22 5 y2 y10

b. x2y3x5y 5 x215y311 5 x7y4

d. 1x2x52 3 5 1x7 2 3 5 x21 1y 2 02

5x2y 2 52 1x22 2y2 25x4y2 5 5 b 1z32 2 z3 z6

Simplify: a. 1y32 2

b. 1a2a42 3

1z 2 02

c. 1x22 3 1x32 2

d. a

3a3b2 3 b c3

1c 2 02

Now Try Exercise 49. If we assume that the rules for natural-number exponents hold for exponents of 0, we can write x0xn 5 x01n 5 xn 5 1xn Since x0xn 5 1xn, it follows that if x 2 0, then x0 5 1.

Zero Exponent

x0 5 1

1x 2 02

If we assume that the rules for natural-number exponents hold for exponents that are negative integers, we can write x2nxn 5 x2n1n 5 x0 5 1

1x 2 02

However, we know that 1 ? xn 5 1 xn

1x 2 02

1 xn n n ? x 5 n , and any nonzero number divided by itself is 1. x x

Since x2nxn 5 x1n ? xn , it follows that x2n 5 x1n 1x 2 02 . Negative Exponents

If n is an integer and x 2 0, then x2n 5

1 xn

and

1 5 xn x2n

Because of the previous definitions, all of the rules for natural-number exponents will hold for integer exponents.

EXAMPLE 3

Simplifying Expressions with Integer Exponents Simplify and write all answers without using negative exponents: 1 b. 3 1x02 c. x24 d. 26 f. 1x24x82 25 e. x23x a. 13x2 0 x

SOLUTION

We will use the definitions of zero exponent and negative exponents to simplify each expression. 1 1 a. 13x2 0 5 1 b. 3 1x02 5 3 112 5 3 c. x24 5 4 d. 26 5 x6 x x

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Section 0.2

5 x22

Self Check 3

19

f. 1x24x82 25 5 1x42 25

e. x23x 5 x2311

5

Integer Exponents and Scientific Notation

5 x220

1 x2

5

1 x20 d. 1a3a272 3

Simplify and write all answers without using negative exponents: a. 7a0

c. a24a2

b. 3a22

Now Try Exercise 59. To develop the Quotient Rule for Exponents, we proceed as follows: 1 xm 5 xm a n b 5 xmx2n 5 xm1 1 2n2 5 xm2n xn x

1x 2 02

This suggests that to divide two exponential expressions with the same nonzero base, we keep the base and subtract the exponent in the denominator from the exponent in the numerator.

Quotient Rule for Exponents

If m and n are integers, then xm 5 xm2n xn

EXAMPLE 4

1x 2 02

Simplifying Expressions with Integer Exponents Simplify and write all answers without using negative exponents: a.

SOLUTION

x8 x5

b.

x2x4 x25

We will apply the Product and Quotient Rules of Exponents. a.

x8 5 x825 x5

b.

x2x4 x6 25 5 25 x x

5 x62 1252

5 x3

5 x11 Self Check 4

Simplify and write all answers without using negative exponents: a.

x26 x2

b.

x4x23 x2

Now Try Exercise 69.

EXAMPLE 5

Simplifying Expressions with Integer Exponents Simplify and write all answers without using negative exponents: a. a

x3y22 22 b x22y3

x 2n b. a b y

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20

Chapter 0

A Review of Basic Algebra

SOLUTION

We will apply the appropriate rules of exponents. x3y22 22 b 5 1x32 1222y22232 22 x22y3

a. a

5 1x5y252 22 5 x210y10

5 x 2n x2n b. a b 5 2n y y

y10 x10

5

x2nxnyn y2nxnyn

Multiply numerator and denominator by 1 in the form

5

x0yn y0xn

x2nxn 5 x0 and y2nyn 5 y0.

yn xn y n 5a b x

xnyn

.

x0 5 1 and y0 5 1.

5

Self Check 5

xnyn

Simplify and write all answers without using negative exponents: a. a

x4y23 2 b x23y2

b. a

2a 23 b 3b

Now Try Exercise 75. Part b of Example 5 establishes the following rule.

A Fraction to a Negative Power

If n is a natural number, then x 2n y n a b 5a b y x

(x 2 0 and

y 2 0)

3. Apply the Rules for Order of Operations to Evaluate Expressions When several operations occur in an expression, we must perform the operations in the following order to get the correct result.

Strategy for Evaluating Expressions Using Order of Operations

If an expression does not contain grouping symbols such as parentheses or brackets, follow these steps: 1. Find the values of any exponential expressions. 2. Perform all multiplications and/or divisions, working from left to right. 3. Perform all additions and/or subtractions, working from left to right. • If an expression contains grouping symbols such as parentheses, brackets, or braces, use the rules above to perform the calculations within each pair of grouping symbols, working from the innermost pair to the outermost pair. • In a fraction, simplify the numerator and the denominator of the fraction separately. Then simplify the fraction, if possible.

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Section 0.2

Comment

Integer Exponents and Scientific Notation

21

Many students remember the Order of Operations Rule with the acronym PEMDAS: • • • • • •

Parentheses Exponents Multiplication Division Addition Subtraction

For example, to simplify

3 3 4 2 16 1 102 4 , we proceed as follows: 22 2 16 1 72

3 3 4 2 16 1 102 4 3 14 2 162 5 2 2 2 2 16 1 72 2 2 16 1 72 3 12122 5 2 2 2 13 3 12122 5 4 2 13 236 5 29

Simplify within the inner parentheses: 6 1 10 5 16.

Simplify within the parentheses: 4 2 16 5 212, and 6 1 7 5 13. Evaluate the power: 22 5 4. 3 12122 5 236. 4 2 13 5 29.

54

EXAMPLE 6

Evaluating Algebraic Expressions If x 5 22, y 5 3, and z 5 24, evaluate a. 2x2 1 y2z

SOLUTION

b.

2z3 2 3y2 5x2

In each part, we will substitute the numbers for the variables, apply the rules of order of operations, and simplify. a. 2x2 1 y2z 5 2 1222 2 1 32 1242 5 2 142 1 9 1242 5 24 1 12362

Do the multiplication.

2z 2 3y 2 1242 2 3 132 5 2 5x 5 1222 2 2 12642 2 3 192 5 5 142 2128 2 27 5 20 2155 5 20 31 52 4 5 240

3

b.

Self Check 6

2

3

Evaluate the powers.

Do the addition.

2

If x 5 3 and y 5 22, evaluate

Evaluate the powers. Do the multiplications. Do the subtraction. Simplify the fraction.

2x2 2 3y2 . x2y

Now Try Exercise 91.

4. Express Numbers in Scientific Notation Scientists often work with numbers that are very large or very small. These numbers can be written compactly by expressing them in scientific notation. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

22

Chapter 0

A Review of Basic Algebra

Scientific Notation

A number is written in scientific notation when it is written in the form N 3 10n

where 1 # 0 N 0 , 10 and n is an integer.

▲

Standard notation

29,980,000,000 5 2.998 3 1010

▲

Light travels 29,980,000,000 centimeters per second. To express this number in scientific notation, we must write it as the product of a number between 1 and 10 and some integer power of 10. The number 2.998 lies between 1 and 10. To get 29,980,000,000, the decimal point in 2.998 must be moved ten places to the right. This is accomplished by multiplying 2.998 by 1010. Scientific notation

0.0006214 5 6214 3 1024

▲

Standard notation

▲

ollirg/Shutterstock.com

One meter is approximately 0.0006214 mile. To express this number in scientific notation, we must write it as the product of a number between 1 and 10 and some integer power of 10. The number 6.214 lies between 1 and 10. To get 0.0006214, the decimal point in 6.214 must be moved four places to the left. This is accomplished by multiplying 6.214 by 101 4 or by multiplying 6.214 by 1024. Scientific notation

To write each of the following numbers in scientific notation, we start to the right of the first nonzero digit and count to the decimal point. The exponent gives the number of places the decimal point moves, and the sign of the exponent indicates the direction in which it moves.

EXAMPLE 7

a. 3 7 2 0 0 0 5 3.72 3 105

5 places to the right.

b. 0 . 0 0 0 5 3 7 5 5.37 3 1024

4 places to the left.

c. 7.36 5 7.36 3 100

No movement of the decimal point.

Writing Numbers in Scientific Notation Write each number in scientific notation: a. 62,000

SOLUTION

b. 20.0027

a. We must express 62,000 as a product of a number between 1 and 10 and some integer power of 10. This is accomplished by multiplying 6.2 by 104. 62,000 5 6.2 3 104 b. We must express 20.0027 as a product of a number whose absolute value is between 1 and 10 and some integer power of 10. This is accomplished by multiplying 22.7 by 1023. 20.0027 5 22.7 3 1023

Self Check 7

Write each number in scientific notation: b. 0.0000087 a. 293,000,000 Now Try Exercise 103.

EXAMPLE 8

Writing Numbers in Standard Notation Write each number in standard notation: a. 7.35 3 102

b. 3.27 3 1025

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Section 0.2

SOLUTION

Integer Exponents and Scientific Notation

23

a. The factor of 102 indicates that 7.35 must be multiplied by 2 factors of 10. Because each multiplication by 10 moves the decimal point one place to the right, we have 7.35 3 102 5 735 b. The factor of 1025 indicates that 3.27 must be divided by 5 factors of 10. Because each division by 10 moves the decimal point one place to the left, we have 3.27 3 1025 5 0.0000327

Self Check 8

Write each number in standard notation: a. 6.3 3 103

b. 9.1 3 1024

Now Try Exercise 111.

5. Use Scientific Notation to Simplify Computations Another advantage of scientific notation becomes evident when we multiply and divide very large and very small numbers.

EXAMPLE 9

Using Scientific Notation to Simplify Computations Use scientific notation to calculate

SOLUTION

13,400,0002 10.000022 . 170,000,000

After changing each number to scientific notation, we can do the arithmetic on the numbers and the exponential expressions separately. 13,400,0002 10.000022 13.4 3 1062 12.0 3 10252 5 170,000,000 1.7 3 108 6.8 5 3 1061 1 252 28 1.7 5 4.0 3 1027 5 0.0000004

Self Check 9

Use scientific notation to simplify

1192,0002 10.00152 . 10.00322 14,5002

Now Try Exercise 119.

ACCENT ON TECHNOLOGy

Scientific Notation Graphing calculators will often give an answer in scientific notation. For example, consider 218.

FIGURE 0-17

We see in the figure that the answer given is 3.782285936E10, which means 3.782285936 3 1010.

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24

Chapter 0

A Review of Basic Algebra

Comment

218 on a scientific calculator, it is necessary to convert 0.000000000061 the denominator to scientific notation because the number has too many digits to fit on the screen. To calculate an expression like

• For scientific notation, we enter these numbers and press these keys: 6.1 EXP 11 1/2 . • To evaluate the expression above, we enter these numbers and press these keys: 21 yx 8 5 4 6.1 EXP 11 1/2 5 The display will read 6.20046874820 . In standard notation, the answer is approximately 620,046,874,800,000,000,000.

Self Check Answers

1. a. 7 ? 7 ? 7 5 343 b. 1232 1232 5 9 c. 5 ? a ? a ? a 1 2 1 2 1 2 1 2 d. 5a 5a 5a 5a 5 625 ? a ? a ? a ? a 2. a. y6 b. a18 27a9b6 c9 x14 5. a. 10 y 8. a. 6,300

d.

3 1 1 c. 2 d. 12 2 a a a 27b3 6 b. 6. 7. a. 29.3 3 107 8a3 5 b. 0.00091 9. 20 3. a. 7

b.

4. a.

c. x12 1 x8

b.

1 x

b. 8.7 3 1026

Exercises 0.2 Getting Ready

15. 252

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

17. 4x3 19. 125x2 4

Fill in the blanks. 1. Each quantity in a product is called a of the product. 2. A number exponent tells how many times a base is used as a factor. 3. In the expression 12x2 3 , is the exponent and is the base. 4. The expression xn is called an expression. notation when it is writ5. A number is in ten in the form N 3 10n , where 1 # 0 N 0 , 10 and n is an . 6. Unless

indicate otherwise, are performed before additions.

Complete each exponent rule. Assume x 2 0. 7. xmxn 5 8. 1xm2 n 5 9. 1xy2 n 5 11. x0 5

10.

xm 5 xn

12. x2n 5

Practice Write each number or expression without using exponents. 13. 132

21. 28x4

16. 1252 2 18. 14x2 3 20. 26x2 22. 128x2 4

Write each expression using exponents. 23. 7xxx 25. 12x2 12x2

24. 28yyyy 26. 12a2 12a2 12a2

27. 13t2 13t2 123t2

28. 2 12b2 12b2 12b2 12b2

29. xxxyy

30. aaabbbb

Use a calculator to simplify each expression. 31. 2.23 33. 20.54

32. 7.14 34. 120.22 4

Simplify each expression. Write all answers without using negative exponents. Assume that all variables are restricted to those numbers for which the expression is defined. 35. x2x3 37. 1z22 3

36. y3y4 38. 1t62 7

39. 1y5y22 3 41. 1z22 3 1z42 5

40. 1a3a62 a4 42. 1t32 4 1t52 2

43. 1a22 3 1a42 2 45. 13x2 3

44. 1a22 4 1a32 3 46. 122y2 4

47. 1x2y2 3

48. 1x3z42 6

14. 103 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.2

49. a

a2 3 b b

50. a

51. 12x2 0 53. 14x2 0

59. 1x3x242 22

63. 65. 67. 69. 71.

62. 64. 66. 68. 70. 72.

r4r26 2 b r3r23

75. a

x5y22 4 b x23y2

79. a

3x y b 6x25y3

74.

5x23y22 22 77. a 2 23 b 3x y

81. 82.

5 23

99. 2177,000,000

60. 1y22y32 24

x7 x3 a21 a17 1x22 2 x2x m3 3 a 2b n 1a32 22 aa2 a23 24 a 21 b b

73. a

97. 372,000

54. 22x0 1 56. 22 t 58. 2m22m3

57. y22y23

22

1822z23y2 21 15y2z222 3 15yz222 21

r5 r2 t13 t4 s9s3 1s22 2 t4 3 a 3b t r9r23 1r222 3 t24 22 a 23 b t

x27y5 3 b x7y24

80. a

12x y z b 4x4y23z5

104. 20.000000089

105. one trillion

106. one millionth

107. 9.37 3 105

108. 4.26 3 109

109. 2.21 3 1025

110. 2.774 3 1022

111. 0.00032 3 104

112. 9,300.0 3 1024

113. 23.2 3 1023

114. 27.25 3 103

117. 3

118.

120. 6 3 3 2 14 2 72 4 25 12 2 422 2

84.

86. 2x2

87. x3 89. 12xz2 3 2 1x2z32 91. 2 z 2 y2

88. 2x3

23x23z22 6x2z23

103. 20.000000693

119.

85. x2

95.

102. 0.00052

116.

Let x 5 22, y 5 0, and z 5 3 and evaluate each expression.

93. 5x2 2 3y3z

101. 0.007

115.

1m22n3p42 22 1mn22p32 4 1mn22p32 24 1mn2p2 21

Simplify each expression. 5 3 62 1 19 2 52 4 83. 2 4 12 2 32 2

90. 2xz3 z2 1x2 2 y22 92. x3z

94. 3 1x 2 z2 2 1 2 1y 2 z2 3

96.

125x2z232 2 5xz22

100. 223,470,000,000

Use the method of Example 9 to do each calculation. Write all answers in scientific notation.

3x2y25 23 78. a 22 26 b 2x y 24 3 25

98. 89,500

Express each number in standard notation.

1x23x22 2 1x2x252 23

76. a

25

Express each number in scientific notation.

52. 4x0

55. z24

61.

x 4 b y3

Integer Exponents and Scientific Notation

165,0002 145,0002 250,000 10.0000000452 10.000000122 45,000,000 10.000000352 1170,0002 0.00000085 10.00000001442 112,0002 600,000 145,000,000,0002 1212,0002 0.00018 10.000000002752 14,7502 500,000,000,000

Applications Use scientific notation to compute each answer. Write all answers in scientific notation. 121. Speed of sound The speed of sound in air is 3.31 3 104 centimeters per second. Compute the speed of sound in meters per minute. 122. Volume of a box Calculate the volume of a box that has dimensions of 6,000 by 9,700 by 4,700 millimeters. 123. Mass of a proton The mass of one proton is 0.00000000000000000000000167248 gram. Find the mass of one billion protons.

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Chapter 0

A Review of Basic Algebra

124. Speed of light The speed of light in a vacuum is approximately 30,000,000,000 centimeters per second. Find the speed of light in miles per hour. 1160,934.4 cm 5 1 mile.2 125. Astronomy The distance d, in miles, of the nth planet from the sun is given by the formula

mass within 1 millimeter (0.04 inch) a year by using a combination of four space-based techniques.

NASA Goddard Space Flight Center

26

d 5 9,275,200 3 3 12n222 1 4 4

To the nearest million miles, find the distance of Earth and the distance of Mars from the sun. Give each answer in scientific notation.

Jupiter

Mars Earth

Venus

The distance from the Earth’s center to the North Pole (the polar radius) measures approximately 6,356.750 km, and the distance from the center to the equator (the equatorial radius ) measures approximately 6,378.135 km. Express each distance using scientific notation. 128. Refer to Exercise 127. Given that 1 km is approximately equal to 0.62 miles, use scientific notation to express each distance in miles.

Mercury

126. License Plates The number of different license plates of the form three digits followed by three letters, as in the illustration, is 10 ? 10 ? 10 ? 26 ? 26 ? 26. Write this expression using exponents. Then evaluate it and express the result in scientific notation. WB COUNTY

UTAH

12

123ABC Discovery and Writing 127. New way to the center of the earth The spectacular “blue marble’ image is the most detailed true-color image of the entire Earth to date. A new NASAdeveloped technique estimates Earth’s center of

Write each expression with a single base. xm 129. xnx2 130. 3 x xmx2 x3m15 131. 132. x3 x2 m11 3 n23 3 133. x x 134. a a 4 135. Explain why 2x and 12x2 4 represent different numbers. 136. Explain why 32 3 102 is not in scientific notation. 137. Graph the interval 122,42 .

Review

138. Graph the interval 12`,23 4 c 3 3,` 2 . 139. Evaluate 0 p 2 5 0 .

140. Find the distance between 27 and 25 on the number line.

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Section 0.3

Rational Exponents and Radicals

27

0.3 Rational Exponents and Radicals In this section, we will learn to 1. Define rational exponents whose numerators are 1. 2. Define rational exponents whose numerators are not 1. 3. Define radical expressions. 4. Simplify and combine radicals. 5. Rationalize denominators and numerators.

Medioimages/Photodisc/Jupiter Images

“Dead Man’s Curve” is a 1964 hit song by the rock and roll duo Jan Berry and Dean Torrence. The song details a teenage drag race that ends in an accident. Today, dead man’s curve is a commonly used expression given to dangerous curves on our roads. Every curve has a “critical speed.” If we exceed this speed, regardless of how skilled a driver we are, we will lose control of the vehicle. The radical expression 3.9"r gives the critical speed in miles per hour when we travel a curved road with a radius of r feet. A knowledge of square roots and radicals is important and used in the construction of safe highways and roads. We will study the topic of radicals in this section.

1. Define Rational Exponents Whose Numerators Are 1 If we apply the rule 1xm2 n 5 xmn to 1251/22 2 , we obtain 1251/22 2 5 2511/222 5 251

Caution

Keep the base and multiply the exponents. 1 2

?251

5 25 1 /n

In the expression a , there is no real-number nth root of a when n is even and a , 0. For example, 12642 1/2 is not a real number, because the square of no real number is 264.

Rational Exponents

Thus, 251/2 is a real number whose square is 25. Although both 52 5 25 and 1252 2 5 25, we define 251/2 to be the positive real number whose square is 25: 251/2 5 5

Read 251/2 as “the square root of 25.”

In general, we have the following definition.

If a $ 0 and n is a natural number, then a1/n (read as “the nth root of a”) is the nonnegative real number b such that bn 5 a

Since b 5 a1/n, we have bn 5 1a1/n2 n 5 a.

EXAMPLE 1

Simplifying Expressions with Rational Exponents In each case, we will apply the definition of rational exponents. a. 161/2 5 4

Because 42 5 16. Read 161/2 as “the square root of 16.”

b. 271/3 5 3

Because 33 5 27. Read 271/3 as “the cube root of 27.”

c. a

1 1/4 1 b 5 81 3

1 1 4 1 1 /4 1 Because a b 5 . Read a b as “the fourth root of .” 3 81 81 81

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28

Chapter 0

A Review of Basic Algebra

d. 2321/5 5 2 1321/52 5 2 122

Read 321/5 as “the fifth root of 32.” Because 25 5 32.

5 22 Self Check 1

Simplify: a. 1001/2

b. 2431/5

Now Try Exercise 15. If n is even in the expression a1/n and the base contains variables, we often use absolute value symbols to guarantee that an even root is nonnegative. 149x22 1/2 5 7 0 x 0

116x42 1/4 5 2 0 x 0

1729x122 1/6 5 3x2

Because 17 0 x 0 2 2 5 49x2 . Since x could be negative, absolute value sym-

bols are necessary to guarantee that the square root is nonnegative.

Because 12 0 x 0 2 4 5 16x4 . Since x could be negative, absolute value sym-

bols are necessary to guarantee that the fourth root is nonnegative.

Because 13x22 6 5 729x12 . Since x2 is always nonnegative, no absolute

value symbols are necessary. Read 1729x122 1/6 as “the sixth root of 729x12 .”

If n is an odd number in the expression a1/n , the base a can be negative.

EXAMPLE 2

Simplifying Expressions with Rational Exponents Simplify by using the definition of rational exponent. a. 1282 1/3 5 22

Because 1222 3 5 28.

c. a2

Because a2

b. 123,1252 1/5 5 25

Self Check 2

1/3 1 1 b 52 1,000 10

Simplify: a. 121252 1/3

Because 1252 5 5 23,125. 1 3 1 . b 52 10 1,000

b. 12100,0002 1/5

Now Try Exercise 19.

Caution

If n is odd in the expression a1/n , we don’t need to use absolute value symbols, because odd roots can be negative. 1227x32 1/3 5 23x

12128a72 1/7 5 22a

Because 123x2 3 5 227x3.

Because 122a2 7 5 2128a7.

We summarize the definitions concerning a1/n as follows.

Summary of Definitions of a1/n

If n is a natural number and a is a real number in the expression a1/n, then If a $ 0, then a1/n is the nonnegative real number b such that bn 5 a. If a , 0 e

and n is odd, then a1/n is the real number b such that bn 5 a. and n is even, then a1/n is not a real number.

The following chart also shows the possibilities that can occur when simplifying a1/n . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.3

Strategy for Simplifying Expressions of the Form a1/n

Rational Exponents and Radicals

n

a1/n

a50

n is a natural number.

01/n is the real number 0 because 0n 5 0.

01/2 5 0 because 02 5 0. 01/5 5 0 because 05 5 0.

a.0

n is a natural number.

a1/n is the nonnegative real number such that 1a1/n2 n 5 a.

161/2 5 4 because 42 5 16. 271/3 5 3 because 33 5 27.

a,0

n is an odd natural number.

a1/n is the real number such that 1a1/n2 n 5 a.

a,0

n is an even natural number.

a1/n is not a real number.

a

29

Examples

12322 1/5 5 22 because 1222 5 5 232. 121252 1/3 5 25 because 1252 3 5 2125. 1292 1/2 is not a real number. 12812 1/4 is not a real number.

2. Define Rational Exponents Whose Rational Exponents Are Not 1 The definition of a1/n can be extended to include rational exponents whose numerators are not 1. For example, 43/2 can be written as either 141/22 3 or 1432 1/2

Because of the Power Rule, 1xm2 n 5 xmn.

This suggests the following rule.

Rule for Rational Exponents

If m and n are positive integers, the fraction mn is in lowest terms, and a1/n is a real number, then am/n 5 1a1/n2 m 5 1am2 1/n

In the previous rule, we can view the expression am/n in two ways:

1. 1a1/n2 m: the mth power of the nth root of a 2. 1am2 1/n: the nth root of the mth power of a

For example, 12272 2/3 can be simplified in two ways: 12272 2/3 5 3 12272 1/3 4 2 5 1232 2 59

or

12272 2/3 5 3 12272 2 4 1/3 5 17292 1/3 59

As this example suggests, it is usually easier to take the root of the base first to avoid large numbers. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

30

Chapter 0

A Review of Basic Algebra

Comment

Negative Rational Exponents

EXAMPLE 3

It is helpful to think of the phrase power over root when we see a rational exponent. The numerator of the fraction represents the power and the denominator represents the root. Begin with the root when simplifying to avoid large numbers.

If m and n are positive integers, the fraction mn is in lowest terms and a1/n is a real number, then 1 1 1a 2 02 and 5 am/n a2m/n 5 m/n 2m/n a a

Simplifying Expressions with Rational Exponents We will apply the rules for rational exponents. a. 253/2 5 1251/22 3

b. a2

5 53

5 a2

5 125

5

1 322/5 1 5 1321/52 2 1 5 2 2 1 5 4

c. 3222/5 5

Self Check 3

x6 2/3 x6 1/3 2 b 5 c a2 b d 1,000 1,000

Simplify: a. 493/2

d.

x2 2 b 10

x4 100

1 5 813/4 8123/4 5 1811/42 3 5 33 5 27

b. 1623/4

c.

1 127x32 22/3

Now Try Exercise 43. Because of the definition, rational exponents follow the same rules as integer exponents.

EXAMPLE 4

Using Exponent Rules to Simplify Expressions with Rational Exponents Simplify each expression. Assume that all variables represent positive numbers, and write answers without using negative exponents. a.

136x2 1/2 5 361/2x1/2 5 6x1/2

b.

1a1/3b2/32 6 a6/3b12/3 5 1y32 2 y6 5

a2b4 y6

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Section 0.3

c.

ax/2ax/4 5 ax/21x/42x/6 ax/6

Rational Exponents and Radicals

d. c

31

2c22/5 5/3 d 5 12c22/524/52 5/3 c4/5

5 3 1212 1c26/52 4 5/3

5 a6x/1213x/1222x/12

5 1212 5/3 1c26/52 5/3

5 a7x/12

5 21c230/15 5 2c22 1 52 2 c Self Check 4

Use the directions for Example 4: 19r2s2 1/2 y2 1/2 b3/7b2/7 a. a b b. c. 49 b4/7 rs23/2 Now Try Exercise 59.

3. Define Radical Expressions Radical signs can also be used to express roots of numbers. Definition of "a n

If n is a natural number greater than 1 and if a1/n is a real number, then "a 5 a1/n n

In the radical expression "a, the symbol " is the radical sign, a is the radicand, and n is the index (or the order) of the radical expression. If the order is 2, the expression is a square root, and we do not write the index. n

2 "a 5 " a

If the index of a radical is 3, we call the radical a cube root.

nth Root of a Nonnegative Number

If n is a natural number greater than 1 and a $ 0, then "a is the nonnegative number whose nth power is a. n

Q"aR 5 a n

Caution

n

In the expression "a, there is no real-number nth root of a when n is even and a , 0. For n

example, "264 is not a real number, because the square of no real number is 264. If 2 is substituted for n in the equation Q"aR 5 a, we have n

n

2 Q" aR 5 Q"aR 5 "a"a 5 a for a $ 0 2

2

This shows that if a number a can be factored into two equal factors, either of those factors is a square root of a. Furthermore, if a can be factored into n equal factors, any one of those factors is an nth root of a.

If n is an odd number greater than 1 in the expression "a, the radicand can be negative. n

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32

Chapter 0

A Review of Basic Algebra

EXAMPLE 5

Finding nth Roots of Real Numbers We apply the definitions of cube root and fifth root. 3 a. "227 5 23

Because 1232 3 5 227.

3 b. " 28 5 22

c.

Because 1222 3 5 28.

3 27 52 Å 1,000 10 3

2

Because a2

3 3 27 . b 52 10 1,000

5 5 d. 2" 2243 5 2Q" 2243R 5 2 1232 53

Self Check 5

3 Find each root: a. " 216

b.

Now Try Exercise 69.

1 Å 32 5

2

We summarize the definitions concerning "a as follows. n

Summary of Definitions of "a n

If n is a natural number greater than 1 and a is a real number, then

If a $ 0, then "a is the nonnegative real number such that Q"aR 5 a. n

n

n

and n is odd, then "a is the real number such that Q"aR 5 a. If a , 0 e n and n is even, then "a is not a real number. n

n

n

The following chart also shows the possibilities that can occur when simplifyn ing "a.

Strategy for Simplifying n Expressions of the Form "a

a a50

a.0

a,0

a,0

"a n

n

n is a natural "0 is the real number number 0 greater than because 0n 5 0. 1. n

n is a natural number greater than 1.

"a is the nonnegative real number such n

that Q"aR 5 a. n

Examples

"0 5 0 because 03 5 0. 5 " 0 5 0 because 05 5 0. 3

"16 5 4 because 42 5 16. 3 " 27 5 3 because 33 5 27.

n

n 5 n is an odd "a is the real " 232 5 22 because 1222 5 5 232. 3 natural num- number such "2125 5 25 because 1252 3 5 2125. n n ber greater that Q"aR 5 a. than 1.

n is an even natural number.

"a is not a real "29 is not a real number. 4 number. " 281 is not a real number. n

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.3

Rational Exponents and Radicals

33

We have seen that if a1/n is real, then am/n 5 1a1/n2 m 5 1am2 1/n . This same fact can be stated in radical notation. am/n 5 1"a2 m 5 "am n

n

Thus, the mth power of the nth root of a is the same as the nth root of the mth power 3 of a. For example, to find " 272, we can proceed in either of two ways: 3 3 3 3 272 5 Q" 27R 5 32 5 9 or " " 272 5 " 729 5 9 2

By definition, "a2 represents a nonnegative number. If a could be negative, we must use absolute value symbols to guarantee that "a2 will be nonnegative. Thus, if a is unrestricted, "a2 5 0 a 0

A similar argument holds when the index is any even natural number. The sym4 bol "a4, for example, means the positive fourth root of a4. Thus, if a is unrestricted, 4 4 " a 5 0a0

EXAMPLE 6 SOLUTION

Simplifying Radical Expressions

6 If x is unrestricted, simplify a. " 64x6

c. "9x8

We apply the definitions of sixth roots, cube roots, and square roots. 6 a. "64x6 5 2 0 x 0 3 3 b. " x 5x

c. "9x 5 3x 8

Self Check 6

3 3 b. " x

Use absolute value symbols to guarantee that the result will be nonnegative. Because the index is odd, no absolute value symbols are needed.

4

Because 3x4 is always nonnegative, no absolute value symbols are needed.

Use the directions for Example 6: 4 a. " 16x4

3 b. " 27y3

4 8 c. " x

Now Try Exercise 73.

4. Simplify and Combine Radicals Many properties of exponents have counterparts in radical notation. For example, 1 /n since a1/nb1/n 5 1ab2 1/n and ab1/n 5 1ab2 1/n and 1b 2 02 , we have the following.

Multiplication and Division Properties of Radicals

"a

If all expressions represent real numbers, "a"b 5 "ab n

n

n

n

"b n

5

a Åb n

1b 2 02

In words, we say The product of two nth roots is equal to the nth root of their product. The quotient of two nth roots is equal to the nth root of their quotient. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

34

Chapter 0

A Review of Basic Algebra

Caution

These properties involve the nth root of the product of two numbers or the nth root of the quotient of two numbers. There is no such property for sums or differences. For example, "9 1 4 2 "9 1 "4, because "9 1 4 5 "13

"9 1 "4 5 3 1 2 5 5

but

and "13 2 5. In general,

"a 1 b 5 "a 1 "b

and

"a 2 b 5 "a 2 "b

Numbers that are squares of positive integers, such as 1, 4, 9, 16, 25, and 36 are called perfect squares. Expressions such as 4x2 and 19x6 are also perfect squares, because each one is the square of another expression with integer exponents and rational coefficients. 2 1 1 4x2 5 12x2 2 and x6 5 a x3 b 9 3

Numbers that are cubes of positive integers, such as 1, 8, 27, 64, 125, and 216 1 9 x are also perfect cubes, are called perfect cubes. Expressions such as 64x3 and 27 because each one is the cube of another expression with integer exponents and rational coefficients.

64x3 5 14x2 3 and

3 1 1 9 x 5 a x3 b 27 3

There are also perfect fourth powers, perfect fifth powers, and so on. We can use perfect powers and the Multiplication Property of Radicals to simplify many radical expressions. For example, to simplify "12x5, we factor 12x5 so that one factor is the largest perfect square that divides 12x5. In this case, it is 4x4. We then rewrite 12x5 as 4x4 ? 3x and simplify. "12x5 5 "4x4 ? 3x

5 "4x4"3x 5 2x2"3x

Factor 12x5 as 4x4 ? 3x.

Use the Multiplication Property of Radicals: "ab 5 "a"b. "4x4 5 2x2

3 To simplify " 432x9y, we find the largest perfect-cube factor of 432x9y (which is 9 216x ) and proceed as follows: 3 3 " 432x9y 5 " 216x9 ? 2y

3 3 216x9 " 2y 5"

5 6x3"2y 3

Factor 432x9y as 216x9 ? 2y.

3 3 3 Use the Multiplication Property of Radicals: " ab 5 " a" b. 3 " 216x9 5 6x3

Radical expressions with the same index and the same radicand are called like or similar radicals. We can combine the like radicals in 3"2 1 2"2 by using the Distributive Property. 3"2 1 2"2 5 13 1 22 "2 5 5"2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.3

Rational Exponents and Radicals

35

This example suggests that to combine like radicals, we add their numerical coefficients and keep the same radical. When radicals have the same index but different radicands, we can often change them to equivalent forms having the same radicand. We can then combine them. For example, to simplify "27 2 "12, we simplify both radicals and combine like radicals. "27 2 "12 5 "9 ? 3 2 "4 ? 3

5 "9"3 2 "4"3 5 3"3 2 2"3 5 "3

EXAMPLE 7 SOLUTION

Factor 27 and 12. "ab 5 "a"b

"9 5 3 and "4 5 2. Combine like radicals.

Adding and Subtracting Radical Expressions

Simplify: a. "50 1 "200

5 5 b. 3z" 64z 2 2" 2z6

We will simplify each radical expression and then combine like radicals. a. "50 1 "200 5 "25 ? 2 1 "100 ? 2

5 "25"2 1 "100"2 5 5"2 1 10"2 5 15"2

5 5 5 5 5 b. 3z" 64z 2 2" 2z6 5 3z" 32 ? 2z 2 2" z ? 2z

5 5 5 5 5 5 3z" 32 " 2z 2 2" z "2z 5 5 5 3z 122 " 2z 2 2z" 2z 5 5 5 6z" 2z 2 2z" 2z 5 5 4z" 2z

Self Check 7

Simplify: a. "18 2 "8

3 3 b. 2" 81a4 1 a" 24a

Now Try Exercise 85.

5. Rationalize Denominators and Numerators By rationalizing the denominator, we can write a fraction such as "5

"3

as a fraction with a rational number in the denominator. All that we must do is multiply both the numerator and the denominator by "3. (Note that "3"3 is the rational number 3.) "5

"3

5

"5"3

"3"3

5

"15 3

To rationalize the numerator, we multiply both the numerator and the denominator by "5. (Note that "5"5 is the rational number 5.) "5

"3

5

"5"5

"3"5

5

"15 5

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36

Chapter 0

A Review of Basic Algebra

EXAMPLE 8

Rationalizing the Denominator of a Radical Expression Rationalize each denominator and simplify. Assume that all variables represent positive numbers. a.

SOLUTION

"7 1

b.

3 Å4 3

c.

3 Åx

d.

3a3 Å 5x5

We will multiply both the numerator and the denominator by a radical that will make the denominator a rational number. a.

"7 1

5

5

"7"7 1"7

b.

"7 7

3 " 3 3 5 3 Å4 "4 3

5

3 3 3" 2 "

5

3 " 6

5

c.

3 "3 5 Åx "x 5

5

d.

"3"x

3 3 " 4" 2 3 " 8

3 " 6 2

3a3 "3a3 5 Å 5x5 "5x5

"x"x

5

"3x x

5

5

5

Self Check 8

Multiply numerator and 3 denominator by " 2, because 3 3 3 "4"2 5 "8 5 2.

"3a3"5x

"5x5"5x "15a3x

Multiply numerator and denominator by "5x, because "5x5 "5x 5 "25x6 5 5x3.

"25x6

"a2"15ax 5x3 a"15ax 5x3

Use the directions for Example 8: a.

"6 6

b.

3 2 Å 5x

Now Try Exercise 101.

EXAMPLE 9

Rationalizing the Numerator of a Radical Expression Rationalize each numerator and simplify. Assume that all variables represent posi3 "x 9x 2" b. tive numbers: a. 7 3

SOLUTION

We will multiply both the numerator and the denominator by a radical that will make the numerator a rational number.

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Section 0.3

a.

"x "x ? "x 5 7 7"x

b.

5

7"x 5

5

Self Check 9

"2x 5

37

3 3 3 2" 9x 9x ? " 3x2 2" 5 3 3 3"3x2

x

5

Rational Exponents and Radicals

3 2" 27x3 3 3" 3x2

2 13x2

3 3" 3x2

"3x2 2x

3

Divide out the 3’s.

Use the directions for Example 9: a.

b.

3 2y2 3" 6

Now Try Exercise 111. After rationalizing denominators, we often can simplify an expression.

EXAMPLE 10

Rationalizing Denominators and Simplifying Simplify:

SOLUTION

1 1 1 . Å2 Å8

We will rationalize the denominators of each radical and then combine like radicals. 1 1 1 "1 1 "1 5 5 ; 5 5 Å2 Å 8 "2 "2 "8 "8

1 1 1 1 1 5 1 Å2 Å8 "2 "8 5

5

5

5

Self Check 10

Simplify:

"2"2 1"2

1

"8"2 1"2

"2 "2 1 2 "16 "2 "2 1 2 4 3"2 4

x x 2 3 . Å2 Å 16 3

Now Try Exercise 115. Another property of radicals can be derived from the properties of exponents. If all of the expressions represent real numbers, #"x 5 "x1/m 5 1x1/m2 1/n 5 x1/ 1mn2 5 "x n m

n

mn

n #"x 5 "x1/n 5 1x1/n2 1/m 5 x1/1nm2 5 "x m

m

mn

These results are summarized in the following theorem (a fact that can be proved).

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38

Chapter 0

A Review of Basic Algebra

Theorem

If all of the expressions involved represent real numbers, then #"x 5 #"x 5 "x m

n

mn

n m

We can use the previous theorem to simplify many radicals. For example, 3 3 # "8 5 #" 8 5 "2

Rational exponents can be used to simplify many radical expressions, as shown in the following example.

EXAMPLE 11

Simplifying Radicals Using the Previously Stated Theorem Simplify. Assume that x and y are positive numbers. 6 a. "4

SOLUTION

b. "x3

9 c. " 8y3

12

In each case, we will write the radical as an exponential expression, simplify the resulting expression, and write the final result as a radical. 6 3 a. "4 5 41/6 5 1222 1/6 5 22/6 5 21/3 5 "2 4 b. "x3 5 x3/12 5 x1/4 5 " x 12

9 3 c. " 8y3 5 123y32 1/9 5 12y2 3/9 5 12y2 1/3 5 "2y

Self Check 11

4 Simplify: a. " 4

9 b. " 27x3

Now Try Exercise 117.

Self Check Answers

1. a. 10

b. 3

c. 9x2 b. 3y 9. a.

4. a.

2. a. 25

y 7

b. b1/7

2x 5"2x

c. 3s2

7. a. "2

c. x2 b.

3 " 4y

y

3. a. 343

b. 210

5. a. 6

10.

3 " 4x 4

b. 2

8. a. "6

3 b. 8a" 3a

11. a. "2

1 8

b. 1 2 b.

6. a. 2 0 x 0

3 50x2 " 5x

3 b. " 3x

Exercises 0.3 Getting Ready

3. If a , 0 and n is an even number, then a1/n is a real number.

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. If a 5 0 and n is a natural number, then a1/n 5 2. If a . 0 and n is a natural number, then a1/n is a number.

5. "a 5

4. 62/3 can be written as n

.

n

8.

9. "x 1 y 10.

"x 1 "y

n a 5 Åb

#"x or #"x can be written as m

n

.

2

7. "a"b 5 n

6. "a 5 or

n

m

.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.3

61. a

Practice Simplify each expression. 11. 91/2 1 1/2 13. a b 25

12. 81/3 16 1/4 14. a b 625

15. 2811/4

16. 2a

17. 110,0002 1/4 19. a2

27 1/3 b 8

21. 12642 1/2

8 1/3 b 27

18. 1,0241/5 20. 2641/3

22. 121252 1/3

Simplify each expression. Use absolute value symbols when necessary. 23. 116a22 1/2 24. 125a42 1/2

25. 116a 2 27. 1232a52 1/5 4 1/4

29. 12216b 2

6 1/3

31. a

16a b 25b2

33. a2

4

1/2

1,000x6 1/3 b 27y3

26. 1264a 2 28. 164a62 1/6

3 1/3

30. 1256t 2

8 1/4

32. 34. a

a2

a5 1/5 b 32b10

49t2 1/2 b 100z4

Simplify each expression. Write all answers without using negative exponents. 35. 43/2 37. 2163/2 39. 21,0002/3

36. 82/3 38. 1282 2/3 40. 1003/2

41. 6421/2

42. 2521/2

43. 6423/2

44. 4923/2

45. 2923/2

4 5/2 47. a b 9 49. a2

27 22/3 b 64

46. 12272 22/3 48. a 50. a

25 3/2 b 81

125 24/3 b 8

Simplify each expression. Assume that all variables represent positive numbers. Write all answers without using negative exponents. 51. 1100s42 1/2 52. 164u6v32 1/3 53. 132y10z52 21/5

55. 1x10y52 3/5

57. 1r8s162 23/4

8a6 2/3 59. a2 b 125b9

54. 1625a4b82 21/4

56. 164a6b122 5/6

58. 128x9y122 22/3

16x4 3/4 60. a b 625y8

63.

Rational Exponents and Radicals

27r6 22/3 b 1,000s12

62. a2

a2/5a4/5 a1/5

64.

Simplify each radical expression. 65. "49 3 67. " 125 3 69. "2125 32 71. 5 2 Å 100,000

39

32m10 22/5 b 243n15

x6/7x3/7 x2/7x5/7

66. "81 3 68. " 264 5 70. "2243 256 72. 4 Å 625

Simplify each expression, using absolute value symbols when necessary. Write answers without using negative exponents. 73. "36x2

75. "9y4 3 8y3 77. "

79.

x4y8 Å z12 4

76. "a4b8

74. 2"25y2 3 78. " 227z9

80.

a10b5 Å c15 5

Simplify each expression. Assume that all variables represent positive numbers so that no absolute value symbols are needed. 81. "8 2 "2 82. "75 2 2"27 83. "200x2 1 "98x2 84. "128a3 2 a"162a 85. 2"48y5 2 3y"12y3

86. y"112y 1 4"175y3

3 3 87. 2" 81 1 3" 24 4 4 5 89. "768z 1 "48z5

4 4 88. 3" 32 2 2" 162 5 5 2 90. 22"64y 1 3" 486y2

91. "8x2y 2 x"2y 1 "50x2y 92. 3x"18x 1 2"2x3 2 "72x3 3 3 3 93. " 16xy4 1 y" 2xy 2 " 54xy4 4 4 4 94. "512x5 2 "32x5 1 "1,250x5

Rationalize each denominator and simplify. Assume that all variables represent positive numbers. 95. 97. 99. 101. 103. 105.

"3 2 3

96.

"x 2

98.

3 " 2 5a

100.

3 " 25a

"3a 2b

4

2

2u4 Å 9v 3

102. 104. 106.

"5 8 6

"y 4d 3 " 9 7

3 " 36c

x Å 2y

3s5 Å 4r2 3

2

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40

Chapter 0

A Review of Basic Algebra

Rationalize each numerator and simplify. Assume that all variables are positive numbers. "5 "y 107. 108. 10 3 "9 3 3

109. 111.

"16b2 16 3

110.

5 16b3 " 64a

112.

3x Å 57

115.

1 1 2 Å 3 Å 27

114.

x x x 2 1 Å8 Å2 Å 32

116.

4 117. "9

1 3 1 1 3 Å 2 Å 16

y y y 1 3 2 3 Å 4 Å 32 Å 500 3

6 118. " 27

6 120. " 27x9

10

Discovery and Writing We often can multiply and divide radicals with different indi3 ces. For example, to multiply "3 by " 5, we first write each radical as a sixth root 6 3 6 "3 5 31/2 5 33/6 5 " 3 5" 27

and then multiply the sixth roots.

"3"5 5 "27"25 5 " 1272 1252 5 "675 6

6

6

123.

124.

"3

3 " 2

"2 "5 4 4 x 5 x? Explain. 125. For what values of x does " 126. If all of the radicals involved represent real numbers and y 2 0, explain why

x "x 5 n Åy "y 127. If all of the radicals involved represent real numbers and there is no division by 0, explain why m x 2m/n n y 5 a b y Å xm n 128. The definition of xm/n requires that "x be a real number. Explain why this is important. (Hint: Consider what happens when n is even, m is odd, and x is negative.)

Review

129. Write 22 , x # 5 using interval notation. 130. Write the expression 0 3 2 x 0 without using absolute value symbols. Assume that x . 4. Evaluate each expression when x 5 22 and y 5 3. xy 1 4y 132. 131. x2 2 y2 x 133. Write 617,000,000 in scientific notation.

3 6 2 6 " 5 5 51/3 5 52/6 5 " 5 5" 25

3

3 122. "3" 5

4

n

Simplify each radical expression. 119. "16x6

3 121. "2" 2

n

Rationalize each denominator and simplify. 113.

Use this idea to write each of the following expressions as a single radical.

134. Write 0.00235 3 104 in standard notation.

6

Division is similar.

0.4 Polynomials In this section, we will learn to 1. Define polynomials. 2. Add and subtract polynomials. 3. Multiply polynomials.

© PCN Photography/Alamy

4. Rationalize denominators. 5. Divide polynomials. Football is one of the most popular sports in the United States. Brett Favre, one of the most talented quarterbacks ever to play the game, holds the record for the most career NFL touchdown passes, the most NFL pass completions, and the most passing yards. He led the Green Bay Packers to the Super Bowl and most recently played for the Minnesota Vikings. His nickname is Gunslinger.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.4

Polynomials

41

An algebraic expression can be used to model the trajectory or path of a football when passed by Brett Favre. Suppose the height in feet, t seconds after the football leaves Brett’s hand, is given by the algebraic expression 20.1t2 1 t 1 5.5 At a time of t 5 3 seconds, we see that the height of the football is 20.1 132 2 1 3 1 5.5 5 7.6 ft.

Algebraic expressions like 20.1t2 1 t 1 5.5 are called polynomials, and we will study them in this section.

1. Define Polynomials A monomial is a real number or the product of a real number and one or more variables with whole-number exponents. The number is called the coefficient of the variables. Some examples of monomials are 3x,

7ab2,

25ab2c4,

x3,

and

212

with coefficients of 3, 7, 25, 1, and 212, respectively. The degree of a monomial is the sum of the exponents of its variables. All nonzero constants (except 0) have a degree of 0. The degree of 3x is 1.

The degree of 7ab2 is 3.

The degree of 25ab2c4 is 7.

The degree of x3 is 3.

The degree of 212 is 0 (since 212 5 212x0).

0 has no defined degree.

A monomial or a sum of monomials is called a polynomial. Each monomial in that sum is called a term of the polynomial. A polynomial with two terms is called a binomial, and a polynomial with three terms is called a trinomial. Monomials

Binomials

Trinomials

2a 1 3b

x2 1 7x 2 4

225xy

4x3 2 3x2

4y4 2 2y 1 12

a 2b 3c

22x3 2 4y2

12x3y2 2 8xy 2 24

3x

2

The degree of a polynomial is the degree of the term in the polynomial with highest degree. The only polynomial with no defined degree is 0, which is called the zero polynomial. Here are some examples. • • • •

3x2y3 1 5xy2 1 7 is a trinomial of 5th degree, because its term with highest degree (the first term) is 5. 3ab 1 5a2b is a binomial of degree 3.

4 5x 1 3y2 1 " 3z4 2 "7 is a polynomial, because its variables have wholenumber exponents. It is of degree 4. 5 27y1/2 1 3y2 1 " 3z is not a polynomial, because one of its variables (y in the first term) does not have a whole-number exponent.

If two terms of a polynomial have the same variables with the same exponents, they are like or similar terms. To combine the like terms in the sum 3x2y 1 5x2y or the difference 7xy2 2 2xy2, we use the Distributive Property: 3x2y 1 5x2y 5 13 1 52 x2y 5 8x2y

7xy2 2 2xy2 5 17 2 22 xy2 5 5xy2

This illustrates that to combine like terms, we add (or subtract) their coefficients and keep the same variables and the same exponents. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

42

Chapter 0

A Review of Basic Algebra

2. Add and Subtract Polynomials Recall that we can use the Distributive Property to remove parentheses enclosing the terms of a polynomial. When the sign preceding the parentheses is +, we simply drop the parentheses: 1 1a 1 b 2 c2 5 11 1a 1 b 2 c2 5 1a 1 1b 2 1c 5a1b2c When the sign preceding the parentheses is 2, we drop the parentheses and the 2 sign and change the sign of each term within the parentheses. 2 1a 1 b 2 c2 5 21 1a 1 b 2 c2

5 21a 1 1212 b 2 1212 c 5 2a 2 b 1 c

We can use these facts to add and subtract polynomials. To add (or subtract) polynomials, we remove parentheses (if necessary) and combine like terms.

EXAMPLE 1 SOLUTION

Adding Polynomials

Add: 13x3y 1 5x2 2 2y2 1 12x3y 2 5x2 1 3x2 . To add the polynomials, we remove parentheses and combine like terms. 13x3y 1 5x2 2 2y2 1 12x3y 2 5x2 1 3x2

5 3x3y 1 5x2 2 2y 1 2x3y 2 5x2 1 3x 5 3x3y 1 2x3y 1 5x2 2 5x2 2 2y 1 3x

Use the Commutative Property to rearrange terms.

5 5x3y 2 2y 1 3x

Combine like terms.

We can add the polynomials in a vertical format by writing like terms in a column and adding the like terms, column by column. 3x3y 1 5x2 2 2y 2x3y 2 5x2 1 3x 3 5x y 2 2y 1 3x Self Check 1

Add: 14x2 1 3x 2 52 1 13x2 2 5x 1 72 . Now Try Exercise 21.

EXAMPLE 2 SOLUTION

Subtracting Polynomials

Subtract: 12x2 1 3y22 2 1x2 2 2y2 1 72 . To subtract the polynomials, we remove parentheses and combine like terms. 12x2 1 3y22 2 1x2 2 2y2 1 72

5 2x2 1 3y2 2 x2 1 2y2 2 7 5 2x2 2 x2 1 3y2 1 2y2 2 7

Use the Commutative Property to rearrange terms.

2

2

5 x 1 5y 2 7

Combine like terms.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.4

Polynomials

43

We can subtract the polynomials in a vertical format by writing like terms in a column and subtracting the like terms, column by column. 2x2 1 3y2

2 1x2 2 2y2 1 72 x2 1 5y2 2 7 Self Check 2

2x2 2 x2 5 12 2 12 x2

3y2 2 1222 y2 5 3y2 1 2y2 5 13 1 22 y2 0 2 7 5 27

Subtract: 14x2 1 3x 2 52 2 13x2 2 5x 1 72 . Now Try Exercise 23. We can also use the Distributive Property to remove parentheses enclosing several terms that are multiplied by a constant. For example, 4 13x2 2 2x 1 62 5 4 13x22 2 4 12x2 1 4 162 5 12x2 2 8x 1 24 This example suggests that to add multiples of one polynomial to another, or to subtract multiples of one polynomial from another, we remove parentheses and combine like terms.

EXAMPLE 3 SOLUTION

Using the Distributive Property and Combining Like Terms

Simplify: 7x 12y2 1 13x22 2 5 1xy2 2 13x32 . 7x 12y2 1 13x22 2 5 1xy2 2 13x32

5 14xy2 1 91x3 2 5xy2 1 65x3 2

2

3

3

5 14xy 2 5xy 1 91x 1 65x 2

3

5 9xy 1 156x Self Check 3

Use the Distributive Property to remove parentheses. Use the Commutative Property to rearrange terms. Combine like terms.

Simplify: 3 12b2 2 3a2b2 1 2b 1b 1 a22 . Now Try Exercise 30.

3. Multiply Polynomials To find the product of 3x2y3z and 5xyz2, we proceed as follows: 13x2y3z2 15xyz22 5 3 ? x2 ? y3 ? z ? 5 ? x ? y ? z2 5 3 ? 5 ? x2 ? x ? y3 ? y ? z ? z2

Use the Commutative Property to rearrange terms.

3 4 3

5 15x y z

This illustrates that to multiply two monomials, we multiply the coefficients and then multiply the variables. To find the product of a monomial and a polynomial, we use the Distributive Property. 3xy2 12xy 1 x2 2 7yz2 5 3xy2 12xy2 1 13xy22 1x22 2 13xy22 17yz2 5 6x2y3 1 3x3y2 2 21xy3z This illustrates that to multiply a polynomial by a monomial, we multiply each term of the polynomial by the monomial. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

44

Chapter 0

A Review of Basic Algebra

To multiply one binomial by another, we use the Distributive Property twice.

EXAMPLE 4

SOLUTION

Multiplying Binomials

Multiply: a. 1x 1 y2 1x 1 y2

b. 1x 2 y2 1x 2 y2

c. 1x 1 y2 1x 2 y2

a. 1x 1 y2 1x 1 y2 5 1x 1 y2 x 1 1x 1 y2 y 5 x2 1 xy 1 xy 1 y2 5 x2 1 2xy 1 y2 b. 1x 2 y2 1x 2 y2 5 1x 2 y2 x 2 1x 2 y2 y 5 x2 2 xy 2 xy 1 y2 5 x2 2 2xy 1 y2 c. 1x 1 y2 1x 2 y2 5 1x 1 y2 x 2 1x 1 y2 y 5 x2 1 xy 2 xy 2 y2 5 x2 2 y2

Self Check 4

Multiply: a. 1x 1 22 1x 1 22 c. 1x 1 42 1x 2 42

b. 1x 2 32 1x 2 32

Now Try Exercise 45.

The products in Example 4 are called special products. Because they occur so often, it is worthwhile to learn their forms.

Special Product Formulas

1x 1 y2 2 5 1x 1 y2 1x 1 y2 5 x2 1 2xy 1 y2

1x 2 y2 2 5 1x 2 y2 1x 2 y2 5 x2 2 2xy 1 y2 1x 1 y2 1x 2 y2 5 x2 2 y2

Caution

Remember that 1x 1 y2 2 and 1x 2 y2 2 have trinomials for their products and that 1x 1 y2 2 2 x2 1 y2

For example,

13 1 52 2 2 32 1 52 82 2 9 1 25 64 2 34

and

and

1x 2 y2 2 2 x2 2 y2

13 2 52 2 2 32 2 52 1222 2 2 9 2 25 4 2 216

We can use the FOIL method to multiply one binomial by another. FOIL is an acronym for First terms, Outer terms, Inner terms, and Last terms. To use this method to multiply 3x 2 4 by 2x 1 5, we write Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.4

First terms

Polynomials

45

Last terms

13x 2 42 12x 1 52 5 3x 12x2 1 3x 152 2 4 12x2 2 4 152 Inner terms

5 6x2 1 15x 2 8x 2 20

Outer terms

5 6x2 1 7x 2 20

In this example, • • • •

the product of the product of the product of the product of

the first terms is 6x2, the outer terms is 15x, the inner terms is 28x, and the last terms is 220.

The resulting like terms of the product are then combined.

EXAMPLE 5

SOLUTION

Using the FOIL Method to Multiply Polynomials

Use the FOIL method to multiply: Q"3 1 xRQ2 2 "3xR. 1"3 1 x2 12 2 "3x2 5 2"3 2 "3"3x 1 2x 2 x"3x 5 2"3 2 3x 1 2x 2 "3x2 5 2"3 2 x 2 "3x2

Self Check 5

Multiply: Q2x 1 "3RQx 2 "3R. Now Try Exercise 63.

To multiply a polynomial with more than two terms by another polynomial, we multiply each term of one polynomial by each term of the other polynomial and combine like terms whenever possible.

EXAMPLE 6

SOLUTION

Multiplying Polynomials

Multiply: a. 1x 1 y2 1x2 2 xy 1 y22

b. 1x 1 32 3

a. 1x 1 y2 1x2 2 xy 1 y22 5 x3 2 x2y 1 xy2 1 yx2 2 xy2 1 y3 5 x3 1 y3

b. 1x 1 32 3 5 1x 1 32 1x 1 32 2 5 1x 1 32 1x2 1 6x 1 92

5 x3 1 6x2 1 9x 1 3x2 1 18x 1 27 5 x3 1 9x2 1 27x 1 27

Self Check 6

Multiply: 1x 1 22 12x2 1 3x 2 12 . Now Try Exercise 67. We can use a vertical format to multiply two polynomials, such as the polynomials given in Self Check 6. We first write the polynomials as follows and draw a line beneath them. We then multiply each term of the upper polynomial by each term of the lower polynomial and write the results so that like terms appear in each column. Finally, we combine like terms column by column.

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46

Chapter 0

A Review of Basic Algebra

2x2 1 3x 2 1 x12 4x2 1 6x 2 2 2x 1 3x2 2 x

Multiply 2x2 1 3x 2 1 by 2.

2x3 1 7x2 1 5x 2 2

In each column, combine like terms.

3

Multiply 2x2 1 3x 2 1 by x.

If n is a whole number, the expressions an 1 1 and 2an 2 3 are polynomials and we can multiply them as follows: 1an 1 12 12an 2 32 5 2a2n 2 3an 1 2an 2 3 5 2a2n 2 an 2 3

Combine like terms.

We can also use the methods previously discussed to multiply expressions that are not polynomials, such as x22 1 y and x2 2 y21. 1x22 1 y2 1x2 2 y212 5 x2212 2 x22y21 1 x2y 2 y121 1 1 x2y 2 y0 x2y 1 5 1 2 2 1 x2y 2 1 xy

5 x0 2

5 x2y 2

x0 5 1 and y0 5 1.

1 x2y

4. Rationalize Denominators If the denominator of a fraction is a binomial containing square roots, we can use the product formula 1x 1 y2 1x 2 y2 to rationalize the denominator. For example, to rationalize the denominator of "7 1 2 6

To rationalize a denominator means to change the denominator into a rational number.

we multiply the numerator and denominator by "7 2 2 and simplify. "7 1 2 6

5

6 1"7 2 22

1"7 1 22 1"7 2 22

6 1"7 2 22 724 6 1"7 2 22 5 3

"7 2 2 "7 2 2

51

5

5 2 1"7 2 22

Here the denominator is a rational number.

In this example, we multiplied both the numerator and the denominator of the given fraction by "7 2 2. This binomial is the same as the denominator of the given fraction "7 1 2, except for the sign between the terms. Such binomials are called conjugate binomials or radical conjugates.

Conjugate Binomials

Conjugate binomials are binomials that are the same except for the sign between their terms. The conjugate of a 1 b is a 2 b, and the conjugate of a 2 b is a 1 b.

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Section 0.4

EXAMPLE 7

Rationalizing the Denominator of a Radical Expression Rationalize the denominator:

SOLUTION

"3x 2 "2

"3x 1 "2

1x . 02 .

We multiply the numerator and the denominator by "3x 2 "2 (the conjugate of "3x 1 "22 and simplify. "3x 2 "2 "3x 1 "2

5

Q"3x 2 "2RQ"3x 2 "2R

5

"3x"3x 2 "3x"2 2 "2"3x 1 "2"2

5

5

Self Check 7

47

Polynomials

"3x 2 "2

Q"3x 1 "2RQ"3x 2 "2R

"3x 2 "2

51

Q"3xR 2 Q"2R 2

2

3x 2 "6x 2 "6x 1 2 3x 2 2 3x 2 2"6x 1 2 3x 2 2

Rationalize the denominator: Now Try Exercise 89.

"x 1 2 "x 2 2

.

In calculus, we often rationalize a numerator.

EXAMPLE 8

Rationalizing the Numerator of a Radical Expression Rationalize the numerator:

SOLUTION

"x 1 h 2 "x . h

To rid the numerator of radicals, we multiply the numerator and the denominator by the conjugate of the numerator and simplify. Q"x 1 h 2 "xRQ"x 1 h 1 "xR "x 1 h 2 "x 5 h hQ"x 1 h 1 "xR 5

5

5

Self Check 8

x1h2x

hQ"x 1 h 1 "xR

"x 1 h 1 "x

"x 1 h 1 "x

Here the numerator has no radicals.

h

hQ"x 1 h 1 "xR

"x 1 h 1 "x

Rationalize the numerator:

1

Divide out the common factor of h.

"4 1 h 2 2 . h

Now Try Exercise 99. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

51

48

Chapter 0

A Review of Basic Algebra

5. Divide Polynomials To divide monomials, we write the quotient as a fraction and simplify by using the rules of exponents. For example, 6x2y3 5 23x223y321 22x3y 5 23x21y2 3y2 x To divide a polynomial by a monomial, we write the quotient as a fraction, write the fraction as a sum of separate fractions, and simplify each one. For example, to divide 8x5y4 1 12x2y5 2 16x2y3 by 4x3y4, we proceed as follows: 52

8x5y4 12x2y5 216x2y3 8x5y4 1 12x2y5 2 16x2y3 5 3 41 3 4 3 4 1 4x y 4x y 4x y 4x3y4 5 2x2 1

3y 4 2 x xy

To divide two polynomials, we can use long division. To illustrate, we consider the division 2x2 1 11x 2 30 x17 which can be written in long division form as x 1 7q2x2 1 11x 2 30 The binomial x 1 7 is called the divisor, and the trinomial 2x2 1 11x 2 30 is called the dividend. The final answer, called the quotient, will appear above the long division symbol. We begin the division by asking “What monomial, when multiplied by x, gives 2x2?” Because x ? 2x 5 2x2, the answer is 2x. We place 2x in the quotient, multiply each term of the divisor by 2x, subtract, and bring down the –30. 2x x 1 7q2x 1 11x 2 30 2x2 1 14x 2 3x 2 30 2

We continue the division by asking “What monomial, when multiplied by x, gives 23x?” We place the answer, 23, in the quotient, multiply each term of the divisor by 23, and subtract. This time, there is no number to bring down. 2x 2 3 x 1 7q2x2 1 11x 2 30 2x2 1 14x 2 3x 2 30 2 3x 2 21 29 Because the degree of the remainder, 29, is less than the degree of the divisor, the division process stops, and we can express the result in the form quotient 1

remainder divisor

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Section 0.4

Polynomials

49

Thus, 2x2 1 11x 2 30 29 5 2x 2 3 1 x17 x17

EXAMPLE 9

Using Long Division to Divide Polynomials Divide 6x3 2 11 by 2x 1 2.

SOLUTION

Comment

We set up the division, leaving spaces for the missing powers of x in the dividend. 2x 1 2q6x3 2 11

In Example 9, we could write the missing powers of x using coefficients of 0. 2x 1 2q6x3 1 0x2 1 0x 2 11

Self Check 9

The division process continues as usual, with the following results: 3x2 2 3x 1 3 2 11 2x 1 2q6x 3 2 6x 1 6x 2 6x2 2 6x2 2 6x 1 6x 2 11 1 6x 1 6 2 17 6x3 2 11 217 Thus, 5 3x2 2 3x 1 3 1 . 2x 1 2 2x 1 2 3

Divide: 3x 1 1q9x2 2 1. Now Try Exercise 113.

EXAMPLE 10

Using Long Division to Divide Polynomials Divide 23x3 2 3 1 x5 1 4x2 2 x4 by x2 2 3.

SOLUTION

The division process works best when the terms in the divisor and dividend are written with their exponents in descending order. x3 2 x2 1 1 x2 2 3qx5 2 x4 2 3x3 1 4x2 2 3 x5 2 3x3 2 x4 1 4x2 4 2x 1 3x2 x2 2 3 x2 2 3 0 Thus,

Self Check 10

23x3 2 3 1 x5 1 4x2 2 x4 5 x3 2 x2 1 1. x2 2 3

Divide: x2 1 1q3x2 2 x 1 1 2 2x3 1 3x4. Now Try Exercise 115.

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50

Chapter 0

A Review of Basic Algebra

Self Check Answers

1. 7x2 2 2x 1 2 4. a. x2 1 4x 1 4

2. x2 1 8x 2 12 b. x2 2 6x 1 9

x 1 4"x 1 4 x24 x 1 1 10. 3x2 2 2x 1 2 x 11

6. 2x3 1 7x2 1 5x 2 2 9. 3x 2 1

3. 8b2 2 7a2b c. x2 2 16

7.

8.

5. 2x2 2 x"3 2 3

"4 1 h 1 2 1

Exercises 0.4 Getting Ready

Practice

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. A is a real number or the product of a real number and one or more . 2. The of a monomial is the sum of the exponents of its . 3. A is a polynomial with three terms. 4. A is a polynomial with two terms. 5. A monomial is a polynomial with term. 6. The constant 0 is called the polynomial. 7. Terms with the same variables with the same exponents are called terms. 8. The of a polynomial is the same as the degree of its term of highest degree. 9. To combine like terms, we add their and keep the same and the same exponents. 10. The conjugate of 3"x 1 2 is

.

Determine whether the given expression is a polynomial. If so, tell whether it is a monomial, a binomial, or a trinomial, and give its degree. 11. x2 1 3x 1 4 12. 5xy 2 x3 3

1/2

13. x 1 y

14. x23 2 5y22

15. 4x2 2 "5x3 16. x2y3

17. "15 5 x 18. 1 1 5 x 5 19. 0 20. 3y3 2 4y2 1 2y 1 2

Perform the operations and simplify. 21. 1x3 2 3x22 1 15x3 2 8x2

22. 12x4 2 5x32 1 17x3 2 x4 1 2x2 23. 1y5 1 2y3 1 72 2 1y5 2 2y3 2 72 24. 13t7 2 7t3 1 32 2 17t7 2 3t3 1 72

25. 2 1x2 1 3x 2 12 2 3 1x2 1 2x 2 42 1 4 26. 5 1x3 2 8x 1 32 1 2 13x2 1 5x2 2 7

27. 8 1t2 2 2t 1 52 1 4 1t2 2 3t 1 22 2 6 12t2 2 82 28. 23 1x3 2 x2 1 2 1x2 1 x2 1 3 1x3 2 2x2 29. y 1y2 2 12 2 y2 1y 1 22 2 y 12y 2 22

30. 24a2 1a 1 12 1 3a 1a2 2 42 2 a2 1a 1 22

31. xy 1x 2 4y2 2 y 1x2 1 3xy2 1 xy 12x 1 3y2

32. 3mn 1m 1 2n2 2 6m 13mn 1 12 2 2n 14mn 2 12 33. 2x2y3 14xy42 34. 215a3b 122a2b32

35. 23m2n 12mn22 a2

mn b 12 3r2s3 2r2s 15rs2 36. 2 a ba b 5 3 2

37. 24rs 1r2 1 s22 38. 6u2v 12uv2 2 y2

39. 6ab2c 12ac 1 3bc2 2 4ab2c2 mn2 14mn 2 6m2 2 82 2 41. 1a 1 22 1a 1 22

42. 1y 2 52 1y 2 52

45. 1x 1 42 1x 2 42

46. 1z 1 72 1z 2 72

40. 2

43. 1a 2 62 2

44. 1t 1 92 2

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Section 0.4

47. 1x 2 32 1x 1 52

48. 1z 1 42 1z 2 62

49. 1u 1 22 13u 2 22

50. 14x 1 12 12x 2 32

53. 13a 2 2b2 2

54. 14a 1 5b2 14a 2 5b2

57. 12y 2 4x2 13y 2 2x2

58. 122x 1 3y2 13x 1 y2

51. 15x 2 12 12x 1 32 55. 13m 1 4n2 13m 2 4n2 59. 19x 2 y2 1x2 2 3y2

52. 14x 2 12 12x 2 72

91.

"2 2 "3

92.

"3 2 "2

93.

"x 2 "y

94.

"2x 1 y

56. 14r 1 3s2 2

60. 18a2 1 b2 1a 1 2b2

95.

62. 1y 2 2x22 1x2 1 3y2

97.

65. 13x 2 12 3

66. 12x 2 32 3

99.

67. 13x 1 12 12x2 1 4x 2 32 68. 12x 2 52 1x2 2 3x 1 22

64. Q"2 1 xRQ3 1 "2xR

69. 13x 1 2y2 12x2 2 3xy 1 4y22 70. 14r 2 3s2 12r2 1 4rs 2 2s22 Multiply the expressions as you would multiply polynomials. 71. 2yn 13yn 1 y2n2

72. 3a2n 12an 1 3an212

73. 25x2nyn 12x2ny2n 1 3x22nyn2 74. 22a3nb2n 15a23nb 2 ab22n2 75. 1xn 1 32 1xn 2 42 76. 1an 2 52 1an 2 32

96.

y 2 "3 y 1 "3

"x 1 3 2 "x 3

98.

100.

"x 2 3 3

"a 2 "b

"a 1 "b "2 1 h 2 "2 h

Perform each division and write all answers without using negative exponents. 36a2b3 245r2s5t3 101. 102. 6 18ab 27r6s2t8 103.

16x6y4z9 224x9y6z0

104.

32m6n4p2 26m6n7p2

105.

5x3y2 1 15x3y4 10x2y3

106.

9m4n9 2 6m3n4 12m3n3

Perform each division. If there is a nonzero remainder, write the answer in quotient 1 remainder divisor form. 2 109. x 1 3q3x 1 11x 1 6 110. 3x 1 2q3x2 1 11x 1 6

82. 1x3/2 1 y1/22 2

111. 2x 2 5q2x2 2 19x 1 37

Rationalize each denominator. 2 1 84. 83. "3 2 1 "5 1 2 3x 14y 85. 86. "7 1 2 "2 2 3 x y 87. 88. x 2 "3 2y 1 "7 y 2 "2

"2 1 1 2

"2x 2 y

24x5y7 2 36x2y5 1 12xy 60x5y4 9a3b4 1 27a2b4 2 18a2b3 108. 18a2b7

79. x1/2 1x1/2y 1 xy1/22 80. ab1/2 1a1/2b1/2 1 b1/22 81. 1a1/2 1 b1/22 1a1/2 2 b1/22

y 1 "2

"x 1 "y

90.

x 2 "3

x 1 "3

51

1 1 "2

107.

77. 12rn 2 72 13rn 2 22 78. 14zn 1 32 13zn 1 12

89.

1 2 "3

Rationalize each numerator.

61. 15z 1 2t2 1z2 2 t2

63. Q"5 1 3xRQ2 2 "5xR

Polynomials

112. x 2 7q2x2 2 19x 1 35 2x3 1 1 113. x21 2x3 2 9x2 1 13x 2 20 114. 2x 2 7 115. x2 1 x 2 1qx3 2 2x2 2 4x 1 3 116. x2 2 3qx3 2 2x2 2 4x 1 5 117.

x5 2 2x3 2 3x2 1 9 x3 2 2

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52

Chapter 0

A Review of Basic Algebra

x5 2 2x3 2 3x2 1 9 x3 2 3 x5 2 32 119. x22 x4 2 1 120. x11

126. Travel Complete the following table, which shows the rate (mph), time traveled (hr), and distance traveled (mi) by a family on vacation.

121. 11x 2 10 1 6x2 q36x4 2 121x2 1 120 1 72x3 2 142x

Discovery and Writing

118.

r

?

t

3x 1 4

122. x 1 6x2 2 12q2121x2 1 72x3 2 142x 1 120 1 36x4

Applications 123. Geometry Find an expression that represents the area of the brick wall.

5

d 3x2 1 19x 1 20

127. Show that a trinomial can be squared by using the formula 1a 1 b 1 c2 2 5 a2 1 b2 1 c2 1 2ab 1 2bc 1 2ac. 128. Show that 1a 1 b 1 c 1 d2 2 5 a2 1 b2 1 c2 1 d 2 1 2ab 1 2ac 1 2ad 1 2bc 1 2bd 1 2cd . 129. Explain the FOIL method. 130. Explain how to rationalize the numerator of "xx1 2 . 131. Explain why 1a 1 b2 2 2 a2 1 b2. 132. Explain why "a2 1 b2 2 "a2 1 "b2.

(x – 2) ft

Review (x + 5) ft

124. Geometry The area of the triangle shown in the illustration is represented as 1x2 1 3x 2 402 square feet. Find an expression that represents its height.

Simplify each expression. Assume that all variables represent positive numbers. 8 22/3 133. 93/2 134. a b 125 4 3/4 625x 135. a 136. "80x4 b 16y8 3 3 137. "16ab4 2 b"54ab

4 4 138. x" 1,280x 1 " 80x5

Height

(x + 8) ft

125. Gift Boxes The corners of a 12 in.-by-12 in. piece of cardboard are folded inward and glued to make a box. Write a polynomial that represents the volume of the resulting box. Crease here and fold inward x

x

x

x

12 in.

x

x x

x 12 in.

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Section 0.5

Factoring Polynomials

53

0.5 Factoring Polynomials In this section, we will learn to 1. 2. 3. 4.

Factor Factor Factor Factor

out a common monomial. by grouping. the difference of two squares. trinomials.

© Peter Coombs/Alamy

5. Factor trinomials by grouping. 6. Factor the sum and difference of two cubes. 7. Factor miscellaneous polynomials. The television network series CSI: Crime Scene Investigation and its spinoffs, CSI: NY and CSI: Miami, are favorite television shows of many people. Their fans enjoy watching a team of forensic scientists uncover the circumstances that led to an unusual death or crime. These mysteries are captivating because many viewers are interested in the field of forensic science. In this section, we will investigate a mathematics mystery. We will be given a polynomial and asked to unveil or uncover the two or more polynomials that were multiplied together to obtain the given polynomial. The process that we will use to solve this mystery is called factoring. When two or more polynomials are multiplied together, each one is called a factor of the resulting product. For example, the factors of 7 1x 1 22 1x 1 32 are 7,

x 1 2,

and

x13

The process of writing a polynomial as the product of several factors is called factoring. In this section, we will discuss factoring where the coefficients of the polynomial and the polynomial factors are integers. If a polynomial cannot be factored by using integers only, we call it a prime polynomial.

1. Factor Out a Common Monomial The simplest type of factoring occurs when we see a common monomial factor in each term of the polynomial. In this case, our strategy is to use the Distributive Property and factor out the greatest common factor.

EXAMPLE 1

Factoring by Removing a Common Monomial Factor: 3xy2 1 6x.

SOLUTION

We note that each term contains a greatest common factor of 3x: 3xy2 1 6x 5 3x 1y22 1 3x 122

We can then use the Distributive Property to factor out the common factor of 3x: 3xy2 1 6x 5 3x 1y2 1 22

Self Check 1

We can check by multiplying: 3x 1y2 1 22 5 3xy2 1 6x.

Factor: 4a2 2 8ab. Now Try Exercise 11.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

54

Chapter 0

A Review of Basic Algebra

EXAMPLE 2

Factoring by Removing a Common Monomial Factor: x2y2z2 2 xyz.

SOLUTION

We factor out the greatest common factor of xyz: x2y2z2 2 xyz 5 xyz 1xyz2 2 xyz 112 5 xyz 1xyz 2 12

We can check by multiplying.

The last term in the expression x2y2z2 2 xyz has an understood coefficient of 1. When the xyz is factored out, the 1 must be written. Self Check 2

Factor: a2b2c2 1 a3b3c3. Now Try Exercise 13.

2. Factor by Grouping A strategy that is often helpful when factoring a polynomial with four or more terms is called grouping. Terms with common factors are grouped together and then their greatest common factors are factored out using the Distributive Property.

EXAMPLE 3

Factoring by Grouping Factor: ax 1 bx 1 a 1 b.

SOLUTION

Although there is no factor common to all four terms, we can factor x out of the first two terms and write the expression as ax 1 bx 1 a 1 b 5 x 1a 1 b2 1 1a 1 b2

We can now factor out the common factor of a 1 b. ax 1 bx 1 a 1 b 5 x 1a 1 b2 1 1a 1 b2

5 x 1a 1 b2 1 1 1a 1 b2 5 1a 1 b2 1x 1 12

Self Check 3

We can check by multiplying.

Factor: x2 1 xy 1 2x 1 2y. Now Try Exercise 17.

3. Factor the Difference of Two Squares A binomial that is the difference of the squares of two quantities factors easily. The strategy used is to write the polynomial as the product of two factors. The first factor is the sum of the quantities and the other factor is the difference of the quantities.

EXAMPLE 4

Factoring the Difference of Two Squares Factor: 49x2 2 4.

SOLUTION

We observe that each term is a perfect square: 49x2 2 4 5 17x2 2 2 22

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Section 0.5

Factoring Polynomials

55

The difference of the squares of two quantities is the product of two factors. One is the sum of the quantities, and the other is the difference of the quantities. Thus, 49x2 2 4 factors as 49x2 2 4 5 17x2 2 2 22

5 17x 1 22 17x 2 22

Self Check 4

We can check by multiplying.

Factor: 9a2 2 16b2. Now Try Exercise 21. Example 4 suggests a formula for factoring the difference of two squares.

Factoring the Difference of Two Squares

EXAMPLE 5

x2 2 y2 5 1x 1 y2 1x 2 y2

Factoring the Difference of Two Squares Twice in One Problem Factor: 16m4 2 n4.

SOLUTION

Caution If you are limited to integer coefficients, the sum of two squares cannot be factored. For example, x2 1 y2 is a prime polynomial.

Self Check 5

The binomial 16m4 2 n4 can be factored as the difference of two squares: 16m4 2 n4 5 14m22 2 2 1n22 2

5 14m2 1 n22 14m2 2 n22

The first factor is the sum of two squares and is prime. The second factor is a difference of two squares and can be factored: 16m4 2 n4 5 14m2 1 n22 3 12m2 2 2 n2 4

5 14m2 1 n22 12m 1 n2 12m 2 n2

We can check by multiplying.

Factor: a4 2 81b4. Now Try Exercise 23.

EXAMPLE 6

Removing a Common Factor and Factoring the Difference of Two Squares Factor: 18t2 2 32.

SOLUTION

We begin by factoring out the common monomial factor of 2. 18t2 2 32 5 2 19t2 2 162

Since 9t2 2 16 is the difference of two squares, it can be factored. 18t2 2 32 5 2 19t2 2 162

5 2 13t 1 42 13t 2 42

Self Check 6

We can check by multiplying.

Factor: 23x2 1 12. Now Try Exercise 61.

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56

Chapter 0

A Review of Basic Algebra

4. Factor Trinomials Trinomials that are squares of binomials can be factored by using the following formulas.

Factoring Trinomial Squares

(1) x2 1 2xy 1 y2 5 1x 1 y2 1x 1 y2 5 1x 1 y2 2 (2) x2 2 2xy 1 y2 5 1x 2 y2 1x 2 y2 5 1x 2 y2 2

For example, to factor a2 2 6a 1 9, we note that it can be written in the form a2 2 2 13a2 1 32

x 5 a and y 5 3

which matches the left side of Equation 2 above. Thus, a2 2 6a 1 9 5 a2 2 2 13a2 1 32 5 1a 2 32 1a 2 32 5 1a 2 32 2

We can check by multiplying.

Factoring trinomials that are not squares of binomials often requires some guesswork. If a trinomial with no common factors is factorable, it will factor into the product of two binomials.

EXAMPLE 7

Factoring a Trinomial with Leading Coefficient of 1 Factor: x2 1 3x 2 10.

SOLUTION

To factor x2 1 3x 2 10, we must find two binomials x 1 a and x 1 b such that x2 1 3x 2 10 5 1x 1 a2 1x 1 b2

where the product of a and b is –10 and the sum of a and b is 3. ab 5 210

and

a1b53

To find such numbers, we list the possible factorizations of –10: 10 1212

5 1222

210 112

25 122

Only in the factorization 5 1222 do the factors have a sum of 3. Thus, a 5 5 and b 5 22, and x2 1 3x 2 10 5 1x 1 a2 1x 1 b2

(3)

x2 1 3x 2 10 5 1x 1 52 1x 2 22

We can check by multiplying.

Because of the Commutative Property of Multiplication, the order of the factors in Equation 3 is not important. Equation 3 can also be written as x2 1 3x 2 10 5 1x 2 22 1x 1 52

Self Check 7

Factor: p2 2 5p 2 6. Now Try Exercise 29.

EXAMPLE 8

Factoring a Trinomial with Leading Coefficient not 1 Factor: 2x2 2 x 2 6.

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Section 0.5

SOLUTION

Factoring Polynomials

57

Since the first term is 2x2, the first terms of the binomial factors must be 2x and x: 2x2 2 x 2 6 5 12x

2 1x

2

The product of the last terms must be 26, and the sum of the products of the outer terms and the inner terms must be 2x. Since the only factorization of 26 that will cause this to happen is 3 1222 , we have 2x2 2 x 2 6 5 12x 1 32 1x 2 22

Self Check 8

We can check by multiplying.

Factor: 6x2 2 x 2 2. Now Try Exercise 33. It is not easy to give specific rules for factoring trinomials, because some guesswork is often necessary. However, the following hints are helpful.

Strategy for Factoring a General Trinomial with Integer Coefficients

EXAMPLE 9

1. Write the trinomial in descending powers of one variable. 2. Factor out any greatest common factor, including 21 if that is necessary to make the coefficient of the first term positive. 3. When the sign of the first term of a trinomial is 1 and the sign of the third term is 1, the sign between the terms of each binomial factor is the same as the sign of the middle term of the trinomial. When the sign of the first term is 1 and the sign of the third term is 2, one of the signs between the terms of the binomial factors is 1 and the other is 2. 4. Try various combinations of first terms and last terms until you find one that works. If no possibilities work, the trinomial is prime. 5. Check the factorization by multiplication.

Factoring a Trinomial Completely Factor: 10xy 1 24y2 2 6x2.

SOLUTION

We write the trinomial in descending powers of x and then factor out the common factor of 22. 10xy 1 24y2 2 6x2 5 26x2 1 10xy 1 24y2

5 22 13x2 2 5xy 2 12y22

Since the sign of the third term of 3x2 2 5xy 2 12y2 is 2, the signs between the binomial factors will be opposite. Since the first term is 3x2, the first terms of the binomial factors must be 3x and x: 22 13x2 2 5xy 2 12y22 5 22 13x

2 1x

2

2

The product of the last terms must be 212y , and the sum of the outer terms and the inner terms must be 25xy. Of the many factorizations of 212y2, only 4y 123y2 leads to a middle term of 25xy. So we have 10xy 1 24y2 2 6x2 5 26x2 1 10xy 1 24y2

5 22 13x2 2 5xy 2 12y22 5 22 13x 1 4y2 1x 2 3y2

We can check by multiplying.

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58

Chapter 0

A Review of Basic Algebra

Self Check 9

Factor: 26x2 2 15xy 2 6y2. Now Try Exercise 69.

5. Factor Trinomials by Grouping Another way of factoring trinomials involves factoring by grouping. This method can be used to factor trinomials of the form ax2 1 bx 1 c. For example, to factor 6x2 1 5x 2 6, we note that a 5 6, b 5 5, and c 5 26, and proceed as follows: 1. Find the product ac: 6 1262 5 236. This number is called the key number. 2. Find two factors of the key number 12362 whose sum is b 5 5. Two such numbers are 9 and 24. 9 1242 5 236

and

9 1 1242 5 5

3. Use the factors 9 and 24 as coefficients of two terms to be placed between 6x2 and 26. 6x2 1 5x 2 6 5 6x2 1 9x 2 4x 2 6 4. Factor by grouping:

6x2 1 9x 2 4x 2 6 5 3x 12x 1 32 2 2 12x 1 32 5 12x 1 32 13x 2 22

EXAMPLE 10

Factor out 2x 1 3.

Factoring a Trinomial by Grouping Factor: 15x2 1 x 2 2.

SOLUTION

Since a 5 15 and c 5 22 in the trinomial, ac 5 230. We now find factors of 230 whose sum is b 5 1. Such factors are 6 and 25. We use these factors as coefficients of two terms to be placed between 15x2 and –2. 15x2 1 6x 2 5x 2 2 Finally, we factor by grouping.

3x 15x 1 22 2 15x 1 22 5 15x 1 22 13x 2 12

Self Check 10

Factor: 15a2 1 17a 2 4. Now Try Exercise 39. We can often factor polynomials with variable exponents. For example, if n is a natural number, a2n 2 5an 2 6 5 1an 1 12 1an 2 62

because 1an 1 12 1an 2 62 5 a2n 2 6an 1 an 2 6 5 a2n 2 5an 2 6

Combine like terms.

6. Factor the Sum and Difference of Two Cubes Two other types of factoring involve binomials that are the sum or the difference of two cubes. Like the difference of two squares, they can be factored by using a formula. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 0.5

Factoring the Sum and Difference of Two Cubes

EXAMPLE 11

Factoring Polynomials

59

x3 1 y3 5 1x 1 y2 1x2 2 xy 1 y22 x3 2 y3 5 1x 2 y2 1x2 1 xy 1 y22

Factoring the Sum of Two Cubes Factor: 27x6 1 64y3.

SOLUTION

We can write this expression as the sum of two cubes and factor it as follows: 27x6 1 64y3 5 13x22 3 1 14y2 3

5 13x2 1 4y2 3 13x22 2 2 13x22 14y2 1 14y2 2 4 5 13x2 1 4y2 19x4 2 12x2y 1 16y22

Self Check 11

We can check by multiplying.

Factor: 8a3 1 1,000b6. Now Try Exercise 41.

EXAMPLE 12

Factoring the Difference of Two Cubes Factor: x3 2 8.

SOLUTION

This binomial can be written as x3 2 23, which is the difference of two cubes. Substituting into the formula for the difference of two cubes gives x3 2 23 5 1x 2 22 1x2 1 2x 1 222 5 1x 2 22 1x2 1 2x 1 42

Self Check 12

We can check by multiplying.

Factor: p3 2 64. Now Try Exercise 43.

7. Factor Miscellaneous Polynomials EXAMPLE 13

Factoring a Miscellaneous Polynomial Factor: x2 2 y2 1 6x 1 9.

SOLUTION

Here we will factor a trinomial and a difference of two squares.

x2 2 y2 1 6x 1 9 5 x2 1 6x 1 9 2 y2 5 1x 1 32 2 2 y2

5 1x 1 3 1 y2 1x 1 3 2 y2

Use the Commutative Property to rearrange the terms. Factor x2 1 6x 1 9. Factor the difference of two squares.

We could try to factor this expression in another way.

x2 2 y2 1 6x 1 9 5 1x 1 y2 1x 2 y2 1 3 12x 1 32

Factor x2 2 y2 and 6x 1 9.

However, we are unable to finish the factorization. If grouping in one way doesn’t work, try various other ways. Self Check 13

Factor: a2 1 8a 2 b2 1 16. Now Try Exercise 97.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

60

Chapter 0

A Review of Basic Algebra

EXAMPLE 14

Factoring a Miscellaneous Trinomial Factor: z4 2 3z2 1 1.

SOLUTION

This trinomial cannot be factored as the product of two binomials, because no combination will give a middle term of 23z2. However, if the middle term were 22z2, the trinomial would be a perfect square, and the factorization would be easy: z4 2 2z2 1 1 5 1z2 2 12 1z2 2 12 5 1z2 2 12 2

We can change the middle term in z4 2 3z2 1 1 to 22z2 by adding z2 to it. However, to make sure that adding z2 does not change the value of the trinomial, we must also subtract z2. We can then proceed as follows. z4 2 3z2 1 1 5 z4 2 3z2 1 z2 1 1 2 z2

Add and subtract z2.

5 z4 2 2z2 1 1 2 z2

Combine 23z2 and z2.

5 1z2 2 1 1 z2 1z2 2 1 2 z2

Factor z4 2 2z2 1 1.

5 1z2 2 12 2 2 z2

Factor the difference of two squares.

In this type of problem, we will always try to add and subtract a perfect square in hopes of making a perfect-square trinomial that will lead to factoring a difference of two squares. Self Check 14

Factor: x4 1 3x2 1 4. Now Try Exercise 103. It is helpful to identify the problem type when we must factor polynomials that are given in random order.

Factoring Strategy

1. Factor out all common monomial factors. 2. If an expression has two terms, check whether the problem type is a. The difference of two squares: x2 2 y2 5 1x 1 y2 1x 2 y2

b. The sum of two cubes:

x3 1 y3 5 1x 1 y2 1x2 2 xy 1 y22

c. The difference of two cubes:

x3 2 y3 5 1x 2 y2 1x2 1 xy 1 y22

3. 4. 5. 6.

Self Check Answers

If an expression has three terms, try to factor it as a trinomial. If an expression has four or more terms, try factoring by grouping. Continue until each individual factor is prime. Check the results by multiplying.

2. a2b2c2 11 1 abc2 3. 1x 1 y2 1x 1 22 1. 4a 1a 2 2b2 5. 1a2 1 9b22 1a 1 3b2 1a 2 3b2 4. 13a 1 4b2 13a 2 4b2 7. 1p 2 62 1p 1 12 8. 13x 2 22 12x 1 12 6. 23 1x 1 22 1x 2 22 10. 13a 1 42 15a 2 12 9. 23 1x 1 2y2 12x 1 y2 11. 8 1a 1 5b22 1a2 2 5ab2 1 25b42 12. 1p 2 42 1p2 1 4p 1 162 13. 1a 1 4 1 b2 1a 1 4 2 b2 14. 1x2 1 2 1 x2 1x2 1 2 2 x2

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Section 0.5

Factoring Polynomials

61

Exercises 0.5 Getting Ready

33. 12x2 2 xy 2 6y2

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

In each expression, factor the trinomial by grouping.

Fill in the blanks. 1. When polynomials are multiplied together, each polynomial is a of the product. 2. If a polynomial cannot be factored using coefficients, it is called a polynomial.

34. 8x2 2 10xy 2 3y2

35. x2 1 10x 1 21

36. x2 1 7x 1 10

37. x2 2 4x 2 12

38. x2 2 2x 2 63

39. 6p2 1 7p 2 3

40. 4q2 2 19q 1 12

In each expression, factor the sum of two cubes.

Complete each factoring formula.

41. t3 1 343

3. ax 1 bx 5

42. r3 1 8s3

4. x2 2 y2 5 In each expression, factor the difference of two cubes. 43. 8z3 2 27 44. 125a3 2 64

5. x2 1 2xy 1 y2 5 6. x2 2 2xy 1 y2 5 7. x3 1 y3 5

Factor each expression completely. If an expression is prime, so indicate.

8. x3 2 y3 5

Practice

45. 3a2bc 1 6ab2c 1 9abc2

In each expression, factor out the greatest common monomial.

46. 5x3y3z3 1 25x2y2z2 2 125xyz

9. 3x 2 6 2

47. 3x3 1 3x2 2 x 2 1

10. 5y 2 15 3

48. 4x 1 6xy 2 9y 2 6

12. 9y3 1 6y2 14. 25y2z 2 15yz2

11. 8x 1 4x 13. 7x2y2 1 14x3y2

49. 2txy 1 2ctx 2 3ty 2 3ct 50. 2ax 1 4ay 2 bx 2 2by 51. ax 1 bx 1 ay 1 by 1 az 1 bz

In each expression, factor by grouping. 16. b 1x 2 y2 1 a 1x 2 y2 15. a 1x 1 y2 1 b 1x 1 y2 17. 4a 1 b 2 12a2 2 3ab

18. x2 1 4x 1 xy 1 4y

52. 6x2y3 1 18xy 1 3x2y2 1 9x 53. x2 2 1y 2 z2 2 54. z2 2 1y 1 32 2 55. 1x 2 y2 2 2 1x 1 y2 2

56. 12a 1 32 2 2 12a 2 32 2

57. x4 2 y4

58. z4 2 81

In each expression, factor the difference of two squares. 19. 4x2 2 9

20. 36z2 2 49

59. 3x2 2 12

60. 3x3y 2 3xy

21. 4 2 9r2

22. 16 2 49x2

61. 18xy2 2 8x

62. 27x2 2 12

23. 81x4 2 1

24. 81 2 x4

63. x2 2 2x 1 15 65. 215 1 2a 1 24a2

64. x2 1 x 1 2 66. 232 2 68x 1 9x2

67. 6x2 1 29xy 1 35y2

68. 10x2 2 17xy 1 6y2

69. 12p2 2 58pq 2 70q2

70. 3x2 2 6xy 2 9y2

71. 26m2 1 47mn 2 35n2

72. 214r2 2 11rs 1 15s2

73. 26x3 1 23x2 1 35x

74. 2y3 2 y2 1 90y

25. 1x 1 z2 2 2 25

26. 1x 2 y2 2 2 9

In each expression, factor the trinomial. 28. a2 2 12a 1 36 27. x2 1 8x 1 16 2

2

29. b 2 10b 1 25 2

30. y 1 14y 1 49 2

31. m 1 4mn 1 4n

2

2

32. r 2 8rs 1 16s

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

62

Chapter 0

A Review of Basic Algebra

75. 6x4 2 11x3 2 35x2

76. 12x 1 17x2 2 7x3

77. x4 1 2x2 2 15

78. x4 2 x2 2 6

79. a2n 2 2an 2 3

80. a2n 1 6an 1 8

81. 6x2n 2 7xn 1 2

82. 9x2n 1 9xn 1 2

83. 4x2n 2 9y2n

84. 8x2n 2 2xn 2 3

110. Movie Stunts The formula that gives the distance a stuntwoman is above the ground t seconds after she falls over the side of a 144-foot tall building is f 5 144 2 16t2. Factor the right side.

144 ft 2n

n

4n

2n

85. 10y 2 11y 2 6

86. 16y 2 25y

87. 2x3 1 2,000

88. 3y3 1 648

89. 1x 1 y2 3 2 64 90. 1x 2 y2 3 1 27 6

Discovery and Writing 111. Explain how to factor the difference of two squares. 112. Explain how to factor the difference of two cubes.

6

91. 64a 2 y 92. a6 1 b6

113. Explain how to factor a2 2 b2 1 a 1 b.

93. a3 2 b3 1 a 2 b 94. 1a2 2 y22 2 5 1a 1 y2

114. Explain how to factor x2 1 2x 1 1.

95. 64x6 1 y6

Factor the indicated monomial from the given expression.

96. z2 1 6z 1 9 2 225y2

115. 3x 1 2; 2

97. x2 2 6x 1 9 2 144y2

116. 5x 2 3; 5

2

98. x2 1 2x 2 9y2 1 1 99. 1a 1 b2 2 2 3 1a 1 b2 2 10 100. 2 1a 1 b2 2 2 5 1a 1 b2 2 3

117. x 1 2x 1 4; 2

118. 3x2 2 2x 2 5; 3

119. a 1 b; a

120. a 2 b; b

1/2

121. x 1 x ; x

103. x4 1 x2 1 1

123. 2x 1 "2y; "2

122. x3/2 2 x1/2 ; x1/2

104. x4 1 3x2 1 4

125. ab3/2 2 a3/2b; ab

126. ab2 1 b; b21

101. x6 1 7x3 2 8

102. x6 2 13x4 1 36x2

4

1/2

124. "3a 2 3b; "3

2

105. x 1 7x 1 16 106. y4 1 2y2 1 9 107. 4a4 1 1 1 3a2

Factor each expression by grouping three terms and two terms.

108. x4 1 25 1 6x2

127. x2 1 x 2 6 1 xy 2 2y

109. Candy To find the amount of chocolate used in the outer coating of one of the malted-milk balls shown, we can find the volume V of the chocolate shell using the formula V 5 43pr31 2 43pr32. Factor the expression on the right side of the formula.

128. 2x2 1 5x 1 2 2 xy 2 2y 129. a4 1 2a3 1 a2 1 a 1 1 130. a4 1 a3 2 2a2 1 a 2 1

Review 131. Which natural number is neither prime nor composite? 132. Graph the interval 3 22,32 .

Outer radius r1

© Istockphoto.com/maureenpr

Inner radius r2

133. Simplify: 1x3x22 4. 1a32 3 1a22 4 . 134. Simplify: 1a2a32 3 136. Simplify: "20x5.

135. a

137. Simplify: "20x 2 "125x. 138. Rationalize the denominator:

3x4x3 0 b 6x22x4

3 3 . ! 3

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Section 0.6

Rational Expressions

63

0.6 Rational Expressions In this section, we will learn to 1. 2. 3. 4.

Define rational expressions. Simplify rational expressions. Multiple and divide rational expressions. Add and subtract rational expressions.

5. Simplify complex fractions.

Photo by Tim Boyle/Getty Images

Abercrombie and Fitch (A&F) is a very successful American clothing company founded in 1892 by David Abercrombie and Ezra Fitch. Today the stores are popular shopping destinations for university students wanting to keep up with the latest styles and trends.

Suppose that a clothing manufacturer finds that the cost in dollars of producing x fleece vintage shirts is given by the algebraic expression 13x 1 1,000. The average cost of producing each shirt could be obtained by dividing the production cost, 13x 1 1,000, by the number of shirts produced, x. The algebraic fraction 13x 1 1,000 x represents the average cost per shirt. We see that the average cost of producing 200 shirts would be $18. 13 12002 1 1,000 5 18 200

An understanding of algebraic fractions is important in solving many real-life problems.

1. Define Rational Expressions

Caution Remember that the denominator of a fraction cannot be zero.

If x and y are real numbers, the quotient xy 1y 2 02 is called a fraction. The number x is called the numerator, and the number y is called the denominator. Algebraic fractions are quotients of algebraic expressions. If the expressions are polynomials, the fraction is called a rational expression. The first two of the following algebraic fractions are rational expressions. The third is not, because the numerator and denominator are not polynomials. 5y2 1 2y y 2 3y 2 7 2

8ab2 2 16c3 2x 1 3

x1/2 1 4x x3/2 2 x1/2

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64

Chapter 0

A Review of Basic Algebra

We summarize some of the properties of fractions as follows:

Properties of Fractions

If a, b, c, and d are real numbers and no denominators are 0, then Equality of Fractions a c 5 if and only if b d

ad 5 bc

Fundamental Property of Fractions ax a 5 bx b Multiplication and Division of Fractions a c ac ? 5 b d bd

a c a d ad 4 5 ? 5 b d b c bc

and

Addition and Subtraction of Fractions a c a1c 1 5 b b b

and

a c a2c 2 5 b b b

The first two examples illustrate each of the previous properties of fractions.

EXAMPLE 1

Illustrating the Properties of Fractions Assume that no denominators are 0. 2a 4a 5 Because 2a 162 5 3 14a2 a. 3 6 b.

6xy 3 12xy2 5 10xy 5 12xy2 5

Self Check 1

a. Is

Factor the numerator and denominator and divide out the common factors.

3 5

3y 15z 5 ? 5 25

b. Simplify:

15a2b . 25ab2

Now Try Exercise 9.

EXAMPLE 2

Illustrating the Properties of Fractions Assume that no denominators are 0. a.

2r 3r 2r ? 3r ? 5 7s 5s 7s ? 5s 5

c.

b.

6r2 35s2

5

2ab ab 2ab 1 ab 1 5 5xy 5xy 5xy 5

3mn 2pq 3mn 7mn 4 5 ? 4pq 7mn 4pq 2pq

3ab 5xy

d.

21m2n2 8p2q2

3uv2 6uv2 2 3uv2 6uv2 2 2 2 5 7w 7w 7w2 5

3uv2 7w2

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Section 0.6

Self Check 2

Perform each operation: 3a 2a 2ab 2rs a. ? b. 4 5b 7b 3rs 4ab

c.

Rational Expressions

5pq 3pq 1 3t 3t

d.

65

5mn2 mn2 2 3w 3w

Now Try Exercise 19.

To add or subtract rational expressions with unlike denominators, we write each expression as an equivalent expression with a common denominator. We can then add or subtract the expressions. For example, 3x 2x 3x 172 2x 152 1 5 1 5 7 5 172 7 152

4a2 3a2 4a2 122 3a2 132 2 5 2 15 10 15 122 10 132

5

21x 10x 1 35 35

5

8a2 9a2 2 30 30

5

21x 1 10x 35

5

8a2 2 9a2 30

5

31x 35

5

2a2 30 a2 52 30

A rational expression is in lowest terms if all factors common to the numerator and the denominator have been removed. To simplify a rational expression means to write it in lowest terms.

2. Simplify Rational Expressions To simplify rational expressions, we use the Fundamental Property of Fractions. This enables us to divide out all factors that are common to the numerator and the denominator.

EXAMPLE 3

Simplifying a Rational Expression Simplify:

SOLUTION

x2 2 9 1x 2 0, 32 . x2 2 3x

We factor the difference of two squares in the numerator, factor out x in the denominator, and divide out the common factor of x 2 3. 1x 1 32 1x 2 32 x2 2 9 5 2 x 2 3x x 1x 2 32 5

Self Check 3

Simplify:

x23 51 x23

x13 x

a2 2 4a 1a 2 4,232 . a 2 a 2 12 2

Now Try Exercise 23. We will encounter the following properties of fractions in the next examples. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

66

Chapter 0

A Review of Basic Algebra

Properties of Fractions

EXAMPLE 4

If a and b represent real numbers and there are no divisions by 0, then •

a 5a 1

•

•

a 2a a 2a 5 52 52 b 2b 2b b

a a 2a 2a • 2 5 5 52 b 2b b 2b

Simplifying a Rational Expression Simplify:

SOLUTION

x2 2 2xy 1 y2 1x 2 y2 . y2x

We factor the trinomial in the numerator, factor 21 from the denominator, and divide out the common factor of x 2 y. 1x 2 y2 1x 2 y2 x2 2 2xy 1 y2 5 y2x 21 1x 2 y2 x2y 5 21 x2y 52 1 5 2 1x 2 y2

Self Check 4

a 51 a

Simplify:

x2y x2y

51

a2 2 ab 2 2b2 12b 2 a 2 02 . 2b 2 a

Now Try Exercise 25.

EXAMPLE 5

Simplifying a Rational Expression Simplify:

SOLUTION

x2 2 3x 1 2 1x 2 2,212 . x2 2 x 2 2

We factor the numerator and denominator and divide out the common factor of x 2 2. 1x 2 12 1x 2 22 x2 2 3x 1 2 5 2 1x 1 12 1x 2 22 x 2x22 5

Self Check 5

Simplify:

x22 51 x22

x21 x11

a2 1 3a 2 4 1a 2 1,232 . a2 1 2a 2 3

Now Try Exercise 27.

3. Multiply and Divide Rational Expressions EXAMPLE 6

Multiplying Rational Expressions Multiply:

x2 2 x 2 2 x2 1 2x 2 3 1x 2 1, 21, 22 . ? x2 2 1 x22

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Section 0.6

SOLUTION

Rational Expressions

67

To multiply the rational expressions, we multiply the numerators, multiply the denominators, and divide out the common factors.

1x2 2 x 2 22 1x2 1 2x 2 32 x2 2 x 2 2 x2 1 2x 2 3 ? 5 1x2 2 12 1x 2 22 x2 2 1 x22 5

1x 2 22 1x 1 12 1x 2 12 1x 1 32 1x 1 12 1x 2 12 1x 2 22

x22 x11 x21 5 1, 5 1, 51 x22 x11 x21

5x13

Self Check 6

Simplify:

x2 2 9 x 2 1 1x 2 0, 1, 32 . ? x2 2 x x2 2 3x

Now Try Exercise 33.

EXAMPLE 7

Dividing Rational Expressions Divide:

SOLUTION

x2 2 2x 2 3 x2 1 2x 2 15 1x 2 2, 22, 3, 252 . 4 2 2 x 24 x 1 3x 2 10

To divide the rational expressions, we multiply by the reciprocal of the second rational expression. We then simplify by factoring the numerator and denominator and dividing out the common factors.

x2 2 2x 2 3 x2 1 2x 2 15 x2 2 2x 2 3 x2 1 3x 2 10 4 5 ? 2 x2 2 4 x2 1 3x 2 10 x2 2 4 x 1 2x 2 15 5 5

5

Self Check 7

Simplify:

1x2 2 2x 2 32 1x2 1 3x 2 102 1x2 2 42 1x2 1 2x 2 152

1x 2 32 1x 1 12 1x 2 22 1x 1 52 1x 1 22 1x 2 22 1x 1 52 1x 2 32

x23 x22 x15 5 1, 5 1, 51 x23 x22 x15

x11 x12

a2 2 a a2 2 2a 1a 2 0, 2, 222 . 4 2 a12 a 24

Now Try Exercise 39.

EXAMPLE 8

Using Multiplication and Division to Simplify a Rational Expression Simplify:

SOLUTION

2x2 2 5x 2 3 3x2 1 2x 2 1 2x2 1 x 1 1 ? 2 4 ax 2 , 21, 3, 0, 2 b. 3x 2 1 x 2 2x 2 3 3x 3 2

We can change the division to a multiplication, factor, and simplify.

2x2 2 5x 2 3 3x2 1 2x 2 1 2x2 1 x ? 2 4 3x 2 1 x 2 2x 2 3 3x 2x2 2 5x 2 3 3x2 1 2x 2 1 3x ? 2 ? 2 3x 2 1 x 2 2x 2 3 2x 1 x 12x2 2 5x 2 32 13x2 1 2x 2 12 13x2 5 13x 2 12 1x2 2 2x 2 32 12x2 1 x2 5

5

1x 2 32 12x 1 12 13x 2 12 1x 1 12 3x 13x 2 12 1x 1 12 1x 2 32 x 12x 1 12

x23 2x 1 1 3x 2 1 x11 x 5 1, 5 1, 5 1, 5 1, 5 1 x23 2x 1 1 3x 2 1 x11 x

53 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

68

Chapter 0

A Review of Basic Algebra

Self Check 8

Simplify:

x2 2 25 x2 2 5x x2 1 2x 1x 2 0, 2, 5, 252 . 4 2 ? x22 x 2 2x x2 1 5x

Now Try Exercise 45.

4. Add and Subtract Rational Expressions To add (or subtract) rational expressions with like denominators, we add (or subtract) the numerators and keep the common denominator.

EXAMPLE 9

Adding Rational Expression with Like Deonominators Add:

SOLUTION

2x 1 5 3x 1 20 1x 2 252 . 1 x15 x15

2x 1 5 3x 1 20 5x 1 25 1 5 x15 x15 x15 5

5 1x 1 52 1 1x 1 52

Add the numerators and keep the common denominator.

Factor out 5 and divide out the common factor of x 1 5.

55

Self Check 9

Add:

3x 2 2 x26 1x 2 22 . 1 x22 x22

Now Try Exercise 49.

To add (or subtract) rational expressions with unlike denominators, we must find a common denominator, called the least (or lowest) common denominator (LCD). Suppose the unlike denominators of three rational expressions are 12, 20, and 35. To find the LCD, we first find the prime factorization of each number. 12 5 4 ? 3 5 22 ? 3

20 5 4 ? 5

35 5 5 ? 7

5 22 ? 5

Because the LCD is the smallest number that can be divided by 12, 20, and 35, it must contain factors of 22, 3, 5, and 7. Thus, the LCD 5 22 ? 3 ? 5 ? 7 5 420

Comment Remember to find the least common denominator, use each factor the greatest number of times that it occurs.

EXAMPLE 10

That is, 420 is the smallest number that can be divided without remainder by 12, 20, and 35. When finding an LCD, we always factor each denominator and then create the LCD by using each factor the greatest number of times that it appears in any one denominator. The product of these factors is the LCD. This rule also applies if the unlike denominators of the rational expressions contain variables. Suppose the unlike denominators are x2 1x 2 52 and x 1x 2 52 3. To find the LCD, we use x2 and 1x 2 52 3. Thus, the LCD is the product x2 1x 2 52 3 .

Adding Rational Expressions with Unlike Denominators Add:

1 2 1x 2 2, 222 . 1 2 x2 2 4 x 2 4x 1 4

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Section 0.6

SOLUTION

Rational Expressions

69

We factor each denominator and find the LCD. x2 2 4 5 1x 1 22 1x 2 22

x2 2 4x 1 4 5 1x 2 22 1x 2 22 5 1x 2 22 2

The LCD is 1x 1 22 1x 2 22 2. We then write each rational expression with its denominator in factored form, convert each rational expression into an equivalent expression with a denominator of 1x 1 22 1x 2 22 2, add the expressions, and simplify.

1 2 1 2 1 2 5 1 1x 1 22 1x 2 22 1x 2 22 1x 2 22 x 24 x 2 4x 1 4 2

5 5

Comment

5

Always attempt to simplify the final result. In this case, the final fraction is already in lowest terms.

5

Self Check 10

Add:

1 1x 2 22 2 1x 1 22 1 1x 1 22 1x 2 22 1x 2 22 1x 2 22 1x 2 22 1x 1 22 1 1x 2 22 1 2 1x 1 22 1x 1 22 1x 2 22 1x 2 22

x22 x12 5 1, 51 x22 x12

x 2 2 1 2x 1 4 1x 1 22 1x 2 22 1x 2 22 3x 1 2 1x 1 22 1x 2 22 2

3 1 1x 2 3,232 . 1 2 x2 2 6x 1 9 x 29

Now Try Exercise 59.

EXAMPLE 11

Combining and Simplifying Rational Expressions with Unlike Denominators Simplify:

SOLUTION

x22 x13 3 1x 2 1, 21, 222 . 2 2 1 2 2 x 21 x 1 3x 1 2 x 1x22

We factor the denominators to find the LCD. x2 2 1 5 1x 1 12 1x 2 12

x2 1 3x 1 2 5 1x 1 22 1x 1 12 x2 1 x 2 2 5 1x 1 22 1x 2 12

The LCD is 1x 1 12 1x 1 22 1x 2 12 . We now write each rational expression as an equivalent expression with this LCD, and proceed as follows: x22 x13 3 2 2 1 2 2 x 21 x 1 3x 1 2 x 1x22 x22 x13 3 5 2 1 1x 1 12 1x 2 12 1x 1 12 1x 1 22 1x 2 12 1x 1 22 1x 2 22 1x 1 22 1x 1 32 1x 2 12 3 1x 1 12 5 2 1 1x 1 12 1x 2 12 1x 1 22 1x 1 12 1x 1 22 1x 2 12 1x 2 12 1x 1 22 1x 1 12 1x2 2 42 2 1x2 1 2x 2 32 1 13x 1 32 5 1x 1 12 1x 1 22 1x 2 12 5

x12 x21 x11 5 1, 5 1, 51 x12 x21 x11

x2 2 4 2 x2 2 2x 1 3 1 3x 1 3 1x 1 12 1x 1 22 1x 2 12

x12 1x 1 12 1x 1 22 1x 2 12 1 5 1x 1 12 1x 2 12 5

Divide out the common factor of x 1 2.

x12 51 x12

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70

Chapter 0

A Review of Basic Algebra

Self Check 11

Simplify:

4y 2 2 1 2 1y 2 1, 212 . y 21 y11 2

Now Try Exercise 61.

5. Simplify Complex Fractions A complex fraction is a fraction that has a fraction in its numerator or a fraction in its denominator. There are two methods generally used to simplify complex fractions. These are stated here for you.

Strategies for Simplifying Complex Fractions

Method 1: Multiply the Complex Fraction by 1 • Determine the LCD of all fractions in the complex fraction. • Multiply both numerator and denominator of the complex fraction by the LCD. Note that when we multiply by LCD LCD we are multiplying by 1. Method 2: Simplify the Numerator and Denominator and then Divide • Simplify the numerator and denominator so that both are single fractions. • Perform the division by multiplying the numerator by the reciprocal of the denominator.

EXAMPLE 12

Simplifying Complex Fractions 1 1 1 x y 1x, y 2 02 . Simplify: x y Method 1: We note that the LCD of the three fractions in the complex fraction is xy. So we multiply the numerator and denominator of the complex fraction by xy and simplify: 1 xy 1 1 xy 1 1 xya 1 b 1 y1x x y x y x y 5 5 5 x x xyx x2 xya b y y y

Method 2: We combine the fractions in the numerator of the complex fraction to obtain a single fraction over a single fraction. 1 1 1 1y2 1 1x2 y1x 1 1 x y x 1y2 y 1x2 xy 5 5 x x x y y y

Then we use the fact that any fraction indicates a division: y1x 1y 1 x2 y xy y1x x y1x y y1x 5 4 5 ? 5 5 x xy y xy x xyx x2 y

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Section 0.6

Self Check 12

Rational Expressions

71

1 1 2 x y 1x, y 2 02 . Simplify: 1 1 1 x y Now Try Exercise 81.

Self Check Answers

1. a. no 3.

a a13

8. x 1 2

b.

3a 5b

2. a.

4. 2 1a 1 b2

9. 4

6a2 35b2

b.

4a2b2 3r2s2

c.

8pq 3t

d.

4mn2 3w

a14 x13 6. 7. a 2 1 a13 x2 4x 1 6 2y y2x 10. 11. 12. 2 1x 1 32 1x 2 32 y21 y1x 5.

Exercises 0.6 Getting Ready

Perform the operations and simplify, whenever possible. Assume that no denominators are 0. 4x 2 25y 4 15. ? 16. ? 7 5a 2z y2 8m 3m 15p 25p 4 18. 4 17. 5n 10n 8q 16q2 3z 2z 7a 3a 19. 1 20. 2 5c 5c 4b 4b

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. In the fraction ab, a is called the . a 2. In the fraction b , b is called the . a c 3. b 5 d if and only if . 4. The denominator of a fraction can never be Complete each formula. a c 5. ? 5 b d 7.

a c 1 5 b b

a c 6. 4 5 b d 8.

a c 2 5 b b

.

21.

11.

25xyz 50a2bc , 12ab2c 24xyz

3x2 12y2 10. , 4y2 16x2 12.

15rs2 37.5a3 , 4rs2 10a3

22.

8rst2 7rst2 1 15m4t2 15m4t2

Simplify each fraction. Assume that no denominators are 0. 2x 2 4 x2 2 16 23. 2 24. 2 x 24 x 2 8x 1 16 25.

4 2 x2 x 2 5x 1 6

26.

25 2 x2 x 1 10x 1 25

27.

6x3 1 x2 2 12x 4x3 1 4x2 2 3x

28.

6x4 2 5x3 2 6x2 2x3 2 7x2 2 15x

29.

x3 2 8 x 1 ax 2 2x 2 2a

30.

xy 1 2x 1 3y 1 6 x3 1 27

Determine whether the fractions are equal. Assume that no denominators are 0. 8x 16x 9. , 3y 6y

15x2y x2y 2 7a2b3 7a2b3

2

2

2

Practice Simplify each rational expression. Assume that no denominators are 0. 7a2b 35p3q2 13. 14. 2 21ab 49p4q

Perform the operations and simplify, whenever possible. Assume that no denominators are 0. x2 2 1 x2 31. ? 2 x x 1 2x 1 1

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72

Chapter 0

A Review of Basic Algebra

32.

y12 y2 2 2y 1 1 ? 2 y y 1y22

54.

33.

3x2 1 7x 1 2 x2 2 x ? 2 x2 1 2x 3x 1 x

55.

34.

2x2 1 x 2 3 x2 1 x ? 2 2x 1 3x x2 2 1

35.

x2 1 x x2 2 1 ? x21 x12 2

36.

x 1 5x 1 6 x 1 2 ? x2 1 6x 1 9 x2 2 4

x2 1 16 2x2 1 32 4 37. 8 2 38.

x2 1 x 2 6 x2 2 4 4 2 2 x 2 6x 1 9 x 29

z2 1 z 2 20 z2 2 25 39. 4 z2 2 4 z25 x2 2 1 ax 1 bx 1 a 1 b 4 40. a2 1 2ab 1 b2 x2 2 2x 1 1 2

a13 a 1 2 a2 1 7a 1 12 a 2 16 a 2 56. 2 1 2 a 1a22 a 2 5a 1 4 x 1 2 57. 2 x 24 x12

58. 59. 60. 61. 62. 63.

2

41.

3x 1 5x 2 2 6x 1 13x 2 5 4 3 2 x 1 2x 2x3 1 5x2

64.

42.

2x2 2 x 2 3 x2 1 13x 1 12 4 8x2 2 6x 2 5 8x2 2 14x 1 5

65.

43.

x2 1 7x 1 12 x2 2 3x 2 10 x3 2 4x2 1 3x ? ? 2 x3 2 x2 2 6x x2 1 2x 2 3 x 2 x 2 20

66.

44. 45.

46. 47. 48. 49. 50. 51. 52. 53.

x 1x 2 22 2 3 x 1x 1 12 2 2 ? x 1x 1 72 2 3 1x 2 12 x 1x 2 72 1 3 1x 1 12

67.

x2 2 2x 2 3 3x 2 8 x2 1 6x 1 5 ? 4 21x2 2 50x 2 16 x 2 3 7x2 2 33x 2 10

68.

x2 1 4x 1 3 x3 1 27 x2 1 x 2 6 4a 4 2 b 2 2 x 24 x 1 2x x 2 3x 1 9 3 x12 1 x13 x13 3 x11 4x x21 6x x22 2 52x 3 x26

1 2 2 1 2

x12 x11 4 x21 3 x22 1 x25 2 62x

3 2 1 x11 x21

3 x 1 x14 x24

b2 4 2 2 2 b 24 b 1 2b 3x 2 2 x 2 2 x2 1 2x 1 1 x 21 2t t11 2 2 2 t 2 25 t 1 5t 2 1 131 2 y 21 y11 4 1 2 21 2 t 24 t22 1 3 3x 2 2 1 2 2 x22 x12 x 24 x 5 3 13x 2 12 2 1 x23 x13 x2 2 9 1 x23 1 1 a b? x22 x23 2x 1 1 1 2 b4 a x11 x22 x22 3x x 3x 1 1 2 2 x24 x14 16 2 x2 7x 3x 3x 2 1 1 1 2 x25 52x x 2 25

69.

1 2 1 2 2 1 2 x 1 3x 1 2 x 1 4x 1 3 x 1 5x 1 6

70.

22 2 2z 2 2y 1 2 1y 2 x2 1z 2 x2 x2y x2z

2

71.

3x 2 2 4x2 1 2 3x2 2 25 2 2 1 2 x 1 x 2 20 x 2 25 x 2 16

72.

3x 1 2 x14 1 1 2 2 8x 2 10x 2 3 6x 2 11x 1 3 4x 1 1

2

2

Simplify each complex fraction. Assume that no denominators are 0. 3a b 73. 6ac b2

3t2 9x 74. t 18x

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Section 0.6

3a b 75. ab 27 x2y ab 77. y2x ab

x2 2 5x 1 6 2x2y 78. x2 2 9 2x2y

1 1 1 x y 79. xy

80.

2

3a 4a 2 b x 83. 1 1 1 b ax

84.

6 x112 x 85. 6 x151 x

86.

87.

89.

3xy 1 12 xy 3x x1

1 x

x 2 2 x12 x21 91. 3 x 1 x12 x21

12

k5

1 1 1 1 k1 k2 Section 1

R5

x y

1 1 1 1 1 1 R1 R2 R3

Discovery and Writing

2z 3 z

x231

Section 2

98. Electronics The combined resistance R of three resistors with resistances of R1, R2, and R3 is given by the following formula. Simplify the complex fraction on the right side of the formula.

x2 21 y2

12

88.

97. Engineering The stiffness k of the shaft shown in the illustration is given by the following formula where k1 and k2 are the individual stiffnesses of each section. Simplify the complex fraction on the right side of the formula.

xy

11 11 1 x y 1 1 2 x y 82. 1 1 1 x y

1 1 1 x y 81. 1 1 2 x y

73

Applications

3u2v 4t 76. 3uv

2

Rational Expressions

1 x

1 2 2x13 x

2x2 1 4 90. 4x 21 5 2x 1 1 x23 x22 92. 3 x 2 x23 x22

Simplify each complex fraction. Assume that no denominators are 0. x ab 100. 99. 1 3 1 1 21 2 1 21 3x 2a 1 y 101. 102. 1 2 11 21 1 2 11 21 x y a c ad 1 bc 1 5 is valid. b d bd a c a d 104. Explain why the formula 4 5 ? is valid. b d b c 105. Explain the Commutative Property of Addition and explain why it is useful. 106. Explain the Distributive Property and explain why it is useful.

103. Explain why the formula

Review Write each expression without using negative exponents, and simplify the resulting complex fraction. Assume that no denominators are 0. y21 1 93. 94. 21 21 11x x 1 y21

3 1x 1 22 21 1 2 1x 2 12 21 95. 1x 1 22 21 96.

2x 1x 2 32 21 2 3 1x 1 22 21 1x 2 32 21 1x 1 22 21

Write each expression without using absolute value symbols. 107. 0 26 0 108. 0 5 2 x 0 , given that x , 0 Simplify each expression. 109. a

x3y22 23 b x21y

111. "20 2 "45 112. 2 1x2 1 42 2 3 12x2 1 52

110. 127x62 2/3

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74

Chapter 0

A Review of Basic Algebra

CHAPTER REVIEW SEctIoN 0.1

Sets of Real Numbers

Definitions and Concepts

Examples

Natural numbers: The numbers that we count with.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, %

Whole numbers: The natural numbers and 0.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, %

Integers: The whole numbers and the negatives of the natural numbers. Rational numbers: {x 0 x can be written in the form ab 1b 2 02 , where a and b are integers.} All decimals that either terminate or repeat. Irrational numbers: Nonrational real numbers. All decimals that neither terminate nor repeat.

% , 25, 24, 23, 22, 21, 0, 1, 2, 3, 4, 5, % 2 5 2 7 1 2 5 , 25 5 2 , , 2 , 0.25 5 1 1 3 3 4 3 13 2 7 5 0.25, 5 2.6, 5 0.666 % 5 0.6, 5 0.63 4 5 3 11

"2, 2"26, p, 0.232232223 % 9 , 0, p, "31, 10.73 13

Real numbers: Any number that can be expressed as a decimal.

26, 2

Prime numbers: A natural number greater than 1 that is divisible only by itself and 1.

2, 3, 5, 7, 11, 13, 17, %

Composite numbers: A natural number greater than 1 that is not prime.

4, 6, 8, 9, 10, 12, 14, 15, %

Even integers: The integers that are exactly divisible by 2.

% , 26, 24, 22, 0, 2, 4, 6, %

Odd integers: The integers that are not exactly divisible by 2.

% , 25, 23, 21, 1, 3, 5, %

Associative Properties: of addition 1a 1 b2 1 c 5 a 1 1b 1 c2

5 1 14 1 72 5 15 1 42 1 7

Commutative Properties: of addition a 1 b 5 b 1 a

4175714

of multiplication 1ab2 c 5 a 1bc2

of multiplication ab 5 ba Distributive Property: a 1b 1 c2 5 ab 1 ac Double Negative Rule: 2 12a2 5 a

Open intervals have no endpoints. Closed intervals have two endpoints. Half-open intervals have one endpoint.

15 ? 42 ? 7 5 5 14 ? 72

1.7 1 2.5 5 2.5 1 1.7

4 ? 7 5 7 ? 4   1.7 12.52 5 2.5 11.72

3 1x 1 62 5 3x 1 3 ? 6

0.2 1y 2 102 5 0.2y 2 0.2 1102

2 1252 5 5   2 12a2 5 a 123, 22

3 23, 2 4 123, 2 4

(

)

–3

2

[

]

–3

2

(

]

–3

2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Review

75

Definitions and Concepts

Examples

Absolute value:

If x $ 0, then 0 x 0 5 x.

7 7 0 7 0 5 7    0 3.5 0 5 3.5    ` ` 5     0 0 0 5 0 2 2

Distance: The distance between points a and b on a number line is d 5 0b 2 a0.

The distance on the number line between points with coordinates of 23 and 2 is d 5 0 2 2 1232 0 5 0 5 0 5 5.

If x , 0, then 0 x 0 5 2x.

Consider the set 5 26, 23, 0, 12, 3, p, "5, 6, 8 6 . List the numbers in this set that are 1. natural numbers.

EXERCISES

2. whole numbers. 3. integers.

7 7 0 27 0 5 7    0 23.5 0 5 3.5    ` 2 ` 5   2 2

16. 12x 1 y2 1 z 5 1y 1 2x2 1 z 17. 2 1262 5 6

Graph each subset of the real numbers: 18. the prime numbers between 10 and 20

4. rational numbers. 5. irrational numbers.

19. the even integers from 6 to 14

6. real numbers.

Consider the set 5 26, 23, 0, 12, 3, p, "5, 6, 8 6 . List the numbers in this set that are 7. prime numbers. 8. composite numbers. 9. even integers. 10. odd integers. Determine which property of real numbers justifies each statement. 11. 1a 1 b2 1 2 5 a 1 1b 1 22

Graph each interval on the number line. 20. 23 , x # 5 21. x $ 0 or x , 21 22. 122, 4 4 23. 12`, 22 d 125,` 2

12. a 1 7 5 7 1 a

24. 12`,242 c 3 6,` 2

14. 3 1a 1 b2 5 3a 1 3b

Write each expression without absolute value symbols. 26. 0 225 0 25. 0 6 0

13. 4 12x2 5 14 ? 22 x 15. 15a2 7 5 7 15a2

27. 0 1 2 "2 0

28. 0 "3 2 1 0

29. On a number line, find the distance between points with coordinates of 25 and 7.

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76

Chapter 0

A Review of Basic Algebra

Integer Exponents and Scientific Notation

SEctIoN 0.2 Definitions and Concepts

Examples

Natural-number exponents: n factors of x           

x5 5 x ? x ? x ? x ? x

xn 5 x ? x ? x ? c ? x

Rules of exponents: If there are no divisions by 0, • 1xm2 n 5 xmn • xmxn 5 xm1n • 1xy2 n 5 xnyn

• x 5 1 1x 2 02 0

m

•

x 5 xm2n xn

x2x5 5 x215 5 x7

x x • a b 5 n y y n

2n

• x

n

1 5 n x

x • a b y

2n

y 5a b x

n

1x22 7 5 x2?7 5 x14

1xy2 5 5 x5y5

x 5 x5 a b 5 5 y y

60 5 1

622 5

x6 5 x624 5 x2 x4

6 3 x 23 63 216 a b 5a b 5 35 3 6 x x x

1 1 5 62 36

Scientific notation: A number is written in scientific notation when it is written in the form N 3 10n, where 1 # 0 N 0 , 10.

Write each number in scientific notation. 386,000 5 3.86 3 105 0.0025 5 2.5 3 1023

EXERCISES

42. a

Write each expression without using exponents. 31. 125a2 2 30. 25a3 Write each expression using exponents. 32. 3ttt 33. 122b2 13b2 Simplify each expression. 34. n2n4

36. 1x3y22 4

38. 1m n 2

23 0 2

a5 40. 8 a

35. 1p32 2 37. a

3x2y22 22 b x2y2 23x3y 22 44. a b xy3

43. a

a23b2 22 b ab23 2m22n0 23 45. a2 2 21 b 4m n

46. If x 5 23 and y 5 3, evaluate 2x2 2 xy2 Write each number in scientific notation.

a b b2 p22q2 3 39. a b 2 a2 22 41. a 3 b b 4

Write each number in standard form. 7.3 3 103 5 7,300 5.25 3 1024 5 0.000525

47. 6,750

48. 0.00023

3

Write each number in standard notation. 49. 4.8 3 102

50. 0.25 3 1023

51. Use scientific notation to simplify

145,0002 1350,0002 . 0.000105

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Chapter Review

Rational Exponents and Radicals

SEctIoN 0.3 Definitions and Concepts

Examples

Summary of a1/n definitions: • If a $ 0, then a1/n is the nonnegative number b such that bn 5 a. • If a , 0 and n is odd, then a1/n is the real number b such that bn 5 a. • If a , 0 and n is even, then a1/n is not a real number. Rule for rational exponents: If m and n are positive integers, mn is in lowest terms, and a1/n is a real number, then am/n 5 1a1/n2 m 5 1am2 1/n

Definition of "a: n "a 5 a1/n

Properties of radicals: If all radicals are real numbers and there are no divisions by 0, then n n n • "ab 5 "a"b n

n

m

n

12272 1/3 5 23 because 1232 3 5 227

12162 1/2 is not a real number because no real number squared is 216.

82/3 5 181/32 2 5 22 5 4 or 1822 1/3 5 641/3 5 4

5 5 5 10 " 32x10 5 " 32" x 5 2x2

4 12 x " x12 x3 x12/4 5 5 4 5 Å 625 5 5 "625 4

#"a 5 #"a 5"a

•

161/2 5 4 because 42 5 16

3 " 125 5 1251/3 5 5

n

a "a • 5 n Åb "b

77

n m

3 3 3 # "64 5 " 8 5 2  #" 64 5 "4 5 2

mn

6 "64 5 " 64 5 2

2#3

EXERCISES Simplify each expression, if possible. 52. 1211/2

54. 132x52 1/5

56. 121,000x62 1/3

58. 1x12y22 1/2

60. a

2c2/3c5/3 1/3 b c22/3

53. a

27 b 125

Simplify each expression. 1/3

55. 181a42 1/4

57. 1225x22 1/2 59. a

61. a

x12 21/2 b y4

a21/4a3/4 21/2 b a9/2

Simplify each expression. 2/3

23/5

63. 32

16 3/4 64. a b 81

32 2/5 65. a b 243

68. 12216x32 2/3

16 23/4 67. a b 625 pa/2pa/3 69. pa/6

71. 2"49

72.

9 Å 25

73.

27 Å 125

76.

m8n4 Å p16

77.

a15b10 Å c5

74. "x2y4

Simplify and combine terms. 78. "50 1 "8

80. "24x 2 "3x "7

4

3

3 3 75. " x

4

3

62. 64

8 22/3 66. a b 27

70. "36

3

4

5

79. "12 1 "3 2 "27

Rationalize each denominator. 81. 83.

"5 1 3 " 2

82. 84.

"8 2 8

3 " 25

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78

Chapter 0

A Review of Basic Algebra

Rationalize each numerator. 85.

"2 5

86.

SEctIoN 0.4

"5 5

87.

"2x 3

88.

3 7x 3" 2

Polynomials

Definitions and Concepts

Examples

Monomial: A polynomial with one term.

2, 3x, 4x2y, 2x3y2z

Binomial: A polynomial with two terms.

2t 1 3, 3r2 2 6r, 4m 1 5n

Trinomial: A polynomial with three terms.

3p2 2 7p 1 8, 3m2 1 2n 2 p

The degree of a monomial is the sum of the exponents on its variables.

The degree of 4x2y3 is 2 1 3 5 5

The degree of a polynomial is the degree of the term in the polynomial with highest degree.

The degree of the first term of 3p2q3 2 6p3q4 1 9pq2 is 2 1 3 5 5, the degree of the second term is 3 1 4 5 7, and the degree of the third term is 1 1 2 5 3. The degree of the polynomial is the largest of these. It is 7.

a 1b 1 c 1 d 1 c 2 5 ab 1 ac 1 ad 1 c

Multiplying a monomial times a polynomial:

Addition and subtraction of polynomials: To add or subtract polynomials, remove parentheses and combine like terms.

3 1x 1 22 5 3x 1 6 2x 13x2 2 2y 1 32 5 6x3 2 4xy 1 6x

Add: 13x2 1 5x2 1 12x2 2 2x2 .

13x2 1 5x2 1 12x2 2 2x2 5 3x2 1 5x 1 2x2 2 2x 5 3x2 1 2x2 1 5x 2 2x 5 5x2 1 3x

Subtract: 14a2 2 5b2 2 13a2 2 7b2 .

14a2 2 5b2 2 13a2 2 7b2 5 4a2 2 5b 2 3a2 1 7b 5 4a2 2 3a2 2 5b 1 7b 5 a2 1 2b

Special products:

1x 1 y2 2 5 x2 1 2xy 1 y2

12m 1 32 2 5 4m2 1 12m 1 9

1x 1 y2 1x 2 y2 5 x2 2 y2

12m 1 n2 12m 2 n2 5 4m2 2 2mn 1 2mn 2 n2 5 4m2 2 n2

1x 2 y2 2 5 x2 2 2xy 1 y2 Multiplying a binomial times a binomial: Use the FOIL method.

14t 2 3s2 2 5 16t2 2 24ts 1 9s2

12a 1 b2 1a 2 b2 5 2a 1a2 1 2a 12b2 1 ba 1 b 12b2 5 2a2 2 2ab 1 ab 2 b2 5 2a2 2 ab 2 b2

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Chapter Review

Definitions and Concepts

Examples

The conjugate of a 1 b is a 2 b. To rationalize the denominator of a radical expression, multiply both the numerator and denominator of the rational expression by the conjugate of the denominator.

Rationalize the denominator: "x 1 2 x

5

x 1"x 2 22

"x 1 2 x

1"x 1 22 1"x 2 22

.

Multiply the numerator and denominator by the conjugate of "x 1 2.

5 Division of polynomials: To divide polynomials, use long division.

x"x 2 2x x24

Simplify.

Divide: 2x 1 3q6x3 1 7x2 2 x 1 3. 3x2 2x 1 3q 6x3 1 7x2 6x3 1 9x2 2 2x2 2 2x2

2 x11 2 x13 2 x 2 3x 1 2x 1 3 1 2x 1 3 0

EXERCISES Give the degree of each polynomial and tell whether the polynomial is a monomial, a binomial, or a trinomial. 89. x3 2 8 90. 8x 2 8x2 2 8 91. "3x2

92. 4x4 2 12x2 1 1 Perform the operations and simplify. 93. 2 1x 1 32 1 3 1x 2 42 94. 3x2 1x 2 12 2 2x 1x 1 32 2 x2 1x 1 22 95. 13x 1 22 13x 1 22 96. 13x 1 y2 12x 2 3y2

97. 14a 1 2b2 12a 2 3b2 98. 1z 1 32 13z2 1 z 2 12 99. 1an 1 22 1an 2 12

100. Q"2 1 xR

Rationalize each denominator. 2 22 104. 103. "3 2 1 "3 2 "2

105.

"x 2 2 2x

Rationalize each numerator. "x 1 2 107. 5 Perform each division. 3x2y2 109. 6x3y

106.

108.

110.

"x 2 "y

"x 1 "y

1 2 "a a

4a2b3 1 6ab4 2b2

2

101. Q"2 1 1RQ"3 1 1R

3 3 3 102. Q"3 2 2RQ"9 1 2"3 1 4R

111. 2x 1 3q2x3 1 7x2 1 8x 1 3 112. x2 2 1qx5 1 x3 2 2x 2 3x2 2 3

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79

80

Chapter 0

A Review of Basic Algebra

Factoring Polynomials

SEctIoN 0.5 Definitions and Concepts

Examples

Factoring out a common monomial:

3p3 2 6p2q 1 9p 5 3p 1p2 2 2pq 1 32

Factoring the difference of two squares:

4x2 2 9 5 12x 1 32 12x 2 32

ab 1 ac 5 a 1b 1 c2

x2 2 y2 5 1x 1 y2 1x 2 y2

Factoring trinomials: • Trinomial squares

x2 1 2xy 1 y2 5 1x 1 y2 2

9a2 1 12ab 1 4b2 5 13a 1 2b2 13a 1 2b2 5 13a 1 2b2 2

x2 2 2xy 1 y2 5 1x 2 y2 2

r2 2 4rs 1 4s2 5 1r 2 2s2 1r 2 2s2 5 1r 2 2s2 2

• To factor general trinomials use trial and error or grouping. Factoring the sum and difference of two cubes: x3 1 y3 5 1x 1 y2 1x2 2 xy 1 y22 x3 2 y3 5 1x 2 y2 1x2 1 xy 1 y22

6x2 2 5x 2 6 5 12x 2 32 13x 1 22 r3 1 8 5 1r 1 22 1r2 2 2r 1 42

27a3 2 8b3 5 13a 2 2b2 19a2 1 6ab 1 4b22

EXERCISES Factor each expression completely, if possible. 113. 3t3 2 3t

114. 5r3 2 5

115. 6x2 1 7x 2 24

116. 3a2 1 ax 2 3a 2 x

117. 8x3 2 125

118. 6x2 2 20x 2 16

SEctIoN 0.6

119. x2 1 6x 1 9 2 t2

120. 3x2 2 1 1 5x

121. 8z3 1 343

122. 1 1 14b 1 49b2

123. 121z2 1 4 2 44z

124. 64y3 2 1,000

125. 2xy 2 4zx 2 wy 1 2zw 126. x8 1 x4 1 1

Algebraic Fractions

Definitions and Concepts

Examples

Properties of fractions: If there are no divisions by 0, then a c • 5 if and only if ad 5 bc. b d

3x 6x 5 because 13x2 8 and 4 16x2 both equal 24x. 4 8

a ax 5 b bx a c ac • ? 5 b d bd •

•

a c ad 4 5 b d bc

6x2 3?2?x?x 3 5 5 8x3 2?4?x?x?x 4x 3p 2p 3p ? 2p 3p ? 2p p2 ? 5 5 5 2 2q 6q 2q ? 6q 2q ? 3 ? 2q 2q 2t 2t 2t 6s 2t ? 6s 2t ? 2 ? 3s 4 5 ? 5 5 52 3s 6s 3s 2t 3s ? 2t 3s2t

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Review

Definitions and Concepts a c a2c 2 5 b b b a a • a?15a • 5a • 51 1 a a 2a a 2a 52 52 • 5 b 2b 2b b a a 2a 2a 5 52 • 2 5 b 2b b 2b •

a c a1c 1 5 b b b

•

Examples x y x1y 3p p 3p 2 p 2p p 2 5 5 5 1 5 4 4 4 2q 2q 2q 2q q 7 7 7?151 57 51 1 7 7 27 7 27 5 52 52 2 22 22 2 7 7 27 27 2 5 5 52 2 22 2 22

Simplifying rational expressions: To simplify a rational expression, factor the numerator and denominator, if possible, and divide out factors that are common to the numerator and denominator.

2 2 x To simplify 2x 2 4 , factor 21 from the numerator and 2 from the denominator to get

Adding or subtracting rational expressions: To add or subtract rational expressions with unlike denominators, find the LCD of the expressions, write each expression with a denominator that is the LCD, add or subtract the expressions, and simplify the result, if possible.

2x 2x 2x 1x 2 32 2x 1x 1 22 2 5 2 1x 1 22 1x 2 32 1x 2 32 1x 1 22 x12 x23 2x 1x 2 32 2 2x 1x 1 22 5 1x 1 22 1x 2 32

21 1x 2 22 21 21 122 1 x2 1 22x 5 5 5 52 2x 2 4 2 1x 2 22 2 1x 2 22 2 2

2x2 2 6x 2 2x2 2 4x 1x 1 22 1x 2 32 210x 5 1x 1 22 1x 2 32 5

Complex fractions: To simplify complex fractions, multiply the numerator and denominator of the complex fraction by the LCD of all the fractions.

y y 2ya1 2 b 2 2 2y 2 y2 y 12 2 y2 5 5 5 1 1 21y 21y 1 1 1 2ya 1 b y 2 y 2

12

EXERCISES Simplify each rational expression. 22x a2 2 9 127. 2 128. 2 x 2 4x 1 4 a 2 6a 1 9 Perform each operation and simplify. Assume that no denominators are 0.

81

133.

x2 1 x 2 6 x2 2 x 2 6 x2 2 4 ? 4 x2 2 x 2 6 x2 1 x 2 2 x2 2 5x 1 6

2x2 2 2x 2 4 x2 2 x 2 2 2x 1 6 4 b 2 x15 x2 2 25 x 2 2x 2 15 2 3x 1 135. x24 x15 134. a

129.

x2 2 4x 1 4 x2 1 5x 1 6 ? x12 x22

130.

2y2 2 11y 1 15 y2 2 2y 2 8 ? 2 y2 2 6y 1 8 y 2y26

136.

5x 3x 1 7 2x 1 1 2 1 x22 x12 x12

131.

2t2 1 t 2 3 10t 1 15 4 2 2 3t 2 7t 1 4 3t 2 t 2 4

137.

x x x 1 1 x21 x22 x23

132.

p2 1 7p 1 12 p2 2 9 4 p3 1 8p2 1 4p p2

138.

x 3x 1 7 2x 1 1 2 1 x11 x12 x12

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82

139. 140.

Chapter 0

A Review of Basic Algebra

3 1x 1 12 5 1x2 1 32 x 2 1 x x2 x11 3x x2 1 4x 1 3 x2 1 x 2 6 1 2 2 x11 x 1 3x 1 2 x2 2 4

Simplify each complex fraction. Assume that no denominators are 0. 5x 2 141. 3x2 8 1 1 1 x y 143. x2y

3x y 142. 6x y2

144.

x21 1 y21 y21 2 x21

CHAPTER TEST Consider the set 5 27, 223, 0, 1, 3, "10, 4 6 . 1. List the numbers in the set that are odd integers. 2. List the numbers in the set that are prime numbers. Determine which property justifies each statement. 3. 1a 1 b2 1 c 5 1b 1 a2 1 c 4. a 1b 1 c2 5 ab 1 ac

Graph each interval on a number line. 6. 12`, 232 c 3 6,` 2 5. 24 , x # 2

Write each number in standard notation. 17. 3.7 3 103

18. 1.2 3 1023

Simplify each expression. Assume that all variables represent positive numbers, and write all answers without using negative exponents. 36 3/2 19. 125a42 1/2 20. a b 81 8t6 22/3 3 21. a 9 b 22. "27a6 27s 23. "12 1 "27

3 3 24. 2" 3x4 2 3x" 24x

"x 2 2 "x 2 "y

25. Rationalize the denominator: Write each expression without using absolute value symbols. 8. 0 x 2 7 0 , when x , 0 7. 0 217 0 Find the distance on a number line between points with the following coordinates. 10. 220 and 212 9. 24 and 12 Simplify each expression. Assume that all variables represent positive numbers, and write all answers without using negative exponents. r2r3s 12. 4 2 11. x4x5x2 rs 1a21a22 22 x0x2 6 13. 14. a b a23 x22 Write each number in scientific notation. 15. 450,000

16. 0.000345

26. Rationalize the numerator: Perform each operation. 27. 1a2 1 32 2 12a2 2 42

x

"x 1 "y

. .

28. 13a3b22 122a3b42 29. 13x 2 42 12x 1 72 30. 1an 1 22 1an 2 32 31. 1x2 1 42 1x2 2 42

32. 1x2 2 x 1 22 12x 2 32 33. x 2 3q6x2 1 x 2 23 34. 2x 2 1q2x3 1 3x2 2 1 Factor each polynomial. 35. 3x 1 6y 36. x2 2 100 37. 10t2 2 19tw 1 6w2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Test

38. 3a3 2 648 39. x4 2 x2 2 12 40. 6x4 1 11x2 2 10 Perform each operation and simplify if possible. Assume that no denominators are 0. x 2 41. 1 x12 x12 42.

x x 2 x11 x21

43.

x2 1 x 2 20 x2 2 25 ? x2 2 16 x25

44.

x12 x2 2 4 4 x2 1 2x 1 1 x11

83

Simplify each complex fraction. Assume that no denominators are 0. 1 1 1 a b x21 46. 21 45. 1 x 1 y21 b

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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

1

Equations and Inequalities

CAREERS AND MATHEMATICS:

Marketing

© Istockphoto.com/Arthur Kwiatkowski

People who work in the field of marketing coordinate their companies’ market research, marketing strategy, sales, advertising, promotion, pricing, product development, and public relations activities. Advertising managers direct a firm’s promotional campaign while marketing managers promote products and services. Promotions managers work on incentives to increase sales while public relations managers plan and direct the public image for the employer. Sales managers plan and direct the distribution of the product to the customer.

Education and Mathematics Required •

•

For marketing, sales, and promotions management positions, employers often prefer a bachelor’s or master’s degree in business administration with an emphasis on marketing. For advertising management positions, some employers prefer a bachelor’s degree in advertising or journalism. For public relations management positions, some employers prefer a bachelor’s or master’s degree in public relations or journalism. College Algebra, Business Calculus I and II, and Economic Statistics are required.

1.1

Linear Equations and Rational Equations

1.2

Applications of Linear Equations

1.3

Quadratic Equations

1.4

Applications of Quadratic Equations

1.5

Complex Numbers

1.6

Polynomial and Radical Equations

1.7

Inequalities

1.8

Absolute Value Chapter Review Chapter Test Cumulative Review Exercises

How Marketing Managers Use Math and Who Employs Them • •

Statistics is used for predicting sales and effectiveness of advertising campaigns. These managers are found in virtually every industry. They are employed in wholesale trade, retail trade, manufacturing, the finance and insurance industries, as well as professional, scientific, and technical services, public and private educational services, and healthcare.

Career Outlook and Salary • •

Overall employment of advertising, marketing, promotions, public relations, and sales managers is expected to increase by 13 percent through 2018. The median annual wages in May 2008 were $80,220 for advertising and promotions managers, $108,580 for marketing managers, $97,260 for sales managers, and $89,430 for public relations managers.

For more information see: www.bls.gov/oco

The main topic of this chapter is equations—one of the most important concepts in algebra. Equations are used in almost every academic discipline and especially in chemistry, physics, medicine, computer science, and business.

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86

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1.1 Linear Equations and Rational Equations In this section, we will learn to 1. Use the properties of equality. 2. Solve linear equations. 3. Solve rational equations. 4. Solve formulas for a specific variable.

stocknadia/Shutterstock.com

It’s been said that “weddings today are as beautiful as they are expensive.” Suppose a couple budgets $3,000 for their wedding reception at a historic home. If $600 is charged for renting the home and there is a $24-per-person fee for food and beverages, how many guests can the couple accommodate at their reception? If we let the variable x represent the number of guests, the expression 600 1 24x represents the cost for the reception. That is, $600 for the home rental plus $24 times the number of guests, x. We want to know the value of x that makes the expression equal $3,000. We can write the statement 600 1 24x 5 3000 to indicate that the two quantities are equal. A statement indicating that two quantities are equal is called an equation. In this section, we will review how to solve equations of this type. If x 5 100, the equation is true because when we substitute 100 for x, we obtain a true statement. 600 1 24 11002 5 3000 600 1 2400 5 3000 3000 5 3000 The couple can accommodate 100 guests at their wedding reception. An equation can be either true or false. For example, the equation 2 1 2 5 4 is true, and the equation 2 1 3 5 6 is false. An equation such as 3x 2 2 5 10 can be true or false depending on the value of x. If x 5 4, the equation is true, because 4 satisfies the equation. 3x 2 2 5 10

3 142 2 2 0 10

Substitute 4 for x. Read 0 as “is possibly equal to.”

12 2 2 0 10 10 5 10 This equation is false for all other values of x. Any number that satisfies an equation is called a solution or root of the equation. The set of all solutions of an equation is called its solution set. We have seen that the solution set of 3x 2 2 5 10 is 5 4 6 . To solve an equation means to find its solution set. There can be restrictions on the values of a variable. For example, in the fraction x2 1 4 x22 we cannot replace x with 2, because that would make the denominator equal to 0.

EXAMPLE 1

Finding the Restrictions on the Values of a Variable Find the restrictions on the values of b in the equation:

7b 2 5 . b16 b21

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Section 1.1

SOLUTION

Self Check 1

Linear Equations and Rational Equations

87

For b 7b 1 6 to be a real number, b cannot be 26 because that would make the denominator 0. For b 22 1 to be a real number, b cannot be 1 because that would make the denominator 0. Thus, the values of b are restricted to the set of all real numbers except 1 or 26.

Find the restrictions on a:

5 3a 5 . a15 a22

Now Try Exercise 13.

There are three types of equations: identities, contradictions, and conditional equations. These are defined and illustrated in the following table.

Type of Equation

Definition

Example

Identity

Every acceptable real number replacement for the variable is a solution.

Contradiction

No real number is a solution.

x5x11 The equation has no solution. No real number can be 1 greater than itself.

Conditional Equation

Solution set contains some but not all real numbers.

3x 2 2 5 10 The equation has one solution, the number 4.

x2 2 9 5 1x 1 32 1x 2 32 Every real number x is a solution.

Two equations with the same solution set are called equivalent equations.

1. Use the Properties of Equality There are certain properties of equality that we can use to transform equations into equivalent but less complicated equations. If we use these properties, the resulting equations will be equivalent and will have the same solution set.

Properties of Equality

The Addition and Subtraction Properties If a, b, and c are real numbers and a 5 b, then a1c5b1c

and

a2c5b2c

The Multiplication and Division Properties If a, b, and c are real numbers and a 5 b, then ac 5 bc

and

a b 5 c c

1c 2 02

The Substitution Property In an equation, a quantity may be substituted for its equal without changing the truth of the equation.

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88

Chapter 1

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2. Solve Linear Equations The easiest equations to solve are the first-degree or linear equations. Since these equations involve first-degree polynomials, they are also called first-degree polynomial equations. Linear Equations

A linear equation in one variable (say, x) is any equation that can be written in the form ax 1 b 5 0

(a and b are real numbers and a 2 0)

To solve the linear equation 2x 1 3 5 0, we subtract 3 from both sides of the equation and divide both sides by 2. 2x 1 3 5 0 2x 1 3 2 3 5 0 2 3

To undo the addition of 3, subtract 3 from both sides.

2x 5 23 2x 3 52 2 2 3 x52 2

To undo the multiplication by 2, divide both sides by 2.

To verify that 232 satisfies the equation, we substitute 232 for x and simplify: 2x 1 3 5 0 3 2a2 b 1 3 0 0 2 23 1 3 0 0

3 Substitute 2 for x. 2 3 2a2 b 5 23 2

050 Since both sides of the equation are equal, the solution checks. In Exercise 74, you will be asked to solve the general linear equation ax 1 b 5 0 for x, thereby showing that every conditional linear equation has exactly one solution.

EXAMPLE 2 SOLUTION

Solving a Linear Equation

Find the solution set: 3 1x 1 22 5 5x 1 2. We proceed as follows:

3 1x 1 22 5 5x 1 2 3x 1 6 5 5x 1 2

3x 1 6 2 3x 5 5x 2 3x 1 2 6 5 2x 1 2 6 2 2 5 2x 1 2 2 2 4 5 2x 4 2x 5 2 2 25x

Use the Distributive Property and remove parentheses. Subtract 3x from both sides. Combine like terms. Subtract 2 from both sides. Simplify. Divide both sides by 2. Simplify.

Since all of the above equations are equivalent, the solution set of the original equation is 5 2 6 . Verify that 2 satisfies the equation. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 1.1

Self Check 2

Linear Equations and Rational Equations

89

Find the solution set: 4 1x 2 32 5 7x 2 3. Now Try Exercise 35.

ACCENT ON TECHNOLOgy

Checking Solutions to Linear Equations Using the table feature on a graphing calculator, we can easily check the solution of a linear equation. We enter each side of the equation into the graph editor; go to the table, and then enter the value of x that we found as the solution. The value of both entries in the table should be the same. These steps are shown in Figure 1-1 below for Example 2. We see that both Y1 and Y2 equal 12 when x 5 2. This verifies the solution. Enter each side of the equation.

go to the table and enter the value x 5 2.

(a)

(b) FIgURE 1-1

EXAMPLE 3

Solving a Linear Equation with Fractions 3 2 1 Find the solution set: y 2 5 y. 2 3 5

SOLUTION

To clear the equation of fractions, we multiply both sides by the least common denominator (LCD) of the three fractions and proceed as follows: 3 2 1 y2 5 y 2 3 5

3 2 1 30a y 2 b 5 30a yb 2 3 5 45y 2 20 5 6y 45y 2 20 1 20 5 6y 1 20 45y 5 6y 1 20 45y 2 6y 5 6y 2 6y 1 20 39y 5 20 39y 20 5 39 39 20 y5 39

The solution set is e

Self Check 3

Multiply both sides by 30, the LCD of

3 2 1 , , and . 2 3 5

Remove parentheses and simplify. Add 20 to both sides. Simplify. Subtract 6y from both sides. Combine like terms. Divide both sides by 39. Simplify.

20 20 satisfies the equation. f . Verify that 39 39

Find the solution set:

2 p p235 . 3 6

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90

Chapter 1

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EXAMPLE 4 SOLUTION

Solving Linear Equations

Solve: a. 3 1x 1 52 5 3 11 1 x2 3 1x 1 52 5 3 11 1 x2 3x 1 15 5 3 1 3x

a.

b. 5 1 5 1x 1 22 2 2x 5 3x 1 15

3x 2 3x 1 15 5 3 1 3x 2 3x 15 5 3

Remove parentheses. Subtract 3x from both sides. Combine like terms.

Since 15 5 3 is false, the equation has no roots. Its solution set is the empty set, which is denoted as [. This equation is a contradiction. b. 5 1 5 1x 1 22 2 2x 5 3x 1 15 5 1 5x 1 10 2 2x 5 3x 1 15 3x 1 15 5 3x 1 15

Remove parentheses. Simplify.

Because both sides of the final equation are identical, every value of x will make the equation true. The solution set is the set of all real numbers. This equation is an identity. Self Check 4

Solve: a. 22 1x 2 42 1 6x 5 4 1x 1 12 b. 2 1x 1 12 1 4 5 2 1x 1 32 Now Try Exercise 51.

3. Solve Rational Equations Rational equations are equations that contain rational expressions. Some examples of rational equations are 2 5 7, x23

x11 3 5 , x22 x22

and

x12 1 1 2 51 x13 x 1 2x 2 3

When solving these equations, we will multiply both sides by a quantity containing a variable. When we do this, we could inadvertently multiply both sides of an equation by 0 and obtain a solution that makes the denominator of a fraction 0. In this case, we have found a false solution, called an extraneous solution. These solutions do not satisfy the equation and must be discarded. The following equation has an extraneous solution. x11 3 5 x22 x22 x11 3 1x 2 22 a b 5 1x 2 22 a b x22 x22 x1153 x52

Multiply both sides by x 2 2. x22 51 x22 Subtract 1 from both sides.

If we check by substituting 2 for x, we obtain 0’s in the denominator. Thus, 2 is not a root. The solution set is [.

EXAMPLE 5

Solving a Rational Equation Solve:

x12 1 1 2 5 1. x13 x 1 2x 2 3

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Section 1.1

SOLUTION

Linear Equations and Rational Equations

91

Note that x cannot be –3, because that would cause the denominator of the first fraction to be 0. To find other restrictions, we factor the trinomial in the denominator of the second fraction. x2 1 2x 2 3 5 1x 1 32 1x 2 12

Since the denominator will be 0 when x 5 23 or x 5 1, x cannot be 23 or 1. x12 1 1 2 51 x13 x 1 2x 2 3 x12 1 1 51 1x 1 32 1x 2 12 x13 1 x12 1x 1 32 1x 2 12 c 1 d 5 1x 1 32 1x 2 12 1 1 2 x13 x 1 3 1x 2 12

1x 1 32 1x 2 12 a

1 x12 5 1x 1 32 1x 2 12 1 b 1 1x 1 32 1x 2 12 1x 1 32 1x 2 12 x13 1x 2 12 1x 1 22 1 1 5 1x 1 32 1x 2 12 x2 1 x 2 2 1 1 5 x2 1 2x 2 3 x 2 1 5 2x 2 3

Factor x2 1 2x 2 3. Multiply both sides by 1x 1 32 1x 2 12 . Remove brackets. Simplify. Multiply the binomials. Subtract x2 from both sides and combine like terms.

25x

Add 3 and subtract x from both sides.

Because 2 is a meaningful replacement for x, it is a root. However, it is a good idea to check it. x12 1 1 2 51 x13 x 1 2x 2 3 212 1 01 1 2 213 2 1 2 122 2 3

Substitute 2 for x.

4 1 1 01 5 5 151 Since 2 satisfies the equation, it is a root.

Self Check 5

Solve:

3 7 1 5 2. 5 x12

Now Try Exercise 61.

4. Solve Formulas for a Specific Variable Many equations, called formulas, contain several variables. For example, the formula that converts degrees Celsius to degrees Fahrenheit is F 5 95C 1 32. If we want to change a large number of Fahrenheit readings to degrees Celsius, it would be tedious to substitute each value of F into the formula and then repeatedly solve it for C. It is better to solve the formula for C, substitute the values for F, and evaluate C directly.

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92

Chapter 1

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EXAMPLE 6

Solving a Formula for a Specific Variable 9 Solve F 5 C 1 32 for C. 5

SOLUTION

We use the same methods as for solving linear equations. 9 F 5 C 1 32 5 9 F 2 32 5 C 5

Subtract 32 from both sides.

5 5 9 1F 2 322 5 a Cb 9 9 5 5 1F 2 322 5 C 9

5 Multiply both sides by . 9 Simplify.

This result can also be written in the form C 5

Self Check 6

5F 2 160 . 9

5 Solve C 5 1F 2 322 for F. 9 Now Try Exercise 79.

EXAMPLE 7

Solving a Formula for a Specific Variable The formula A 5 P 1 Prt is used to find the amount of money in a savings account at the end of a specified time. A represents the amount, P represents the principal (the original deposit), r represents the rate of simple interest per unit of time, and t represents the number of units of time. Solve this formula for P.

SOLUTION

We factor P from both terms on the right side of the equation and proceed as follows: A 5 P 1 Prt

A 5 P 11 1 rt2 A 5P 1 1 rt A P5 1 1 rt Self Check 7

Factor out P. Divide both sides by 1 1 rt.

Solve pq 5 fq 1 fp for f. Now Try Exercise 85.

Self Check Answers

1. all real numbers except 2 or 25 4. a. no solution pq 7. f 5 q1p

2. 5 23 6

b. all real numbers

5. 3

3. 5 6 6

9 6. F 5 C 1 32 5

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Section 1.1

Exercises 1.1 getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. If a number satisfies an equation, it is called a or a of the equation. 2. If an equation is true for all values of its variable, it is called an . 3. A contradiction is an equation that is true for values of its variable. equation is true for some values of its 4. A variable and is not true for others. 5. An equation of the form ax 1 b 5 0 is called a equation. 6. If an equation contains rational expressions, it is called a equation. 7. A conditional linear equation has root. 8. The of a fraction can never be 0.

Practice Each quantity represents a real number. Find any restrictions on x. 1 9. x 1 3 5 1 10. x 2 7 5 14 2 1 3 11. 5 12 12. 5 9x x x22 8 5 x 4 13. 5 14. 52 x23 x12 x23 x14 15.

1 5x 5 2 x23 x 2 16

16.

1 5 5 12 x2 2 3x 2 4 x

18. 3x 1 2 5 x 1 8 19. 2 1n 1 22 2 5 5 2n

20. 3 1m 1 22 5 2 1m 1 32 1 m x17 21. 57 2 x 22. 2 7 5 14 2 23. 2 1a 1 12 5 3 1a 2 22 2 a 24. x2 5 1x 1 42 1x 2 42 1 16 25. 3 1x 2 32 5

6x 2 18 2

93

26. x 1x 1 22 5 1x 1 12 2 27.

3 51 b23

28. x2 2 8x 1 15 5 1x 2 32 1x 1 52

29. 2x2 1 5x 2 3 5 12x 2 12 1x 1 32

30. 2x2 1 5x 2 3 5 2xax 1

19 b 2

Solve each equation. If an equation has no solution, so indicate. 31. 2x 1 7 5 10 2 x 32. 9a 2 3 5 15 1 3a 33. 5 1x 2 22 5 2 1x 1 42

35. 7 12x 1 52 2 6 1x 1 82 5 7

34. 5 1r 2 42 5 25 1r 2 42

36. 6 1x 2 52 2 4 1x 1 22 5 21

5 z2857 3 z 39. 1 2 5 4 5 3x 2 2 7 5 2x 1 41. 3 3 3x 1 1 1 5 43. 20 2 37.

45. 46. 47.

Solve each equation, if possible. Classify each one as an identity, a conditional equation, or a contradiction. 17. 2x 1 5 5 15

Linear Equations and Rational Equations

48. 49.

4 y 1 12 5 24 3 3p 40. 2 p 5 24 7 7 15 42. x 1 5 5 x 1 2 2 7 x 4x 1 3 44. 2x 2 1 5 6 6 6 38.

31x x17 1 5 4x 1 1 3 2 23 14 1 x2 3x 5 2 12x 1 12 2 2 2 3 13x 2 22 2 10x 2 4 5 0 2 1a 2 12 2 a 1a 2 32 1 5 5 7 7 1y 1 22 2 y2 5y121 3 3 1t 1 12 1t 2 12 5 1t 1 22 1t 2 32 1 4

51. x 1x 1 22 5 1x 1 12 2 2 1 50.

52. 1x 2 22 1x 2 32 5 1x 1 32 1x 1 42

53. 2 1s 1 22 1 1s 1 32 2 5 s 1s 1 52 1 2a 3 1 4 1 5 x 2 x 2 1 1 55. 1 5 x11 3 x11

17 1 sb 2

54.

56.

3 1 3 1 5 x22 x x22

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94

57.

Chapter 1

Equations and Inequalities

9t 1 6 7 5 t 1t 1 32 t13

79. Pn 5 L 1

3x2 2 122x 1 12 5 3x 1 5 3x 1 5 2 4 59. 5 1a 2 72 1a 1 22 1a 1 32 1a 1 22 2 1 1 60. 1 5 2 n22 n11 n 2n22

si ;s f

80. Pn 5 L 1

si ;f f

58. x 1

61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72.

mMg ;m r2

82.

1 1 1 5 1 ;f f p q

83.

x y 1 5 1; y a b

84.

x y 2 5 1; a a b

85.

1 1 1 5 1 ;r r r1 r2

86.

1 1 1 5 1 ; r1 r r1 r2

81. F 5

2x 1 3 3x 2 2 5x 2 2 1 2 5 2 x2 1 5x 1 6 x 1x26 x 24 3x 2x x12 2 2 5 2 2 x 1x x 1 5x x 1 6x 1 5 3x 1 5 3 2 13x 2 22 1 2 5 1x 2 22 1x2 2 2x 1 42 x3 1 8 x 24 1 3n 2 4 1 2 2 5 n18 5n 1 42n 1 16 5n 1 2 1 2 13n 2 12 1 2 5 2 11 2 n 27n 1 74n 1 33 7n 1 3 4 2 1 2 2 5 2 a2 2 13a 2 48 a 2 18a 1 32 a 1a26 5 2 6 1 1 5 2 2 y14 y12 y12 y 1 6y 1 8 6 3 1 2 5 2 2a 2 6 3 2 3a a 2 4a 1 3 3y 2y 8 1 5 6 2 3y 2y 1 4 4 2 y2 3 1 2a 2 2 3a 5a 2 2 2 2 5 2 2 a 1 6 1 5a a 261a a 24 a 3a 1 2 2152 2 a12 a 1 4a 1 4 x21 x22 1 2 2x 1 5 x13 x23 32x

Solve each formula for the specified variable. 73. k 5 2.2p; p

74. ax 1 b 5 0; x

75. P 5 2l 1 2w; w

1 76. V 5 pr2h; h 3

1 77. V 5 pr2h; r2 3

78. z 5

x2m ;m s

87. l 5 a 1 1n 2 12 d; n 89. a 5 1n 2 22

91. R 5

92. R 5

180 ;n n

88. l 5 a 1 1n 2 12 d; d

90. S 5

a 2 lr ;a 12r

1 ; r1 1 1 1 1 1 r1 r2 r3 1 ; r3 1 1 1 1 1 r1 r2 r3

Discovery and Writing 93. Explain why a conditional linear equation always has exactly one root. 94. Define an extraneous solution and explain how such a solution occurs.

Review Simplify each expression. Use absolute value symbols when necessary. 95. 125x22 1/2

96. a

97. a

98. a2

125x3 22/3 b 8y6

99. "25y2 101.

a4b12 Å z8 4

25p2 1/2 b 16q4

27y3 1/3 b 1,000x6

3 100. " 2125y9

102.

x10y5 Å z15 5

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Section 1.2

Applications of Linear Equations

95

1.2 Applications of Linear Equations In this section, we will learn to 1. Solve number problems. 2. Solve geometric problems. 3. Solve investment problems. 4. Solve break-point analysis problems. 5. Solve shared-work problems. 6. Solve mixture problems. 7. Solve uniform motion problems.

John Shearer/WireImage

Tim McGraw is one of the most successful country music singers. He is married to country singer Faith Hill and is the son of former baseball player Tug McGraw. His albums sales have exceeded 40 million copies and his concert tours have been attended by thousands of fans. Suppose you hear that for one day only Tim McGraw concert tickets can be purchased for 30% off the original price. Knowing this is a bargain, you immediately purchase a ticket for $28, the selling price. Later that day, this question comes to mind: What was the original price of the concert ticket? A linear equation can be used to model this problem. We can let x represent the original price of the concert ticket and subtract 30% of x (the discount) and we will get the selling price $28. The linear equation is x 2 0.3x 5 28. We can solve this equation to determine the original price of the concert ticket. x 2 0.3x 5 28 0.7x 5 28 28 0.7 x 5 40 x5

Combine like terms. Divide both sides by 0.7.

The original price of the concert ticket is $40. In this section, we will use the equation-solving techniques discussed in the previous section to solve applied problems (often called word problems). To solve these problems, we must translate the verbal description of the problem into an equation. The process of finding the equation that describes the words of the problem is called mathematical modeling. The equation itself is often called a mathematical model of the situation described in the word problem. The following list of steps provides a strategy to follow when we try to find the equation that models an applied problem.

Strategy for Modeling Equations

1. Analyze the problem to see what you are to find. Often, drawing a diagram or making a table will help you visualize the facts. 2. Pick a variable to represent the quantity that is to be found, and write a sentence telling what that variable represents. Express all other quantities mentioned in the problem as expressions involving this single variable. 3. Find a way to express a quantity in two different ways. This might involve a formula from geometry, finance, or physics. 4. Form an equation indicating that the two quantities found in Step 3 are equal. 5. Solve the equation. 6. Answer the questions asked in the problem. 7. Check the answers in the words of the problem.

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96

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This list does not apply to all situations, but it can be used for a wide range of problems with only slight modifications.

1. Solve Number Problems EXAMPLE 1

Solving a Number Problem A student has scores of 74%, 78%, and 70% on three exams. What score is needed on a fourth exam for the student to earn an average grade of 80%?

SOLUTION

To find an equation that models the problem, we can let x represent the required grade on the fourth exam. The average grade will be one-fourth of the sum of the four grades. We know this average is to be 80. The average of the four grades 74 1 78 1 70 1 x 4

equals

the required average grade.

5

80

We can solve this equation for x. 222 1 x 5 80 4 222 1 x 5 320 x 5 98

74 1 78 1 70 5 222 Multiply both sides by 4. Subtract 222 from both sides.

To earn an average of 80%, the student must score 98% on the fourth exam. Self Check 1

A student scores 82%, 96%, 91%, and 92% on four college algebra exams. What score is needed on a fifth exam for the student to earn an average grade of 90%? Now Try Exercise 9.

2. Solve geometric Problems EXAMPLE 2

Solving a geometric Problem A city ordinance requires a man to install a fence around the swimming pool shown in Figure 1-2. He wants the border around the pool to be of uniform width. If he has 154 feet of fencing, find the width of the border. x

16 ft x

x 20 ft x FIgURE 1-2

SOLUTION

We can let x represent the width of the border. The distance around the large rectangle, called its perimeter, is given by the formula P 5 2l 1 2w, where l is the length, 20 1 2x, and w is the width, 16 1 2x. Since the man has 154 feet of fencing, the perimeter will be 154 feet. To find an equation that models the problem, we substitute these values into the formula for perimeter.

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Section 1.2

P 5 2l 1 2w

154 5 2 120 1 2x2 1 2 116 1 2x2

Applications of Linear Equations

97

The formula for the perimeter of a rectangle. Substitute 154 for P, 20 1 2x for l, and 16 1 2x for w.

154 5 40 1 4x 1 32 1 4x

Use the Distributive Property to remove parentheses.

154 5 72 1 8x

Combine like terms.

82 5 8x

Subtract 72 from both sides.

1 10 5 x 4

Divide both sides by 8.

This border will be 1014 feet wide. Self Check 2

In Example 2, if 168 feet of fencing is available, find the border’s width. Now Try Exercise 13.

3. Solve Investment Problems EXAMPLE 3

Solving an Investment Problem A woman invested $10,000, part at 9% and the rest at 14%. If the annual income from these investments is $1,275, how much did she invest at each rate?

SOLUTION

We can let x represent the amount invested at 9%. Then 10,000 2 x represents the amount invested at 14%. Since the annual income from any investment is the product of the interest rate and the amount invested, we have the following information.

Type of investment

Rate

Amount invested

Interest earned

9% investment

0.09

x

0.09x

14% investment

0.14

10,000 2 x

0.14 110,000 2 x2

The total income from these two investments can be expressed in two ways: as $1,275 and as the sum of the incomes from the two investments. The income from the 9% investment

plus

0.09x

1

the income from the 14% investment 0.14 110,000 2 x2

equals

the total income.

5

1,275

We can solve this equation for x.

0.09x 1 0.14 110,000 2 x2 5 1,275

9x 1 14 110,000 2 x2 5 127,500 9x 1 140,000 2 14x 5 127,500 25x 1 140,000 5 127,500 25x 5 212,500 x 5 2,500

To eliminate the decimal points, multiply both sides by 100. Use the Distributive Property to remove parentheses. Combine like terms. Subtract 140,000 from both sides. Divide both sides by 25.

The amount invested at 9% was $2,500, and the amount invested at 14% was $7,500 1$10,000 2 $2,5002 . These amounts are correct, because 9% of $2,500 is $225, 14% of $7,500 is $1,050, and the sum of these amounts is $1,275.

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Self Check 3

A man invests $12,000, part at 7% and the rest at 9%. If the annual income from these investments is $965, how much was invested at each rate? Now Try Exercise 21.

4. Solve Break-Point Analysis Problems Running a machine involves two costs—setup costs and unit costs. Setup costs include the cost of installing a machine and preparing it to do a job. Unit cost is the cost to manufacture one item, which includes the costs of material and labor.

EXAMPLE 4

Solving a Break-Point Analysis Problem Suppose that one machine has a setup cost of $400 and a unit cost of $1.50, and a second machine has a setup cost of $500 and a unit cost of $1.25. Find the break point (the number of units manufactured at which the cost on each machine is the same).

SOLUTION

We can let x represent the number of items to be manufactured. The cost C1 of using machine 1 is C1 5 400 1 1.5x and the cost C2 of using machine 2 is C2 5 500 1 1.25x The break point occurs when these two costs are equal. The cost of using machine 1

equals

the cost of using machine 2.

400 1 1.5x

5

500 1 1.25x

We can solve this equation for x. 400 1 1.5x 5 500 1 1.25x 1.5x 5 100 1 1.25x 0.25x 5 100 x 5 400

Subtract 400 from both sides. Subtract 1.25x from both sides. Divide both sides by 0.25.

The break point is 400 units. This result is correct, because it will cost the same amount to manufacture 400 units with either machine.

C1 5 $400 1 $1.5 14002 5 $1,000 Self Check 4

and

C2 5 $500 1 $1.25 14002 5 $1,000

An ATM has a setup cost of $3,000 and operating costs averaging $1 per transaction. Another ATM machine has a setup cost of $3,500 and an operating cost of $0.50 per transaction. Find the number of transactions at which the costs for each ATM is the same. Now Try Exercise 29.

5. Solve Shared-Work Problems EXAMPLE 5

Solving a Shared-Work Problem The Toll Way Authority needs to pave 100 miles of interstate highway before freezing temperatures come in about 60 days. Sjostrom and Sons has estimated that it can do the job in 110 days. Scandroli and Sons has estimated that it can do the job in 140 days. If the authority hires both contractors, will the job get done in time?

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Section 1.2

SOLUTION

Applications of Linear Equations

99

1 Since Sjostrom can do the job in 110 days, they can do 110 of the job in one day. 1 Since Scandroli can do the job in 140 days, they can do 140 of the job in one day. If we let n represent the number of days it will take to pave the highway if both contractors work together, they can do 1n of the job in one day. The work that they can do together in one day is the sum of what each can do in one day.

The part Sjostrom can pave in one day

plus

the part Scandroli can pave in one day

equals

1 110

1

1 140

5

the part they can pave together in one day. 1 n

We can solve this equation for n. 1 1 1 1 5 110 140 n 1 1 1 11102 11402 na 1 b 5 11102 11402 na b 110 140 n

Multiply both sides by 11102 11402 n to eliminate the fractions.

11102 11402 n 11102 11402 n 11102 11402 n 1 5 110 140 n

Use the Distributive Property to remove parentheses. 110 140 n 5 1, 5 1, and 5 1. 110 140 n

140n 1 110n 5 15,400 250n 5 15,400

Combine like terms.

n 5 61.6

Divide both sides by 250.

It will take the contractors about 62 days to pave the highway. With any luck, the job will be done in time. Self Check 5

John and Eric both work at Firestone Auto Care. John can install a new set of tires in 45 minutes. Eric is faster and can install a set in 30 minutes. If John and Eric work together, how long will it take them to install one set of tires? Now Try Exercise 33.

6. Solve Mixture Problems EXAMPLE 6

Solving a Mixture Problem A container is partially filled with 20 liters of whole milk containing 4% butterfat. How much 1% milk must be added to obtain a mixture that is 2% butterfat?

SOLUTION

Since the first container shown in Figure 1-3 contains 20 liters of 4% milk, it contains 0.04(20) liters of butterfat. To this amount, we will add the contents of the second container, which holds 0.01(l) liters of butterfat. The sum of these two amounts will equal the number of liters of butterfat in the third container, which is 0.02 120 1 l2 liters of butterfat. This information is presented in table form in Figure 1-4. (20 + l) liters

20 liters

4%

+

l liters

1%

=

2%

FIgURE 1-3 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

100

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Equations and Inequalities

The butterfat in the 4% milk

plus

the butterfat in the 1% milk

equals

4% of 20 liters

1

1% of l liters

5

Percentage of butterfat

?

Amount of milk

the butterfat in the 2% milk.

2% of 120 1 l2 liters 5

Amount of butterfat

4 % milk

0.04

20

0.04(20)

1% milk

0.01

l

0.01(l)

2% milk

0.02

20 1 l

0.02 120 1 l2

FIgURE 1-4

We can solve this equation for l.

0.04 1202 1 0.01 1l2 5 0.02 120 1 l2 4 1202 1 l 5 2 120 1 l2

Multiply both sides by 100.

80 1 l 5 40 1 2l

Remove parentheses.

40 5 l

Subtract 40 and l from both sides.

To dilute the 20 liters of 4% milk to a 2% mixture, 40 liters of 1% milk must be added. To check, we note that the final mixture contains 0.02 1602 5 1.2 liters of pure butterfat, and that this is equal to the amount of pure butterfat in the 4% milk and the 1% milk; 0.04 1202 1 0.01 1402 5 1.2 liters. Self Check 6

Milk containing 4% butterfat is mixed with 8 gallons of milk containing 1% butterfat to make a low-fat cottage cheese mixture containing 2% butterfat. How many gallons of the richer milk is used? Now Try Exercise 39.

7. Solve Uniform Motion Problems EXAMPLE 7

Solving a Uniform Motion Problem A man leaves home driving his Ford F-150 truck at the rate of 50 mph. When his daughter discovers than he has forgotten his wallet, she drives her Ford Mustang after him at a rate of 65 mph. How long will it take her to catch her dad if he had a 15-minute head start?

Patrick Mezirka/Shutterstock.com

SOLUTION

Uniform motion problems are based on the formula d 5 rt, where d is the distance, r is the rate, and t is the time. We can draw a diagram and organize the information given in the problem in a chart as shown in Figures 1-5 and 1-6. In the chart, t represents the number of hours the daughter must drive to overtake her father. Because

the father has a 15-minute, or 14 hour, head start, he has been on the road for Qt 1 14R hours. 50 mph

65 mph

FIgURE 1-5

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Section 1.2

Applications of Linear Equations

d

5

50Qt 1 14R

Man

65t

Daughter

r

?

101

t

50

t 1 14

65

t

FIgURE 1-6

We can set up the following equation and solve it for t. The distance the man drives

equals

the distance the daughter drives.

1 50at 1 b 4

5

65t

We can solve this equation for t. 1 50at 1 b 5 65t 4 50t 1

25 5 65t 2 25 5 15t 2 5 5t 6

Remove parentheses.

Subtract 50t from both sides.

Divide both sides by 15 and simplify.

It will take the daughter 56 hour, or 50 minutes, to overtake her dad. Self Check 7

On a reality television show, an officer traveling in a police cruiser at 90 mph pursues Jennifer who has a 3-minute head start. If the officer overtakes Jennifer in 12 minutes, how fast is Jennifer traveling? Now Try Exercise 53.

Self Check Answers

1. 89 2. 12 feet 3. $5,750 at 7%; $6,250 at 9% 5. 18 minutes 6. 4 7. 72 mph

4. 1,000

Exercises 1.2 getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. To average n scores, by n. 2. The formula for the P 5 2l 1 2w.

4. The number of units manufactured at which the cost on two machines is equal is called the . 5. Distance traveled is the product of the and the . 6. 5% of 30 liters is liters.

the scores and divide

Practice of a rectangle is

3. The simple annual interest earned on an investment is the product of the interest rate and the invested.

Solve each problem. 7. Test scores Tate scored 5 points higher on his midterm and 13 points higher on his final than he did on his first exam. If his mean (average) score was 90, what was his score on the first exam?

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102

Chapter 1

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8. Test scores Courtney took four tests in science class. On each successive test, her score improved by 3 points. If her mean score was 69.5%, what did she score on the first test? 9. Teacher certification On the Illinois certification test for teachers specializing in learning disabilities, a teacher earned the scores shown in the accompanying table. What was the teacher’s score in program development? Human development with special needs

82

Assessment

90

Program development and instruction

?

Professional knowledge and legal issues

78

AVERAGE SCORE

86

10. Golfing Par on a golf course is 72. If a golfer shot rounds of 76, 68, and 70 in a tournament, what will she need to shoot on the final round to average par? 11. Replacing locks A locksmith at Pop-A-Lock charges $40 plus $28 for each lock installed. How many locks can be replaced for $236? 12. Delivering ads A University of Florida student earns $20 per day delivering advertising brochures door-to-door, plus 75¢ for each person he interviews. How many people did he interview on a day when he earned $56? 13. Electronic LED billboard An electronic LED billboard in Times Square is 26 feet taller than it is wide. If its perimeter is 92 feet, find the dimensions of the billboard. 14. Hockey rink A National Hockey League rink is 115 feet longer than it is wide. If the perimeter of the rink is 570 feet, find the dimensions of the rink? 15. Width of a picture frame The picture frame with the dimensions shown in the illustration was built with 14 feet of framing material. Find x its width.

24 ft

x ft

x ft

17. Wading pool dimensions The area of the triangular swimming pool shown in the illustration is doubled by adding a rectangular wading pool. Find the dimensions of the wading pool. (Hint: The area of a triangle 5 12bh, and the area of a rectangle 5 lw.)

Wading pool 20 ft Swimming pool x ft

16 ft

18. House construction A builder wants to install a triangular window with the angles shown in the illustration. What angles will he have to cut to make the window fit? (Hint: The sum of the angles in a triangle equals 180°.)

x°

(x + 30)°

(x + 30)°

19. Length of a living room If a carpenter adds a porch with dimensions shown in the illustration to the living room, the living area will be increased by 50%. Find the length of the living room.

(x + 2) ft

12 ft x ft

Porch

x ft

16. Fencing a garden If a gardener fences in the total rectangular area shown in the illustration instead of just the square area, he will need twice as much fencing to enclose the garden. How much fencing will he need?

12 ft

Living room

(x + 10) ft

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Section 1.2

20. Depth of water in a trough The trough in the illustration has a cross-sectional area of 54 square inches. Find the depth, d, of the trough. (Hint: Area of a trapezoid 5 12h 1b1 1 b22 .) 12 in.

d

8 in.

21. Investment problem An executive invests $22,000, some at 7% and some at 6% annual interest. If he receives an annual return of $1,420, how much is invested at each rate? 22. Financial planning After inheriting some money, a woman wants to invest enough to have an annual income of $5,000. If she can invest $20,000 at 9% annual interest, how much more will she have to invest at 7% to achieve her goal? (See the table.) Type

Rate

Amount

Income

9% investment

0.09

20,000

.09(20,000)

7% investment

0.07

x

.07x

23. Investment problem Equal amounts are invested at 6%, 7%, and 8% annual interest. If the three investments yield a total of $2,037 annual interest, find the total investment. 24. Investment problem A woman invests $37,000, part at 8% and the rest at 912% annual interest. If the 912% investment provides $452.50 more income than the 8% investment, how much is invested at each rate? 25. Ticket sales A full-price ticket for a college basketball game costs $2.50, and a student ticket costs $1.75. If 585 tickets were sold, and the total receipts were $1,217.25, how many tickets were student tickets? 26. Ticket sales Of the 800 tickets sold to a movie, 480 were full-price tickets costing $7 each. If the gate receipts were $4,960, what did a student ticket cost? 27. Discounts After being discounted 20%, a weather radio sells for $63.96. Find the original price. 28. Markups A merchant increases the wholesale cost of a Maytag washing machine by 30% to determine the selling price. If the washer sells for $588.90, find the wholesale cost. 29. Break-point analysis A machine to mill a brass plate has a setup cost of $600 and a unit cost of $3 for each plate manufactured. A bigger machine has a setup cost of $800 but a unit cost of only $2 for each plate manufactured. Find the break point.

Applications of Linear Equations

103

30. Break-point analysis A machine to manufacture fasteners has a setup cost of $1,200 and a unit cost of $0.005 for each fastener manufactured. A newer machine has a setup cost of $1,500 but a unit cost of only $0.0015 for each fastener manufactured. Find the break point. 31. Computer sales A computer store has fixed costs of $8,925 per month and a unit cost of $850 for every computer it sells. If the store can sell all the computers it can get for $1,275 each, how many must be sold for the store to break even? (Hint: The breakeven point occurs when costs equal income.) 32. Restaurant management A restaurant has fixed costs of $137.50 per day and an average unit cost of $4.75 for each meal served. If a typical meal costs $6, how many customers must eat at the restaurant each day for the owner to break even? 33. Roofing houses Kyle estimates that it will take him 7 days to roof his house. A professional roofer estimates that it will take him 4 days to roof the same house. How long will it take if they work together? 34. Sealing asphalt One crew can seal a parking lot in 8 hours and another in 10 hours. How long will it take to seal the parking lot if the two crews work together? 35. Mowing lawns Julie can mow a lawn with a lawn tractor in 2 hours, and her husband can mow the same lawn with a push mower in 4 hours. How long will it take to mow the lawn if they work together? 36. Filling swimming pools A garden hose can fill a swimming pool in 3 days, and a larger hose can fill the pool in 2 days. How long will it take to fill the pool if both hoses are used? 37. Filling swimming pools An empty swimming pool can be filled in 10 hours. When full, the pool can be drained in 19 hours. How long will it take to fill the empty pool if the drain is left open? 38. Preparing seafood Kevin stuffs shrimp in his job as a seafood chef. He can stuff 1,000 shrimp in 6 hours. When his sister helps him, they can stuff 1,000 shrimp in 4 hours. If Kevin gets sick, how long will it take his sister to stuff 500 shrimp? 39. Diluting solutions How much water should be added to 20 ounces of a 15% solution of alcohol to dilute it to a 10% solution?

20 oz

15%

+

Water x oz

0%

=

(20 + x) oz

10%

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40. Increasing concentrations The beaker shown below contains a 2% saltwater solution. a. How much water must be boiled away to increase the concentration of the salt solution from 2% to 3%? b. Where on the beaker would the new water level be?

300 ml 200 ml 100 ml

41. Winterizing cars A car radiator has a 6-liter capacity. If the liquid in the radiator is 40% antifreeze, how much liquid must be replaced with pure antifreeze to bring the mixture up to a 50% solution? 42. Mixing milk If a bottle holding 3 liters of milk contains 312% butterfat, how much skimmed milk must be added to dilute the milk to 2% butterfat? 43. Preparing solutions A nurse has 1 liter of a solution that is 20% alcohol. How much pure alcohol must she add to bring the solution up to a 25% concentration? 44. Diluting solutions If there are 400 cubic centimeters of a chemical in 1 liter of solution, how many cubic centimeters of water must be added to dilute it to a 25% solution? (Hint: 1,000 cc 5 1 liter.) 45. Cleaning swimming pools A swimming pool contains 15,000 gallons of water. How many gallons of chlorine must be added to “shock the pool” and 3 bring the water to a 100 % solution? 46. Mixing fuels An automobile engine can run on a mixture of gasoline and a substitute fuel. If gas costs $3.50 per gallon and the substitute fuel costs $2 per gallon, what percent of a mixture must be substitute fuel to bring the cost down to $2.75 per gallon? 47. Evaporation How many liters of water must evaporate to turn 12 liters of a 24% salt solution into a 36% solution? 48. Increasing concentrations A beaker contains 320 ml of a 5% saltwater solution. How much water should be boiled away to increase the concentration to 6%?

50. Dairy foods How many gallons of cream that is 22% butterfat must be mixed with milk that is 2% butterfat to get 20 gallons of milk containing 4% butterfat? 51. Mixing solutions How many gallons of a 5% alcohol solution must be mixed with 90 gallons of 1% solution to obtain a 2% solution? 52. Preparing medicines A doctor prescribes an ointment that is 2% hydrocortisone. A pharmacist has 1% and 5% concentrations in stock. How much of each should the pharmacist use to make a 1-ounce tube? 53. Driving rates John drove to Daytona Beach, Florida, in 5 hours. When he returned, there was less traffic, and the trip took only 3 hours. If John averaged 26 mph faster on the return trip, how fast did he drive each way? 54. Distance problem Allison drove home at 60 mph, but her brother Austin, who left at the same time, could drive at only 48 mph. When Allison arrived, Austin still had 60 miles to go. How far did Allison drive? 55. Distance problem Two cars leave Hinds Community College traveling in opposite directions. One car travels at 60 mph and the other at 64 mph. In how many hours will they be 310 miles apart? 56. Bank robbery Some bank robbers leave town, speeding at 70 mph. Ten minutes later, the police give chase, traveling at 78 mph. How long, after the robbery, will it take the police to overtake the robbers? 57. Jogging problem Two Michigan State University cross-country runners are 440 yards apart and are running toward each other, one at 8 mph and the other at 10 mph. In how many seconds will they meet? 58. Driving rates One morning, Justin drove 5 hours before stopping to eat lunch at Pizza Hut. After lunch, he increased his speed by 10 mph. If he completed a 430-mile trip in 8 hours of driving time, how fast did he drive in the morning? 59. Boating problem A Johnson motorboat goes 5 miles upstream in the same time it requires to go 7 miles downstream. If the river flows at 2 mph, find the speed of the boat in still water. 60. Wind velocity A plane can fly 340 mph in still air. If it can fly 200 miles downwind in the same amount of time it can fly 140 miles upwind, find the velocity of the wind.

49. Lowering fat How many pounds of extra-lean hamburger that is 7% fat must be mixed with 30 pounds of hamburger that is 15% fat to obtain a mixture that is 10% fat? Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 1.3

Quadratic Equations

61. Feeding cattle A cattleman wants to mix 2,400 pounds of cattle feed that is to be 14% protein. Barley (11.7% protein) will make up 25% of the mixture. The remaining 75% will be made up of oats (11.8% protein) and soybean meal (44.5% protein). How many pounds of each will he use?

105

h ft

62. Feeding cattle If the cattleman in Exercise 61 wants only 20% of the mixture to be barley, how many pounds of each should he use?

6 ft

Discovery and Writing Use a calculator to help solve each problem. 63. Machine tool design 712.51 cubic millimeters of material was removed by drilling the blind hole as shown in the illustration. Find the depth of the hole. (Hint: The volume of a cylinder is given by V 5 pr2h.)

65. Consider the strategy you use to solve investment and uniform motion problems. Describe any similarities you observe in these problem types. 66. Which type of application was hardest for you to solve? Why? What strategy or approach works best for you when approaching solving this problem?

Review Factor each expression. 67. x2 2 2x 2 63

68. 2x2 1 11x 2 21

69. 9x2 2 12x 2 5

70. 9x2 2 2x 2 7

71. x2 1 6x 1 9

72. x2 2 10x 1 25

73. x3 1 8

74. 27a3 2 64

d mm

9 mm

64. Architecture The Norman window with dimensions as shown is a rectangle topped by a semicircle. If the area of the window is 68.2 square feet, find its height h.

1.3 Quadratic Equations In this section, we will learn to 1. Solve quadratic equations using factoring and the Square Root Property. 2. Solve quadratic equations using completing the square. 3. Solve quadratic equations using the Quadratic Formula. 4. Determine the easiest strategy to use to solve a quadratic equation.

Eric Broder Van Dyke/Shutterstock.com

5. Solve formulas for a variable that is squared. 6. Define and use the discriminant. 7. Write rational equations in quadratic form and solve the equations. Fenway Park, America’s most beloved ballpark, is home to the Boston Red Sox baseball club. The park opened in 1912 and is the oldest major league baseball stadium. In baseball, the distance between home plate and first base is 90 feet and the distance between first base and second base is 90 feet. To find the distance between home plate and second base, we can use the Pythagorean Theorem, which states that The sum of the squares of the two legs of a right triangle is equal to the square of its hypotenuse. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

106

Chapter 1

Equations and Inequalities

Because home plate, first base, and second base form a right triangle, we can let x represent the distance between home plate and second base (the hypotenuse of the right triangle) and 90 feet represent the length of each leg. We can then apply the Pythagorean Theorem and write the equation 902 1 902 5 x2. To find the distance between home plate and second base, we must solve this equation. 902 1 902 5 x2 8,100 1 8,100 5 x2

Square 90 two times.

16,200 5 x2

Simplify.

To find x, we must determine what positive number squared gives 16,200. From Chapter 0, we know that this number is the square root of 16,200. "16,200 < 127.3

Use a calculator and round to the nearest tenth.

To the nearest tenth, the distance between home plate and second base is 127.3 feet. Since this equation contains the term x2, it is an example of a new type of equation, called a quadratic equation. In this section, we will learn several strategies for solving these equations.

1. Solve Quadratic Equations Using Factoring and the Square Root Property Polynomial equations such as 2x2 1 11x 2 21 5 0 and 3x2 2 x 1 2 5 0 are called quadratic or second-degree equations.

Quadratic Equation

A quadratic equation is an equation that can be written in the form ax2 1 bx 1 c 5 0, where a, b, and c are real numbers and a 2 0.

To solve quadratic equations by factoring, we can use the following theorem.

Zero-Factor Theorem

If a and b are real numbers, and if ab 5 0, then a50

PROOF

or

b50

Suppose that ab 5 0. If a 5 0, we are finished, because at least one of a or b is 0. If a 2 0, then a has a reciprocal 1a, and we can multiply both sides of the equation ab 5 0 by 1a to obtain ab 5 0 1 1 1ab2 5 102 a a

1 a ? abb 5 0 a 1b 5 0

1 Multiply both sides by . a Use the Associative Property to group

1 and a together. a

1 ?a51 a

b50 Thus, if a 2 0, then b must be 0, and the theorem is proved.

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Section 1.3

EXAMPLE 1

Quadratic Equations

107

Solving a Quadratic Equation by Factoring Solve: 2x2 2 9x 2 35 5 0.

SOLUTION

The left side can be factored and written as 12x 1 52 1x 2 72 5 0

This product can be 0 if and only if one of the factors is 0. So we can use the Zero-Factor Theorem and set each factor equal to 0. We can then solve each equation for x. 2x 1 5 5 0

or

2x 5 25 x52

x2750 x57

5 2

Because 12x 1 52 1x 2 72 5 0 only if one of its factors is zero, 252 and 7 are the only solutions of the equation. Verify that each one satisfies the equation. Self Check 1

Solve: 6x2 1 7x 2 3 5 0. Now Try Exercise 17.

Comment The Zero-Factor Theorem can be used only when there is a constant term of 0 on the right side of the equation.

In many quadratic equations, the quadratic expression does not factor over the set of integers. For example, the left side of x2 2 5x 1 3 5 0 is a prime polynomial and cannot be factored over the set of integers. To develop a method to solve these equations, we consider the equation x2 5 c. If c is positive, it has two real roots that can be found by adding –c to both sides, factoring x2 2 c over the set of real numbers, setting each factor equal to 0, and solving for x. x2 5 c x2 2 c 5 0

Subtract c from both sides.

x2 2 Q"cR 5 0

Q"cR 5 c

2

2

Qx 2 "cRQx 1 "cR 5 0 x 2 "c 5 0

x 5 "c

or x 1 "c 5 0

Factor the difference of two squares. Set each factor equal to 0.

x 5 2"c

The roots of x2 5 c are x 5 "c and x 5 2"c. This fact is summarized in the Square Root Property.

Square Root Property

EXAMPLE 2

If c . 0, the equation x2 5 c has two real roots: x 5 "c

or

x 5 2"c

Solving a Quadratic Equation by Using the Square Root Property Solve: x2 2 8 5 0.

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108

Chapter 1

Equations and Inequalities

SOLUTION

We solve for x2 and use the Square Root Property. x2 2 8 5 0 x 5 "8

x2 5 8 or x 5 2"8

x 5 2"2

"8 5 "4"2 5 2"2

x 5 22"2

Verify that each root satisfies the equation. Self Check 2

Solve: x2 2 12 5 0. Now Try Exercise 21.

EXAMPLE 3 SOLUTION

Solve: 1x 1 42 2 5 27. Again, we will use the Square Root Property. x 1 4 5 "27

1x 1 42 2 5 27 or

x 1 4 5 3"3

x 1 4 5 2"27

"27 5 "9"3 5 3"3

x 1 4 5 23"3

x 5 24 1 3"3

x 5 24 2 3"3

Verify that each root satisfies the equation. Self Check 3

Solve: 12x 1 52 2 5 45. Now Try Exercise 33.

2. Solve Quadratic Equations Using Completing the Square Another way to solve quadratic equations is called completing the square. This method is based on the following products: x2 1 2ax 1 a2 5 1x 1 a2 2

and

x2 2 2ax 1 a2 5 1x 2 a2 2

The trinomials x2 1 2ax 1 a2 and x2 2 2ax 1 a2 are perfect-square trinomials, because each one factors as the square of a binomial. In each case, the coefficient of the first term is 1. If we take one-half of the coefficient of x in the middle term and square it, we obtain the third term. 2 1 c 12a2 d 5 a2 2

and

2 1 c 122a2 d 5 12a2 2 5 a2 2

This suggests that to make x2 1 bx a perfect-square trinomial, we find one-half of b, square it, and add the result to the binomial. For example, to make x2 1 10x a perfect-square trinomial, we find one-half of 10 to get 5, square 5 to get 25, and add 25 to x2 1 10x. 2 1 x2 1 10x 1 c 1102 d 5 x2 1 10x 1 152 2 2

5 x2 1 10x 1 25

Note that x2 1 10x 1 25 5 1x 1 52 2.

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Section 1.3

Quadratic Equations

109

To make x2 2 11x a perfect-square trinomial, we find one-half of 211 to get 121 121 2 square 211 2 to get 4 , and add 4 to x 2 11x.

211 2,

2 11 2 1 x2 2 11x 1 c 12112 d 5 x2 2 11x 1 a2 b 2 2 121 5 x2 2 11x 1 4

Note that x2 2 11x 1

121 11 2 5 ax 2 b . 4 2

To solve a quadratic equation in x by completing the square, we follow these steps.

Strategy for Completing the Square

1. If the coefficient of x2 is not 1, make it 1 by dividing both sides of the equation by the coefficient of x2. 2. If necessary, add a number to both sides of the equation to get the constant on the right side of the equation. 3. Complete the square on x: a. Identify the coefficient of x, take one-half of it, and square the result. b. Add the number found in part a to both sides of the equation. 4. Factor the perfect-square trinomial and combine like terms. 5. Solve the resulting quadratic equation by using the Square Root Property.

To use completing the square to solve x2 2 10x 1 24 5 0, we note that the coefficient of x2 is 1. We move on to Step 2 and subtract 24 from both sides to get the constant term on the right side of the equal sign. x2 2 10x 5 224

We can then complete the square by adding S12 12102 T 5 25 to both sides. 2

x2 2 10x 1 25 5 224 1 25 x2 2 10x 1 25 5 1

Simplify on the right side.

We then factor the perfect square trinomial on the left side. 1x 2 52 2 5 1

Finally, we use the Square Root Property to solve this equation. x2551 x56

EXAMPLE 4

or

x 2 5 5 21 x54

Solving a Quadratic Equation by Completing the Square Use completing the square to solve x2 1 4x 2 6 5 0.

SOLUTION

Here the coefficient of x2 is already 1. We move to Step 2 and add 6 to both sides to isolate the binomial x2 1 4x. x2 1 4x 5 6 We then find the number to add to both sides by completing the square. Since one-half of 4 (the coefficient of x) is 2 and 22 5 4, we add 4 to both sides.

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110

Chapter 1

Equations and Inequalities

x2 1 4x 1 4 5 6 1 4

Add 4 to both sides.

x2 1 4x 1 4 5 10 x 1 2 5 "10

1x 1 22 2 5 10

x 5 22 1 "10

or

Factor x2 1 4x 1 4.

x 1 2 5 2"10

x 5 22 2 "10

Use the Square Root Property.

Verify that each root satisfies the original equation. Self Check 4

Solve: x2 2 2x 2 9 5 0. Now Try Exercise 49.

EXAMPLE 5 SOLUTION

Solving a Quadratic Equation by Completing the Square

Use completing the square to solve x 1x 1 32 5 9. We remove parentheses to get x2 1 3x 5 9

Since the coefficient of x2 is 1 and the constant is on the right side, we move to Step 3 and find the number to be added to both sides to complete the square. Since onehalf of 3 (the coefficient of x) is 32 and the square of 32 is 94, we add 94 to both sides. 9 9 591 4 4 2 3 45 ax 1 b 5 2 4

x2 1 3x 1

x1 x1

"45 3 5 2 2

3 3"5 5 2 2 23 1 3"5 x5 2

or x 1 or x 1

Add

9 to both sides. 4

9 and 4 combine terms on right. Factor x2 1 3x 1

"45 3 52 2 2

Use the Square Root Property.

3 3"5 52 2 2

"45 5 "9"5 5 3"5

23 2 3"5 x5 2

3 from both sides. 2 and combine. Subtract

Verify that each root satisfies the original equation.

Self Check 5

Solve: x 1x 1 52 5 1. Now Try Exercise 51.

EXAMPLE 6

Solving a Quadratic Equation by Completing the Square Use completing the square to solve 6x2 1 5x 2 6 5 0.

SOLUTION

We begin by dividing both sides of the equation by 6 to make the coefficient of x2 equal to 1. Then we proceed as follows: 6x2 1 5x 2 6 5 0 5 x2 1 x 2 1 5 0 6 5 x2 1 x 5 1 6

Divide both sides by 6. Add 1 to both sides.

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Section 1.3

5 25 25 x2 1 x 1 511 6 144 144 ax 1

5 2 169 b 5 12 144

Quadratic Equations

111

1 5 2 25 Add a ? b , or , to both sides. 2 6 144 5 25 Factor x2 1 x 1 . 6 144

We now apply the Square Root Property. 5 169 5 12 Å 144 5 13 x1 5 12 12 8 x5 12 x1

x5

169 5 52 12 Å 144 5 13 x1 52 12 12 18 x52 12

or x 1

2 3

x52

3 2

Verify that each root satisfies the original equation. Self Check 6

Solve: 2x2 2 5x 2 3 5 0. Now Try Exercise 57.

3. Solve Quadratic Equations Using the Quadratic Formula

We can solve the equation ax2 1 bx 1 c 5 0 1a 2 02 by completing the square. The result will be a formula that we can use to solve quadratic equations. ax2 1 bx 1 c 5 0 ax2 b c 0 1 x1 5 a a a a b c x2 1 x 5 2 a a b b2 b2 4ac x2 1 x 1 2 5 2 2 a 4aa 4a 4a ax 1

b 2 b2 2 4ac b 5 2a 4a2

Divide both sides by a.

Simplify and subtract Add

c from both sides. a

b2

to both sides and multiply the numerator and 4a2 c denominator of by 4a. a

Factor the left side and add the fractions on the right side.

We can now use the Square Root Property. x1

b b2 2 4ac 5 2a Å 4a2 x52

x5

b "b2 2 4ac 1 2a 2a

2b 1 "b2 2 4ac 2a

or x 1

b b2 2 4ac 52 2a Å 4a2 x52

x5

b "b2 2 4ac 2 2a 2a

2b 2 "b2 2 4ac 2a

These values of x are the two roots of the equation ax2 1 bx 1 c 5 0. They are usually combined into a single expression, called the Quadratic Formula. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

112

Chapter 1

Equations and Inequalities

Quadratic Formula

The solutions of the general quadratic equation, ax2 1 bx 1 c 5 0, are x5

2b 6 "b2 2 4ac 2a

1a 2 02

The Quadratic Formula should be read twice, once using the 1 sign and once using the 2 sign. The Quadratic Formula implies that x5

Caution

2b 1 "b2 2 4ac 2a

2b 2 "b2 2 4ac 2a

Be sure to write the Quadratic Formula correctly. Do not write the Quadratic Formula as x 5 2b 6

EXAMPLE 7

x5

or

"b2 2 4ac 2a

Solving a Quadratic Equation Using the Quadratic Formula Use the Quadratic Formula to solve x2 2 5x 1 3 5 0.

SOLUTION

In this equation a 5 1, b 5 25, and c 5 3. We will substitute these values into the Quadratic Formula. x5 x5

2b 6 "b2 2 4ac 2a

2 1252 6 " 1252 2 2 4 112 132 2 112

5 6 "13 2 Both values satisfy the original equation. x5

Self Check 7

This is the Quadratic Formula.

Substitute 1 for a, 25 for b, and 3 for c. 1252 2 2 4 112 132 5 25 2 12 5 13

Solve: 3x2 2 5x 1 1 5 0. Now Try Exercise 65.

EXAMPLE 8

Solving a Quadratic Equation Using the Quadratic Formula Use the Quadratic Formula to solve 2x2 1 8x 2 7 5 0.

SOLUTION

In this equation, a 5 2, b 5 8, and c 5 27. We will substitute these values into the Quadratic Formula. x5 x5 x5 x5

2b 6 "b2 2 4ac 2a

28 6 "82 2 4 122 1272 2 122

28 6 "120 4

28 6 2"30 4

This is the Quadratic Formula.

Substitute 2 for a, 8 for b, and –7 for c. 82 2 4 122 1272 5 64 1 56 5 120 "120 5 "4 ? 30 5 2"30

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Section 1.3

x5

2Q24 6 "30R 4

113

Quadratic Equations

Factor out 2 in the numerator.

24 6 "30 Simplify. 2 Both values satisfy the original equation. x5

Self Check 8

Solve: 4x2 1 16x 2 13 5 0. Now Try Exercise 69.

4. Determine the Easiest Strategy to Use To Solve a Quadratic Equation So far, we have solved quadratic equations by factoring, by the square root method, by completing the square, and by the Quadratic Formula. With so many methods available, it is useful to think about which one will be the easiest way to solve a specific quadratic equation. Although we have used completing the square to develop the Quadratic Formula, it is usually the most complicated way to solve a quadratic equation. Therefore, unless specified, we will usually not use this method. However, we will complete the square again later in the book to write certain equations in specific forms. The following chart summarizes the different types of quadratic equations that can occur, a suggested method for solving them, and an example.

Type of Quadratic Equation

Easiest Strategy to Solve It

Example

Equations of the form ax2 1 bx 1 c 5 0 where the left side factors easily.

Use factoring and the Zero-Factor Theorem.

Equations of the form ax2 1 bx 5 0 where the constant term is missing and the left side factors easily.

Use factoring and the Zero-Factor Theorem.

Equations of the form ax2 2 c 5 0 where the term involving x is missing and the left side factors easily.

Use factoring and the Zero-Factor Theorem.

Solve:

Equations of the form ax2 2 c 5 0 or x2 5 k where k is a contant.

Use the Square Root Property.

Solve: 2x2 2 5 5 0 5 x2 5 2

Solve: 6x2 2 11x 1 3 5 0 12x 2 32 13x 2 12 5 0 2x 2 3 5 0 or 3x 2 1 5 0 3 1 x5 x5 2 3 Solve:

9x2 1 6x 5 0 3x 13x 1 22 5 0 3x 5 0 or 3x 1 2 5 0 x50

x52

2 3

4x2 2 9 5 0 12x 1 32 12x 2 32 5 0 2x 1 3 5 0 or 2x 2 3 5 0 3 3 x5 x52 2 2

x56 x56

5 Å2

"10 2 (continued)

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114

Chapter 1

Equations and Inequalities

Type of Quadratic Equation

Easiest Strategy to Solve It

Example

Equations of the form ax2 1 bx 1 c 5 0 where the left side cannot be factored easily or cannot be factored at all.

Use the Quadratic Formula.

Solve: 3x2 2 x 2 5 5 0.

2b 6 "b2 2 4ac 2a 1 2 2 21 6 " 1212 2 2 4 132 1252 x5 2 132 x5

1 6 "1 1 60 6 1 6 "61 x5 6

a53 b 5 21 c 5 25

x5

5. Solve Formulas for a Variable That Is Squared Many formulas involve quadratic equations. For example, if an object is fired straight up into the air with an initial velocity of 88 feet per second, its height is given by the formula h 5 88t 2 16t2, where h represents its height (in feet) and t represents the elapsed time (in seconds) since it was fired. To solve this formula for t, we use the Quadratic Formula. h 5 88t 2 16t2 16t2 2 88t 1 h 5 0 t5 t5

2 12882 6 " 12882 2 2 4 1162 1h2 2 1162 88 6 "7,744 2 64h 32

Add 16t2 and 288t to both sides.

Substitute into the Quadratic Formula.

Simplify.

6. Define and Use the Discriminant We can predict what type of roots a quadratic equation will have before we solve it. Suppose that the coefficients a, b, and c in the equation ax2 1 bx 1 c 5 0 1a 2 02 are real numbers. Then the two roots of the equation are given by the Quadratic Formula x5

2b 6 "b2 2 4ac 2a

1a 2 02

The value of b2 2 4ac, called the discriminant, determines the nature of the roots. The possibilities are summarized in the table as follows.

EXAMPLE 9

Discriminant

Number of Roots and Type of Roots

0

One repeated rational root

Positive and a perfect square

Two different rational roots

Positive and not a perfect square

Two different irrational roots

Negative

No real number roots

Using the Discriminant to Determine the Number and Type of Roots of a Quadratic Equation Determine the number and type of roots of 3x2 1 4x 1 1 5 0.

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Section 1.3

SOLUTION

Quadratic Equations

115

We calculate the discriminant b2 2 4ac. b2 2 4ac 5 42 2 4 132 112

Substitute 4 for b, 3 for a, and 1 for c.

5 16 2 12 54 Since a, b, and c are real numbers and the discriminant is positive and a perfect square, there will be two different rational roots. Self Check 9

Determine the number and type of the roots of 4x2 2 3x 2 2 5 0. Now Try Exercise 87.

EXAMPLE 10

Using the Discriminant to Find the Constant k If k is a constant, many quadratic equations are represented by the equation 1k 2 22 x2 1 1k 1 12 x 1 4 5 0

Find the values of k that will give an equation with roots that are equal rational numbers. SOLUTION

We calculate the discriminant b2 2 4ac and set it equal to 0. b2 2 4ac 5 1k 1 12 2 2 4 1k 2 22 142

0 5 k2 1 2k 1 1 2 16k 1 32 0 5 k2 2 14k 1 33

0 5 1k 2 32 1k 2 112 k2350

or k 2 11 5 0

k53

k 5 11

When k 5 3 or k 5 11, the equation will have equal roots. As a check, we let k 5 3 and note that the equation 1k 2 22 x2 1 1k 1 12 x 1 4 5 0 becomes 13 2 22 x2 1 13 1 12 x 1 4 5 0 x2 1 4x 1 4 5 0 The roots of this equation are equal rational numbers, as expected: x2 1 4x 1 4 5 0

1x 1 22 1x 1 22 5 0 x1250 x 5 22

or

x1250 x 5 22

Similarly, k 5 11 will give an equation with equal rational roots. Self Check 10

Find k such that 1k 2 22 x2 2 1k 1 32 x 1 9 5 0 will have equal roots. Now Try Exercise 93.

7. Write Rational Equations in Quadratic Form and Solve the Equations If an equation can be written in quadratic form, it can be solved with the techniques used for solving quadratic equations. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

116

Chapter 1

Equations and Inequalities

EXAMPLE 11

Solving a Rational Equation Solve:

SOLUTION

1 3 1 5 2. x21 x11

Since neither denominator can be zero, x 2 1 and x 2 21. If either number appears as a root, it must be discarded.

1 3 1 52 x21 x11 3 1 1x 2 12 1x 1 12 c 1 d 5 1x 2 12 1x 1 12 2 x21 x11 1x 1 12 1 3 1x 2 12 5 2 1x2 2 12 4x 2 2 5 2x2 2 2

Multiply both sides by 1x 2 12 1x 1 12 . Remove brackets and simplify. Remove parentheses and simplify.

2

0 5 2x 2 4x

Add 2 2 4x to both sides.

The resulting equation is a quadratic equation that we can solve by factoring. 2x2 2 4x 5 0

2x 1x 2 22 5 0

Factor 2x2 2 4x.

2x 5 0 or x 2 2 5 0 x50

x52

Verify these results by checking each root in the original equation. Self Check 11

Solve:

1 2 1 5 1. x21 x11

Now Try Exercise 109.

Self Check Answers

1.

1 3 ,2 3 2

2. 2"3, 22"3

3.

25 1 3"5 25 2 3"5 , 2 2

1 25 1 "29 25 2 "29 , 6. 3, 2 2 2 2 24 6 "29 5 6 "13 8. 9. two different irrational roots 7. 6 2 10. 3, 27 11. 0, 3 4. 1 1 "10, 1 2 "10

5.

Exercises 1.3 getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. A quadratic equation is an equation that can be written in the form , where a 2 0. 2. If a and b are real numbers and a 5 0 or b 5 0.

, then

3. If c . 0, the equation x2 5 c has two roots. They and x 5 . are x 5 4. The Quadratic Formula is 1a 2 02 . 5. If a, b, and c are real numbers and if b2 2 4ac 5 0, the two roots of the quadratic equation are repeated . 6. If a, b, and c are real numbers and b2 2 4ac , 0, the two roots of the quadratic equation are .

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Section 1.3

Quadratic Equations

117

Practice

Solve each equation by completing the square.

Solve each equation by factoring. 7. x2 2 x 2 6 5 0 8. x2 1 8x 1 15 5 0

47. x2 2 8x 1 15 5 0

48. x2 1 10x 1 21 5 0

49. x2 1 12x 5 28

50. x2 2 6x 5 21

51. x2 1 5 5 25x

52. x2 1 1 5 24x

53. 2x2 2 20x 5 249

54. 4x2 1 8x 5 7

55. 3x2 5 1 2 4x

56. 3x2 1 4x 5 5

57. 2x2 5 3x 1 1

58. 2x2 1 5x 5 14

9. x2 2 144 5 0 2

11. 2x 1 x 2 10 5 0 2

13. 5x 2 13x 1 6 5 0 2

10. x2 1 4x 5 0 2

12. 3x 1 4x 2 4 5 0 2

14. 2x 1 5x 2 12 5 0 2

15. 15x 1 16x 5 15

16. 6x 2 25x 5 225

17. 12x2 1 9 5 24x

18. 24x2 1 6 5 24x

Use the Square Root Property to solve each equation. 19. x2 5 9

20. x2 5 64

Use the Quadratic Formula to solve each equation. 59. x2 2 12 5 0 60. x2 2 60 5 0

21. y2 2 50 5 0

22. x2 2 75 5 0

61. x2 2 25x 5 0

62. x2 1 x 5 0

23. 2x2 5 40

24. 5x2 5 400

63. 2x2 2 x 2 15 5 0

64. 6x2 1 x 2 2 5 0

25. 4x2 5 7

26. 16x2 5 11

65. 3x2 5 25x 2 1

66. 2x2 5 5x 1 11

27. 2x2 2 13 5 0

28. 23x2 5 211

67. x2 1 1 5 27x

68. 13x2 1 1 5 210x

29. 1x 2 12 5 4

30. 1y 1 22 2 49 5 0

2

31. 1x 1 12 2 2 8 5 0

33. 12x 1 12 2 5 27

32. 1y 1 22 2 2 98 5 0

34. 15y 1 22 2 2 48 5 0

Complete the square to make each a perfect-square trinomial. 36. x2 1 8x 35. x2 1 6x 37. x2 2 4x

38. x2 2 12x

39. a2 1 5a

40. t2 1 9t

41. r2 2 11r

42. s2 2 7s

3 43. y 1 y 4

3 44. p 1 p 2

1 45. q2 2 q 5

2 46. m2 2 m 3

2

69. 3x2 1 6x 5 21

2

2

1 71. 5xax 1 b 5 3 5

70. 2x 1x 1 32 5 21 72. 7x2 5 2x 1 2

Solve each formula for the indicated variable. 1 73. h 5 gt2; t 74. x2 1 y2 5 r2; x 2 75. h 5 64t 2 16t2; t

76. y 5 16x2 2 4; x

77.

x2 y2 2 1 2 5 1; y a b

78.

x2 y2 2 2 2 5 1; x a b

79.

x2 y2 2 2 2 5 1; a a b

80.

x2 y2 2 2 2 5 1; b a b

81. x2 1 xy 2 y2 5 0; x 82. x2 2 3xy 1 y2 5 0; y

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118

Chapter 1

Equations and Inequalities

Use the discriminant to determine the number and type of roots. Do not solve the equation. 2

2

83. x 1 6x 1 9 5 0

84. 23x 1 2x 5 21

85. 3x2 2 2x 1 5 5 0

86. 9x2 1 42x 1 49 5 0

87. 10x2 1 29x 5 21

88. 10x2 1 x 5 21

89. x2 2 5x 1 2 5 0

90. 28x2 2 2x 5 13

91. Does 1,492x2 1 1,984x 2 1,776 5 0 have any roots that are real numbers? 92. Does 2,004x2 1 10x 1 1,994 5 0 have any roots that are real numbers? 93. Find two values of k such that x2 1 kx 1 3k 2 5 5 0 will have two roots that are equal. 94. For what value(s) of b will the solutions of x2 2 2bx 1 b2 5 0 be equal? Change each rational equation to quadratic form and solve it by the most efficient method. 12 15 95. x 1 1 5 96. x 2 2 5 x x 3 4 97. 8x 2 5 10 98. 15x 2 5 4 x x 5 4 6 1 99. 5 2 2 6 100. 2 1 5 12 x x x x 101. xa30 2

13 10 b5 x x

102. xa20 2

1 3 1 52 x x12 1 1 5 104. 1 5 x21 x24 4 103.

1 5 1 51 x11 2x 2 4 10 x 12x 1 12 5 106. x22 x22

105.

x12 3 5 x21 x21 1 1 1 108. 5 1 42y 4 y12

107. x 1 1 1

24 212 2 11 5 a a11 1a 2 22 1a 1 42 a 1a 2 32 5 110. 10 5 41a a22 111. 5 2a 3 36 224 112. 2 17 5 b b11

109.

17 10 b5 x x

Discovery and Writing 113. If r1 and r2 are the roots of ax2 1 bx 1 c 5 0, show that r1 1 r2 5 2ba. 114. If r1 and r2 are the roots of ax2 1 bx 1 c 5 0, show that r1r2 5 ac . In Exercises 115 and 116, a stone is thrown straight upward, higher than the top of a tree. The stone is even with the top of the tree at time t1 on the way up and at time t2 on the way down. If the height of the tree is h feet, both t1 and t2 are solutions of h 5 v0t 2 16t2. 115. Show that the tree is 16t1t2 feet tall. 116. Show that v0 is 16 1t1 1 t22 feet per second.

117. Explain why the Zero-Factor Theorem is true. 118. Explain how to complete the square on x2 2 17x. Quadratic equations can be solved automatically by using a computer program, such as Excel. 119. Solve: 2x2 2 3x 2 4 5 0. a. Open an Excel spreadsheet. In cell B1, enter the left side of the equation as 5 2*A1^2 2 3*A1 2 4 (Cell A1 is reserved for the value of x.) b. After pressing ENTER, you will see the number 24 in cell B1. Since cell A1 is empty, its value is considered to be 0 and the value of the quadratic is 24 when x 5 0. c. To solve the equation, enter a guess for the solution in cell A1. To find a positive solution, enter 1 as a guess in cell A1. d. Click inside cell B1. On the Menu bar, look under the Tools menu for SOLVER. (If there is no SOLVER, choose Add-Ins and then, inside the dialog box, select Solver Add-In and click OK.) After choosing SOLVER, a parameters window will appear. Inside the window, in front of Equal To, select Value Of, and be sure it is followed by the number 0. In the GUESS box, enter $A$1. Then click Solve. When the Solver Results dialog box opens, be sure that Keep Solver Solution is selected, and click OK. The solution of the equation appears in cell A1. e. Find the negative solution of the equation. 120. Use Excel to solve 6x2 1 13x 2 5 5 0.

Review

Simplify each expression. 121. 5x 1x 2 22 2 x 13x 2 22

122. 1x 1 32 1x 2 92 2 x 1x 2 52 123. 1m 1 32 2 2 1m 2 32 2 124. 3 1y 1 z2 1y 2 z2 4 2

125. "50x3 2 x"8x 2x 126. "5 2 2

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Section 1.4

Applications of Quadratic Equations

119

1.4 Applications of Quadratic Equations In this section, we will learn to 1. 2. 3. 4.

Solve Solve Solve Solve

geometric problems. uniform motion problems. falling body problems. business problems.

5. Solve shared-work problems.

© Bennie Thornton/Alamy

The Grand Canyon Skywalk is a tourist attraction located along the Colorado River in the state of Arizona. The glass walkway is shaped like a horseshoe and is 4,000 feet above the floor of the canyon. The Grand Canyon, known for its overwhelming size and beautiful landscape, is awe-inspiring and one of our nation’s most astounding natural wonders. If a Clif energy bar is accidentally dropped over the side of the skywalk, how long will it take it to hit the canyon floor? If t represents the time in seconds, the quadratic equation 216t2 1 4,000 5 0 models the time it takes the energy bar to fall to the canyon floor. We can solve this equation by using the Square Root Property 216t2 1 4,000 5 0 216t2 5 24,000 t2 5 250

Subtract 4,000 from both sides. Divide both sides by 216.

t 5 6"250

Use the Square Root Property.

t < 615.8

Round to the nearest tenth.

Because time cannot be negative, we disregard the negative answer. The time it will take the energy bar to reach the canyon floor is about 15.8 seconds. As this example illustrates, the solutions of many problems involve quadratic equations.

1. Solve geometric Problems EXAMPLE 1

Solving an Area Problem The length of a rectangle exceeds its width by 3 feet. If its area is 40 square feet, find its dimensions.

SOLUTION

To find an equation that models the problem, we can let w represent the width of the rectangle. Then, w 1 3 will represent its length (see Figure 1-7). Since the formula for the area of a rectangle is A 5 lw 1area 5 length 3 width2 , the area of the rectangle is 1w 1 32 w, which is equal to 40.

w ft

(w + 3) ft FIgURE 1-7

The length of the rectangle 1w 1 32

times

the width of the rectangle

equals

?

w

5

the area of the rectangle. 40

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120

Chapter 1

Equations and Inequalities

We can solve this equation for w. 1w 1 32 w 5 40

w2 1 3w 5 40 w2 1 3w 2 40 5 0

1w 2 52 1w 1 82 5 0 w2550

Subtract 40 from both sides. Factor.

or w 1 8 5 0

w55

w 5 28

When w 5 5, the length is w 1 3 5 8. The solution −8 must be discarded, because a rectangle cannot have a negative width. We can verify that this solution is correct by observing that a rectangle with dimensions of 5 feet by 8 feet has an area of 40 square feet. Self Check 1

The length of a rectangle exceeds its width by 10 feet. If its area is 375 square feet, find its dimensions. Now Try Exercise 5.

EXAMPLE 2

Solve a Right Triangle Problem On a college campus, a sidewalk 85 meters long (represented by the red lines in Figure 1-8) joins a dormitory building D with the student center C. However, the students prefer to walk directly from D to C. If segment DC is 65 meters long, how long is each piece of the existing sidewalk? C

65 m

D

lm

(85 – l) m

FIgURE 1-8

SOLUTION

We note that the triangle shown in the figure is a right triangle, with a hypotenuse that is 65 meters long. If we let the shorter leg of the triangle be l meters long, the length of the longer leg will be 185 2 l 2 meters. By the Pythagorean Theorem, we know that the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. Thus, we can form the equation l 2 1 185 2 l 2 2 5 652

which we can solve as follows. l 2 1 7,225 2 170l 1 l 2 5 4,225 2l 2 2 170l 1 3,000 5 0 l 2 2 85l 1 1,500 5 0

In a right triangle, a2 1 b2 5 c2. Expand 185 2 l 2 2. Combine like terms and subtract 4,225 from both sides. Divide both sides by 2.

Since the left side is difficult to factor, we will solve this equation using the Quadratic Formula.

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Section 1.4

Applications of Quadratic Equations

121

2b 6 "b2 2 4ac 2a 1 2 2 285 6 " 12852 2 2 4 112 11,5002 l5 2 112 l5

85 6 "1,225 2 85 6 35 l5 2 85 1 35 85 2 35 l5 or l 5 2 2 5 60 5 25 l5

The length of the shorter leg is 25 meters. The length of the longer leg is 185 2 252 meters, or 60 meters. Self Check 2

The length of a video screen is 21 feet shorter than its height. If the diagonal of the screen is 39 feet, find the height of the screen. Now Try Exercise 15.

2. Solve Uniform Motion Problems EXAMPLE 3

Solving a Uniform Motion Problem A man drives 600 miles to a convention. On the return trip, he is able to increase his speed by 10 mph and save 2 hours of driving time. How fast did he drive in each direction?

SOLUTION

We can let s represent the car’s speed (in mph) driving to the convention. On the return trip, his speed was s 1 10 mph. Recall that the distance traveled by an object moving at a constant rate for a certain time is given by the formula d 5 rt. If we divide both sides of this formula by r, we will have a formula for time. t5

d r

We can organize the information given in this problem as shown in the following table. d

r

5

t

?

Outbound trip

600

s

600 s

Return trip

600

s 1 10

600 s 1 10

Although neither the outbound nor the return travel time is given, we know the difference of those times. The longer time of the outbound trip

minus

the shorter time of the return trip

equals

the difference in travel times.

600 s

2

600 s 1 10

5

2

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122

Chapter 1

Equations and Inequalities

We can solve this equation for s. 600 600 2 52 s s 1 10 600 600 s 1s 1 102 a 2 b 5 s 1s 1 102 122 s s 1 10 600 1s 1 102 2 600s 5 2s 1s 1 102

Simplify.

2

600s 1 6,000 2 600s 5 2s 1 20s

Remove parentheses.

6,000 5 2s2 1 20s

Combine like terms.

0 5 2s2 1 20s 2 6,000 2

0 5 s 1 10s 2 3,000 0 5 1s 2 502 1s 1 602 s 2 50 5 0

or s 1 60 5 0

s 5 50

Multiply both sides by s 1s 1 102 to clear the equation of fractions.

Subtract 6,000 from both sides. Divide both sides by 2. Factor. Set each factor equal to 0.

s 5 260

The solution s 5 260 must be discarded. The man drove 50 mph to the convention and 50 1 10, or 60, mph on the return trip. These answers are correct, because a 600-mile trip at 50 mph would take 600 50 , or 12 hours. At 60 mph, the same trip would take only 10 hours, which is 2 hours less time. Self Check 3

Terrence drives his motorcycle 600 miles to Key West, Florida. On the return trip, he is able to increase his speed by 10 mph and save 3 hours of driving time. How fast did he drive in each direction? Now Try Exercise 19.

3. Solve Falling Body Problems EXAMPLE 4

Solving a Falling Body Problem If an object is thrown straight up into the air with an initial velocity of 144 feet per second, its height is given by the formula h 5 144t 2 16t2, where h represents its height (in feet) and t represents the time (in seconds) since it was thrown. How long will it take for the object to return to the point from which it was thrown?

SOLUTION

When the object returns to its starting point, its height is again 0. Thus, we can set h equal to 0 and solve for t. h 5 144t 2 16t2 0 5 144t 2 16t2 0 5 16t 19 2 t2

16t 5 0

or

92t50

t50

Let h 5 0. Factor. Set each factor equal to 0.

t59

At t 5 0, the object’s height is 0, because it was just released. When t 5 9, the height is again 0, and the object has returned to its starting point. Self Check 4

How long does it take the ball in Example 4 to reach a height of 324 feet? Now Try Exercise 23.

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Section 1.4

Applications of Quadratic Equations

123

4. Solve Business Problems EXAMPLE 5

Solving a Business Problem A bus company shuttles 1,120 passengers daily between Rockford, Illinois, and O’Hare Airport. The current one-way fare is $10. For each 25¢ increase in the fare, the company predicts that it will lose 48 passengers. What increase in fare will produce daily revenue of $10,208?

SOLUTION

Let q represent the number of quarters the fare will be increased. Then the new fare will be 110 1 0.25q2 . Since the company will lose 48 passengers for each 25¢ increase, 48q passengers will be lost when the rate increases by q quarters. The passenger load will then be 11,120 2 48q2 passengers. Since the daily revenue of $10,208 will be the product of the rate and the number of passengers, we have

110 1 0.25q2 11,120 2 48q2 5 10,208 11,200 2 480q 1 280q 2 12q2 5 10,208

Remove parentheses.

212q2 2 200q 1 992 5 0

Combine like terms and subtract 10,208 from both sides.

2

3q 1 50q 2 248 5 0

Divide both sides by 24.

Since the left side is difficult to factor, we will solve this equation with the Quadratic Formula. q5 q5

q5

q5

2b 6 "b2 2 4ac 2a

250 6 "502 2 4 132 122482 2 132 250 6 "2,500 1 2,976 6 250 6 "5,476 6

q5

250 6 74 6

q5

250 1 74 6

5

Substitute 3 for a, 50 for b, and 2248 for c.

or

24 6

54

q5 5

250 2 74 6 2124 6

52

62 3

Since the number of riders cannot be negative, the result of 262 3 must be discarded. To generate $10,208 in daily revenues, the company should raise the fare by 4 quarters, or $1, to $11. Self Check 5

A rock band has been drawing average crowds of 400 people. It is projected that for every $1 increase in the $10 ticket price, the average attendance will decrease by 20. At what ticket price will nightly receipts be $4,500? Now Try Exercise 31.

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124

Chapter 1

Equations and Inequalities

5. Solve Shared-Work Problems EXAMPLE 6

Solving a Shared-Work Problem One environmental company can clean up an oil spill on a beach in 2 days less time than its competitor. Working together they were able to clean up the spill in 10 days. How long would it have taken the first company to clean up the spill if it worked alone?

SOLUTION

Suppose the first company can clean up the spill in x days. Then the first company can do x1 of the job each day. Because the first company can do the work in 2 days less than its competitor, it will take the competitor 1x 1 22 days to clean up the spill. The competitor can do x 11 2 of the job each day. Working together, they can clean up the spill in 10 days. So together they can 1 of the job each day. The sum of the work each can do in one day is equal to do 10 the work that they can do together in one day.

The part the first company can clean up in one day 1 x

plus

the part the second company can clean up in one day

equals

the part they can clean up together in one day.

1

1 x12

5

1 10

We can solve this equation for x. 1 1 1 1 5 x x12 10 1 1 1 10x 1x 1 22 a 1 b 5 10x 1x 1 22 a b x x12 10

10x 1x 1 22 10x 1x 1 22 10x 1x 1 22 1 5 x x12 10 10 1x 1 22 1 10x 5 x 1x 1 22 10x 1 20 1 10x 5 x2 1 2x

0 5 x2 2 18x 2 20

Multiply both sides by 10x 1x 1 22 to eliminate the fractions. Distribute the multiplication by 10x 1x 1 22 . x x12 10 5 1, 5 1, and 5 1. x x12 10 Use the Distributive Property to remove parentheses. Subtract 20x and 20 from both sides.

Since the right side cannot be factored over the integers, we will solve the equation with the Quadratic Formula. x5 x5

x5

x5 x5 x5

18 1 20.09975124 2

< 19.05

or

2b 6 "b2 2 4ac 2a

2 12182 6 " 12182 2 2 4 112 12202 2 112

Substitute 1 for a, 218 for b, and 220 for c.

18 6 "324 1 80 2 18 6 "404 2

18 6 20.09975124 2 x5

18 2 20.09975124 2

< 21.05

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Section 1.4

Applications of Quadratic Equations

125

Since the work cannot be completed in a negative number of days, we discard the solution of 21.05. Thus, the first company can complete the job working alone in a little over 19 days. Self Check 6

A hose can fill a swimming pool in 7 hours. Another hose needs 2 more hours to fill the pool than the two hoses combined. How long would it take the second hose to fill the pool? Now Try Exercise 39.

Self Check Answers

1. 15 feet by 25 feet 2. 36 feet 3. 40 mph to Key West and 50 mph returning 4. 4.5 seconds 5. $15 6. 5

) −5

( 0

5

FIgURE 1-35

In general, the inequality 0 x 0 . k 1k . 02 indicates that a point with coordinate x is more than k units from the origin. (See Figure 1-36.) x < −k or x > k

) −k

( 0

k

FIgURE 1-36

Similarly, the inequality 0 x 0 $ k 1k . 02 indicates that a point with coordinate x is k or more units from the origin and thus, x # 25 or x $ 5.

Inequalities of the Forms 0 x 0 . k and 0 x 0 $ k

EXAMPLE 5

SOLUTION

If k . 0, then

• 0 x 0 . k is equivalent to x , 2k or x . k

• 0 x 0 $ k is equivalent to x # 2k or x $ k

Solving an Absolute Value Inequality of the Form 0 x 0 $ k

Solve: `

2x 1 3 ` 1 7 $ 12. 2

We begin by subtracting 7 from both sides of the inequality to isolate the absolute value on the left side. `

2x 1 3 ` $5 2

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170

Chapter 1

Equations and Inequalities

This result is equivalent to two inequalities that can be solved separately. 2x 1 3 # 25 2 2x 1 3 # 210

or

2x 1 3 $5 2 2x 1 3 $ 10

2x # 213 x#2

2x $ 7

13 2

x$

7 2

7 The solution set is the union of the intervals Q2`, 213 2 T and S2 , `R. Its graph appears in Figure 1-37.

]

[

–13/2

0

7/2

FIgURE 1-37

Self Check 5

Solve: `

3x 2 6 ` 1 2 $ 12. 3

Now Try Exercise 59.

6. Solve Compound Inequalities with Absolute Value EXAMPLE 6 SOLUTION

Solving a Compound Inequality with Absolute Value

Solve: 0 , 0 x 2 5 0 # 3.

The inequality 0 , 0 x 2 5 0 # 3 consists of two inequalities that can be solved separately. The solution will be the intersection of the inequalities 0 , 0x 2 50

and

0x 2 50 # 3

The inequality 0 , 0 x 2 5 0 is true for all x except 5. The inequality 0 x 2 5 0 # 3 is equivalent to the inequality 23 # x 2 5 # 3 2#x#8

Add 5 to each part.

The solution set is the intersection of these two solutions, which is the interval 3 2, 8 4 , except 5. This is the union of the intervals 3 2, 52 and 15, 8 4 , as shown in Figure 1-38.

[

)(

]

2

5

8

FIgURE 1-38

Self Check 6

Solve: 0 , 0 x 1 2 0 # 5. Now Try Exercise 63.

7. Solve Inequalities with Two Absolute Values

In Example 9b of Section 1.5, we saw that 0 a 0 could be defined as 0 a 0 5 "a2

We will use this fact in the next example. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 1.8

EXAMPLE 7 SOLUTION

171

Absolute Value

Solving an Inequality with Two Absolute Values

Solve 0 x 1 2 0 . 0 x 1 1 0 and give the result in interval notation. 0x 1 20 . 0x 1 10

" 1x 1 22 2 . " 1x 1 12 2 1x 1 22 2 . 1x 1 12 2

2

Use 0 a 0 5 "a2. Square both sides.

2

x 1 4x 1 4 . x 1 2x 1 1

Expand each binomial.

4x . 2x 2 3

Subtract x2 and 4 from both sides.

2x . 23 Subtract 2x from both sides. 3 x.2 Divide both sides by 2. 2 The solution set is the interval Q232, `R. Check several numbers in this interval to verify that this interval is the solution. Self Check 7

Solve 0 x 2 3 0 # 0 x 1 2 0 and give the result in interval notation. Now Try Exercise 77. Three other properties of absolute value are sometimes useful.

Properties of Absolute Value

0a0 a 1b 2 02 2. ` ` 5 0b0 b

If a and b are real numbers, then 1. 0 ab 0 5 0 a 0 0 b 0

3. 0 a 1 b 0 # 0 a 0 1 0 b 0

Properties 1 and 2 above indicate that the absolute value of a product (or a quotient) is the product (or the quotient) of the absolute values. Property 3 indicates that the absolute value of a sum is either equal to or less than the sum of the absolute values. Self Check Answers

1. a. 0

b. 17

0 x 1 5 0 5 2x 2 5

−5

4. 1212, 62

5. 12`, 28 4 c 3 12, ` 2

(

)

]

−12

6

−8

Exercises 1.8 getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. If x $ 0, then 0 x 0 5 2. If x , 0, then 0 x 0 5

c. if x 1 5 $ 0, 0 x 1 5 0 5 x 1 5; if x 1 5 , 0, 1 2. 25, 2 3. 2, 4

3. 0 x 0 4. 0 a 0 5. 0 x 0 6. 0 x 0

0

0

2

6. 3 27, 222 c 122, 3 4

[

[

)(

]

12

−7

−2

3

5 k is equivalent to 5 0 b 0 is equivalent to a 5 b or , k is equivalent to

. k is equivalent to 7. 0 x 0 $ k is equivalent to

. .

8. "a2 5

1 7. c , `b 2

. . . . .

.

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172

Chapter 1

Equations and Inequalities

Practice

Write each expression without absolute value symbols. 10. 0 29 0 9. 0 7 0 11. 0 0 0 13. 0 5 0 2 0 23 0 15. 0 p 2 2 0 17. 0 x 2 5 0 and x $ 5 19. 0 x3 0

12. 0 3 2 5 0 14. 0 23 0 1 0 5 0 16. 0 p 2 4 0

18. 0 x 2 5 0 and x # 5

20. 0 2x 0

Solve each absolute value equation for x. 21. 0 x 1 2 0 5 2 22. 0 2x 1 5 0 5 3 23. 0 3x 2 1 0 2 7 5 22

24. 0 7x 2 5 0 1 5 5 8

3x 2 4 25. ` ` 55 2

9 10x 1 1 26. ` ` 5 2 2

27. `

2x 2 4 ` 1658 5

28. `

3x 1 11 ` 2 15 5 214 7

29. `

x23 ` 5 22 4

30. `

x15 ` 1352 2

31. `

x25 ` 50 3

32. `

x17 ` 50 9

33. `

4x 2 2 ` 53 x

34. `

35. 0 x 0 5 x

2 1x 2 32 ` 56 3x

36. 0 x 0 1 x 5 2

49. 0 x 1 3 0 . 6

50. 0 x 1 2 0 # 4

51. 0 2x 1 4 0 $ 10

52. 0 5x 2 2 0 , 7

53. 0 3x 1 5 0 1 1 # 9

54. 0 2x 2 7 0 2 3 . 2

55. 0 x 1 3 0 . 0

56. 0 x 2 3 0 # 0

5x 1 2 ` ,1 3

57. `

59. 3 `

61.

3x 2 1 ` .5 2

0x 2 10 . 23 22

58. `

3x 1 2 ` .2 4

60. 2 `

62.

8x 1 2 ` #1 5

0 2x 2 3 0 , 21 23

37. 0 x 1 3 0 5 0 x 0

38. 0 x 1 5 0 5 0 5 2 x 0

Solve each compound inequality with absolute value. Express the solution set in interval notation, and graph it. 64. 0 , 0 2x 2 3 0 , 1 63. 0 , 0 2x 1 1 0 , 3

41. 0 x 1 2 0 5 0 x 2 2 0

42. 0 2x 2 3 0 5 0 3x 2 5 0

65. 8 . 0 3x 2 1 0 . 3

66. 8 . 0 4x 2 1 0 . 5

67. 2 , `

68. 3 , `

x23 ` ,5 2

70. 5 $ `

x12 ` .1 3

39. 0 x 2 3 0 5 0 2x 1 3 0

43. `

x13 ` 5 0 2x 2 3 0 2

3x 2 1 2x 1 3 45. ` ` 5 ` ` 2 3

40. 0 x 2 2 0 5 0 3x 1 8 0

44. `

x22 ` 5 06 2 x0 3

5x 1 2 x21 46. ` ` 5 ` ` 3 4

Solve each absolute value inequality. Express the solution set in interval notation, and graph it. 48. 0 x 2 2 0 $ 4 47. 0 x 2 3 0 , 6

x25 ` ,4 3

69. 10 . `

x22 ` .4 2

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Section 1.8

71. 2 # `

x11 ` ,3 3

72. 8 . `

3x 1 1 ` .2 2

Solve each inequality and express the solution using interval notation. 74. 0 x 1 1 0 , 0 x 1 2 0 73. 0 x 1 1 0 $ 0 x 0 75. 0 2x 1 1 0 , 0 2x 2 1 0

76. 0 3x 2 2 0 $ 0 3x 1 1 0

77. 0 x 1 1 0 , 0 x 0

78. 0 x 1 2 0 # 0 x 1 1 0

79. 0 2x 1 1 0 $ 0 2x 2 1 0

80. 0 3x 2 2 0 , 0 3x 1 1 0

Absolute Value

173

86. Light bulbs A light bulb is expected to last h hours, where 0 h 2 1,500 0 # 200. Express this range without using absolute value symbols. 87. Error analysis In a lab, students measured the percent of copper p in a sample of copper sulfate. The students know that copper sulfate is actually 25.46% copper by mass. They are to compare their results to the actual value and find the amount of experimental error. a. Which measurements shown in the illustration satisfy the absolute value inequality 0 p 2 25.46 0 # 1.00?

b. What can be said about the amount of error for each of the trials listed in part a? Section A Lab 4 Title: “Percent copper (CU) in copper sulfate (CuSO4• 5H2O)” Results

Applications

81. Finding temperature ranges The temperatures on a summer day satisfy the inequality 0 t 2 78° 0 # 8°, where t is the temperature in degrees Fahrenheit. Express this range without using absolute value symbols. 82. Finding operating temperatures A car CD player has an operating temperature of 0 t 2 40° 0 , 80°, where t is the temperature in degrees Fahrenheit. Express this range without using absolute value symbols.

Trial #1: Trial #2: Trial #3 Trial #4:

% Copper 22.91% 26.45% 26.49% 24.76%

88. Error analysis See Exercise 87. a. Which measurements satisfy the absolute value inequality 0 p 2 25.46 0 . 1.00? b. What can be said about the amount of error for each of the trials listed in part a?

Discovery and Writing 83. Range of camber angles The specifications for a certain car state that the camber angle c of its wheels should be 0.6° 6 0.5°. Express this range with an inequality containing an absolute value.

89. Explain how to find the absolute value of a number. 90. Explain why the equation 0 x 0 1 9 5 0 has no solution. 91. Explain the use of parentheses and brackets when graphing inequalities.

84. Tolerance of a sheet of steel A sheet of steel is to be 0.25 inch thick, with a tolerance of 0.015 inch. Express this specification with an inequality containing an absolute value.

92. If k . 0, explain the differences between the solution sets of 0 x 0 , k and 0 x 0 . k. 93. If k , 0, explain why the solution set of 0 x 0 , k has no solution. 94. If k , 0, explain why the solution set of of 0 x 0 . k is all real numbers.

85. Humidity level A Steinway piano should be placed in an environment where the relative humidity h is between 38% and 72%. Express this range with an inequality containing an absolute value.

Review Write each number in scientific notation.

© Istockphoto.com/iLexx

95. 37,250

96. 0.0003725

Write each number in standard notation. 97. 5.23 3 105 98. 7.9 3 1024 Simplify each expression. 99. 1x 2 y2 2 2 1x 1 y2 2

100. 1p 1 q2 2 1 1p 2 q2 2

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174

Chapter 1

Equations and Inequalities

CHAPTER REVIEW SECTION 1.1

Linear Equations and Rational Equations

Definitions and Concepts

Examples

An equation is a statement indicating that two quantities are equal.

Equations: 2x 2 5 5 10,

2 1x 2 22 7x 1 3 5 x23 x12

There can be restrictions on the variable in an equation.

In the equation 2x 2 5 5 10, x can be any real number. 1 2 1 3 In the equation 2 xx 22 32 5 7x x 1 2 , x cannot be 3 or 22, because this would give a 0 in the denominator.

Properties of equality:

If a 5 b, then

If a 5 b and c is a number, then

a175b17

a1c5b1c

and

a2c5b2c

ac 5 bc

and

a b 5 c c

1c 2 02

A linear equation is an equation that can be written in the form ax 1 b 5 0 1a 2 02 . To solve a linear equation, use the properties of equality to isolate x on one side of the equation.

7a 5 7b

and

Solve 3x 2 5 5 4. 3x 2 5 5 4 3x 2 5 1 5 5 4 1 5

A contradiction is an equation that is false for all acceptable replacements for its variable. Rational equations are equations that contain rational expressions. To solve rational equations, multiply both sides of the equation by an expression that will remove the denominators and solve the resulting equation. Be sure to check the answers to identify any extraneous solutions.

Add 5 to both sides.

3x 5 9

Combine like terms.

3x 9 5 3 3

Divide both sides by 3.

x53 An identity is an equation that is true for all acceptable replacements for its variable.

and a275b27 a b 5 7 7

Identities: x 1 x 5 2x,

2 1x 1 12 5 2x 1 2

Contradictions: x 1 1 5 x, 2 1x 1 12 5 2x 1 3 Solve

2x 6 5 . x23 x23

1x 2 32 a

2x 6 5 x23 x23

2x 6 b 5 1x 2 32 a b x23 x23 2x 5 6 x53

Multiply both sides by x 2 3. Simplify. Divide both sides by 2 and simplify.

The result of 3 is extraneous because when you substitute 3 into the original equation, you get a denominator of 0.

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175

Chapter Review

Definitions and Concepts

Examples

Formulas can be solved for a specific variable.

1 Solve A 5 bh for h. 2 1 A 5 bh 2 2A 5 bh

Multiply both sides by 2.

2A bh 5 b b

Divide both sides by b.

2A 5h b

Simplify.

EXERCISES Find the restrictions on x, if any. 1 1. 3x 1 7 5 4 2. x 1 5 2 x 1 1 2 3. 54 4. 5 x21 x22 x23 Solve each equation and classify it as an identity, a conditional equation, or a contradiction. 3 5. 3 19x 1 42 5 28 6. a 5 7 1a 1 112 2 7. 8 13x 2 52 2 4 1x 1 32 5 12 x13 x13 1 52 8. x14 x12 3 1 8x2 1 72x 9. 5 10. 5 8x x21 2 91x

SECTION 1.2

1 2x2 5 2x 2 3 2x 2 3

11.

3x 5 2 53 x21 x13

13.

4 1 2 2 2 5 2 x 2 13x 2 48 x 1x26 x 2 18x 1 32

14.

a21 2a 2 1 22a 1 5 a13 32a a23

12. x 1

2

Solve each formula for the indicated variable. 5 si 15. C 5 1F 2 322 ; F 16. Pn 5 l 1 ; f 9 f

17.

1 1 1 5 1 ; f1 f f1 f2

18. S 5

a 2 lr ;l 12r

Application of Linear Equations

Definitions and Concepts

Examples

Use the following steps to solve an application problem:

Two students leave their dorm in two cars traveling in opposite directions. If one student drives at a rate of 55 mph and the other at a rate of 50 mph, how long will it take for them to be 210 miles apart?

1. Analyze the problem. 2. Pick a variable to represent the quantity to be found. 3. Form an equation. 4. Solve the equation. 5. Check the solution in the words of the problem.

Analyze the problem and pick a variable. We can organize the facts of the problem in the following chart. Since each student drives the same amount of time, let t represent that time. d

5

r

?

t

Student 1

55t

55

t

Student 2

50t

50

t

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176

Chapter 1

Equations and Inequalities

Definitions and Concepts

Examples Form and solve an equation. Since the students are driving in opposite directions, the distance they are apart in t hours is the sum of the distances they drive, a total of 210 miles. We can form and solve the following equation: 55t 1 50t 5 210 105t 5 210 t52

Combine like terms. Divide both sides by 105.

Check. In 2 hours, student 1 drives 55(2) miles and student 2 drives 50(2) miles, or 110 miles plus 100 miles. At this time, they will be 210 miles apart.

EXERCISES 19. Test scores Carlos took four tests in an English class. On each successive test, his score improved by 4 points. If his mean score was 66%, what did he score on the first test? 20. Fencing a garden A homeowner has 100 ft of fencing to enclose a rectangular garden. If the garden is to be 5 ft longer than it is wide, find its dimensions. 21. Travel Two women leave a shopping center by car traveling in opposite directions. If one car averages 45 mph and the other 50 mph, how long will it take for the cars to be 285 miles apart? 22. Travel Two taxis leave an airport and travel in the same direction. If the average speed of one taxi is 40 mph and the average speed of the other taxi is 46 mph, how long will it take before the cars are 3 miles apart? 23. Preparing a solution A liter of fluid is 50% alcohol. How much water must be added to dilute it to a 20% solution?

SECTION 1.3

24. Washing windows Scott can wash 37 windows in 3 hours, and Bill can wash 27 windows in 2 hours. How long will it take the two of them to wash 100 windows? 25. Filling a tank A tank can be filled in 9 hours by one pipe and in 12 hours by another. How long will it take both pipes to fill the empty tank? 26. Producing brass How many ounces of pure zinc must be alloyed with 20 ounces of brass that is 30% zinc and 70% copper to produce brass that is 40% zinc? 27. Lending money A bank lends $10,000, part of it at 11% annual interest and the rest at 14%. If the annual income is $1,265, how much was lent at each rate? 28. Producing oriental rugs An oriental rug manufacturer can use one loom with a setup cost of $750 that can weave a rug for $115. Another loom, with a setup cost of $950, can produce a rug for $95. How many rugs are produced if the costs are the same on each loom?

Quadratic Equations

Definitions and Concepts

Examples

A quadratic equation is an equation that can be written in the form ax2 1 bx 1 c 5 0, where a, b, and c are real numbers and a 2 0.

3x2 2 5x 2 7 5 0,  5x2 2 25 5 0,  7x2 1 14x 5 0

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Chapter Review

177

Definitions and Concepts

Examples

Zero-Factor Theorem:

Solve x2 2 x 2 6 5 0 using the Zero-Factor Theorem.

If ab 5 0, then a 5 0 or b 5 0.

x2 2 x 2 6 5 0

1x 1 22 1x 2 32 5 0 x1250

Factor x2 2 x 2 6.

or x 2 3 5 0 x53

x 5 22 If x2 5 32, then

Square Root Property:

x 5 "32

If c . 0, x2 5 c has two real roots: x 5 "c or x 5 2"c

5 "16 ? 2

or

x 5 2"32 5 2"16 ? 2

5 4"2

5 24"2

Steps to complete the square:

Solve x2 2 x 2 6 5 0 by completing the square:

1. Make the coefficient of x2 equal to 1.

1. Since the coefficient of x2 5 1, we go to Step 2.

2. Get the constant on the right side of the equation.

2. Add 6 to both sides to get the constant on the right side: x2 2 x 5 6.

3. Complete the square on x. Take one-half the coefficient of x, square it, and add it to both sides of the equation.

1 2 1 2 3. x2 2 x 1 a2 b 5 6 1 a2 b 2 2

4. Factor the resulting perfect-square trinomial and combine like terms.

4.

5. Solve the resulting quadratic equation by using the Square Root Property.

5.

1 2 25 ax 2 b 5 2 4 x2

25 1 56 2 Å4 x5

x5 2b 6 "b2 2 4ac 2a

Quadratic Formula: x5

1a 2 02

6 5 3 or 2

1 5 6 2 2 4 x 5 2 5 22 2

If 3x2 2 5x 1 1 5 0, then a 5 3, b 5 25, and c 5 1. So x5

5

2 1252 6 " 1252 2 2 4 132 112 2b 6 "b2 2 4ac 5 2a 2 132

5 6 "25 2 12 5 6 "13 5 6 6

Discriminant: The value b2 2 4ac is the discriminant. • If b2 2 4ac 5 0, the roots of ax2 1 bx 1 c 5 0 are repeated rational numbers.

In 4x2 2 12x 1 9 5 0, the discriminant is b2 2 4ac 5 12122 2 2 4 142 192 5 0. So the roots are repeated rational numbers.

• If b2 2 4ac . 0 and a perfect square, the roots of ax2 1 bx 1 c 5 0 are two different rational numbers.

In x2 2 x 2 6 5 0, the discriminant is b2 2 4ac 5 1212 2 2 4 112 1262 5 25. Since 25 is positive and a perfect square, the roots are two different rational numbers.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

178

Chapter 1

Equations and Inequalities

Definitions and Concepts

Examples

• If b2 2 4ac . 0 and not a perfect square, the roots of ax2 1 bx 1 c 5 0 are two different irrrational numbers.

In x2 2 x 2 5 5 0, the discriminant is b2 2 4ac 5 1212 2 2 4 112 1252 5 21. Since 21 is positive and not a perfect square, the roots are two different irrational numbers.

• If b2 2 4ac , 0, the roots of ax2 1 bx 1 c 5 0 are two different nonreal numbers.

In 3x2 2 2x 1 1 5 0, the discriminant is b2 2 4ac 5 1222 2 2 4 132 112 5 28 , 0. So the roots are two different nonreal numbers.

EXERCISES Solve each equation by factoring.

43. 5x2 5 1 2 x

29. 2x2 2 x 2 6 5 0

30. 12x2 1 13x 5 4

31. 5x2 2 8x 5 0

32. 27x2 5 30x 2 8

45. Calculate the discriminant associated with the equation 6x2 1 5x 1 1 5 0. 46. Determine the number and nature of the roots of the equation in Exercise 45.

Solve each equation by using the Square Root Property. 33. 2x2 5 16

35. 14z 2 52 2 5 32

34. 12x2 5 60

36. 15x 2 72 2 5 45

Solve each equation by completing the square. 38. 3x2 1 18x 5 224 37. x2 2 8x 1 15 5 0 39. 5x2 2 x 2 1 5 0

47. Find the value of k that will make the roots of kx2 1 4x 1 12 5 0 equal. 48. Find the values of k that will make the roots of 4y2 1 1k 1 22 y 5 1 2 k equal. 49. Solve:

1 1 3 2 5 . a 5 2a

50. Solve:

4 4 1 5 5. a24 a21

40. 5x2 2 x 5 0

Use the Quadratic Formula to solve each equation. 41. x2 1 5x 2 14 5 0

SECTION 1.4

44. 5 5 a2 1 2a

42. 3x2 2 25x 5 18

Applications of Quadratic Equations

Definitions and Concepts

Examples

Many real-life problems are modeled by quadratic equations.

If a missile is launched straight up into the air with an initial velocity of 128 feet per second, its height will be given by the formula h 5 216t2 1 128t, where h represents its height (in feet) and t represents the time (in seconds) since it was launched. How long will it take the missile to return to its starting point? When the missile returns to its starting point, its height will again be 0. So we let h 5 0 and solve for t. 0 5 216t2 1 128t 0 5 216t2 1t 2 82 216t 5 0 or t 2 8 5 0 t50

t58

The missile will leave its starting point at 0 seconds and return at 8 seconds. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Review

179

EXERCISES 51. Fencing a field A farmer wishes to enclose a rectangular garden with 300 yards of fencing. A river runs along one side of the garden, so no fencing is needed there. Find the dimensions of the rectangle if the area is 10,450 square yards. 52. Flying rates A jet plane, flying 120 mph faster than a propeller-driven plane, travels 3,520 miles in 3 hours less time than the propeller plane requires to fly the same distance. How fast does each plane fly?

SECTION 1.5

53. Flight of a ball A ball thrown into the air reaches a height h (in feet) according to the formula h 5 216t2 1 64t, where t is the time elapsed since the ball was thrown. Find the shortest time it will take the ball to reach a height of 48 feet. 54. Width of a walk A man built a walk of uniform width around a rectangular pool. If the area of the walk is 117 square feet and the dimensions of the pool are 16 feet by 20 feet, how wide is the walk?

Complex Numbers

Definitions and Concepts

Examples

Complex numbers: Numbers that can be written in the form a 1 bi, where a and b are real numbers and i 5 "21, are complex numbers.

"236 5 " 1212 1362 5 "21"36 5 1i2 162 5 6i

Equality of complex numbers: a 1 bi 5 c 1 di if and only if a 5 c and b 5 d Adding, subtracting, and multiplying complex numbers:

7 2 2i, 9 1 5i, and 2 1 "7i are complex numbers.

3 1 "4i 5 62 1 2i because 3 5 62 and "4 5 2.

• 1a 1 bi2 1 1c 1 di2 5 1a 1 c2 1 1b 1 d2 i

123 1 4i2 1 12 1 7i2 5 123 1 22 1 14 1 72 i 5 21 1 11i

• 1a 1 bi2 1c 1 di2 5 1ac 2 bd2 1 1ad 1 bc2 i or multiply them as if they were binomials.

123 1 4i2 12 1 7i2 5 23 122 2 3 17i2 1 4i 122 1 4i 17i2

• 1a 1 bi2 2 1c 1 di2 5 1a 2 c2 1 1b 2 d2 i

123 1 4i2 2 12 1 7i2 5 123 2 22 1 14 2 72 i 5 25 2 3i 5 26 2 21i 1 8i 1 28i 2 5 26 2 21i 1 8i 2 28 5 234 2 13i

The complex conjugate of a 1 bi is a 2 bi.

3 1 4i and 3 2 4i are complex conjugates.

Division of complex numbers:

Divide 2 1 i by 2 2 i.

To divide complex numbers, rationalize the denominator.

12 1 i2 12 1 i2 21i 5 12 2 i2 12 1 i2 22i 5

4 1 2i 1 2i 1 i2 4 1 2i 2 2i 2 i2

5

4 1 4i 2 1 4 2 1212

5

3 1 4i 5

5

3 4 1 i 5 5

21i 51 21i

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180

Chapter 1

Equations and Inequalities

Definitions and Concepts

Examples

Powers of i:

i 2 5 21, i 3 5 2i, i 4 5 1, i 5 5 i, i 6 5 2i, i 7 5 2i, i 8 5 1, . . .

If n is a natural number that has a remainder of r when divided by 4, then i n 5 i r. Absolute value of a complex number: 0 a 1 bi 0 5 "a2 1 b2

0 5 2 7i 0 5 "52 1 1272 2 5 "25 1 49 5 "74

EXERCISES Perform all operations and express all answers in a 1 bi form. 56. 12 2 3i2 2 14 1 2i2 55. 12 2 3i2 1 124 1 2i2 57. 13 2 "2362 1 1"216 1 22

62.

2i 22i

63.

31i 32i

64.

3 2 2i 11i

66. Simplify: i 103.

67. 0 3 2 i 0

2 60. 2 3 i

SECTION 1.6

3 11i

65. Simplify: i 53.

58. 13 1 "292 12 2 "2252 3 59. i

61.

68. `

11i ` 12i

69. Solve: 3x2 2 2x 1 1 5 0. 70. Solve: 3x2 1 4 5 2x.

Polynomials and Radical Equations

Definitions and Concepts

Examples

Polynomial equations:

Solve x3 2 5x2 1 6x 5 0.

Many polynomial equations of higher degree can be solved by factoring.

x3 2 5x2 1 6x 5 0

x 1x2 2 5x 1 62 5 0

x 1x 2 22 1x 2 32 5 0

Factor out x. Factor x2 2 5x 1 6.

x 5 0 or x 2 2 5 0 or

x2350

x52

x53

The solution set is 5 0, 2, 3 6 . Power property of real numbers:

If x 5 5, then x2 5 52 or x2 5 25.

If a 5 b, then a2 5 b2. Factoring can be used to solve certain nonpolynomial equations. Check all solutions because extraneous roots can be introduced.

Solve 2x 2 5x1/2 1 3 5 0. 2x 2 5x1/2 1 3 5 0

12x1/2 2 32 1x1/2 2 12 5 0 2x1/2 2 3 5 0 3 x1/2 5 2 9 x5 4

or

x1/2 2 1 5 0 x1/2 5 1 x51

Both roots check. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter Review

Definitions and Concepts

181

Examples

Radical equations: To solve radical equations, use the Power Property of Real Numbers.

Solve "2x 2 3 5 x 2 1. "2x 2 3 5 x 2 1

Q"2x 2 3R2 5 1x 2 12 2

Square both sides.

2x 2 3 5 x2 2 2x 1 1 0 5 x2 2 4x 1 4

0 5 1x 2 22 1x 2 22 x 2 2 5 0 or x 2 2 5 0 x52

x52

Since 2 checks, it is a root.

EXERCISES Solve each equation. 3x 2x 71. 2 5x23 2 x21 73. x4 2 2x2 1 1 5 0

75. a 2 a1/2 2 6 5 0

72.

12 x 2 5x23 x 2

77. "x 2 1 1 x 5 7

76. x2/3 1 x1/3 2 6 5 0

78. "a 1 9 2 "a 5 3

79. "5 2 x 1 "5 1 x 5 4 80. "y 1 5 1 "y 5 1

74. x4 1 36 5 37x2

SECTION 1.7

Inequalities

Definitions and Concepts

Examples

Addition, Subtraction, Multiplication, and Division Properties of Inequalities: If a, b, and c are real numbers: If a , b, a 1 c , b 1 c and a 2 c , b 2 c. If a , b and c . 0, ac , bc and

a b , . c c

If a , b and c , 0, ac . bc and

a b . . c c

If x , 10, then x 1 6 , 10 1 6 and x 2 6 , 10 2 6. If x , 12, then 3x , 3 1122 and

x 12 , . 3 3

If x # 12, then 23x $ 23 1122 and

x 12 $ . 23 23

Trichotomy Property: a , b, a 5 b, or a . b

Either x , 3, x 5 3, or x . 3.

Transitive Property: If a , b and b , c, then a , c.

If x , 4 and 4 , y, then x , y.

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182

Chapter 1

Equations and Inequalities

Definitions and Concepts

Examples

Types of inequalities:

24 1x 2 32 $ 7

• Linear inequality • Compound inequality

25 # 2x 1 1 , 3

• Quadratic inequality

x2 2 2x 1 3 # 0

• Rational inequality

x11 .0 x22

Solving a linear inequality: Use the same steps to solve a linear inequality as you would use to solve a linear equation. However, remember to reverse the order of the inequality when you multiply (or divide) both sides of an inequality by a negative number.

Solve the linear inequality: 24 1x 2 32 $ 7. 24 1x 2 32 $ 7 24x 1 12 $ 7 24x $ 25 5 x# 4

Remove parentheses. Subtract 12 from both sides. Divide both sides by 24.

] 5– 4

In interval notation, the solution is Q2`, 54T. Solving a compound inequality:

Solve the compound inequality: 25 # 2x 1 1 , 3.

For a compound inequality, isolate x in the middle, if possible. If not, solve each inequality separately. The intersection of the intervals is the solution.

25 # 2x 1 1 , 3 26 # 2x , 2

Subtract 1 from all three parts.

23 # x , 1

Divide each part by 2.

[

)

−3

1

In interval notation, the solution is 3 23, 12 . Solving quadratic and rational inequalities: Quadratic and rational inequalities can be solved by constructing a table and testing values or by constructing a sign graph.

1 1 Solve the rational inequality: xx 2 2 . 0.

First note that the solution of x 1 1 5 0 1x 5 212 and the solution of x 2 2 5 0 1x 5 22 form three intervals: 12`, 212 , 121, 22 , and 12, ` 2 . To determine which intervals are solutions, we test a number in each interval to see whether it satisfies the inequality.

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Chapter Review

Definitions and Concepts

183

Examples

Interval

12`, 212

Inequality x11 +0 x22

Test Value 22

121, 22 12, ` 2

Result

22 1 1 1 5 .0 22 2 2 4

The numbers in this interval are solutions.

True

0

011 1 52 .0 022 2

The numbers in this interval are not solutions.

False

3

311 54.0 322

The numbers in this interval are solutions.

True

The solution is the union of two intervals: 12`, 212 c 12, ` 2

EXERCISES Solve each inequality; graph the solution set and write the answer in interval notation. 81. 2x 2 9 , 5 82. 5x 1 3 $ 2 5 1x 2 12 ,x 2 1 2 1 1 84. x 1 x 2 x . 1 1x 1 12 4 3 2 2

88. 1x 2 12 1x 1 42 , 0

89. x2 2 2x 2 3 , 0

90. 2x2 1 x 2 3 . 0

SECTION 1.8 Definitions and Concepts Definition of absolute value of x: 0x0 5 e

91.

x12 $0 x23

92.

x21 #0 x14

93.

x2 1 x 2 2 $0 x23

94.

5 ,2 x

31x ,4 2

86. 2 1 a , 3a 2 2 # 5a 1 2

( 2

87. 1x 1 22 1x 2 42 . 0

83.

85. 0 #

) −1

Absolute Value Examples 050 5 5

0 27 0 5 7

2 0 210 0 5 210

x when x $ 0 2x when x , 0

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184

Chapter 1

Equations and Inequalities

Definitions and Concepts

Examples

Absolute value equations:

Solve: 0 x 2 2 0 5 6.

• If k $ 0, then 0 x 0 5 k is equivalent to x 5 k or x 5 2k. • If a and b are algebraic expressions, 0 a 0 5 0 b 0 is equivalent to a 5 b or a 5 2b.

x2256

or x 2 2 5 26

x58

x 5 24

Solve: 0 3x 0 5 0 x 2 2 0 . 3x 5 x 2 2

or

2x 5 22

3x 5 2 1x 2 22

3x 5 2x 1 2 4x 5 2

x 5 21

x5

1 2

Both solutions check. Absolute value inequality properties:

• If k . 0, then 0 x 0 , k is equivalent to 2k , x , k.

• If k . 0, then 0 x 0 . k is equivalent to x . k or x , 2k. These two properties also hold for # and $.

Solve: 0 x 2 2 0 , 6. 26 , x 2 2 , 6 24 , x , 8

Add 2 to all three parts.

The solution in interval notation is 124, 82 .

Solve: 0 x 2 2 0 . 6. x 2 2 . 6 or

x 2 2 , 26

x.8 Thus, x , 24 or

x , 24

Add 2 to both parts.

x.8

The solution in interval notation is 12`, 242 c 18, ` 2 . Properties of absolute value:

0 23x 0 5 0 23 0 0 x 0 5 3 0 x 0

1. 0 ab 0 5 0 a 0 0 b 0 0a0 a 2. ` ` 5 0b0 b

1b 2 02

`

3. 0 a 1 b 0 # 0 a 0 1 0 b 0

0 23 0 23 3 1x 2 02 5 ` 5 0x0 0x0 x

0x 1 30 # 0x0 1 030

EXERCISES Solve each equation or inequality. 96. 0 2x 2 1 0 5 0 2x 1 1 0 95. 0 x 1 1 0 5 6 97. `

3x 1 11 ` 2150 7

99. 0 x 1 3 0 , 3

98. `

2a 2 6 ` 2650 3a

101. `

x12 ` ,1 3

103. 1 , 0 2x 1 3 0 , 4

102. `

x23 ` .8 4

104. 0 , 0 3x 2 4 0 , 7

100. 0 3x 2 7 0 $ 1

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Chapter Test

185

CHAPTER TEST Find all restrictions on x. x 1. 52 x 1x 2 12

Find each absolute value.

2.

4 1357 3x 2 2

3. 7 12a 1 52 2 7 5 6 1a 1 82

23. z4 2 13z2 1 36 5 0

6. Solve for a:

1 ` 31i

24. 2p2/5 2 p1/5 2 1 5 0

25. "x 1 5 5 12

3 4 5 2 x 2 5x 2 14 x 1 5x 1 6 2

5. Solve for x: z 5

22. `

Solve each equation.

Solve each equation.

4.

21. 0 5 2 12i 0

26. "2z 1 3 5 1 2 "z 1 1

x2m . s

1 1 1 5 1 . a b c

7. Test scores A student’s average on three tests is 75. If the final is to count as two one-hour tests, what grade must the student make to bring the average up to 80? 8. Investment A woman invested part of $20,000 at 6% interest and the rest at 7%. If her annual interest is $1,260, how much did she invest at 6%?

Solve each inequality; graph the solution set and write the answer using interval notation. x13 2x 2 4 27. 5x 2 3 # 7 28. . 4 3

29. 5 # 2x 2 1 , 7

Solve each equation. 9. 4x2 2 8x 1 3 5 0

10. 2b2 2 12 5 25b

30. 1 1 x , 3x 2 3 , 4x 2 2

11. Write the Quadratic Formula. 12. Use the Quadratic Formula to solve 3x2 2 5x 2 9 5 0.

31. x2 2 7x 2 8 $ 0

14. Height of a projectile The height of a projectile shot up into the air is given by the formula h 5 216t2 1 128t. Find the time t required for the projectile to return to its starting point.

Solve each equation.

13. Find k such that x2 1 1k 1 12 x 1 k 1 4 5 0 will have two equal roots.

Perform each operation and write all answers in a 1 bi form. 15. 14 2 5i2 2 123 1 7i2

17.

2 22i

16. 14 2 5i2 13 2 7i2

18.

11i 12i

33. `

3x 1 2 ` 54 2

32.

x12 #0 x21

34. 0 x 1 3 0 5 0 x 2 3 0

Solve each inequality; graph the solution set and write the answer using interval notation. 2x 1 3 35. 0 2x 2 5 0 . 2 36. ` ` #5 3

Simplify each expression. 19. i13

20. i 0

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

186

Chapter 1

Equations and Inequalities

CUMULATIVE REVIEW EXERCISES Consider the set 5 25, 23, 22, 0, 1, "2, 2, 52, 5, 6, 11 6 . 1. Which numbers are even integers?

25. 13x 2 52 12x 1 72 26. 1z 1 22 1z2 2 z 1 22

2. Which numbers are prime numbers?

27. 3x 1 2q6x3 1 x2 1 x 1 2

Write each inequality as an interval and graph it. 3. 24 # x , 7

28. x2 1 2q3x4 1 7x2 2 x 1 2 Factor each polynomial. 29. 3t2 2 6t 30. 3x2 2 10x 2 8 31. x8 2 2x4 1 1

4. x $ 2 or x , 0

32. x6 2 1

Determine which property of the real numbers justifies each expression. 5. 1a 1 b2 1 c 5 c 1 1a 1 b2

6. If x , 3 and 3 , y, then x , y. Simplify each expression. Assume that all variables represent positive numbers. Give all answers with positive exponents. 8. 81 1a42 1/2

7. 181a42 1/2 9. 1a b 2

4x4 22 10. a b 12x2y

23 22 22

11. a

4x0y2 22 b x2y

13. 1a1/2b2 2 1ab1/22 2

12. a

4x25y2 2 b 6x22y23

14. 1a1/2b1/2c2 2

Rationalize each denominator and simplify. 15.

17.

Perform the operations and simplify. x2 2 2x 2 15 x2 2 4 ? 2 33. 2 x 1 5x 1 6 x 1 3x 2 10 34.

6x3 1 x2 2 x 3x2 2 x 4 2 x12 x 1 4x 1 4

35.

2 5x 1 x13 x23

36.

x22 x13 2 1b a x 1 3 x2 2 4

1 1 1 a b 37. 1 ab

38.

x21 2 y21 x2y

Solve each equation. 3x x 39. 5 x15 x25

40. 8 12x 2 32 2 3 15x 1 22 5 4

"3

16.

"4x

Solve each formula for the indicated variable. 1 1 1 41. 5 1 ;R R R1 R2

y 2 "3

18.

"x 2 1

42. S 5

3

3

2

3

3x

Simplify each expression and combine like terms. 19. "75 2 3"5

21. Q"2 2 "3R

2

20. "18 1 "8 2 2"2 22. Q3 2 "5RQ3 1 "5R

23. 13x2 2 2x 1 52 2 3 1x2 1 2x 2 12

Perform the operations and simplify when necessary. 24. 5x2 12x2 2 x2 1 x 1x2 2 x32

a 2 lr ;r 12r

43. Gardening A gardener wishes to enclose her rectangular raspberry patch with 40 feet of fencing. The raspberry bushes are planted along the garage, so no fencing is needed on that side. Find the dimensions if the total area is to be 192 square feet. 44. Financial planning A college student invested part of a $25,000 inheritance at 7% interest and the rest at 6%. If his annual interest is $1,670, how much did he invest at 6%?

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Cumulative Review Exercises

Perform the operations. If the result is not real, express the answer in a 1 bi form. 21i i 13 2 i2 46. 45. 1 22i 1 1 i2 11 1 i2 47. 0 3 1 4i 0

54. x2 2 8x 1 15 . 0

5 48. 7 1 5i i

55.

x2 1 4x 1 3 $0 x22

50. x4 1 36 5 13x2

56.

9 .x x

Solve each equation. 49.

x13 6 2 51 x21 x

51. "y 1 2 1 "11 2 y 5 5

52. z2/3 2 13z1/3 1 36 5 0

Graph the solution set of each inequality and write the answer using interval notation. 53. 5x 2 7 # 4

57. 0 2x 2 3 0 $ 5 58. `

3x 2 5 ` ,2 2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

187

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

2

The Rectangular Coordinate System and Graphs of Equations

© Stock Connection Blue/Alamy

CAREERS AND MATHEMATICS:

Cartographer

Cartographers collect, analyze, interpret, and map geographic information about the earth’s surface. Their work involves geographical research and compiling data to produce maps. They analyze latitude and longitude, elevation and distance, as well as population density, land-use patterns, precipitation levels, and demographic characteristics. Their maps are prepared in either digital or graphic form, using information provided by geodetic surveys, aerial cameras, satellites, lightimaging detection, and Geographic Information Systems (GIS).

2.1

The Rectangular Coordinate System

2.2

The Slope of a Nonvertical Line

2.3

Writing Equations of Lines

2.4

Graphs of Equations

2.5

Proportion and Variation Chapter Review Chapter Test

Education and Mathematics Required • •

Cartographers usually have a bachelor’s degree in cartography, geography, surveying, engineering, forestry, computer science, or a physical science. College Algebra, Trigonometry, Calculus I and II, Elementary Statistics, and Spatial Statistics are usually required.

How Cartographers Use Math and Who Employs Them •

•

Math helps cartographers with map scale, coordinate systems, and map projection. Map scale is the relationship between distances on a map and the corresponding distances on the earth’s surface. Coordinate systems are numeric methods of representing locations on the earth’s surface. Map projection is a function or transformation that relates coordinates of points on a curved surface to coordinates of points on a plane. Most cartographers work with engineering, architectural, and surveying firms. Government agencies hire cartographers to work in highway departments and in areas such as land management, natural resources planning, and national defense.

Career Outlook and Salary • •

Employment of cartographers is expected to grow 19 percent from 2008 to 2018. The median annual wage in May 2008 was $51,180.

For more information see: www.bls.gov/oco

Mathematical expressions often indicate relationships between two variables. To visualize these relationships, we draw graphs of their equations.

189 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

190

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

2.1 The Rectangular Coordinate System In this section, we will learn to 1. Plot points in the rectangular coordinate system. 2. Graph linear equations. 3. Graph vertical and horizontal lines.

We often say that a picture is worth a thousand words. In fact, pictures and graphs are an effective way to present information. For this reason, they appear frequently in newspapers and magazines. For example, the graph shown in Figure 2-1 provides a visual representation of the manatee population in Florida after the year 2003. From the graph we can note many facts about manatees. Among them are: • • •

The number of manatees in Florida declined from 2005 to 2007. In 2005, the population was about 3,100 animals. In this time period, the lowest population of manatees occurred in 2004. Florida Manatees

Manatees

Liquid Productions, LLC/Shutterstock.com

4. Solve applications using linear equations. 5. Find the distance between two points. 6. Find the midpoint of a line segment.

6,000 5,000 4,000 3,000 2,000 1,000 0

1

2

3 4 5 6 Years after 2003

7

FIGURE 2-1

In mathematics, graphs are also an effective way to present information. In this chapter, we will draw graphs of equations containing two variables and then discuss the information that we can derive from graphs. The solutions of an equation with variables x and y such as y 5 212x 1 4 are ordered pairs of real numbers 1x, y2 that satisfy the equation. To find some ordered pairs that satisfy the equation, we substitute input values of x into the equation and find the corresponding output values of y. For example, if we substitute 2 for x, we obtain 1 y52 x14 2 1 y 5 2 122 1 4 2 5 21 1 4

Substitute 2 for x.

53

Since y 5 3 when x 5 2, the ordered pair 12, 32 is a solution of the equation. The first coordinate, 2, of the ordered pair is usually called the x-coordinate. The second coordinate, 3, is usually called the y-coordinate. The solution 12, 32 and several other solutions are listed in the table of values shown in Figure 2-2.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.1

The Rectangular Coordinate System

191

1 y52 x14 2 1x, y2 x y

Comment To complete the table of solutions, we first pick values for x. Next we compute each y-value. Then we write each solution as an ordered pair. Note that we choose x-values that are multiples of the denominator, 2. This makes the computations easier when multiplying the x-value by 212 to find the corresponding y-value.

24

6

22

5

0

4

2

3

4

2

▲

▲

124, 62 122, 52 10, 42 12, 32 14, 22 ▲

Pick values for x.

Write each solution as an ordered pair. Compute each y-value. FIGURE 2-2

ACCENT ON TECHNOlOGy

Generating Table Values

© Texas Instruments images used with permission

If an equation in x and y is solved for y, we can use a graphing calculator to generate a table of solutions. The instructions in this discussion are for a TI-84 Plus graphing calculator. For details about other brands, please consult the owner’s manual. To construct a table of solutions for x 1 2y 5 8, we first solve the equation for y. x 1 2y 5 8 2y 5 2x 1 8 1 y52 x14 2 • •

•

Subtract x from both sides. Divide both sides by 2 and simplify.

To construct a table of values for y 5 212x 1 4, we first enter the equation. We STAT PLOT F1 11/22 x 1 4, as shown in Figure 2-3(a). press Y= and enter 2 TBLSET F2 WINDOW and enter one value for x on the line labeled Next, we press TblStart5. In Figure 2-3(b), 24 has been entered on this line. Other values for x that will appear in the table are determined by setting an increment value on the line labeled D Tbl5. In Figure 2-3(b), an increment value of 2 has been entered. This means that each x-value in the table will be 2 units larger than the previous one. GRAPH to obtain the table of values shown in Figure Finally, we press 2-3(c). This table contains all of the solutions listed in Figure 2-2, plus the two additional solutions 16, 12 and 18, 02 .

To see other values, we simply scroll up and down the screen by pressing the up and down arrow keys.

(a)

(b)

(c)

FIGURE 2-3

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

2

192

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Before we can present the table of solutions shown in Figure 2-2 in graphical form, we need to discuss the rectangular coordinate system.

1. Plot Points in the Rectangular Coordinate System The rectangular coordinate system consists of two perpendicular number lines that divide the plane into four quadrants, numbered as shown in Figure 2-4. The horizontal number line is called the x-axis, and the vertical number line is called the y-axis. These axes intersect at a point called the origin, which is 0 on each axis. The positive direction on the x-axis is to the right, the positive direction on the y-axis is upward, and the same unit distance is used on both axes, unless otherwise indicated. y Quadrant II

Quadrant I

3 2

Origin

1 –5 –4 –3 –2 –1

1 2

–1

–2 –3

Quadrant III

3

4

5

x

Quadrant IV

FIGURE 2-4

To plot (or graph) the point associated with the pair x 5 2 and y 5 3, denoted as 12, 32 , we start at the origin, count 2 units to the right, and then count 3 units up. (See Figure 2-5.) Point P (which lies in the first quadrant) is the graph of the ordered pair 12, 32 . The ordered pair 12, 32 gives the coordinates of point P. To plot point Q with coordinates 124, 62 , we start at the origin, count 4 units to the left, and then count 6 units up. Point Q lies in the second quadrant. Point R with coordinates 16, 242 lies in the fourth quadrant. y Q(−4, 6)

6 5 4

P(2, 3)

3 2 1 x

Caution The ordered pairs 124, 62 and 16, 242 represent different points. 124, 62 is in the second quadrant and 16, 242 is in the fourth quadrant.

–4 –3 –2 –1 –1 –2 –3 –4

1

2

3

4

5

R(6, –4)

FIGURE 2-5

2. Graph linear Equations The graph of the equation y 5 212x 1 4 is the graph of all points 1x, y2 on the rectangular coordinate system whose coordinates satisfy the equation. To graph y 5 212x 1 4, we plot the pairs listed in the table of solutions shown in Figure 2-6. These points lie on the line shown in the figure. This line is the graph of the equation.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.1

The Rectangular Coordinate System

193

y

1 y52 x14 2 1x, y2 x y

(−4, 6) (−2, 5) 1 y = − –x + 4 2

(0, 4) (2, 3) (4, 2)

x

24

6

22

5

0

4

2

3

4

2

124, 62

122, 52 10, 42 12, 32 14, 22

FIGURE 2-6

Comment

When we say that the graph of an equation is a line, we imply two things: 1. Every point with coordinates that satisfy the equation will lie on the line. 2. Every point on the line will have coordinates that satisfy the equation.

When the graph of an equation is a line, we call the equation a linear equation. These equations are often written in standard form as Ax 1 By 5 C, where A, B, and C are specific numbers (called constants) and x and y are variables. Either A or B can be 0, but A and B cannot both be 0. Standard Form of an Equation of a line

The standard form of an equation of a line is Ax 1 By 5 C where A, B, and C are real numbers and A and B are not both 0.

Here are four examples of linear equations written in standard form.

EXAMPlE 1

Linear Equation

Values of A, B, and C

3x 1 2y 5 6

A 5 3, B 5 2, C 5 6

5x 2 2y 5 210

A 5 5, B 5 22, C 5 210

2y 5 7

A 5 0, B 5 2, C 5 7

x 5 24

A 5 1, B 5 0, C 5 24

Graphing a linear Equation Graph: x 1 2y 5 5.

SOlUTION

We will solve the equation for y and form a table of solutions by picking values for x, substituting them into the equation, and solving for the other variable y. We will plot the points represented in the table of solutions and draw a line through the points. Solve the equation for y. x 1 2y 5 5 x 2 x 1 2y 5 5 2 x 2y 5 2x 1 5 1 5 y52 x1 2 2

Subtract x from both sides. Simplify. Divide both sides by 2.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

194

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Pick values for x and solve for y. If we pick x 5 0, we can find y as follows: 1 5 y52 x1 2 2 1 5 y 5 2 102 1 2 2 5 y5 2

Substitute 0 in for x. Simplify.

5 The ordered pair a0, b satisfies the equation. 2 To find another ordered pair, we pick x 5 1 and find y. 1 5 y52 x1 2 2 1 5 y 5 2 112 1 2 2 y52

Substitute 1 in for x. Simplify.

The ordered pair 11, 22 satisfies the equation. These pairs and others that satisfy the equation are shown in Figure 2-7. We plot the points and join them with a line to get the graph of the equation. x 1 2y 5 5

y

(0, 5–2) (1, 2) (3, 1)

Comment

(5, 0) (7, –1)

Even though there are infinitely many points that lie on a line, only two are required to graph a line. However, it is a good idea to find a third point as a check.

Self Check 1

x + 2y = 5 x

x

y

1x, y2

0

5 2

5 a0, b 2

1

2

11, 22

3

1

5

0

7

21

13, 12 15, 02

17, 212

FIGURE 2-7

Graph: 3x 2 2y 5 6. Now Try Exercise 35.

EXAMPlE 2 SOlUTION

Graphing a linear Equation

Graph: 3 1y 1 22 5 2x 2 3.

We will solve the equation for y and find ordered pairs 1x, y2 that satisfy the equation. Then, we will plot the points and graph the line. Solve the equation for y. 3 1y 1 22 5 2x 2 3 3y 1 6 5 2x 2 3 3y 5 2x 2 9 2 y5 x23 3

Use the Distributive Property to remove parentheses. Subtract 6 from both sides. Divide both sides by 3.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.1

The Rectangular Coordinate System

195

Pick values for x, solve for y. We now substitute numbers for x to find the corresponding values of y. If we let x 5 0 and find y, we get 2 y5 x23 3 2 y 5 102 2 3 3 y 5 23

Substitute 0 for x. Simplify.

The point 10, 232 lies on the graph. If we let x 5 3, we get 2 y5 x23 3 2 y 5 132 2 3 3 y5223

Substitute 3 for x. Simplify.

y 5 21 The point 13, 212 lies on the graph. We plot these points and others, as in Figure 2-8, and draw the line that passes through the points.

y

3 1y 1 22 5 2x 2 3

(6, 1) x (3, –1) (0, –3)

3(y + 2) = 2x – 3

(–3, –5)

x

y

23

25

0

23

3

21

6

1

1x, y2

123, 252 10, 232 13, 212 16, 12

FIGURE 2-8

Self Check 2

Graph: 2 1x 2 12 5 6 2 8y. Now Try Exercise 39.

ACCENT ON TECHNOlOGy

Graphing Equations We can graph equations with a graphing calculator. To see a graph, we must choose the minimum and maximum values of the x- and y-coordinates that will appear on the calculator’s window. A window with standard settings of Xmin 5 210

Xmax 5 10

Ymin 5 210

Ymax 5 10

will produce a graph where the value of x is in the interval 3 210, 10 4 , and the value of y is in the interval 3 210, 10 4 .

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

196

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

•

To use a graphing calculator to graph 3x 1 2y 5 12, we first solve the equation for y. 2y 5 23x 1 12 3 y52 x16 2

• •

Subtract 3x from both sides. Divide both sides by 2.

Next, we press Y= and enter the right side of the equation. The screen is shown in Figure 2-9(a). We then press GRAPH to obtain the graph shown in Figure 2-9(b).

(a)

(b) FIGURE 2-9

In Figure 2-9, the graph intersects the y-axis at the point 10, 62 , which is called the y-intercept. It intersects the x-axis at the point 14, 02 , which is called the x-intercept.

Intercepts of a line

The y-intercept of a line is the point 10, b2 , where the line intersects the y-axis. To find b, substitute 0 for x in the equation of the line and solve for y.

The x-intercept of a line is the point 1a, 02 , where the line intersects the x-axis. To find a, substitute 0 for y in the equation of the line and solve for x.

EXAMPlE 3

Graphing a line by Finding the Intercepts Use the x- and y-intercepts to graph the equation 3x 1 2y 5 12.

SOlUTION

To find the y-intercept, we substitute 0 for x and solve for y. To find the x-intercept, we substitute 0 for y and solve for x. We will also find a third point as a check and then plot the points and draw the graph. Find the y-intercept. To find the y-intercept, we substitute 0 for x and solve for y. 3x 1 2y 5 12

3 102 1 2y 5 12 2y 5 12 y56

Substitute 0 for x. Simplify. Divide both sides by 2.

The y-intercept is the point 10, 62 .

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.1

The Rectangular Coordinate System

197

Find the x-intercept. To find the x-intercept, we substitute 0 for y and solve for x. 3x 1 2y 5 12

3x 1 2 102 5 12

Substitute 0 for y.

3x 5 12

Simplify.

x54

Divide both sides by 3.

The x-intercept is the point 14, 02 . Find a third point as a check. If we let x 5 2, we will find that y 5 3. 3x 1 2y 5 12

3 122 1 2y 5 12

Substitute 2 for x.

6 1 2y 5 12

Simplify.

2y 5 6

Subtract 6 from both sides.

y53

Divide both sides by 2.

The point 12, 32 satisfies the equation. We plot each pair (as in Figure 2-10) and join them with a line to get the graph of the equation. y

(0, 6)

3x 1 2y 5 12 3x + 2y = 12 (2, 3)

(4, 0)

x

x

y

0

6

2

3

4

0

1x, y2 10, 62

12, 32 14, 02

FIGURE 2-10

Self Check 3

Graph: 2x 2 3y 5 12. Now Try Exercise 47.

ACCENT ON TECHNOlOGy

Finding Intercepts Using Zoom and Trace We can use the trace feature on a graphing calculator to find the approximate coordinates of any point on a graph. When we press TRACE , a flashing cursor will appear on the screen. The coordinates of the cursor will also appear at the bottom of the screen. •

To find the y-intercept of the graph of 2y 5 25x 2 7 1or y 5 252x 2 722 , we graph the equation, using 3 210, 10 4 for x and 3 210, 10 4 for y, and press TRACE to get Figure 2-11(a). We see from the figure that the y-intercept is (0, 23.5).

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

198

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

•

We can approximate the x-intercept by using the left arrow key and moving the cursor toward the x-intercept until we arrive at a point with the coordinates shown in Figure 2-11(b). We see from the graph that the x-coordinate of the x-intercept is approximately 21.489362.

To get better results, we can zoom in to get a magnified picture, trace again, and move the cursor to the point with coordinates shown in Figure 2-11(c). Since the y-coordinate is almost 0, we now have a good approximation for the x-intercept. The x-coordinate of the x-intercept is approximately 21.382979.We can achieve better results with repeated zooms.

(a)

(b)

(c)

FIGURE 2-11

3. Graph Horizontal and Vertical lines In the next example, we will graph a horizontal and a vertical line.

EXAMPlE 4

Graphing Horizontal and Vertical lines Graph: a. y 5 2

SOlUTION

b. x 5 23

In each case, we will plot a few ordered pairs that satisfy the equation and then draw the graph of the line. a. In the equation y 5 2, the value of y is always 2. Any value can be used for x. If we pick x-values of 23, 0, 2, and 4, we get the ordered pairs: 123, 22 , 10, 22 , 12, 22 , and (4, 22 . Plotting the pairs shown in Figure 2-12, we see that the graph is a horizontal line, parallel to the x-axis and having a y-intercept of 10, 22 . The line has no x-intercept. b. In the equation x 5 23, the value of x is always 23. Any value can be used for y. If we pick y-values of 22, 0, 2 and 3, we get the ordered pairs: 123, 222 , 123, 02 , 123, 22 and 123, 32 . After plotting the pairs shown in Figure 2-12, we see that the graph is a vertical line, parallel to the y-axis and having an x-intercept of 123, 02 . The line has no y-intercept. y

y52 x

y

23

2

0

2

2

2

4

2

1x, y2

123, 22 10, 22 12, 22 14, 22

x 5 23 (–3, 3) (–3, 2)

y=2 (0, 2)

(–3, 0) (–3, –2) x = −3

(2, 2) (4, 2) x

x

y

23

22

23

0

23

2

23

3

1x, y2

123, 222 123, 02

123, 22 123, 32

FIGURE 2-12

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Section 2.1

Self Check 4

The Rectangular Coordinate System

199

Graph: a. x 5 2 b. y 5 23 Now Try Exercise 51. Example 4 suggests the following facts.

Equations of Vertical and Horizontal lines

If a and b are real numbers, then • •

The graph of the equation x 5 a is a vertical line with x-intercept of 1a, 02 . If a 5 0, the line x = 0 is the y-axis. The graph of the equation y 5 b is a horizontal line with y-intercept of 10, b2 . If b 5 0, the line y = 0 is the x-axis.

4. Solve Applications Using linear Equations EXAMPlE 5

Solving an Application Problem A computer purchased for $2,750 is expected to depreciate according to the formula y 5 2550x 1 $2,750, where y is the value of the computer after x years. When will the computer be worth nothing?

SOlUTION

The computer will have no value when its value (y) is 0. To find x when y 5 0, we substitute 0 for y and solve for x. y 5 2550x 1 2,750 0 5 2550x 1 2,750 22,750 5 2550x 55x

Subtract 2,750 from both sides. Divide both sides by 2550.

The computer will have no value in 5 years. Self Check 5

When will the value of the computer be $1,650? Now Try Exercise 97.

5. Find the Distance between Two Points To derive the formula used to find the distance between two points on a rectangular coordinate system, we use subscript notation and denote the points as P 1x1, y12

Q 1x2, y22

y

Q(x2, y2) d

|y2 – y1|

P(x1, y1) |x2 – x1|

R(x2, y1)

FIGURE 2-13

Read as “point Q with coordinates of x sub 2 and y sub 2.”

If P 1x1, y12 and Q 1x2, y22 are two points in Figure 2-13 and point R has coordinates 1x2, y12 , triangle PQR is a right triangle. By the Pythagorean Theorem, the square of the hypotenuse of right triangle PQR is equal to the sum of the squares of the two legs. Because leg RQ is vertical, the square of its length is 1y2 2 y12 2. Since leg PR is horizontal, the square of its length is 1x2 2 x12 2. Thus, we have (1)

x

Read as “point P with coordinates of x sub 1 and y sub 1.”

d 2 5 1x2 2 x12 2 1 1y2 2 y12 2

Because equal positive numbers have equal positive square roots, we can take the positive square root of both sides of Equation 1 to obtain the Distance Formula.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

200

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

The Distance Formula

The distance d between points 1x1, y12 and 1x2, y22 is given by d 5 " 1x2 2 x12 2 1 1y2 2 y12 2

EXAMPlE 6 SOlUTION

Finding the Distance between Two Points

Find the distance between P 121, 222 and Q 127, 82 .

We use the Distance Formula, d 5 " 1x2 2 x12 2 1 1y2 2 y12 2, to find the distance between P 121, 222 and Q 127, 82 . If we let P 121, 222 5 P 1x1, y12 and Q 127, 82 5 Q 1x2, y22 , we can substitute –1 for x1, –2 for y1, –7 for x2, and 8 for y2 into the formula and simplify. d 1PQ2 5 " 1x2 2 x12 2 1 1y2 2 y12 2

Read d 1PQ2 as “the length of segment PQ.”

d 1PQ2 5 " 3 27 2 1212 4 2 1 3 8 2 1222 4 2 5 " 1262 2 1 1102 2 5 "36 1 100 5 "136

5 "4 ? 34

"4 ? 34 5 "4"34 5 2"34

5 2"34 Self Check 6

Find the distance between P 122, 252 and Q 13, 72 . Now Try Exercise 73.

6. Find the Midpoint of a line Segment

If point M in Figure 2-14 lies midway between points P 1x1, y12 and Q 1x2, y22 , point M is called the midpoint of segment PQ. To find the coordinates of M, we find the average of the x-coordinates and the average of the y-coordinates of P and Q. The Midpoint Formula

The midpoint of the line segment with endpoints at P 1x1, y12 and Q 1x2, y22 is the point M with coordinates of M5a

x1 1 x2 y1 1 y2 , b 2 2

y

x1 + x2 y1 + y2 M –––––– , –––––– 2 2

(

)

Q(x2, y2)

P(x1, y1) x

FIGURE 2-14 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.1

201

The Rectangular Coordinate System

You will be asked to prove this formula in Exercise 107 by using the Distance Formula to show that d 1PM2 1 d 1MQ2 5 d 1PQ2 .

EXAMPlE 7

SOlUTION

Finding the Midpoint of a line Segment

Find the midpoint of the segment joining P 127, 22 and Q 11, 242 . We use the Midpoint Formula, M 5 a

x1 1 x2 y1 1 y2 , b to find the midpoint of 2 2 the line segment joining P 127, 22 and Q 11, 242 . To do so, we substitute P 127, 22 for P 1x1, y12 and Q 11, 242 for Q 1x2, y22 into the Midpoint Formula to get x1 1 x2 2 27 1 1 5 2 26 5 2 5 23

y1 1 y2 2 2 1 1242 5 2 22 5 2 5 21

and

xM 5

yM 5

The midpoint is M 123, 212 . Self Check 7

Find the midpoint of the segment joining P 127, 282 and Q 122, 102 . Now Try Exercise 83.

EXAMPlE 8

SOlUTION

Using Midpoint Formula to Find Coordinates

The midpoint of the segment joining P 123, 22 and Q 1x2, y22 is M 11, 42 . Find the coordinates of Q. We can let P 1x1, y12 5 P 123, 22 and M 1xM, yM2 5 M 11, 42 , and then find the coordinates x2 and y2 of point Q 1x2, y22 . x1 1 x2 2 23 1 x2 15 2 2 5 23 1 x2

xM 5

y1 1 y2 2 2 1 y2 45 2 8 5 2 1 y2

and

yM 5

5 5 x2

Multiply both sides by 2.

6 5 y2

The coordinates of point Q are 15, 62 . Self Check 8

If the midpoint of a segment PQ is M 12, 252 and one endpoint is Q 16, 92 , find P. Now Try Exercise 87.

Self Check Answers

1.

2.

y

x

3.

y

2(x − 1) = 6 – 8y

4.

y

y x=2

2x − 3y = 12

x

x

x

y = –3

3x − 2y = 6

5. 2 years

6. 13

9 7. Ma2 , 1b 2

8. 122, 2192

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202

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Exercises 2.1 Getting Ready

Graph each point. Indicate the quadrant in which the point lies, or the axis on which it lies. 24. 123, 42 23. 12, 52

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. The coordinate axes divide the plane into four . 2. The coordinate axes intersect at the . 3. The positive direction on the x-axis is

25. 124, 252 27. 15, 22 29. 14, 02

.

4. The positive direction on the y-axis is . 5. The x-coordinate is the coordinate in an ordered pair. coordinate in an 6. The y-coordinate is the ordered pair. 7. A equation is an equation whose graph is a line. 8. The point where a line intersects the is called the y-intercept. 9. The point where a line intersects the x-axis is called the . 10. The graph of the equation x 5 a will be a line. 11. The graph of the equation y 5 b will be a line. 12. Complete the Distance Formula: d5 . 13. If a point divides a segment into two equal segments, the point is called the of the segment. 14. The midpoint of the segment joining P 1x1, y12 and Q 1x2, y22 is

26. 16, 22 28. 13, 242 30. 10, 22

Solve each equation for y and graph the equation. Then check your graph with a graphing calculator. 31. y 2 2x 5 7

32. y 1 3 5 24x y

y x

x

33. y 1 5x 5 5

34. y 2 3x 5 6

y

y

x

x

35. 6x 2 3y 5 10

36. 4x 1 8y 2 1 5 0

y

y

x

. x

Practice Refer to the illustration and determine the coordinates of each point.

37. 3x 5 6y 2 1

38. 2x 1 1 5 4y

y 5 4 A 3 2 1 E F x –5 –4 –3–2–1–1 1 2 3 4 5 –2 H C –3 –4 –5 G D

y

y

B

15. A

16. B

17. C

18. D

19. E

20. F

21. G

x

x

22. H Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.1

39. 2 1x 1 y 1 12 5 x 1 2

40. 5 1x 1 22 5 3y 2 x

y

203

The Rectangular Coordinate System

Graph each equation. 49. y 5 3

y

50. x 5 24 y

y

x

x

x

x

Find the x- and y-intercepts and use them to graph each equation. 41. x 1 y 5 5 42. x 2 y 5 3 y

51. 3x 1 5 5 21

52. 7y 2 1 5 6

y

y

y

x

x

x

x

43. 2x 2 y 5 4

53. 3 1y 1 22 5 y 44. 3x 1 y 5 9

y

54. 4 1 3y 5 3 1x 1 y2 y

y

y

x

x x

x

45. 3x 1 2y 5 6

46. 2x 2 3y 5 6

55. 3 1y 1 2x2 5 6x 1 y y

y

56. 5 1y 2 x2 5 x 1 5y y

y

x

x

x

x

47. 4x 2 5y 5 20

48. 3x 2 5y 5 15

y

y

x

x

Use a graphing calculator to graph each equation and then find the x-coordinate of the x-intercept to the nearest hundredth. 3 5 57. y 5 3.7x 2 4.5 58. y 5 x 1 5 4 60. 0.3x 1 y 5 7.5 59. 1.5x 2 3y 5 7 Find the distance between P and O(0, 0). 62. P 125, 122 61. P 14, 232 63. P 123, 22

65. P 11, 12

67. PQ"3, 1R

64. P 15, 02

66. P 16, 282

68. PQ"7, "2R

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Find the distance between P and Q. 70. P 14, 92 ; Q 19, 212 69. P 13, 72 ; Q 16, 32 72. P 10, 52 ; Q 16, 232 71. P 14, 262 ; Q 121, 62

73. P 122, 2152 ; Q 126, 2212 74. P 127, 112 ; Q 1211, 72 75. P 13, 232 ; Q 125, 52

76. P 16, 232 ; Q 123, 22

77. P 1p, 222 ; Q 1p, 52

78. PQ"5, 0R; Q 10, 22

Find the midpoint of the line segment PQ. 80. P 13, 262 ; Q 121, 262 79. P 12, 42 ; Q 16, 82 81. P 12, 252 ; Q 122, 72

82. P 10, 32 ; Q 1210, 2132

83. P 128, 52 ; Q 16, 242

84. P 13, 222 ; Q 12, 232

95. In the illustration, point M is the midpoint of the hypotenuse of right triangle AOB. Show that the area of rectangle OLMN is one-half of the area of triangle AOB. y

B(0, b)

O

L

x A(a, 0)

96. Rectangle ABCD in the illustration is twice as long as it is wide, and its sides are parallel to the coordinate axes. If the perimeter is 42, find the coordinates of point C. y D

85. P 10, 02 ; QQ"5, "5R

M

N

C

86. PQ"3, 0R; QQ0, 2"5R x A(–3, –2)

89. P 15, 252 ; M 15, 52

90. P 127, 32 ; M 10, 02

91. Show that a triangle with vertices at 113, 222 , 19, 282 , and 15, 222 is isosceles. 92. Show that a triangle with vertices at 121, 22 , 13, 12 , and 14, 52 is isosceles. 93. In the illustration, points M and N are the midpoints of AC and BC, respectively. Find the length of MN.

Applications 97. House appreciation A house purchased for $225,000 is expected to appreciate according to the formula y 5 17,500x 1 225,000, where y is the value of the house after x years. Find the value of the house 5 years later. jcpjr/Shutterstock.com

One endpoint P and the midpoint M of line segment PQ are given. Find the coordinates of the other endpoint, Q. 88. P 12, 272 ; M 125, 62 87. P 11, 42 ; M 13, 52

B

98. Car depreciation A Chevy Cruze car purchased for $17,000 is expected to depreciate according to the formula y 5 21,360x 1 17,000, where y is the value of the car after x years. When will the car be worthless?

C(6, 10) N M B(4, 6) A(2, 4)

94. In the illustration, points M and N are the midpoints of AC and BC, respectively. Show that d 1MN2 5 12 3 d 1AB2 4 .

Olga Besnard/Shutterstock.com

204

C(b, c)

M

A(0, 0)

N

B(a, 0)

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Section 2.1

99. Demand equations The number of photo scanners that consumers buy depends on price. The higher the price, the fewer photo scanners people will buy. The equation that relates price to the number of photo scanners sold at that price is called a demand equation. If the demand equation for a photo 1 q 1 170, where p is the price scanner is p 5 210 and q is the number of photo scanners sold at that price, how many photo scanners will be sold at a price of $150? 100. Supply equations The number of television sets that manufacturers produce depends on price. The higher the price, the more TVs manufacturers will produce. The equation that relates price to the number of TVs produced at that price is called a supply equation. If the supply equation for a 1 q 1 130, where p is the price 25-inch TV is p 5 10 and q is the number of TVs produced for sale at that price, how many TVs will be produced if the price is $150? 101. Meshing gears The rotational speed V of a large gear (with N teeth) is related to the speed v of the smaller gear (with n teeth) by the equation V 5 nv N. If the larger gear in the illustration is making 60 revolutions per minute, how fast is the smaller gear spinning?

The Rectangular Coordinate System

205

72 mi 47 mi

Pigeon Cove 23 mi

84 mi

Home port

104. Engineering Two holes are to be drilled at locations specified by the engineering drawing shown in the illustration. Find the distance between the centers of the holes. 19.3 mm

5.0 mm 4.3 mm 11.7 mm 5.0 mm 3.7 mm

Discovery and Writing 105. Explain how to graph a line using the intercept method.

V v

106. Explain how to determine the quadrant in which the point P 1a, b2 lies. 107. In Figure 2-14, show that d 1PM2 1 d 1MQ2 5 d 1PQ2 .

102. Crime prevention The number n of incidents of family violence requiring police response appears to be related to d, the money spent on crisis intervention, by the equation n 5 430 2 0.005d. What expenditure would reduce the number of incidents to 350? 103. Navigation See the illustration. An ocean liner is located 23 miles east and 72 miles north of Pigeon Cove Lighthouse, and its home port is 47 miles west and 84 miles south of the lighthouse. How far is the ship from port?

108. Use the result of Exercise 107 to explain why point M is the midpoint of segment PQ.

Review Graph each interval on the number line. 110. 121, 42 d 3 22, 2 4 109. 3 23, 22 c 122, 3 4 111. 3 23, 222 d 12, 3 4

112. 3 24, 232 c 12, 3 4

Solve each equation. 3 4 113. 5 y16 y14 114.

z14 z11 8 2 2 5 2 z2 1 z z 1 2z z 1 3z 1 2

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206

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

2.2 The Slope of a Nonvertical Line In this section, we will learn to 1. 2. 3. 4.

Find the slope of a line. Use slope to solve applications. Find slopes of horizontal and vertical lines. Find slopes of parallel and perpendicular lines.

The world’s steepest passenger railway is the Lookout Mountain Incline Railway in Chattanooga, Tennessee. Passengers experience breathtaking views of the city and surrounding mountains as the trolley-style railcars travel up Lookout Mountain. The grade or steepness of the track is 72.7% near the top. Mathematicians use the term slope to represent the measure of the steepness of a line. We will explore the topic of slope in this section because it has many real-life applications.

1. Find the Slope of a line © dk/Alamy

Suppose that a college student rents a room for $300 per month, plus a $200 nonrefundable deposit. The table shown in Figure 2-15(b) gives the cost (y) for different numbers of months (x). If we construct a graph from these data, we get the line shown in Figure 2-15(a). y

Total cost (y)

$2,000 $1,500 (4, 1,400) $1,000

(3, 1,100) (2, 800)

$500

(1, 500) (0, 200) 0

1 2 3 4 5 Number of months (x)

x

(a)

Time in Months

Total Cost

x

y

0

200

1

500

2

800

3

1,100

4

1,400

(b) FIGURE 2-15

From the graph, we can see that if x changes from 0 to 1, y changes from 200 to 500. As x changes from 1 to 2, y changes from 500 to 800, and so on. The ratio of the change in y divided by the change in x is the constant 300. Change in y 500 2 200 800 2 500 1,100 2 800 1,400 2 1,100 300 5 5 5 5 5 5 300 Change in x 120 221 322 423 1 The ratio of the change in y divided by the change in x between any two points on any line is always a constant. This constant rate of change is called the slope of the line.

The Slope of a Nonvertical line

The slope of the nonvertical line (see Figure 2-16) passing through points P 1x1, y12 and Q 1x2, y22 is m5

change in y y2 2 y1 5 change in x x2 2 x1

1x2 2 x12

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Section 2.2

The Slope of a Nonvertical Line

207

y

l Q(x2, y2)

Comment

y2 – y1

Slope is often considered to be a measure of the steepness or tilt of a line. Note that we can use the coordinates of any two points on a line to compute the slope of the line.

P(x1, y1)

R(x2, y1)

x2 – x1

x FIGURE 2-16

EXAMPlE 1

SOlUTION

Finding the Slope of a line Given Two Points

Find the slope of the line passing through P 121, 222 and Q 17, 82 . We will substitute the points P 121, 222 and Q 17, 82 into the slope formula, change in y y2 2 y1 m5 5 , to find the slope of the line. change in x x2 2 x1 Let P 1x1, y12 5 P 121, 222 and Q 1x2, y22 5 Q 17, 82 . Then we substitute 21 for x1, 22 for y1, 7 for x2, and 8 for y2 to get change in y change in x y 2 y1 m5 2 x2 2 x1 m5

8 2 1222 7 2 1212 10 5 8 5 5 4 The slope of the line is 54. See Figure 2-17. 5

y Q(7, 8)

Change in y = 10 x P(–1, –2) Change in x = 8 R(7, –2) FIGURE 2-17

We would have obtained the same result if we had let P 1x1, y12 5 P 17, 82 and Q 1x2, y22 5 Q 121, 222 . Self Check 1

Find the slope of the line passing through P 123, 242 and Q 15, 92 . Now Try Exercise 13.

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208

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Caution

When calculating slope, always subtract the y-values and the x-values in the same order. m5

y2 2 y1 x2 2 x1

or

m5

y1 2 y2 x1 2 x2

Otherwise, we will obtain an incorrect result.

A slope can be a positive real number, 0, or a negative real number. If the denominator of the slope formula is 0, slope is not defined. The change in y (often denoted as Dy) is the rise of the line between points P and Q. The change in x (often denoted as Dx) is the run. Using this terminology, we can define slope to be the ratio of the rise to the run: m5

EXAMPlE 2

Dy y2 2 y1 rise 5 5 x2 2 x1 Dx run

1Dx 2 02

Finding the Slope of a line Given Its Equation in Standard Form Find the slope of the line determined by 5x 1 2y 5 10. (See Figure 2-18.)

SOlUTION y

We will find the coordinates of the x- and y-intercepts and substitute into slope change in y y2 2 y1 formula, m 5 5 , to find the slope of the line change in x x2 2 x1 • •

5x + 2y = 10

If y 5 0, then x 5 2, and the point 12, 02 lies on the line. If x 5 0, then y 5 5, and the point 10, 52 lies on the line.

We then find the slope of the line between P 12, 02 and Q 10, 52 . change in y change in x y 2 y1 m5 2 x2 2 x1

x

m5

FIGURE 2-18

520 022 5 52 2 The slope is 252 . 5

Self Check 2

Find the slope of the line determined by 3x 2 2y 5 9. Now Try Exercise 27.

2. Use Slope to Solve Applications EXAMPlE 3

Using Slope to Solve an Application Problem If carpet costs $25 per square yard plus a delivery charge of $30, the total cost C of n square yards is given by the formula

Total cost

equals

cost per square yard

times

the number of square yards purchased

plus

the delivery charge.

C

5

25

?

n

1

30

Graph the equation C 5 25n 1 30 and interpret the slope of the line. SOlUTION

We will complete a table of solutions and graph the equation on a coordinate system with a vertical C-axis and a horizontal n-axis. Figure 2-19 shows a table of ordered pairs and the graph.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.2

The Slope of a Nonvertical Line

209

C 1,300

Cost

1,200 1,100 1,000 900 800

C 5 25n 1 30 n

C

(n, C)

10

280

(10, 280)

20

530

(20, 530)

500

30

780

(30, 780)

400

40

1,030

(40, 1,030)

300 200

50

1,280

(50, 1,280)

700

C = 25n + 30

600

100 n 0

30 40 50 60 10 20 Number of square yards purchased FIGURE 2-19

If we pick the points 130, 7802 and 150, 1,2802 to find the slope, we have DC Dn C2 2 C1 5 n2 2 n1

m5

5

1,280 2 780 50 2 30

5

500 20

Substitute 1,280 for C2, 780 for C1, 50 for n2, and 30 for n1.

5 25 The slope of 25 (in dollars/square yard) is the cost per square yard of the carpet. Self Check 3

If the cost of the carpet in Example 3 increases to $35 per square yard, find the slope of the line. Now Try Exercise 81.

EXAMPlE 4

Solving an Application Problem It takes a skier 25 minutes to complete the course shown in Figure 2-20. Find his average rate of descent in feet per minute. y (in ft)

(0, 12,000)

Altitude

12,000

8,500 (25, 8,500)

Time

25

x (in min)

FIGURE 2-20 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

210

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

SOlUTION

To find the average rate of descent, we will find the ratio of the change in altitude to the change in time. To find this ratio, we will calculate the slope of the line passing through the points (0, 12,000) and (25, 8,500). Average rate 12,000 2 8,500 5 of descent 0 2 25 5

3, 500 225

5 2140 The average rate of descent is 140 ft/min. Self Check 4

If it takes the skier 20 minutes to complete the course, find his average rate of descent in feet per minute. Now Try Exercise 83.

3. Find Slopes of Horizontal and Vertical lines

If P 1x1, y12 and Q 1x2, y22 are points on the horizontal line shown in Figure 2-21(a), then y1 5 y2, and the numerator of the fraction is 0. y2 2 y1 x2 2 x1

On a horizontal line, x2 2 x1 .

Thus, the value of the fraction is 0, and the slope of the horizontal line is 0. If P 1x1, y12 and Q 1x2, y22 are points on the vertical line shown in Figure 2-21(b), then x1 5 x2, and the denominator of the fraction is 0. y2 2 y1 x2 2 x1

On a vertical line, y2 2 y1 .

Since the denominator of a fraction cannot be 0, the slope of a vertical line is not defined. y

y

Q(x2, y2) P(x1, y1)

Q(x2, y2) P(x1, y1)

x

x

(b)

(a) FIGURE 2-21

Slopes of Horizontal and Vertical lines

The slope of a horizontal line (a line with an equation of the form y 5 b) is 0. The slope of a vertical line (a line with an equation of the form x 5 a) is not defined.

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Section 2.2

211

The Slope of a Nonvertical Line

Here are a few facts about slope. See Figure 2-22. • • • •

If a line rises as we follow it from left to right, as in Figure 2-22(a), its slope is positive. If a line drops as we follow it from left to right, as in Figure 2-22(b), its slope is negative. If a line is horizontal, as in Figure 2-22(c), its slope is 0. If a line is vertical, as in Figure 2-22(d), it has no defined slope. Slope Concepts

y

y

y

y

∆y > 0 ∆x > 0

∆y > 0

∆y = 0

∆x > 0 ∆y < 0 x

∆x = 0

∆x > 0 x

x

x

Positive slope

Negative slope

Zero slope

Undefined slope

(a)

(b)

(c)

(d)

FIGURE 2-22

4. Find Slopes of Parallel and Perpendicular lines To see a relationship between parallel lines and their slopes, we refer to the parallel lines l1 and l2 shown in Figure 2-23, with slopes of m1 and m2, respectively. Because right triangles ABC and DEF are similar, it follows that m1 5

Dy of l1 Dy of l2 5 5 m2 Dx of l1 Dx of l2 y

l1 l2

C ∆y of l1 Slope = m1

A ∆x of l1

F ∆y of l2

B D ∆x of l2

E

x

Slope = m2 FIGURE 2-23

This shows that if two nonvertical lines are parallel, they have the same slope. It is also true that when two lines have the same slope, they are parallel.

Slopes of Parellel lines

Nonvertical parallel lines have the same slope, and lines having the same slope are parallel. Since vertical lines are parallel, lines with undefined slopes are parallel.

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212

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EXAMPlE 5

Solving a Slope Problem Involving Parallel lines The lines in Figure 2-24 are parallel. Find y. y R(−2, 5) Q(−3, 4)

T(x, 0)

P(1, −2)

x

S(3, y)

FIGURE 2-24

SOlUTION

Since the lines are parallel, their slopes are equal. To find y, we will find the slope of each line, set them equal, and solve the resulting equation. Slope of PQ 5 slope of RS 22 2 4 y25 5 1 2 1232 3 2 1222 26 y25 5 4 5 230 5 4 1y 2 52

Simplify. Multiply both sides by 20.

230 5 4y 2 20

Remove parentheses and simplify.

210 5 4y

Add 20 to both sides.

5 2 5y 2

Divide both sides by 4 and simplify.

Thus, y 5 252. Self Check 5

Find x in Figure 2-24. Now Try Exercise 59.

Comment If the product of two numbers is 21, the numbers are called negative reciprocals.

Slopes of Perpendicular lines

The following theorem relates perpendicular lines and their slopes.

If two nonvertical lines are perpendicular, the product of their slopes is 21. If the product of the slopes of two lines is 21, the lines are perpendicular.

PROOF

Suppose l1 and l2 are lines with slopes of m1 and m2 that intersect at some point. See Figure 2-25. Then superimpose a coordinate system over the lines so that the intersection point is the origin. Let P 1a, b2 be a point on l1, and let Q 1c, d2 be a point on l2. Neither point P nor point Q can be the origin.

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Section 2.2

The Slope of a Nonvertical Line

213

y

l1

l2

P(a, b) Q(c, d)

Slope = m1

Slope = m2

x

O(0, 0) FIGURE 2-25

First, we suppose that l1 and l2 are perpendicular. Then triangle POQ is a right triangle with its right angle at O. By the Pythagorean Theorem, d 1OP2 2 1 d 1OQ2 2 5 d 1PQ2 2

1a 2 02 2 1 1b 2 02 2 1 1c 2 02 2 1 1d 2 02 2 5 1a 2 c2 2 1 1b 2 d2 2 a2 1 b2 1 c2 1 d 2 5 a2 2 2ac 1 c2 1 b2 2 2bd 1 d 2 0 5 22ac 2 2bd bd 5 2ac b d ? 5 21 Divide both sides by ac. a c The coordinates of P are 1a, b2 , and the coordinates of O are 10, 02 . Using the definition of slope, we have (1)

b20 b 5 a20 a Similarly, we have m1 5

d c We substitute m1 for ba and m2 for dc in Equation 1 to obtain m2 5

m1m2 5 21

Comment It is also true that a horizontal line is perpendicular to a vertical line.

EXAMPlE 6

Hence, if lines l1 and l2 are perpendicular, the product of their slopes is 21. Conversely, we suppose that the product of the slopes of lines l1 and l2 is 21. Because the steps in the previous discussion are reversible, we have d 1OP2 2 1 d 1OQ2 2 5 d 1PQ2 2. By the Pythagorean Theorem, triangle POQ is a right triangle. Thus, l1 and l2 are perpendicular.

Solving a Slope Problem Involving Perpendicular lines Are the lines shown in Figure 2-26 perpendicular? y Q(9, 4)

R(3, 3)

O(0, 0)

x S(7, 0)

P(3, −4) FIGURE 2-26

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214

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

SOlUTION

We will determine the slopes of the lines and see whether their product is –1. Dy Dx y 2 y1 5 2 x2 2 y1

Dy Dx y 2 y1 5 2 x2 2 y1

Slope of OP 5

Slope of PQ 5

4 2 1242 923 8 5 6 4 5 3

24 2 0 320 4 52 3 5

5

Since the product of the slopes is 216 9 and not 21, the lines are not perpendicular. Self Check 6

Is either line in Figure 2-26 perpendicular to the line passing through R and S? Now Try Exercise 55.

Self Check Answers

1.

13 8

2.

3 2

3. 35

4. 175 ft/min

5.

4 3

6. yes

Exercises 2.2 Getting Ready

Practice

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

Find the slope of the line passing through each pair of points, if possible. 12. P 13, 212 ; Q 15, 32 11. P 12, 22 ; Q 121, 212

Fill in the blanks. 1. The slope of a nonvertical line is defined to be the change in y by the change in x. 2. The change in is often called the rise. . 3. The change in x is often called the 4. When computing the slope from the coordinates of two points, always subtract the y-values and the x-values in the . 5. The symbol Dy means y. 6. The slope of a line is 0. 7. The slope of a line is undefined. 8. If the slopes of two lines are equal, the lines are .

y

y

Q(5, 3) P(2, 2) x Q(–1, –1)

x P(3, –1)

13. P 126, 32 ; Q 16, 222

14. P 12, 52 ; Q 13, 102 y Q(3, 10)

y P(–6, 3) P(2, 5)

9. If the product of the slopes of two lines is 21, the lines are . 10. If two lines are perpendicular, the product of their slopes is .

x Q(6, –2)

x

15. P 13, 222 ; Q 121, 52 16. P 13, 72 ; Q 16, 162 17. P 18, 272 ; Q 14, 12

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Section 2.2

The Slope of a Nonvertical Line

18. P 15, 172 ; Q 117, 172 19. P 124, 32 ; Q 124, 232

47. m1 5 2"2; m2 5

3 2 5 7 21. Pa , b; Qa , b 2 3 2 3 2 1 3 5 22. Pa2 , b; Qa , 2 b 5 3 5 3

50. m1 5 0.125; m2 5

215

"2 2 48. m1 5 2"7; m2 5 "28 49. m1 5 20.125; m2 5 8

20. PQ2, "7R; QQ"7, 2R

1 8 51. m1 5 ab21; m2 5 2a21b

1a 2 0, b 2 02

23. P 1a 1 b, c2 ; Q 1b 1 c, a2 assume c 2 a 24. P 1b, 02 ; Q 1a 1 b, a2 assume a 2 0

a 21 b 52. m1 5 a b ; m2 5 2 b a

Find two points on the line and use slope formula to find the slope of the line. 25. y 5 3x 1 2 26. y 5 5x 2 8

Determine whether the line through the given points and the line through R 123, 52 and S 12, 72 are parallel, perpendicular, or neither. 54. P 123, 82 ; Q 1213, 42 53. P 12, 42 ; Q 17, 62

27. 5x 2 10y 5 3 29. 3 1y 1 22 5 2x 2 3

31. 3 1y 1 x2 5 3 1x 2 12

28. 8y 1 2x 5 5 30. 4 1x 2 22 5 3y 1 2 32. 2x 1 5 5 2 1y 1 x2

Find the slope of the line, if possible. 33. y 5 7 34. 2y 5 5 1 35. x 5 2 36. x 2 7 5 0 2

39.

x

40.

y

41.

x

57. P 1a, a2 ; Q 13a, 6a2 1a 2 02 58. P 1b, b2 ; Q 12b, 6b2 1b 2 02

60. Parallel: P 12, 232 ; Q 15, 72 ; R 13, 212 ; S 16, y2 61. Perpendicular: P 12, 272 ; Q 11, 02 ; R 129, 52 ; S 122, y2 62. Perpendicular: P 11, 222 ; Q 13, 42 ; R 1x, 62 ; S 16, 52

Find the slopes of lines PQ and PR, and determine whether points P, Q, and R lie on the same line. 63. P 122, 82 ; Q 126, 92 ; R 12, 52

66. P 1a, a 1 b2 ; Q 1a 1 b, b2 ; R 1a 2 b, a2

x

42.

y

56. P 10, 292 ; Q 14, 12

64. P 11, 212 ; Q 13, 222 ; R 123, 02 65. P 12a, a2 ; Q 10, 02 ; R 1a, 2a2

y

x

55. P 124, 62 ; Q 122, 12

Lines PQ and RS are either parallel or perpendicular. Find x or y. 59. Parallel: P 123, 72 ; Q 12, 92 ; R 110, 242 ; S 1x, 262

Determine whether the slope of the line is positive, negative, 0, or undefined. y y 37. 38.

x

1a 2 0, b 2 0, a 2 b2

Determine which, if any, of the three lines PQ, PR, and QR are perpendicular. 67. P 15, 42 ; Q 12, 252 ; R 18, 232 68. P 18, 222 ; Q 14, 62 ; R 16, 72 69. P 11, 32 ; Q 11, 92 ; R 17, 32

y

x

70. P 12, 232 ; Q 123, 22 ; R 13, 82 71. P 10, 02 ; Q 1a, b2 ; R 12b, a2

72. P 1a, b2 ; Q 12b, a2 ; R 1a 2 b, a 1 b2 Determine whether the lines with the given slopes are parallel, perpendicular, or neither. 1 2 3 44. m1 5 ; m2 5 43. m1 5 3; m2 5 2 3 3 2 45. m1 5 "8; m2 5 2"2

46. m1 5 1; m2 5 21

73. Right triangles Show that the points A 121, 212 , B 123, 42 , and C 14, 12 are the vertices of a right triangle. 74. Right triangles Show that the points D 10, 12 , E 121, 32 , and F 13, 52 are the vertices of a right triangle. 75. Squares Show that the points A 11, 212 , B 13, 02 , C 12, 22 , and D 10, 12 are the vertices of a square.

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216

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The Rectangular Coordinate System and Graphs of Equations

76. Squares Show that the points E 121, 212 , F 13, 02 , G 12, 42 , and H 122, 32 are the vertices of a square. 77. Parallelograms Show that the points A 122, 222 , B 13, 32 , C 12, 62 , and D 123, 12 are the vertices of a parallelogram. (Show that both pairs of opposite sides are parallel.) 78. Trapezoids Show that points E 11, 222 , F 15, 12 , G 13, 42 , and H 123, 42 are the vertices of a trapezoid. (Show that only one pair of opposite sides is parallel.) 79. Geometry In the illustration, points M and N are midpoints of CB and BA, respectively. Show that MN is parallel to AC. C(5, 9) M

84. Hospital costs The table shows the changing mean daily cost for a hospital room. For the ten-year period, find the rate of change per year of the portion of the room cost that is absorbed by the hospital.

Year

Total Cost to the Hospital

Amount Passed on to Patient

2000

$459

$212

2005

$670

$295

2010

$812

$307

B(7, 5)

N A(1, 3)

80. Geometry In the illustration, d 1AB2 5 d 1AC2 . Show that AD is perpendicular to BC. C(b, c)

83. Rate of decrease The price of computers has been dropping steadily for the past ten years. If a desktop PC cost $6,700 ten years ago, and the same computing power cost $2,200 three years ago, find the rate of decrease per year. (Assume a straight-line model.)

85. Charting temperature changes The following Fahrenheit temperature readings were recorded over a four-hour period. Time

D(a + b, c)

12:00

1:00

2:00

3:00

4:00

47°

53°

59°

65°

71°

Temperature

A(0, 0)

Let t represent the time (in hours), with 12:00 corresponding to t 5 0. Let T represent the temperature. Plot the points 1t, T 2 , and draw the line through those points. Explain the meaning of DT Dt .

B(a, 0)

Applications 81. Rate of growth When a college started an aviation program, the administration agreed to predict enrollments using a straight-line method. If the enrollment during the first year was 12, and the enrollment during the fifth year was 26, find the rate of growth per year (the slope of the line). See the illustration.

T 75 70 65 60 55 50 45 t 1

Enrollment

26

ENROLL IN THE AVIATION PROGRAM

1

3 4

86. Tracking the Dow The Dow Jones Industrial Averages at the close of trade on three consecutive days were as follows:

FLY WITH US!

12

2

5 Years

82. Rate of growth A small business predicts sales according to a straight-line method. If sales were $50,000 in the first year and $110,000 in the third year, find the rate of growth in dollars per year (the slope of the line).

Day

Monday

Tuesday

Wednesday

Close

12,981

12,964

12,947

Let d represent the day, with d 5 0 corresponding to Monday, and let D represent the Dow Jones average. Plot the points 1d, D2 , and draw the graph. Explain the meaning of DD Dd .

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Section 2.3

D

Writing Equations of Lines

217

A (amount, in dollars)

12,990 12,980 12,970 12,960 12,950 12,940

125 100 75 50 25

d 1

2 1

Ilja Mas˘ík/Shutterstock.com

87. Speed of an airplane A pilot files a flight plan indicating her intention to fly at a constant speed of 590 mph. Write an equation that expresses the distance traveled in terms of the flying time. Then graph the equation and interpret the slope of the line. (Hint: d 5 rt.)

d (mi)

2

3

4 5

n (number of deposits)

Discovery and Writing 89. Explain why the slope of a vertical line is undefined. 90. Explain how to determine whether two lines are parallel, perpendicular, or neither.

Review Solve each equation for y and simplify. 91. 3x 1 7y 5 21

93.

x y 1 51 5 2

92. y 2 3 5 5 1x 1 22

94. x 2 5y 5 15

1,200

Factor each expression. 600

95. 6p2 1 p 2 12 1

2

t (hr)

96. b3 2 27 97. mp 1 mq 1 np 1 nq

88. Growth of savings A student deposits $25 each month in a Holiday Club account at her bank. The account pays no interest. Write an equation that expresses the amount A in her account in terms of the number of deposits n. Then graph the line, and interpret the slope of the line.

98. x4 1 x2 2 2

2.3 Writing Equations of Lines In this section, we will learn to 1. 2. 3. 4.

Use point-slope form to write an equation of a line. Use slope-intercept form to write an equation of a line. Graph linear equations using the slope and y-intercept. Determine whether linear equations represent lines that are parallel, perpendicular, or neither. 5. Write equations of parallel and perpendicular lines. 6. Recognize and use the standard form of the equation of a line. 7. Write an equation of a line that models a real-life problem. 8. Use linear curve fitting to solve problems.

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218

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

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Suppose we purchase a Kawasaki motorcycle for $10,000 and know that it depreciates $1,200 in value each year. We can use the facts given to write a linear equation that represents the value y of the motorcycle x years after it was purchased. Because the motorcycle’s value decreases $1,200 each year, the slope of the line’s graph is 21,200. Because the purchase price is $10,000, we know that when we let x 5 0, the value of y will equal 10,000. A linear equation that satisfies these two conditions is y 5 21,200x 1 10,000. This equation represents the straight-line depreciation of the motorcycle. In this section, we will write equations of lines given specific characteristics or features of the line.

1. Use Point-Slope Form to Write an Equation of a line Suppose that line l in Figure 2-27 has a slope of m and passes through the point P 1x1, y12 . If Q 1x, y2 is any other point on line l, we have m5

y 2 y1 x 2 x1 y l Q(x, y) Slope = m P(x1, y1)

∆y = y – y1

∆x = x – x1

x

FIGURE 2-27

Comment y 2 y1 5 m 1x 2 x12 is often referred to as point-slope formula because we substitute a point and slope into it to write an equation of a line.

Point-Slope Form of an Equation of a line

EXAMPlE 1

If we multiply both sides by x 2 x1, we have (1)

y 2 y1 5 m 1x 2 x12

Since Equation 1 displays the coordinates of the point 1x1, y12 on the line and the slope m of the line, it is called the point-slope form of the equation of a line.

The equation of a line passing through P 1x1, y12 and with slope m is y 2 y1 5 m 1x 2 x12 .

Writing an Equation of the line with a Given Slope Passing through a Given Point

Write an equation of the line with slope 253 and passing through P 13, 212 . SOlUTION

We will substitute 253 for m, 3 for x1, and 21 for y1 in the point-slope form and simplify. y 2 y1 5 m 1x 2 x12

This is the point-slope form.

5 5 y 2 1212 5 2 1x 2 32 Substitute 2 for m, 3 for x1, and 21 for y1. 3 3 5 y1152 x15 Remove parentheses. 3 5 y52 x14 Subtract 1 from both sides. 3 An equation of the line is y 5 253x 1 4. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3

Self Check 1

Writing Equations of Lines

219

Write an equation of the line with slope 223 and passing through P 124, 52 . Now Try Exercise 11.

EXAMPlE 2 SOlUTION

Writing an Equation of the line that Passes through Two Given Points

Find an equation of the line passing through P 13, 72 and Q 125, 32 .

We will find the slope of the line and then choose either point P or point Q and substitute both the slope and coordinates of the point into the point-slope form. First we find the slope of the line. y 2 y1 m5 2 This is the slope formula. x2 2 x1 327 5 Substitute 3 for y2, 7 for y1, 25 for x2, and 3 for x1. 25 2 3 24 5 28 1 5 2 We can choose either point P or point Q and substitute its coordinates into the point-slope form. If we choose P 13, 72 , we substitute 12 for m, 3 for x1, and 7 for y1. y 2 y1 5 m 1x 2 x12

1 y 2 7 5 1x 2 32 2 1 3 y5 x2 17 2 2 1 11 y5 x1 2 2

This is the point-slope form. Substitute

1 for m, 3 for x1, and 7 for y1. 2

Remove parentheses and add 7 to both sides. 3 3 14 11 2 1752 1 5 2 2 2 2

An equation of the line is y 5 12x 1 11 2. Self Check 2

Find an equation of the line passing through P 125, 42 and Q 18, 262 . Now Try Exercise 23.

2. Use Slope-Intercept Form to Write an Equation of a line y

l Slope = m

Since the y-intercept of the line shown in Figure 2-28 is the point P 10, b2 , we can write an equation of the line by substituting 0 for x1 and b for y1 into the point-slope form and simplifying. y 2 y1 5 m 1x 2 x12

P(0, b) x

FIGURE 2-28

(2)

This is the point-slope form.

y 2 b 5 m 1x 2 02

Substitute 0 for x1 and b for y1.

y 2 b 5 mx

x205x

y 5 mx 1 b

Add b to both sides.

Because Equation 2 displays the slope m and the y-coordinate b of the y-intercept, it is called the slope-intercept form of the equation of a line.

Slope-Intercept Form of an Equation of a line

An equation of the line with slope m and y-intercept 10, b2 is y 5 mx 1 b

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220

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Three examples of linear equations written in slope-intercept form are shown below.

Comment Note that when a line is written in slopeintercept form the coefficient of x is the slope and the constant term is the y-coordinate of the y-intercept.

Example

Slope

y-Intercept

y 5 2x 1 7

m52

b 5 7; (0, 7)

2 y5 x25 3

2 m5 3

b 5 25; (0, 25)

m 5 24

1 1 b 5 ; a0, b 5 5

y 5 24x 1

EXAMPlE 3

1 5

Using Slope-Intercept Form to Write an Equation of a line Use slope-intercept form to write an equation of the line with slope 4 that passes through P 15, 92 .

SOlUTION

Since we know that m 5 4 and that the ordered pair 15, 92 satisfies the equation, we substitute 4 for m, 5 for x, and 9 for y in the equation y 5 mx 1 b and solve for b. y 5 mx 1 b

Comment When we are given a point and slope we can determine an equation of the line by substituting into point-slope form or slope-intercept form.

Self Check 3

This is the slope-intercept form.

9 5 4 152 1 b

Substitute 4 for m, 5 for x, and 9 for y.

9 5 20 1 b

Simplify.

211 5 b

Subtract 20 from both sides.

Because m 5 4 and b 5 211, an equation is y 5 4x 2 11. Use slope-intercept form to write an equation of the line with slope 73 and passing through 13, 12 . Now Try Exercise 37.

3. Graph linear Equations Using the Slope and y-Intercept. It is easy to graph a linear equation when it is written in slope-intercept form. For example, to graph y 5 43x 2 2 we note that b 5 22 and that the y-intercept is 10, b2 5 10, 222 . (See Figure 2-29.) Dy 5 43, we can locate another point Q on the line by startBecause the slope is Dx ing at point P and counting 3 units to the right and 4 units up. The change in x from point P to point Q is Dx 5 3, and the corresponding change in y is Dy 5 4. The line joining points P and Q is the graph of the equation. y

4 y = –x − 2 3

Q(3, 2)

∆y = 4 units

P(0, −2)

x

∆x = 3 units

FIGURE 2-29

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Section 2.3

EXAMPlE 4

SOlUTION

Writing Equations of Lines

221

Finding the Slope and y-Intercept of a line and Graphing the line

Find the slope and the y-intercept of the line with equation 3 1y 1 22 5 6x 2 1 and graph it. We will write the equation in the form y 5 mx 1 b to find the slope m and the y-intercept 10, b2 . Then we will use m and b to graph the line. 3 1y 1 22 5 6x 2 1 3y 1 6 5 6x 2 1 3y 5 6x 2 7 7 y 5 2x 2 3

Remove parentheses. Subtract 6 from both sides. Divide both sides by 3.

The slope of the graph is 2, and the y-intercept is Q0, 273R. We plot the y-intercept. Then we find a second point on the line by moving 1 unit to the right and 2 units up to the point Q1, 213R. To get the graph, we draw a line through the two points, as shown in Figure 2-30. y

3(y + 2) = 6x – 1 1 (1, − – ) 3 x 7 (0, − –) 3

2 units 1 unit

FIGURE 2-30

Self Check 4

Find the slope and the y-intercept of the line with equation 2 1x 2 32 5 23 1y 1 52 . Then graph it. Now Try Exercise 45.

4. Determine Whether linear Equations Represent lines that Are Parallel, Perpendicular, or Neither EXAMPlE 5

Determining Whether lines are Parallel, Perpendicular, or Neither Determine whether the lines represented by 4x 1 8y 5 10 and 2x 5 12 2 4y are parallel, perpendicular, or neither.

SOlUTION

We will find the slope of each line and compare their slopes. If their slopes are the same, the lines are parallel. If the product of their slopes is 21, the lines are perpendicular. Otherwise, they are neither parallel nor perpendicular.

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222

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

We solve each equation for y and write each equation in slope-intercept form. 4x 1 8y 5 10

2x 5 12 2 4y

8y 5 24x 1 10

4y 5 22x 1 12

10 12 24x 22x 1 y5 1 8 8 4 4 1 5 1 y52 x1 y52 x13 2 4 2 Since the values of b are different, the lines are distinct. Since each slope is 212, the lines are parallel. y5

Self Check 5

Are the lines represented by y 5 3x 1 2 and 6x 2 2y 5 5 parallel, perpendicular, or neither? Now Try Exercise 57.

EXAMPlE 6

Determining Whether lines are Parallel, Perpendicular, or Neither Determine whether the lines represented by 4x 1 8y 5 10 and 4x 2 2y 5 21 are parallel, perpendicular, or neither.

SOlUTION

We will find the slope of each line and compare their slopes. If their slopes are the same, the lines are parallel. If the product of their slopes is 21, the lines are perpendicular. Otherwise, they are neither parallel nor perpendicular. We solve each equation for y and write each equation in slope-intercept form. 4x 1 8y 5 10

4x 2 2y 5 21

8y 5 24x 1 10

22y 5 24x 1 21

24x 10 24x 21 1 y5 1 8 8 22 22 1 5 21 y52 x1 y 5 2x 2 2 4 2 Since the product of their slopes 1212 and 22 is 21, the lines are perpendicular. y5

Self Check 6

Are the lines represented by 3x 1 2y 5 7 and y 5 23x 1 3 parallel, perpendicular, or neither? Now Try Exercise 59.

5. Write Equations of Parallel and Perpendicular lines EXAMPlE 7

SOlUTION

Writing an Equation of a Parallel line

Write an equation of the line passing through P 122, 52 and parallel to the line y 5 8x 2 3.

We will substitute the coordinates of P 122, 52 and the slope of the line parallel to y 5 8x 2 3 into point-slope form and simplify the results to write an equation of the parallel line. The slope of the line given by y 5 8x 2 3 is 8, the coefficient of x. Since the graph of the desired equation is to be parallel to the graph of y 5 8x 2 3, its slope must also be 8.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3

Writing Equations of Lines

223

We will substitute 22 for x1, 5 for y1, and 8 for m into the point-slope form and simplify. y 2 y1 5 m 1x 2 x12

y 2 5 5 8 3 x 2 1222 4

Substitute 5 for y1, 8 for m, and 22 for x1.

y 2 5 5 8x 1 16

Use the Distributive Property to remove parentheses.

y 2 5 5 8 1x 1 22 y 5 8x 1 21

2 1222 5 2

Add 5 to both sides.

An equation of the desired line is y 5 8x 1 21. Self Check 7

Write an equation of the line passing through Q 11, 22 and parallel to the line y 5 8x 2 3. Now Try Exercise 67.

EXAMPlE 8

SOlUTION

Writing an Equation of a Perpendicular line

Write an equation of the line passing through P 122, 52 and perpendicular to the line y 5 8x 2 3.

We will substitute the coordinates of P 122, 52 and the slope of the line perpendicular to y 5 8x 2 3 into point-slope form and simplify to write an equation of the parallel line. Because the slope of the given line is 8, the slope of the desired perpendicular line must be 218. We substitute 22 for x1, 5 for y1, and 218 for m into the point-slope form and simplify. y 2 y1 5 m 1x 2 x12

1 1 y 2 5 5 2 3 x 2 1222 4 Substitute 5 for y1, 2 for m, and 22 for x1. 8 8 1 y 2 5 5 2 1x 1 22 2 1222 5 2 8 1 1 y52 x2 15 Remove parentheses and add 5 to both sides. 8 4 1 19 1 1 20 19 y52 x1 2 1552 1 5 4 4 4 4 8 4 1 19 An equation of the line is y 5 28x 1 4 . Self Check 8

Write an equation of the line passing through Q 11, 22 and perpendicular to y 5 8x 2 3. Now Try Exercise 73.

6. Recognize and Use the Standard Form of the Equation of a line We have shown that the graph of any equation of the form y 5 mx 1 b is a line with slope m and y-intercept 10, b2 . In Section 2.1, we saw that the graph of any equation of the form Ax 1 By 5 C (where A and B are not both zero) is also a line. We consider three possibilities. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

224

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•

If A 2 0 and B 2 0, the equation Ax 1 By 5 C can be written in slopeintercept form. Ax 1 By 5 C By 5 2Ax 1 C A C y52 x1 B B

Subtract Ax from both sides. Divide both sides by B.

This is an equation of a line with slope 2AB and y-intercept Q0, CBR. •

If A 5 0 and B 2 0, the equation Ax 1 By 5 C can be written in the form

•

If A 2 0 and B 5 0, the equation Ax 1 By 5 C can be written in the form

y 5 CB. This is an equation of a horizontal line with y-intercept Q0, CBR.

C x5C A. This is an equation of a vertical line with x-intercept at QA, 0R.

Recall that Ax 1 By 5 C is called the standard form of the equation of a line.

Standard Form of an Equation on a line

The standard form of an equation of a line is Ax 1 By 5 C where A, B, and C are real numbers, and both A and B are not zero. • If B 2 0, the slope of the nonvertical line is 2AB and the y-intercept is Q0, CBR.

• If B 5 0 the graph is a vertical line with x-intercept of QC A, 0R.

Comment

When writing equations in Ax 1 By 5 C form, we usually clear the equation of fractions and make A positive. For example, the equation 2x 1 52 y 5 2 can be changed to 2x 2 5y 5 24 by multiplying both sides by 22. We will also divide out any common integer factors of A, B, and C. For example, we would write 4x 1 8y 5 12 as x 1 2y 5 3.

Comment

Sometimes linear equations are written with a constant term of 0 on the right side of the equation. This form is referred to as general form. The line 2x 2 5y 1 3 5 0 is written in general form.

EXAMPlE 9

Using Formulas to Find the Slope and y-Intercept Find the slope and the y-intercept of the graph of 3x 2 2y 5 5.

SOlUTION

The equation 3x 2 2y 5 5 is in standard form, with A 5 3, B 5 22, and C 5 5. We will use 2AB to determine the slope of the line and Q0, CBR to determine the y-intercept. The slope of the graph is m52

A 3 3 52 5 B 22 2

and the y-intercept is a0,

C 5 b 5 a0, b B 22

The slope is 32 and the y-intercept is 10, 2522 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3

Self Check 9

Writing Equations of Lines

225

Find the slope and the y-intercept of the graph of 3x 2 4y 5 12. Now Try Exercise 77. We summarize the various forms of an equation of a line as follows.

Forms of an Equation of a line Standard form

Ax 1 By 5 C A and B cannot both be 0.

Slope-intercept form

y 5 mx 1 b The slope is m, and the y-intercept is 10, b2 .

y 2 y1 5 m 1x 2 x12 The slope is m, and the line passes through 1x1, y12 .

Point-slope form A horizontal line

y5b The slope is 0, and the y-intercept is 10, b2 .

A vertical line

x5a There is no defined slope, and the x-intercept is 1a, 02 .

7. Write an Equation of the line that Models a Real-life Problem For tax purposes, many businesses use straight-line depreciation to find the declining value of aging equipment.

EXAMPlE 10

Solving an Application Problem

mpanch/Shutterstock.com

A business purchases a digital multimedia projector for $1,970 and expects it to last for ten years. It can then be sold as scrap for a salvage value of $270. If y is the value of the projector after x years of use, and y and x are related by the equation of a line, a. Find an equation of the line. b. Find the value of the projector after 212 years. c. Find the economic meaning of the y-intercept of the line. d. Find the economic meaning of the slope of the line. SOlUTION

a. We will find the slope and use the point-slope form to write an equation of the line. (See Figure 2-31.) y (0, 1,970)

Value

1,970

(10, 270)

270

x 0

Age

10

FIGURE 2-31 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

226

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The Rectangular Coordinate System and Graphs of Equations

When the projector is new, its age x is 0, and its value y is $1,970. When the projector is 10 years old, x 5 10 and y 5 $270. Since the line passes through the points (0, 1,970) and (10, 270) the slope of the line is y2 2 y1 x2 2 x1

This is the slope formula.

5

270 2 1,970 10 2 0

Substitute 270 for y2, 1,970 for y1, 10 for x2, and 0 for x1.

5

21,700 10

m5

5 2170 To find an equation of the line, we substitute 2170 for m, 0 for x1, and 1,970 for y1 in the point-slope form and simplify. y 2 y1 5 m 1x 2 x12 y 2 1,970 5 2170 1x 2 02

(3)

y 5 2170x 1 1,970 The value y of the projector is related to its age x by the equation y 5 2170x 1 1,970.

b. To find the value after 212 years, we will substitute 2.5 for x in Equation 3 and solve for y. y 5 2170x 1 1,970

5 2170 12.52 1 1,970

Substitute 2.5 for x.

5 2425 1 1,970 5 1,545 In 212 years, the projector will be worth $1,545.

c. The y-intercept of the graph is 10, b2 , where b is the value of y when x 5 0. y 5 2170x 1 1,970

y 5 2170 102 1 1,970

Substitute 0 for x.

y 5 1,970 The y-coordinate b of the y-intercept is the value of a 0-year-old projector, which is the projector’s original cost, $1,970. d. Each year, the value decreases by $170, because the slope of the line is 2170. The slope of the depreciation line is called the annual depreciation rate. Problems that have an annual appreciation rate can be worked similarly. Self Check 10

A business purchases a Canon copier for $2,700 and expects it to last for ten years. It can then be sold for $300. Write a straight-line depreciation line for the copier. Now Try Exercise 91.

8. Use linear Curve Fitting to Solve Problems In statistics, the process of using one variable to predict another is called regression. For example, if we know a woman’s height, we can make a good prediction about her weight, because taller women usually weigh more than shorter women. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3

227

Writing Equations of Lines

Figure 2-32 shows the result of sampling ten women and finding their heights and weights. The graph of the ordered pairs 1h, w2 is called a scattergram.

w (lbs)

Woman

Height (h) in inches

Weight (w) in pounds

1

60

100

2

61

105

3

62

120

4

62

130

5

63

135

6

64

120

7

64

125

8

65

155

9

67

155

10

69

160

170 160 Q(67, 155)

150 140 130 120 110 100

P(60, 100) 65

60

70

h (in.) FIGURE 2-32

To write a prediction equation (sometimes called a regression equation), we must find an equation of the line that comes closer to all of the points in the scattergram than any other possible line. There are statistical methods to find this equation, but we can only approximate it here. To write an approximation of the regression equation, we place a straightedge on the scattergram shown in Figure 2-32 and draw the line joining two points that seems to best fit all the points. In the figure, line PQ is drawn, where point P has coordinates of (60, 100) and point Q has coordinates of (67, 155). Our approximation of the regression equation will be an equation of the line passing through points P and Q. To find an equation of this line, we first find its slope. y2 2 y1 This is the slope formula. m5 x2 2 x1 155 2 100 67 2 60 55 5 7 5

Substitute 155 for y2, 100 for y1, 67 for x2, and 60 for x1.

We can then use point-slope form to find an equation of the line. y 2 y1 5 m 1x 2 x12 55 1x 2 602 7

Choose (60, 100) for 1x1, y12 .

y5

55 3,300 x2 1 100 7 7

Remove parentheses and add 100 to both sides.

y5

55 2,600 x2 7 7

Simplify.

y 2 100 5

(4)

This is the point-slope form.

2,600 Our approximation of the regression equation is y 5 55 7x 2 7 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

228

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

To predict the weight of a woman who is 66 inches tall, for example, we substitute 66 for x in Equation 4 and simplify. 55 2,600 x2 7 7 55 2,600 y 5 1662 2 7 7 y < 147.1428571 We would predict that a 66-inch-tall woman chosen at random will weigh about 147 pounds. y5

ACCENT ON TECHNOlOGy

linear Regression We can use the linear regression feature on a graphing calculator to find an equation of the line that best fits a given set of data points. We will do that for the data given in Figure 2-32. •

to enter the statistics menu on the calculator. This screen First, we press is shown in Figure 2-33(a). Next we press ENTER to input our data. We can input our heights into the L1 column and our weights into the L2 column. This is shown in Figure 2-33(b).

(a)

(b) FIGURE 2-33

•

and then the rightTo obtain an equation of the regression line, we press arrow key once to access the calculate menu. This screen is shown in Figure 2-34(a). to select LinReg 1ax 1 b2 To calculate a linear regression equation, we press and then ENTER to obtain an equation. The screen is shown in Figure 2-34(b).

(a)

(b) FIGURE 2-34

Note that the regression line is of the form y 5 ax 1 b, where a is the slope and b is the y-coordinate of the y-intercept. If we substitute the values shown in Figure 2-34(b) for a and b and round to hundredths, we can write an equation of the line that best fits the data. The regression line is y 5 6.78x 2 301.18.

Self Check Answers

2 7 1. y 5 2 x 1 3 3

2. y 5 2

10 2 x1 13 13

7 3. y 5 x 2 6 3

2 4. 2 , 10, 232 3 y x 3 (0, −3)

−2 (3, −5)

2(x − 3) = −3(y + 5) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3

5. parallel

6. perpendicular

3 9. , 10, 232 4

Writing Equations of Lines

7. y 5 8x 2 6

229

1 17 8. y 5 2 x 1 8 8

10. y 5 2240x 1 2,700

Exercises 2.3 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. The formula for the point-slope form of a line is . 2. In the equation y 5 mx 1 b, is the slope of the graph of the line, and 10, b2 is the . 3. The equation y 5 mx 1 b is called the form of the equation of a line. 4. The standard form of an equation of a line is . 5. The slope of the graph of Ax 1 By 5 C is .

Write an equation of a line that passes through the two given points. Your answer should be written in slopeintercept form. 22. P 125, 252 , Q 10, 02 21. P 10, 02 , Q 14, 42 23. P 13, 42 , Q 10, 232

24. P 14, 02 , Q 16, 282

Write an equation in slope-intercept form of each line shown. y y 26. 25.

x x

6. The y-intercept of the graph of Ax 1 By 5 C is .

Practice Write an equation of the line with the given properties. Your answer should be written in standard form. 7. m 5 2 passing through P 12, 42 8. m 5 23 passing through P 13, 52

Use slope-intercept form to write an equation of the line with the given properties. 1 2 28. m 5 2 ; b 5 27. m 5 3; b 5 22 3 3

9. m 5 2 passing through PQ232, 12R

10. m 5 26 passing through PQ14, 22R

29. m 5 5; b 5 2

11. m 5 passing through P 121, 12 12. m 5 215 passing through P 122, 232 13. m 5 0 passing through P 126, 232 14. m 5 0 passing through P 17, 52 15. m is undefined passing through P 126, 232 16. m is undefined passing through P 16, 212 2 5

31. m 5 a; b 5

Find an equation of each line shown. Your answer should be written in standard form. 20.

y P(2, 5)

y

x

30. m 5 "2; b 5 "2 32. m 5 a; b 5 2a 1 34. m 5 ; b 5 a a

Use slope-intercept form to write an equation of a line passing through the given point and having the given slope. Express the answer in standard form. 3 2 35. P 10, 02 ; m 5 36. P 123, 272 ; m 5 2 2 3 37. P 123, 52 ; m 5 23

P(−3, 2) x

1 a

33. m 5 a; b 5 a

17. m 5 p passing through P 1p, 02 18. m 5 p passing through P 10, p2

19.

1 5

38. P 125, 12 ; m 5 1

39. P Q0, "2R; m 5 "2

40. P Q2"3, 0R; m 5 2"3

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

230

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Write each equation in slope-intercept form to determine the slope and y-intercept. Then use the slope and y-intercept to graph the line. 41. x 2 y 5 1

42. x 1 y 5 2

y

y

x

3 43. x 5 y 2 3 2

x

66. P 10, 02 , x 5 23y 2 12

y

x

x

45. 3 1y 2 42 5 22 1x 2 32

58. 2x 1 3y 5 9, 3x 2 2y 5 5 59. x 5 3y 1 4, y 5 23x 1 7 1 60. 3x 1 6y 5 1, y 5 x 2 61. y 5 3, x 5 4 62. y 5 23, y 5 27 y22 63. x 5 , 3 1y 2 32 1 x 5 0 3 64. 2y 5 8, 3 12 1 x2 5 3 1y 1 22 Write an equation of the line that passes through the given point and is parallel to the given line. Your answer should be written in slope-intercept form. 65. P 10, 02 , y 5 4x 2 7

4 44. x 5 2 y 1 2 5

y

57. y 5 3x 1 7, 2y 5 6x 2 9

46. 24 12x 1 32 5 3 13y 1 82 y

y

67. P 12, 52 , 4x 2 y 5 7 68. P 126, 32 , y 1 3x 5 212 5 69. P 14, 222 , x 5 y 2 2 4 3 70. P 11, 252 , x 5 2 y 1 5 4 Write an equation of the line that passes through the given point and is perpendicular to the given line. Your answer should be written in slope-intercept form. 71. P 10, 02 , y 5 4x 2 7

x

72. P 10, 02 , x 5 23y 2 12 73. P 12, 52 , 4x 2 y 5 7

Find the slope and the y-intercept of the lines determined by the given equations. 48. 22x 1 4y 5 12 47. 3x 2 2y 5 8

74. P 126, 32 , y 1 3x 5 212 5 75. P 14, 222 , x 5 y 2 2 4 3 76. P 11, 252 , x 5 2 y 1 5 4

49. 22 1x 1 3y2 5 5

Use the method of Example 9 to find the slope and the y-intercept of the graph of each equation. 78. 9x 2 12y 5 17 77. 4x 1 5y 5 20

x

51. x 5

2y 2 4 7

50. 5 12x 2 3y2 5 4

52. 3x 1 4 5 2

2 1y 2 32 5

Determine whether the graphs of each pair of equations are parallel, perpendicular, or neither. 53. y 5 3x 1 4, y 5 3x 2 7 1 54. y 5 4x 2 13, y 5 x 1 13 4 55. x 1 y 5 2, y 5 x 1 5 56. x 5 y 1 2, y 5 x 1 3

79. 2x 1 3y 5 12

80. 5x 1 6y 5 30

81. Find an equation of the line perpendicular to the line y 5 3 and passing through the midpoint of the segment joining 12, 42 and 126, 102 .

82. Find an equation of the line parallel to the line y 5 28 and passing through the midpoint of the segment joining 124, 22 and 122, 82 . 83. Find an equation of the line parallel to the line x 5 3 and passing through the midpoint of the segment joining 12, 242 and 18, 122 .

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Section 2.3

@erics/Shutterstock.com

84. Find an equation of the line perpendicular to the line x 5 3 and passing through the midpoint of the segment joining 122, 22 and 14, 282 .

Applications In Exercises 85-95, assume straight-line depreciation or straight-line appreciation. 85. Depreciation A Toyota Tundra truck was purchased for $24,300. Its salvage value at the end of its 7-year useful life is expected to be $1,900. Find a depreciation equation. 86. Depreciation A small business purchases the laptop computer shown. It will be depreciated over a 4-year period, when its salvage value will be $300. Find a depreciation equation.

$2,700

231

Writing Equations of Lines

93. Value of an antique An antique table is expected to appreciate $40 each year. If the table will be worth $450 in 2 years, what will it be worth in 13 years? 94. Value of an antique An antique clock is expected to be worth $350 after 2 years and $530 after 5 years. What will the clock be worth after 7 years? 95. Purchase price of real estate A cottage that was purchased 3 years ago is now appraised at $47,700. If the property has been appreciating $3,500 per year, find its original purchase price. 96. Computer repair A computer repair company charges a fixed amount, plus an hourly rate, for a service call. Use the information in the illustration to find the hourly rate

AAA Computer Repair 87. Appreciation A condominium in San Diego was purchased for $475,000. The owners expect the condominium to double in value in 10 years. Find an appreciation equation. 88. Appreciation A house purchased for $112,000 is expected to double in value in 12 years. Find an appreciation equation. 89. Depreciation Find a depreciation equation for the TV in the following want ad.

For Sale: 3-year-old 54-inch TV, $1,900 new. Asking $1,190. Call 875-5555. Ask for Mike. 90. Depreciation A Bose Wave Radio cost $555 when new and is expected to be worth $80 after 5 years. What will it be worth after 3 years? 91. Salvage value A copier cost $1,050 when new and will be depreciated at the rate of $120 per year. If the useful life of the copier is 8 years, find its salvage value. 92. Rate of depreciation A ski boat that cost $27,600 when new will have no salvage value after 12 years. Find its annual rate of depreciation.

Typical Charges 2 hours $ 70 4 hours $105 97. Automobile repair An auto repair shop charges an hourly rate, plus the cost of parts. If the cost of labor for a 112-hour radiator repair is $69, find the cost of labor for a 5-hour transmission overhaul. 98. Printer charges A printer charges a fixed setup cost, plus $1 for every 100 copies. If 700 copies cost $52, how much will it cost to print 1,000 copies? 99. Predicting fires A local fire department recognizes that city growth and the number of reported fires are related by a linear equation. City records show that 300 fires were reported in a year when the local population was 57,000 persons, and 325 fires were reported in a year when the population was 59,000 persons. How many fires can be expected in the year when the population reaches 100,000 persons? 100. Estimating the cost of rain gutter A neighbor tells you that an installer of rain gutter charges $60, plus a dollar amount per foot. If the neighbor paid $435 for the installation of 250 feet of gutter, how much will it cost you to have 300 feet installed?

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

232

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

101. Converting temperatures Water freezes at 32° F, or 0° C. Water boils at 212° F, or 100° C. Find a formula for converting a temperature from degrees Fahrenheit to degrees Celsius. 102. Converting units A speed of 1 mile per hour is equal to 88 feet per minute, and of course, 0 miles per hour is 0 feet per minute. Find an equation for converting a speed x, in miles per hour, to the corresponding speed y, in feet per minute. 103. Smoking The percent y of 18-to-25-year-old smokers in the United States has been declining at a constant rate since 1974. If about 47% of this group smoked in 1974 and about 29% smoked in 1994, find a linear equation that models this decline. If this trend continues, estimate what percent will smoke in 2014. 104. Forensic science Scientists believe there is a linear relationship between the height h (in centimeters) of a male and the length f (in centimeters) of his femur bone. Use the data in the table to find a linear equation that expresses the height h in terms of f. Round all constants to the nearest thousandth. How tall would you expect a man to be if his femur measures 50 cm? Round to the nearest centimeter.

108. Waste management The corrosive waste in industrial sewage limits the useful life of the piping in a waste processing plant to 12 years. The piping system was originally worth $137,000, and it will cost the company $33,000 to remove it at the end of its 12-year useful life. Find a depreciation equation. 109. Crickets The table shows the approximate chirping rate at various temperatures for one type of cricket. Temperature (° F)

Chirps per minute

50

20

60

80

70

115

80

150

100

250

a. Construct a scattergram below. Chirps/min 250 225 200 175

Person

Length of femur (f )

Height (h)

A

62.5 cm

200 cm

150 125 100

B

40.2 cm

150 cm

75 50 25

105. Predicting stock prices The value of the stock of ABC Corporation has been increasing by the same fixed dollar amount each year. The pattern is expected to continue. Let 2010 be the base year corresponding to x 5 0 with x 5 1, 2, 3, % corresponding to later years. ABC stock was selling at $3712 in 2010 and at $45 in 2012. If y represents the price of ABC stock, find the equation y 5 mx 1 b that relates x and y, and predict the price in the year 2014. 106. Estimating inventory Inventory of unsold goods showed a surplus of 375 units in January and 264 in April. Assume that the relationship between inventory and time is given by the equation of a line, and estimate the expected inventory in March. Because March lies between January and April, this estimation is called interpolation. 107. Oil depletion When a Petroland oil well was first brought on line, it produced 1,900 barrels of crude oil per day. In each later year, owners expect its daily production to drop by 70 barrels. Find the daily production after 312 years.

50 60 70 80 90 100

Temp (°F)

b. Assume a linear relationship and write a regression equation. c. Estimate the chirping rate at a temperature of 90° F. 110. Fishing The table shows the lengths and weights of seven muskies captured by the Department of Natural Resources in Catfish Lake in Eagle River, Wisconsin. Musky

Length (in.)

Weight (lb)

1

26

5

2

27

8

3

29

9

4

33

12

5

35

14

6

36

14

7

38

19

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.3

a. Construct a scattergram for the data. b. Assume a linear relationship and write a regression equation.

Length of femur bone (cm)

Height (cm)

x

y

49.3

180.8

47.4

176.7

47.1

176.1

48.5

176.5

45.2

170.6

47.8

176.8

49.4

178.6

49.6

179.6

50.9

185.2

47.8

176.2

c. Estimate the weight of a musky that is 32 inches long. 111. Use the linear regression feature on a graphing calculator to determine an equation of the line that best fits the data given in Exercise 109. Round to the hundredths. 112. Use the linear regression feature on a graphing calculator to determine an equation of the line that best fits the data given in Exercise 110. Round to the hundredths.

Discovery and Writing 113. Explain how to find an equation of a line passing through two given points. 114. In straight-line depreciation, explain why the slope of the line is called the rate of depreciation.

a. Enter the data in an Excel spreadsheet. Enter the femur values in Column A and the height values in Column B. b. To plot the data, click and drag to highlight the two columns, select Insert on the menu bar, and then select Chart. When a window pops up, select XY (Scatter) as the chart type. When you click Finish, a scattergram will appear on the spreadsheet. c. Observe that the points are approximately linear. To draw the regression line, select Chart on the menu bar and select Add Trendline. When the dialog box appears, be sure that the Linear Regression type is selected, and then click OK. The regression line will appear. d. To find an equation of the regression line, estimate the coordinates of two points on the regression line. Then use the two points to write an equation of the line. e. Use the equation of the regression line to predict the height of a male whose femur bone measures 46 cm.

115. Prove that an equation of a line with x-intercept of 1a, 02 and y-intercept of 10, b2 can be written in the form x y 1 51 a b 116. Find the x- and y-intercepts of the line bx 1 ay 5 ab. Investigate the properties of slope and the y-intercept by experimenting with the following problems. 117. Graph y 5 mx 1 2 for several positive values of m. What do you notice? 118. Graph y 5 mx 1 2 for several negative values of m. What do you notice? 119. Graph y 5 2x 1 b for several increasing positive values of b. What do you notice? 120. Graph y 5 2x 1 b for several decreasing negative values of b. What do you notice? 121. How will the graph of y 5 12x 1 5 compare to the graph of y 5 12x 2 5? Using Excel to solve a linear regression problem. 122. The following table shows the length of a femur bone and the height for ten Caucasian males.

233

Writing Equations of Lines

Review Simplify each expression. 123. x7x3x25 125. a

81 23/2 b 25

127. "27 2 2"12 129.

5

"x 1 2

124.

y3y24 y25

3 126. " 27x7

128.

5

"5

130. Q"x 2 2R

2

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234

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The Rectangular Coordinate System and Graphs of Equations

2.4 Graphs of Equations In this section, we will learn to 1. 2. 3. 4.

Find the x- and y-intercepts of a graph. Use symmetry to help graph equations. Identify the center and radius of a circle. Write equations of circles.

Chrislofoto/Shutterstock.com

5. Graph circles. 6. Solve equations by using a graphing calculator. The London Eye opened in 2000 and is one of the world’s tallest and most beautiful observation wheels in the world. It stands 443 feet high and was the vision of the architects David Marks and Julia Barfield. Its circular design was used as a metaphor for the turning of the century. The Eye has been described as a breathtaking feat of design and engineering. The graphs of many equations are curves and circles. In this section, we will plot several points 1x, y2 that satisfy such an equation and join them with a smooth curve. Usually the shape of the graph will become evident.

1. Find the x- and y-Intercepts of a Graph

In Figure 2-35(a), the x-intercepts of the graph are 1a, 02 and 1b, 02 , the points where the graph intersects the x-axis. In Figure 2-35(b), the y-intercept is 10, c2 , the point where the graph intersects the y-axis. y

y (0, c) (b, 0)

(a, 0)

x

x

(b)

(a) FIGURE 2-35

To graph the equation y 5 x2 2 4, we first find the x- and y-intercepts. To find the x-intercepts, we let y 5 0 and solve for x. y 5 x2 2 4

x1250 x 5 22

0 5 x2 2 4

Substitute 0 for y.

0 5 1x 1 22 1x 2 22

Factor x2 2 4.

or x 2 2 5 0

Set each factor equal to 0.

x52

Since y 5 0 when x 5 22 and x 5 2, the x-intercepts are 122, 02 and 12, 02 . (See Figure 2-36.) To find the y-intercept, we let x 5 0 and solve for y. y 5 x2 2 4 y 5 02 2 4

Substitute 0 for x.

y 5 24 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.4

Graphs of Equations

235

Since y 5 24 when x 5 0 the y-intercept is 10, 242 . We can find other pairs 1x, y2 that satisfy the equation by substituting numbers for x and finding the corresponding values of y. For example, if x 5 23, then y 5 1232 2 2 4 5 5 and the point 123, 52 lies on the graph. The coordinates of the intercepts and other points appear in Figure 2-36. If we plot the points and draw a curve through them, we obtain the graph of the equation. y

y 5 x2 2 4 (–3, 5)

(3, 5) y=

(–2, 0)

x2

–4

(2, 0) x

(–1, –3)

(1, –3) V (0, –4)

x

y

23

5

22

0

21

23

0

24

1

23

2

0

3

5

1x, y2

123, 52 122, 02

121, 232 10, 242 11, 232 12, 02 13, 52

FIGURE 2-36

This graph is called a parabola. Its lowest point, V 10, 242 , is called the vertex. Because the y-axis divides the parabola into two congruent halves, it is called an axis of symmetry. We say that the parabola is symmetric about the y-axis.

ACCENT ON TECHNOlOGy

Graphing an Equation A graphing calculator can be used to draw the graph of the equation y 5 x2 2 4. When using a graphing calculator, we must enter the equation and then set the window. • •

•

To enter the equation, press Y= , which opens the window shown in Figure 2-37(a) and enter the right side of the equation. To set the window, press WINDOW as shown in Figure 2-37(b). We can experiment with different windows, or we can use the ZOOM menu to select some default windows. Press GRAPH , and the equation is graphed as shown in Figure 2-37(c).

(a)

(b)

(c)

FIGURE 2-37

2. Use Symmetry to Help Graph Equations Symmetry is a tool that we can use to help graph equations. When we look at the McDonald’s logo shown on the next page, we can see that the left and right sides of the logo are symmetric (mirror images of each other.) Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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There are several ways in which a graph can have symmetry.

© Istockphoto.com/Micah Young

1. If the point 1x, 2y2 lies on the graph whenever the point 1x, y2 does, the graph is symmetric about the x-axis. See Figure 2-38(a). This implies that a graph is symmetric about the x-axis if we get the same x-coordinate when we evaluate its equation at y or at 2y. 2. If the point 12x, y2 lies on a graph whenever the point 1x, y2 does, the graph is symmetric about the y-axis. See Figure 2-38(b). This implies that a graph is symmetric about the y-axis if we get the same y-coordinate when we evaluate its equation at x or at –x. 3. If the point 12x, 2y2 lies on the graph whenever the point 1x, y2 does, the graph is symmetric about the origin. See Figure 2-38(c). This implies that the graph is symmetric about the origin if we get opposite values of y when we evaluate its equation at x or at 2x. y

y

y

(x, y) (–x, y)

(x, y)

x

(x, y) x

x

(–x, –y)

(x, –y)

Comment If a graph is symmetric with respect to the x-axis and y-axis, then it is symmetric about the origin.

Symmetry Tests

(a)

(b)

(c)

FIGURE 2-38

•

•

•

Test for x-Axis Symmetry To test for x-axis symmetry, replace y with 2y. If the resulting equation is equivalent to the original one, the graph is symmetric about the x-axis. Test for y-Axis Symmetry To test for y-axis symmetry, replace x with 2x. If the resulting equation is equivalent to the original one, the graph is symmetric about the y-axis. Test for Origin Symmetry To test for symmetry about the origin, replace x with 2x and y with 2y. If the resulting equation is equivalent to the original one, the graph is symmetric about the origin.

To graph various equations, we will use the following strategy.

Strategy for Graphing Equations

To graph various equations, we will use the following three steps: 1. Find the x- and y-intercepts. 2. Test for symmetries. 3. Graph the equation by plotting points, making use of symmetry, and joining the points with a smooth curve.

EXAMPlE 1 SOlUTION

Graphing an Equation Using Intercepts and Symmetry

Graph: y 5 0 x 0 .

To graph y 5 0 x 0 , we will find the x- and y-intercepts, test for symmetries, plot points, and join the points with a smooth curve.

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Section 2.4

Step 1: Find the x- and y-intercepts. for x. y 5 0x0 0 5 0x0

Graphs of Equations

237

To find the x-intercepts, we let y 5 0 and solve

Substitute 0 for y.

0x0 5 0 x50

The x-intercept is 10, 02 (See Figure 2-39.) To find the y-intercepts, we let x 5 0 and solve for y. y 5 0x0 y 5 000

Substitute 0 for x.

y50

The y-intercept is 10, 02 . Step 2: Test for symmetries. To test for x-axis symmetry, we replace y with 2y. (1) (2)

y 5 0x0

2y 5 0 x 0

This is the original equation. Replace y with 2y.

Since Equations 1 and 2 are different, the graph is not symmetric about the x-axis. To test for y-axis symmetry, we replace x with 2x. (1)

(3)

y 5 0x0

y 5 0 2x 0

This is the original equation. Replace x with 2x.

y 5 0x0

0 2x 0 5 0 x 0

Since Equations 1 and 3 are the same, the graph is symmetric about the y-axis. To test for symmetry about the origin, we replace x with 2x and y with 2y. (1)

y 5 0x0

This is the original equation.

2y 5 0 2x 0

(4)

Replace x with 2x and y with 2y.

2y 5 0 x 0

0 2x 0 5 0 x 0

Since Equations 1 and 4 are different, the graph is not symmetric about the origin. Step 3: Graph the equation. To graph the equation, we plot the x- and y-intercepts and several other pairs 1x, y2 with positive values of x. We can use the y-axis symmetry to draw the graph for negative values of x. See Figure 2-39. y

y 5 0x0 y = |x|

x

x

y

0

0

1

1

2

2

3

3

4

4

1x, y2 10, 02 11, 12 12, 22

13, 32 14, 42

FIGURE 2-39

Self Check 1

Graph: y 5 2 0 x 0 . Now Try Exercise 53.

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238

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

ACCENT ON TECHNOlOGy

Graphing an Absolute Value Equation

A graphing calculator can be used to draw the graph of the equation y 5 0 0.5x 0 . •

• •

To enter the equation, press Y= , which opens the window shown in Figure 2-40(a). To enter the absolute value equation into Y1, we need to access the absolute value function on our calculator. It is #1 in the MATH NUM menu as shown in Figure 2-40(b). Press ENTER and input 0.5x, and the equation will appear in the graph window. Figure 2-40(c). To set the window, press ZOOM , scroll down to 6: [see Figure 2-40(d)], and then press ENTER . The equation is graphed as shown in Figure 2-40(e).

(a)

(b)

(c)

(d)

(e) FIGURE 2-40

EXAMPlE 2

Graphing an Equation Using Intercepts and Symmetry Graph: y 5 x3 2 x.

SOlUTION

To graph y 5 x3 2 x, we will find the x- and y-intercepts, test for symmetries, plot points, and join the points with a smooth curve. Step 1: Find the x- and y-intercepts. To find the x-intercepts, we let y 5 0 and solve for x. y 5 x3 2 x 0 5 x3 2 x

Substitute 0 for y.

0 5 x 1x2 2 12

Factor out x.

0 5 x 1x 1 12 1x 2 12 x 5 0 or

x1150 x 5 21

Factor x2 2 1.

or x 2 1 5 0

Set each factor equal to 0.

x51

The x-intercepts are 10, 02 , 121, 02 , and 11, 02 . To find the y-intercepts, we let x 5 0 and solve for y. y 5 x3 2 x y 5 03 2 0

Substitute 0 for x.

y50

The y-intercept is 10, 02 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.4

239

Graphs of Equations

Step 2: Test for symmetries. We test for symmetry about the x-axis by replacing y with –y. (1)

y 5 x3 2 x

This is the original equation.

2y 5 x3 2 x (2)

Replace y with 2y.

3

y 5 2x 1 x

Multiply both sides by 21.

Since Equations 1 and 2 are different, the graph is not symmetric about the x-axis. To test for y-axis symmetry, we replace x with –x. (1)

y 5 x3 2 x

This is the original equation.

y 5 12x2 2 12x2

Replace x with 2x.

y 5 2x3 1 x

Simplify.

3

(3)

Since Equations 1 and 3 are different, the graph is not symmetric about the y-axis. To test for symmetry about the origin, we replace x with –x and y with –y. (1)

y 5 x3 2 x

This is the original equation.

2y 5 12x2 2 12x2

Replace x with 2x and y with 2y.

2y 5 2x3 1 x

Simplify.

3

(4)

3

y5x 2x

Multiply both sides by 21.

Since Equations 1 and 4 are the same, the graph is symmetric about the origin. Step 3: Graph the equation. To graph the equation, we plot the x- and y-intercepts and several other pairs 1x, y2 with positive values of x. We can use the property of symmetry about the origin to draw the graph for negative values of x. See Figure 2-41. y

y 5 x3 2 x

(2, 6)

x y = x3 – x

x

y

21

0

0

0

1 2

2

3 8

1

0

2

6

1x, y2

121, 02 10, 02

1 3 a ,2 b 2 8 11, 02 12, 62

FIGURE 2-41

Self Check 2

Graph: y 5 x3 2 9x. Now Try Exercise 49.

ACCENT ON TECHNOlOGy

Finding the Intercepts of a Graph Using the Zero Feature A graphing calculator can find the intercepts of an equation, such as y 5 2x3 2 3x. The easiest way to do this is to use the graph of the equation. The x-coordinate of the x-intercept is also called a zero of the equation. • •

Enter the function into Y1. See Figure 2-42(a). Set the window. Press ZOOM , scroll down to 4:, and then then press this graph. See Figures 2-42(a), (b), and (c).

ENTER

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for

240

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

• • •

• • •

To find the zeros, we need to access the CALC menu, which is found by pressTRACE . Select 2:ZERO. See Figure 2-42(d). ing Move the cursor until it is to the left of the zero we are trying to find and press ENTER . See Figure 2-42(e). Repeat the previous step by moving the cursor to the right of the zero. Press ENTER two times. Do not stop when the screen shows GUESS. See Figures 2-42 (f), (g), and (h). The zero is x 5 21.224745. Repeat the steps to find the zero x 5 1.224745. See Figures 2-42(i), (j), and (k). We can see that another zero is x 5 0.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k) FIGURE 2-42

The x-intercepts are (21.224745, 0), (1.224745, 0), and (0, 0).

EXAMPlE 3

Graphing an Equation Using Intercepts and Symmetry Graph: y 5 "x.

SOlUTION

To graph y 5 "x, we will find the x- and y-intercepts, test for symmetries, plot points, and join the points with a smooth curve. Step 1: Find the x- and y-intercepts. We can see that the x- and y-intercepts of y 5 "x are both 10, 02 . The graph passes through the origin.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 2.4

Graphs of Equations

241

Step 2: Test for symmetries. We test for symmetry about the x-axis by replacing y with 2y. (1) y 5 "x This is the original equation. 2y 5 "x

(2)

Replace y with 2y.

y 5 2"x

Multiply both sides by 21.

Since Equations 1 and 2 are different, the graph is not symmetric about the x-axis. To test for y-axis symmetry, we replace x with 2x. (1) (3)

y 5 "x

This is the original equation.

y 5 "2x

Replace x with 2x.

Since Equations 1 and 3 are different, the graph is not symmetric about the y-axis. To test for symmetry about the origin, we replace x with 2x and y with 2y. (1)

y 5 "x

This is the original equation.

2y 5 "2x (4)

Replace x with 2x and y with 2y.

y 5 2"2x

Multiply both sides by 21.

Since Equations 1 and 4 are different, the graph is not symmetric about the origin. The graph of this equation has no symmetries. Step 3: Graph the equation. We plot several points to obtain the graph in Figure 2-43. y 4

–1

–1

y 5 "x

y = √x

3 2 1 1

2 3

4

5 6

7

8

9

x

y

0

0

1

1

4

2

9

3

x

–2 –3 –4

1x, y2 10, 02 11, 12 14, 22 19, 32

FIGURE 2-43

Self Check 3

Graph: y 5 2"x. Now Try Exercise 59.

ACCENT ON TECHNOlOGy

Graphing a Radical Equation. To draw the graph of y 5 "3.5x using a graphing calculator, enter the equation into the graph menu. See Figure 2-44(a); set the window using the values shown in Figure 2-44(b), and press GRAPH to graph the equation. See Figure 2-44(c).

(a)

(b)

(c)

FIGURE 2-44

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242

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

EXAMPlE 4

Graphing an Equation Using Intercepts and Symmetry Graph: y2 5 x.

SOlUTION

To graph y2 5 x, we will find the x- and y-intercepts, test for symmetries, plot points, and join the points with a smooth curve. Step 1: Find the x- and y-intercepts. We can see that the x- and y-intercepts of y2 5 x are both 10, 02 . The graph passes through the origin. Step 2: Test for symmetries. We test for symmetry about the x-axis by replacing y and –y. (1)

y2 5 x

This is the original equation.

12y2 2 5 x

Replace y with 2y.

y2 5 x

(2)

Simplify.

Since Equations 1 and 2 are the same, the graph is symmetric about the x-axis. We test for symmetry about the y-axis by replacing x with 2x. (1) (3)

y2 5 x

This is the original equation.

2

y 5 2x

Replace x with 2x.

Since Equations 1 and 3 are different, the graph is not symmetric about the y-axis. To test for symmetry about the origin, we replace x with 2x and y with 2y. (1)

(4)

y2 5 x

This is the original equation.

12y2 2 5 2x

Replace x with 2x and y with 2y.

2

y 5 2x

Simplify.

Since Equations 1 and 4 are different, the graph is not symmetric about the origin. Step 3: Graph the equation. To graph the equation, we plot the x- and y-intercepts and several other pairs 1x, y2 with positive values of y. We can use the property of x-axis symmetry to draw the graph for negative values of y. See Figure 2-45.

y 4

–1

y2 5 x

y2 = x

3 2 1

x

y

0

0

–1

1

1

–2

4

2

–3

9

3

1

2 3

4

5 6

7

8

9

x

–4

1x, y2 10, 02 11, 12 14, 22 19, 32

FIGURE 2-45

Self Check 4

Graph: y2 5 2x. Now Try Exercise 57.

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Section 2.4

ACCENT ON TECHNOlOGy

Graphs of Equations

243

Graphing an Equation with a y 2 Term A graphing calculator is programmed to graph equations with y defined in terms of x. To graph the equation y2 5 2x, it is necessary to solve for y and then enter the resulting equations to see the graph. y2 5 2x y 5 6"2x y 5 "2x • •

or

y 5 2"2x

We enter the two equations as shown in Figure 2-46(a). Next we set the WINDOW as shown in Figure 2-46(b). Then we

GRAPH

the equation(s) as shown in Figure 2-46(c).

(a)

(b)

(c)

FIGURE 2-46

The graphs in Examples 3 and 4 are related. We have solved the equation y2 5 x for y, and two equations resulted. y 5 "x and

y 5 2"x

The equation, y 5 "x was graphed in Example 3. It is the top half of the parabola shown in Example 4. The equation, y 5 2"x is the bottom half.

3. Identify the Center and Radius of a Circle Circles

A circle is the set of all points in a plane that are a fixed distance from a point called its center. The fixed distance is the radius of the circle. To find an equation of a circle with radius r and center at C 1h, k2 , we must find all points P 1x, y2 in the xy-plane such that the length of line segment PC is r. (See Figure 2-47.) y

C(h, k)

r P(x, y) x FIGURE 2-47

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Chapter 2

The Rectangular Coordinate System and Graphs of Equations

We can use the Distance Formula to find the length of CP, which is r: r 5 " 1x 2 h2 2 1 1y 2 k2 2 After squaring both sides, we get r2 5 1x 2 h2 2 1 1y 2 k2 2 This equation is called the standard form of an equation of a circle.

The Standard Form of an Equation of a Circle with Center at 1h, k2 and Radius r

The graph of any equation that can be written in the standard form 1x 2 h2 2 1 1y 2 k2 2 5 r2

is a circle with radius r and center at point 1h, k2 . If r 5 0, the circle is a single point called a point circle. If the center of a circle is the origin, then 1h, k2 5 10, 02 and we have the following result.

The Standard Form of an Equation of a Circle with Center at 10, 02 and Radius r

The graph of any equation that can be written in the standard form x2 1 y2 5 r2 is a circle with radius r and center at the origin.

The equations of four circles written in standard form and their graphs are shown in Figure 2-48. y

4

r=

r=

r=

2 (0, 0)

x

y

y

3

y

r=

244

x

(0, 3) (–4, 0)

3

(–1, 1) x

x x2 + y2 = 4 center: (0, 0) radius: 2 (a)

(x + 4)2 + y 2 = 16 center: (–4, 0) radius: 4 (c)

x2 + (y – 3)2 = 9 center: (0, 3) radius: 3 (b)

(x + 1)2 + (y – 1)2 = 9 center: (–1, 1) radius: 3 (d)

FIGURE 2-48

If we are given an equation of a circle in standard form, we can easily determine the coordinates of its center and the length of its radius.

EXAMPlE 5 SOlUTION

Finding the Center and Radius of a Circle in Standard Form

Find the center and radius of the circle with the equation 1x 2 32 2 1 1y 1 22 2 5 36. Since the equation is written in standard form, we can identify h, k, and r by comparing the equation to the equation 1x 2 h2 2 1 1y 2 k2 2 5 r2. Standard Form: Given Form:

1x 2 h2 2 1 1y 2 k2 2 5 r2

1x 2 32 2 1 3 y 2 1222 4 2 5 62

We see that h 5 3, k 5 22, and r 5 6. The center 1h, k2 of the circle is at 13, 222 and the radius is 6.

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Section 2.4

Self Check 5

Graphs of Equations

245

Identify the center and radius of the circle with equation of 1x 2 42 2 1 y2 5 49. Now Try Exercise 69.

4. Write Equations of Circles EXAMPlE 6 SOlUTION

Writing an Equation of a Circle Given the Center and Radius

Write an equation of the circle with center at 122, 42 and radius 1.

We will substitute the coordinates of the center and the radius into the standard form 1x 2 h2 2 1 1y 2 k2 2 5 r2. 1x 2 h2 2 1 1y 2 k2 2 5 r 2

3 x 2 1222 4 2 1 1y 2 42 2 5 12

Substitute 22 for h, 4 for k, and 1 for r.

1x 1 22 1 1y 2 42 5 1 2

2

Simplify.

An equation of the circle with center at 122, 42 and radius 1 is 1x 1 22 2 1 1y 2 42 2 5 1. Self Check 6

Write an equation in standard form of the circle with center 12, 242 and radius "5. Now Try Exercise 79. If we square the binomials in the equation of the circle found in Example 6, we get another form of the equation, called the general equation of a circle. 1x 1 22 2 1 1y 2 42 2 5 1

Comment We can easily recognize the equation of a circle. A circle’s equation will always contain both x2 and y2 terms, and the coefficients of both terms will be equal.

The General Form of an Equation of a Circle

x2 1 4x 1 4 1 y2 2 8y 1 16 5 1 2

2

x 1 y 1 4x 2 8y 1 19 5 0

This is the equation of Example 6. Remove parentheses. Subtract 1 from both sides and simplify.

The general form is x2 1 y2 1 4x 2 8y 1 19 5 0.

The general form of an equation of a circle is x2 1 y2 1 cd 1 dy 1 e 5 0 where c, d, and e are real numbers.

EXAMPlE 7

Finding the General Form of an Equation of a Circle Given the Center and Radius

Find the general form of an equation of the circle with radius 5 and center at 13, 22 . SOlUTION

We will substitute 5 for r, 3 for h, and 2 for k in the standard form of the equation of a circle and simplify. 1x 2 h2 2 1 1y 2 k2 2 5 r2 1x 2 32 1 1y 2 22 5 5 2

2

2

x2 2 6x 1 9 1 y2 2 4y 1 4 5 25 2

2

x 1 y 2 6x 2 4y 2 12 5 0

This is the standard equation. Substitute. Remove parentheses. Subtract 25 from both sides and simplify.

The general form is x2 1 y2 2 6x 2 4y 2 12 5 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

246

Chapter 2

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Self Check 7

Find the general form of an equation of a circle with radius 6 and center at 122, 52 . Now Try Exercise 83.

EXAMPlE 8

Finding the General Form of an Equation of a Circle Given the Endpoints of Its Diameter Find the general form of an equation of the circle with endpoints of its diameter at 18, 232 and 124, 132 .

SOlUTION

We will find the center and the radius of the circle, substitute into the standard equation, and simplify. To determine the center, we will find the midpoint of the diameter. To find the radius, we will find the distance from the center to one endpoint of the diameter. Step 1: Find the center of the circle. We will find the midpoint of its diameter. Since 1x1, y12 5 18, 232 and 1x2, y22 5 124, 132 , we know that x1 5 8, x2 5 24, y1 5 23, and y2 5 13. x1 1 x2 2 8 1 1242 h5 2 4 5 2 52 h5

y1 1 y2 2 23 1 13 k5 2 10 5 2 55 k5

Use the Midpoint Formula.

The center of the circle is at 1h, k2 5 12, 52 .

Step 2: Find the radius of the circle. To find the radius, we find the distance between the center and one endpoint of the diameter. The center is at 12, 52 and one endpoint is 18, 232 . r 5 " 1x2 2 x12 2 1 1y2 2 y12 2

r 5 " 12 2 82 2 1 3 5 2 1232 4 2

Use the Distance Formula. Substitute 8 for x1, 23 for y1, 2 for x2, and 5 for y2.

5 " 1262 2 1 182 2 5 "36 1 64 5 10

"36 1 64 5 "100 5 10

The radius of the circle is 10 units.

Step 3: Substitute and simplify. To find an equation of a circle with center at 12, 52 and radius 10, we substitute 2 for h, 5 for k, and 10 for r in the standard equation of the circle and simplify: 1x 2 h2 2 1 1y 2 k2 2 5 r2

This is the standard equation.

x2 2 4x 1 4 1 y2 2 10y 1 25 5 100

Remove parentheses.

1x 2 22 2 1 1y 2 52 2 5 102 2

2

x 1 y 2 4x 2 10y 2 71 5 0

Subtract 100 from both sides and simplify.

The general form of the equation is x2 1 y2 2 4x 2 10y 2 71 5 0. Self Check 8

Find an equation of a circle with endpoints of its diameter at 122, 22 and 16, 82 . Now Try Exercise 87.

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Section 2.4

Graphs of Equations

247

5. Graph Circles We can convert the general form of the equation of a circle into standard form by completing the square on x and y.

EXAMPlE 9

Graphing a Circle Whose Equation is in General Form Graph the circle whose equation is 2x2 1 2y2 2 8x 1 4y 2 40 5 0.

SOlUTION

We will convert the general equation of the circle into standard form by completing the square on x and y. We will then use the coordinates of the center of the circle and its radius to draw its graph. Step 1: Complete the square on x and y. First, we divide both sides of the equation by 2 to make the coefficients of x2 and y2 equal to 1. 2x2 1 2y2 2 8x 1 4y 2 40 5 0 x2 1 y2 2 4x 1 2y 2 20 5 0 To find the coordinates of the center and the radius, we add 20 to both sides and write the equation in standard form by completing the square on both x and y: x2 1 y2 2 4x 1 2y 5 20 x2 2 4x 1 y2 1 2y 5 20 x2 2 4x 1 4 1 y2 1 2y 1 1 5 20 1 4 1 1 1x 2 22 2 1 1y 1 12 2 5 25

Add 4 and 1 to both sides to complete the square. Factor x2 2 4x 1 4 and y2 1 2y 1 1.

1x 2 22 2 1 3 y 2 1212 4 2 5 52

Step 2: Graph the circle. From the standard form of the equation of the circle, we see that its radius is 5 and that the coordinates of its center are h 5 2 and k 5 21. Thus, the center of the circle is at is 12, 212 . To graph the circle, we plot its center 12, 212 and locate points on the circle that are 5 units from its center. The graph of the circle is shown in Figure 2-49. Because the radius of the circle is 5, the easiest points to locate on the circle to draw the graph are the points that are 5 units to the right, 5 units to the left, 5 units above, and 5 units below the center. Those points are 17, 212 , 123, 212 , 12, 42 , and 12, 262 .

Ra

di 5 us

y

x (2, –1)

2x 2 + 2y 2 – 8x + 4y – 40 = 0

FIGURE 2-49

Self Check 9

Graph: 2x2 1 2y2 1 4x 2 8y 1 2 5 0. Now Try Exercise 107.

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248

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

ACCENT ON TECHNOlOGy

Graphing a Circle When we graphed y2 5 2x using a graphing calculator, we saw that it is necessary to solve the equation for y in terms of x. The same is true when we graph a circle on a graphing calculator. •

To graph 1x 2 22 2 1 1y 1 12 2 5 25, we must solve the equation for y: 1x 2 22 2 1 1y 1 12 2 5 25

1y 1 12 2 5 25 2 1x 2 22 2

y 1 1 5 6"25 2 1x 2 22 2

y 5 21 6 "25 2 1x 2 22 2

•

This last expression represents two equations: y 5 21 1 "25 2 1x 2 22 2 and y 5 21 2 "25 2 1x 2 22 2. We graph both of these equations separately on the same coordinate axes by entering the first equation as Y1 and the second as Y2, as shown in Figure 2-50(a). Depending on the window setting of the maximum and minimum values of x and y, the graph may not appear to be a circle. However, if we use the ZOOM 5: ZSquare window, the graph will be circular. See Figures 2-50(b) and (c). If it appears that there are gaps in the graph, that is due to the way the calculator draws graphs by darkening pixels.

(a)

(b)

(c)

FIGURE 2-50

Note that in this example, it is easier to graph the circle by hand than with a calculator.

6. Solve Equations by Using a Graphing Calculator We can solve many equations using the graphing concepts discussed in this chapter and a graphing calculator. For example, the solutions of x2 2 x 2 3 5 0 will be the numbers x that will make y 5 0 in the equation y 5 x2 2 x 2 3. These numbers will be the x-coordinates of the x-intercepts of the graph of y 5 x2 2 x 2 3.

ACCENT ON TECHNOlOGy

Solving Equations To use a graphing calculator to solve x2 2 x 2 3 5 0, we graph the equation y 5 x2 2 x 2 3, and find the x-coordinates of the x-intercepts. These will be the solutions of the equation. The graph of y 5 x2 2 x 2 3 is shown in Figure 2-51(c). •

Use the ZERO command shown earlier in this section to solve the equation. Notice the screen in Figures 2-51(h) and (i) looks different than in Figures 2-51(e) and (f). On some calculators we can just enter the value we want as the

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Section 2.4

•

Graphs of Equations

left and right bound rather than using the cursor to get to a bound. We see that the two solutions to the equation are x < 21.302776 and x < 2.3027756. Every equation does not have a solution over the real numbers. For example, to try and solve the equation x2 1 2x 1 2 5 0, graph the equation y 5 x2 1 2x 1 2. We can see that it does not intersect the x-axis, and there is no solution to the equation. See Figure 2-51(k).

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k) FIGURE 2-51

Self Check Answers

1.

2.

y

y 10 8 6 4 2

x −4

y = –|x|

3.

249

−2 −2 −4 −6 −8 −10

4.

y

x

y = x3 − 9x

y

x

x y = – √x

1 2 3 4 5

y2 = –x

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250

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

5. 14, 02 , 7 6. 1x 2 22 2 1 1y 1 42 2 5 5 2 2 8. x 1 y 2 4x 2 10y 1 4 5 0 9.

7. x2 1 y2 1 4x 2 10y 2 7 5 0 y

2

Ra

diu s

(–1, 2)

x 2x 2 + 2y 2 + 4x – 8y + 2 = 0

Exercises 2.4 Getting Ready

21. y 5 x4 2 1

22. y 5 x4 2 25x2

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. The point where a graph intersects the x-axis is . called the 2. The y-intercept is the point where a graph intersects the . 3. If a line divides a graph into two congruent halves, we call the line an . 1 2 4. If the point 2x, y lies on a graph whenever 1x, y2 does, the graph is symmetric about the . 5. If the point 1x, 2y2 lies on a graph whenever 1x, y2 does, the graph is symmetric about the . 6. If the point 12x, 2y2 lies on a graph whenever 1x, y2 does, the graph is symmetric about the . 7. A is the set of all points in a plane that are fixed distance from a point called its . 8. A is the distance from the center of a circle to a point on the circle. 9. The standard form of an equation of a circle with center at the origin and radius r is . 10. The standard form of an equation of a circle with center at 1h, k2 and radius r is

Graph each equation. Check your graph with a graphing calculator. 23. y 5 x2

24. y 5 2x2 y

y

x

x

25. y 5 2x2 1 2

26. y 5 x2 2 1 y

y

x x

27. y 5 x2 2 4x

28. y 5 x2 1 2x y

y

. x

Practice

x

Find the x- and y-intercepts of each graph. Do not graph the equation. 11. y 5 x2 2 4

12. y 5 x2 2 9

13. y 5 4x2 2 2x

14. y 5 2x 2 4x2

15. y 5 x2 2 4x 2 5

16. y 5 x2 2 10x 1 21

17. y 5 x2 1 x 2 2

18. y 5 x2 1 2x 2 3

3

3

1 29. y 5 x2 2 2x 2

1 30. y 5 x2 1 3 2

y

y

x x

19. y 5 x 2 9x

20. y 5 x 1 x

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Section 2.4

31. y 5 x2 1 2

32. y 5 3x 1 2

33. y2 1 1 5 x

34. y2 1 y 5 x

35. y2 5 x2

36. y 5 3x 1 7

37. y 5 3x2 1 7

38. x2 1 y2 5 1

54. y 5 3 0 x 0

53. y 5 3 2 0 x 0

Find the symmetries, if any, of the graph of each equation. Do not graph the equation.

251

Graphs of Equations

y

y

x

55. y2 5 2x

x

56. y2 5 4x y

y

39. y 5 3x3 1 7

40. y 5 3x3 1 7x

41. y2 5 3x

42. y 5 3x4 1 7

43. y 5 0 x 0

44. y 5 0 x 1 1 0

45. 0 y 0 5 x

46. 0 y 0 5 0 x 0

x

x

57. y2 5 9x

58. y2 5 24x y

y

Graph each equation. Be sure to find any intercepts and symmetries. Check your graph with a graphing calculator. 47. y 5 x2 1 4x y

x

x

48. y 5 x2 2 6x y x

59. y 5 "x 2 1

60. y 5 1 2 "x

y

y

x

x

x

49. y 5 x3

50. y 5 x3 1 x y

y

61. xy 5 4

62. xy 5 29 y

51. y 5 0 x 2 2 0

y

x

x

x

52. y 5 0 x 0 2 2

y

x

y

x x

Identify the center and radius of each circle written in standard form. 63. x2 1 y2 5 100 65. x2 1 1y 2 52 2 5 49

64. x2 1 y2 5 81 66. x2 1 1y 1 32 2 5 8

67. 1x 1 62 2 1 y2 5

68. 1x 2 52 2 1 y2 5

1 4

69. 1x 2 42 2 1 1y 2 12 2 5 9

70. 1x 1 112 2 1 1x 1 72 2 5 121

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16 25

252

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

1 2 71. ax 2 b 1 1y 1 22 2 5 45 4

72. Qx 1 "5R 1 1y 2 32 2 5 1

Graph each circle. 97. x2 1 y2 2 25 5 0

2

98. x2 1 y2 2 8 5 0 y

y

Write an equation in standard form of the circle with the given properties. x

x

73. Center at the origin; r 5 5 74. Center at the origin; r 5 "3 75. Center at 10, 262 ; r 5 6

76. Center at 10, 72 ; r 5 9 1 77. Center at 18, 02 ; r 5 5 2 1 78. Center at 210, 0 ; r 5 "11

99. 1x 2 12 2 1 1y 1 22 2 5 4 100. 1x 1 12 2 1 1y 2 22 2 5 9 y

y

x

79. Center at 122, 122 , r 5 13 2 80. Center at a , 25b; r 5 7 7

Write an equation in general form of the circle with the given properties.

x

101. x2 1 y2 1 2x 2 24 5 0

81. Center at the origin; r 5 1

102. x2 1 y2 2 4y 5 12

y

y

82. Center at the origin; r 5 4 83. Center at 16, 82 ; r 5 4

84. Center at 15, 32 ; r 5 2 85. Center at 13, 242 ; r 5 "2

x x

86. Center at 129, 82 ; r 5 2"3 87. Ends of diameter at 13, 222 and 13, 82

103. x2 1 y2 1 4x 1 2y 2 11 5 0

88. Ends of diameter at 15, 92 and 125, 292

y

89. Center at 123, 42 and passing through the origin

x

90. Center at 122, 62 and passing through the origin Convert the general form of each circle given into standard form. 91. x2 1 y2 2 6x 1 4y 1 4 5 0

104. x2 1 y2 2 6x 1 2y 1 1 5 0 y

92. x2 1 y2 1 4x 2 8y 2 5 5 0 93. x2 1 y2 2 10x 2 12y 1 57 5 0

x

94. x2 1 y2 1 2x 1 18y 1 57 5 0 95. 2x2 1 2y2 2 8x 2 16y 1 22 5 0 96. 3x2 1 3y2 1 6x 2 30y 1 3 5 0

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Section 2.4

105. 9x2 1 9y2 2 12y 5 5

106. 4x2 1 4y2 1 4y 5 15

y

115. x3 2 3 5 0

Graphs of Equations

253

116. 3x3 2 x2 2 x 5 0

y

x

x

107. 4x2 1 4y2 2 4x 1 8y 1 1 5 0 y

x

Applications 117. Golfing Phil Mickelson’s tee shot follows a path given by y 5 64t 2 16t2, where y is the height of the ball (in feet) after t seconds of flight. How long will it take for the ball to strike the ground? 118. Golfing Halfway through its flight, the golf ball of Exercise 117 reaches the highest point of its trajectory. How high is that? 119. Stopping distances The stopping distance D (in feet) for a Ford Fusion car moving V miles per hour is given by D 5 0.08V 2 1 0.9V . Graph the equation for velocities between 0 and 60 mph. D

108. 9x2 1 9y2 2 6x 1 18y 1 1 5 0 y

V x

120. Stopping distances See Exercise 119. How much farther does it take to stop at 60 mph than at 30 mph?

109. y 5 2x2 2 x 1 1

110. y 5 x2 1 5x 2 6

111. y 5 7 1 x 2 x2

112. y 5 2x2 2 3x 1 2

Use a graphing calculator to solve each equation. Round to the nearest hundredth. 113. x2 2 7 5 0

ulisse/Shutterstock.com

Use a graphing calculator to graph each equation. Then find the coordinates of the vertex of the parabola to the nearest hundredth.

121. Basketball court The center circle of the Kansas Jayhawks basketball court is a circle with a 12-ft diameter. If the center of the circle is located at the origin, find an equation in standard form that models the circle.

122. Oil spill Oil spills from a tanker in the Gulf of Mexico and surfaces continuously at coordinates 10, 02 . If oil spreads in a circular pattern for ten hours and the circle’s radius increases at a rate of 2 inches per hour, write an equation of the circle that models the range of the spill’s effect.

114. x2 2 3x 1 2 5 0

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254

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

123. Super Loop The Fire Ball Super Loop is a rollercoaster ride that is shaped like a circle. Find an equation of the loop in standard form if it is positioned 5 feet off of the ground, has a diameter of 60 feet, and its center is at coordinates 10, 352 .

y

Kenneth William Caleno/Shutterstock.com

x

Discovery and Writing The solution of the inequality y , 0 consists of those numbers x for which the graph of y lies below the x-axis. To solve y , 0, we graph y and trace to find numbers x that produce negative values of y. Solve each inequality.

124. Hurricane As a hurricane strengthens an eye begins to form at the center of the storm. At a wind speed of 80 mph the eye of a hurricane is circular when viewed from above and is 30 miles in diameter. If the eye is located at map coordinates 15, 102 , find an equation, in standard form, of the circle that models the eye of the hurricane. 125. CB radios The CB radio of a trucker covers the circular area shown in the illustration. Find an equation of that circle, in general form.

127. x2 1 x 2 6 , 0

128. x2 2 3x 2 10 . 0

When converting a circle’s equation from general to standard form, it is possible to obtain a constant term on the right side that is zero or negative. If the constant term is zero the graph is a single point. If the constant term is negative the graph is nonexistent. Determine whether the graph of the equation is a single point or nonexistent. 129. x2 2 4x 1 y2 2 6y 1 13 5 0 130. x2 2 12x 1 y2 1 4y 1 43 5 0

Review

y

Solve each equation. 131. 3 1x 1 22 1 x 5 5x

N

E T(7, 4)

Route 73 x

Meadville

(10, 0)

126. Firestone tires Two 24-inch-diameter Firestone tires stand against a wall, as shown in the illustration. Find equations in general form of the circular boundaries of the tires.

132. 12b 1 6 13 2 b2 5 b 1 3 5 12 2 x2 133. 215x15 3 r21 r12 134. 5 12 3 6 135. Mixing an alloy In 60 ounces of alloy for watch cases, there are 20 ounces of gold. How much copper must be added to the alloy so that a watch case weighing 4 ounces, made from the new alloy, will contain exactly 1 ounce of gold? 136. Mixing coffee To make a mixture of 80 pounds of coffee worth $272, a grocer mixes coffee worth $3.25 a pound with coffee worth $3.85 a pound. How many pounds of cheaper coffee should the grocer use?

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Section 2.5

Proportion and Variation

255

2.5 Proportion and Variation In this section, we will learn to 1. 2. 3. 4.

Solve proportions. Use direct variation to solve problems. Use inverse variation to solve problems. Use joint variation to solve problems.

5. Use combined variation to solve problems.

© Istockphoto.com/Aleksander Trankov

Skydiving is an adventurous sport. Jumping out of an aircraft at a height of 14,000 feet and free falling delivers a rush of adrenaline that is an exhilarating experience and one that skydivers never forget. The unique experience allows skydivers to fall approximately 1,300 feet every five seconds and reach speeds between 120 and 150 miles per hour. For a free-falling object, we can calculate the distance fallen given the amount of time. In fact, the distance of the fall is proportional to the square of the time. In this section we will consider how variables are related. These ways include direct variation, inverse variation, and joint variation and combinations of these.

1. Solve Proportions The quotient of two numbers is often called a ratio. For example, the fraction 32 (or the expression 3:2) can be read as “the ratio of 3 to 2.” Some examples are 3 , 5

x11 , 9

7:9,

a , b

and

x2 2 4 x15

An equation indicating that two ratios are equal is called a proportion. Some examples of proportions are 2 4 5 , 3 6

x 3 5 , y 5

and

x2 1 8 17 1x 1 32 5 2 1x 1 32 2

In the proportion ab 5 dc , the numbers a and d are called the extremes, and the numbers b and c are called the means. To develop an important property of proportions, we suppose that a c 5 b d and multiply both sides by bd to get a c bda b 5 bda b b d bda bdc 5 b d da 5 bc Thus, if

Property of Proportions

a b

5 dc then ad 5 bc. This proves the following statement.

In any proportion, the product of the extremes is equal to the product of the means.

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256

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

We can use this property to solve proportions.

EXAMPlE 1

Solving a Proportion x 2 5 . 5 x13

Solve the proportion: SOlUTION

We will use the property of proportions to solve the proportion. x 2 5 5 x13

x 1x 1 32 5 5 ? 2

The product of the extremes equals the product of the means.

x2 1 3x 5 10

Remove parentheses and simplify.

x2 1 3x 2 10 5 0

Subtract 10 from both sides.

1x 2 22 1x 1 52 5 0 x 2 2 5 0 or x52

Factor the trinomial.

x1550

Set each factor equal to 0.

x 5 25

Thus, x 5 2 or x 5 25. Verify each solution. Self Check 1

Solve:

2 3 5 . 5 x24

Now Try Exercise 13.

EXAMPlE 2

Solving an Application Problem Involving a Proportion Gasoline and oil for a Nissan outboard boat motor are to be mixed in a 50:1 ratio. How many ounces of oil should be mixed with 6 gallons of gasoline?

SOlUTION

We will first express 6 gallons in terms of ounces. 6 gallons 5 6 ? 128 ounces 5 768 ounces Then, we let x represent the number of ounces of oil needed, set up the proportion, and solve it. 50 768 5 1 x 50x 5 768 x5

768 50

The product of the extremes equals the product of the means. Divide both sides by 50.

x 5 15.36 Approximately 15 ounces of oil should be added to 6 gallons of gasoline. Self Check 2

How many ounces of oil should be mixed with 6 gallons of gas if the ratio is to be 40 parts of gas to 1 part of oil? Now Try Exercise 15.

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Section 2.5

Proportion and Variation

257

2. Use Direct Variation to Solve Problems Two variables are said to vary directly or be directly proportional if their ratio is a constant. The variables x and y vary directly when y 5 k or, equivalently, y 5 kx (k is a constant) x Direct Variation

The words “y varies directly with x,” or “y is directly proportional to x,” mean that y 5 kx for some real-number constant k. The number k is called the constant of proportionality.

EXAMPlE 3

Solving a Direct Variation Problem Distance traveled in a given time varies directly with the speed. If a car travels 70 miles at 30 mph, how far will it travel in the same time at 45 mph?

SOlUTION

We will use direct variation to solve the problem. The phrase distance varies directly with speed translates into the formula d 5 ks, where d represents the distance traveled and s represents the speed. The constant of proportionality k can be found by substituting 70 for d and 30 for s in the equation d 5 ks. d 5 ks

70 5 k 1302 7 k5 3 To evaluate the distance d traveled at 45 mph, we substitute 73 for k and 45 for s into the formula d 5 ks. 7 d5 s 3 7 5 1452 3 5 105 In the time it takes to go 70 miles at 30 mph, the car could travel 105 miles at 45 mph. Self Check 3

How far will the car travel in the same time if its speed is 60 mph? Now Try Exercise 39.

y m=2 m=1

x 1 m=–– 2 FIGURE 2-52

The statement y varies directly with x is equivalent to the equation y 5 mx, where m is the constant of proportionality. Because the equation is in the form y 5 mx 1 b with b 5 0, its graph is a line with slope m and y-intercept at (0, 0). The graphs of y 5 mx for several values of m are shown in Figure 2-52. The graph of the relationship of direct variation is always a line that passes through the origin.

3. Use Inverse Variation to Solve Problems Two variables are said to vary inversely or be inversely proportional if their product is a constant. xy 5 k

or, equivalently,

y5

k (k is a constant) x

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Inverse Variation

EXAMPlE 4

The words “y varies inversely with x,” or “y is inversely proportional to x,” mean that y 5 xk for some real-number constant k.

Solving an Inverse Variation Problem Intensity of illumination from a light source varies inversely with the square of the distance from the source. If the intensity of a light source is 100 lumens at a distance of 20 feet, find the intensity at 30 feet.

SOlUTION

We will use inverse variation to solve the problem. If I is the intensity and d is the distance from the light source, the phrase intensity varies inversely with the square of the distance translates into the formula k d2 We can evaluate k by substituting 100 for I and 20 for d in the formula and solving for k. k I5 2 d k 100 5 2 20 k 5 40,000 I5

To find the intensity at a distance of 30 feet, we substitute 40,000 for k and 30 for d in the formula k I5 2 d 40,000 I5 302 400 5 9 At 30 feet, the intensity of light would be 400 9 lumens per square centimeter. Self Check 4

Find the intensity at 50 feet. Now Try Exercise 43. The statement y is inversely proportional to x is equivalent to the equation y 5 xk , where k is a constant. Figure 2-53 shows the graphs of y 5 xk 1x . 02 for three values of k. In each case, the equation determines one branch of a curve called a hyperbola. Verify these graphs with a graphing calculator. y

y

y 9 y= – x

1 y= – x (1, 1) x

(a)

x

(3, 3)

(2, –2) x

(b)

–4 y = –– x (c)

FIGURE 2-53

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Section 2.5

Proportion and Variation

259

4. Use Joint Variation to Solve Problems Joint Variation

EXAMPlE 5

The words “y varies jointly with w and x” mean that y 5 kwx for some realnumber constant k.

Solving a Joint Variation Problem Kinetic energy of an object varies jointly with its mass and the square of its velocity. A 25-gram mass moving at the rate of 30 centimeters per second has a kinetic energy of 11,250 dyne-centimeters. Find the kinetic energy of a 10-gram mass that is moving at 40 centimeters per second.

SOlUTION

We will use joint variation to solve the problem. If we let E, m, and v represent the kinetic energy, mass, and velocity, respectively, the phrase energy varies jointly with the mass and the square of its velocity translates into the formula E 5 kmv2 The constant k can be evaluated by substituting 11,250 for E, 25 for m, and 30 for v in the formula. E 5 kmv2 11,250 5 k 1252 1302 2 11,250 5 22,500k k5

1 2

We can now substitute 12 for k, 10 for m, and 40 for v in the formula and evaluate E. E 5 kmv2 1 5 1102 1402 2 2 5 8,000 A 10-gram mass that is moving at 40 centimeters per second has a kinetic energy of 8,000 dyne-centimeters. Self Check 5

Find the kinetic energy of a 25-gram mass that is moving at 100 centimeters per second. Now Try Exercise 45.

5. Use Combined Variation to Solve Problems The preceding terminology can be used in various combinations. In each of the following statements shown in the following table, the formula on the left translates into the words on the right.

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Formula y5

Words

kx z

y varies directly with x and inversely with z.

3 z y 5 kx2"

y5

y5

EXAMPlE 6

y varies jointly with the square of x and the cube root of z.

kx"z

y varies jointly with x and the square root of z and inversely with the cube root of t.

"t 3

k xz

y varies inversely with the product of x and z.

Using Combined Variation to Solve a Problem The time it takes to build a highway varies directly with the length of the road but inversely with the number of workers. If it takes 100 workers 4 weeks to build 2 miles of the George Washington Memorial Parkway, how long will it take 80 workers to build 10 miles of the parkway?

SOlUTION

We will use combined variation to solve the problem. We can let t represent the time in weeks, l represent the length in miles, and w represent the number of workers. Because time varies directly with the length of the parkway but inversely with the number of workers, the relationship between these variables can be expressed by the equation t5

kl w

We substitute 4 for t, 100 for w, and 2 for l to find k: k 122 100 400 5 2k

Multiply both sides by 100.

200 5 k

Divide both sides by 2.

45

We now substitute 80 for w, 10 for l, and 200 for k in the equation t 5 kl w and simplify: kl w 200 1102 t5 80 5 25

t5

It will take 25 weeks for 80 workers to build 10 miles of parkway. Self Check 6

How long will it take 100 workers to build 20 miles of parkway? Now Try Exercise 47.

Self Check Answers

23 2. 19.2 oz 3. 140 mi 4. 16 lumens per cm2 2 5. 125,000 dyne-centimeters 6. 40 weeks

1.

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Section 2.5

261

Proportion and Variation

Exercises 2.5 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. A ratio is the of two numbers. 2. A proportion is a statement that two are equal. 3. In the proportion ab 5 dc , b and c are called the . 4. In the proportion ab 5 dc , a and d are called the . is 5. In a proportion, the product of the equal to the product of the . 6. Direct variation translates into the equation . 7. The equation y 5 xk indicates variation. 8. In the equation y 5 kx, k is called the of proportionality. 9. The equation y 5 kxz represents variation. kx2 10. In the equation y 5 z , y varies directly with and inversely with .

22. z is directly proportional to the sum of x and y. If x 5 2 and y 5 5, then z 5 28. Solve each problem. 23. y is directly proportional to x. If y 5 15 when x 5 4, find y when x 5 75. 24. w is directly proportional to z. If w 5 26 when z 5 2, find w when z 5 23. 25. w is inversely proportional to z. If w 5 10 when z 5 3, find w when z 5 5. 26. y is inversely proportional to x. If y 5 100 when x 5 2, find y when x 5 50. 27. P varies jointly with r and s. If P 5 16 when r 5 5 and s 5 28, find P when r 5 2 and s 5 10. 28. m varies jointly with the square of n and the square root of q. If m 5 24 when n 5 2 and q 5 4, find m when n 5 5 and q 5 9. Determine whether the graph could represent direct variation, inverse variation, or neither. 29. y 30. y

Practice 5 x 5 2 6 x15 7 14. 5 6 82x

12.

Set up and solve a proportion to answer each question. 15. The ratio of women to men in a mathematics class is 3:5. How many women are in the class if there are 30 men? 16. The ratio of lime to sand in mortar is 3:7. How much lime must be mixed with 21 bags of sand to make mortar? Find the constant of proportionality. 17. y is directly proportional to x. If x 5 30, then y 5 15. 18. z is directly proportional to t. If t 5 7, then z 5 21. 19. I is inversely proportional to R. If R 5 20, then I 5 50. 20. R is inversely proportional to the square of I. If I 5 25, then R 5 100. 21. E varies jointly with I and R. If R 5 25 and I 5 5, then E 5 125.

31.

32.

y

y x

x

Applications Set up and solve the required proportion. 33. Caffeine Many convenience stores sell supersize 44-ounce soft drinks in refillable cups. For each of the products listed in the table, find the amount of caffeine contained in one of the large cups. Round to the nearest milligram.

Joe Raedle/Getty Images

Solve each proportion. 4 2 11. 5 x 7 x 3 13. 5 2 x11

x

x

Soft drink, 12 oz

Caffeine (mg)

Mountain Dew

55

Coca-Cola Classic

47

Pepsi

37

Based on data from the Los Angeles Times

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34. Cellphones A country has 221 mobile cellular telephones per 250 inhabitants. If the country’s population is about 280,000, how many mobile cellular telephones does the country have? 35. Wallpapering Read the instructions on the label of wallpaper adhesive. Estimate the amount of adhesive needed to paper 500 square feet of kitchen walls if a heavy wallpaper will be used. COVERAGE: One-half gallon will hang approximately 4 single rolls (140 sq ft), depending on the weight of the wall covering and the condition of the wall 36. Recommended dosages The recommended child’s dose of the sedative hydroxine is 0.006 gram per kilogram of body mass. Find the dosage for a 30-kg child in milligrams. 37. Gas laws The volume of a gas varies directly with the temperature and inversely with the pressure. When the temperature of a certain gas is 330°C, the pressure is 40 pounds per square inch and the volume is 20 cubic feet. Find the volume when the pressure increases 10 pounds per square inch and the temperature decreases to 300°C. 38. Hooke’s Law The force f required to stretch a spring a distance d is directly proportional to d. A force of 5 newtons stretches a spring 0.2 meter. What force will stretch the spring 0.35 meter? 39. Free-falling objects The distance that an object will fall in t seconds varies directly with the square of t. An object falls 16 feet in 1 second. How long will it take the object to fall 144 feet? 40. Heat dissipation The power, in watts, dissipated as heat in a resistor varies directly with the square of the voltage and inversely with the resistance. If 20 volts are placed across a 20-ohm resistor, it will dissipate 20 watts. What voltage across a 10-ohm resistor will dissipate 40 watts? 41. Period of a pendulum The time required for one complete swing of a pendulum is called the period of the pendulum. The period varies directly with the square of its length. If a 1-meter pendulum has a period of 1 second, find the length of a pendulum with a period of 2 seconds. 42. Frequency of vibration The pitch, or frequency, of a vibrating string varies directly with the square root of the tension. If a string vibrates at a frequency of 144 hertz due to a tension of 2 pounds, find the frequency when the tension is 18 pounds. 43. Illumination Intensity of illumination from a light source varies inversely with the square of the distance from the source. If the intensity of a light source is 60 lumens at a distance of 10 feet, find the intensity at 20 feet.

44. Illumination Intensity of illumination from a light source varies inversely with the square of the distance from the source. If the intensity of a light source is 100 lumens at a distance of 15 feet, find the intensity at 25 feet. 45. Kinetic energy The kinetic energy of an object varies jointly with its mass and the square of its velocity. What happens to the energy when the mass is doubled and the velocity is tripled? 46. Heat dissipation The power, in watts, dissipated as heat in a resistor varies jointly with the resistance, in ohms, and the square of the current, in amperes. A 10-ohm resistor carrying a current of 1 ampere dissipates 10 watts. How much power is dissipated in a 5-ohm resistor carrying a current of 3 amperes? 47. Gravitational attraction The gravitational attraction between two massive objects varies jointly with their masses and inversely with the square of the distance between them. What happens to this force if each mass is tripled and the distance between them is doubled? 48. Gravitational attraction In Problem 47, what happens to the force if one mass is doubled and the other tripled and the distance between them is halved? 49. Plane geometry The area of an equilateral triangle varies directly with the square of the length of a side. Find the constant of proportionality. 50. Solid geometry The diagonal of a cube varies directly with the length of a side. Find the constant of proportionality.

Discovery and Writing Explain the terms extremes and means. Distinguish between a ratio and a proportion. Explain the term joint variation. Explain why xy 5 k indicates that y varies directly with x. 55. Explain why xy 5 k indicates that y varies inversely with x. 51. 52. 53. 54.

56. As temperature increases on the Fahrenheit scale, it also increases on the Celsius scale. Is this direct variation? Explain.

Review Perform each operation and simplify. 1 2 1 57. x12 x11 58.

x2 2 1 x21 ? x 1 1 x2 2 2x 1 1

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Chapter Review

59.

x2 1 3x 2 4 x21 4 2 x2 2 5x 1 4 x 2 3x 2 4

60.

x12 4 12x 1 42 3x 2 3

263

x2 1 4 2 1x 1 22 2 4x2 1 1 2 x 3 62. 1 21 x 61.

CHAPTER REVIEW SECTION 2.1

The Rectangular Coordinate System

Definitions and Concepts

Examples

The rectangular coordinate system divides the plane into four quadrants.

y Quadrant II

Quadrant I

3 2

Origin

1 –5 –4 –3 –2 –1

Quadrant III

–1

1 2

–2 –3

3

4

5

x

Quadrant IV

The graph of an equation in x and y is the set of all points 1x, y2 that satisfy the equation.

Use the x- and y-intercepts to graph the equation 6x 1 4y 5 24.

The y-intercept of a line is the point 10, b2 , where the line intersects the y-axis. To find b, substitute 0 for x in the equation of the line and solve for y.

Find the y-intercept. To find the y-intercept, we substitute 0 for x and solve for y. 6x 1 4y 5 24

6 102 1 4y 5 24 4y 5 24 y56

Substitute 0 in for x. Simplify. Divide both sides by 4.

The y-intercept is the point 10, 62 . The x-intercept of a line is the point 1a, 02 , where the line intersects the x-axis. To find a, substitute 0 for y in the equation of the line and solve for x.

Find the x-intercept. To find the x-intercept, we substitute 0 for y and solve for x. 6x 1 4y 5 24

6x 1 4 102 5 24 6x 5 24 x54

Substitute 0 for y. Simplify. Divide both sides by 6.

The x-intercept is the point 14, 02 .

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264

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Definitions and Concepts

Examples Find a third point as a check. If we let x 5 2, we will find that y 5 3. 6x 1 4y 5 24

6 122 1 4y 5 24

Substitute 2 for x.

12 1 4y 5 24

Simplify.

4y 5 12

Subtract 12 from both sides.

y53

Divide both sides by 4.

The point 12, 32 satisfies the equation. We plot each pair and join them with a line to get the graph of the equation. y (0, 6) 6x + 4y = 24 (2, 3)

(4, 0)

Equation of a vertical line through 1a, b2 :

x

y

x5a

y=2

Equation of a horizontal line through 1a, b2 :

x

y5b x = −3

The Distance Formula: The distance d between points 1x1, y12 and 1x2, y22 is given by d 5 " 1x2 2 x12 2 1 1y2 2 y12 2

Find the distance between P 125, 22 and Q 123, 42 . We can use the Distance Formula d 5 " 1x2 2 x12 2 1 1y2 2 y12 2 to find the distance between P 125, 22 and Q 123, 42 . If we let P 125, 22 5 P 1x1, y12 and Q 123, 42 5 Q 1x2, y22 , we can substitute 25 for x1, 2 for y1, 23 for x2, and 4 for y2 into the formula and simplify. d 1PQ2 5 " 1x2 2 x12 2 1 1y2 2 y12 2

d 1PQ2 5 " 3 23 2 1252 4 2 1 3 4 2 122 4 2 5 " 122 2 1 122 2

5 "4 1 4 5 "8 5 "4 ? 2 5 2"2 The distance between the two points is 2"2. The Midpoint Formula: The midpoint of the line segment joining 1x1, y12 and 1x2, y22 is the point M with coordinates M5a

x1 1 x2 y1 1 y2 , b 2 2

To find the midpoint of the segment with endpoints at 124, 52 and 16, 72 , average the x-coordinates and average the y-coordinates: The midpoint is Q24 21 6, 5

1 7 2 R

5 Q22, 12 2 R 5 11, 62 .

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Chapter Review

265

EXERCISES 14. x 2 5y 5 5

13. x 1 y 5 27 Refer to the illustration and find the coordinates of each point. 1. A

2. B

3. C

4. D

y

y x x

y B A

x

Graph each equation.

D

C

15. y 5 4

16. x 5 22 y

Graph each point. Indicate the quadrant in which the point lies or the axis on which it lies. 6. 15, 232 5. 123, 52 7. 10, 272

y

x

1 8. a2 , 0b 2

x

y

17. Depreciation A Ford Mustang purchased for $18,750 is expected to depreciate according to the formula y 5 22,200x 1 18,750. Find its value after 3 years. 18. House appreciation A house purchased for $250,000 is expected to appreciate according to the formula y 5 16,500x 1 250,000, where y is the value of the house after x years. Find the value of the house 5 years later.

x

Find the length of the segment PQ. Solve each equation for y and graph the equation. Then check your graph with a graphing calculator. 9. 2x 2 y 5 6 10. 2x 1 5y 5 210

19. P 123, 72 ; Q 13, 212

20. P 128, 62 ; Q 1212, 102

21. PQ"3, 9R; QQ"3, 7R

22. P 1a, 2a2 ; Q 12a, a2

y

y

Find the midpoint of the segment PQ.

x x

23. P 123, 72 ; Q 13,212

24. P 10, 52 ; Q 1212, 102

25. PQ"3, 9R; QQ"3, 7R

26. P 1a, 2a2 ; Q 12a, a2

Use the x- and the y-intercepts to graph each equation. 11. 3x 2 5y 5 15

12. x 1 y 5 7 y

y

x

x Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

266

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SECTION 2.2

The Slope of a Nonvertical Line

Definitions and Concepts The slope of a nonvertical line passing through points P 1x1, y12 and Q 1x2, y22 is m5

change in y y 2 y1 1x 2 x12 5 2 change in x x2 2 x1 2

Examples Find the slope of the line passing through P 121, 232 and Q 17, 92 . We will substitute the points P 121, 232 and Q 17, 92 into the slope formula, m5

change in y y 2 y1 5 2 change in x x2 2 x1

to find the slope of the line. Let P 1x1, y12 5 P 121, 232 and Q 1x2, y22 5 Q 17, 92 . Then we substitute 21 for x1, 23 for y1, 7 for x2, and 9 for y2 to get m5

change in y change in x

m5

y2 2 y1 x2 2 x1

5

9 2 1232 12 3 5 5 7 2 1212 8 2

The slope of the line is 32 . Slopes of horizontal and vertical lines: The slope of a horizontal line (a line with an equation of the form y 5 b) is 0.

The slope of the graph of y 5 7 is 0.

The slope of a vertical line (a line with an equation of the form x 5 a) is not defined.

The slope of the line x 5 6 is not defined.

Slopes of parallel lines: Nonvertical parallel lines have the same slope.

Determine whether the lines with the given slopes are parallel, perpendicular, or neither.

Slopes of perpendicular lines: The product of the slopes of two perpendicular lines is 21, provided neither line is vertical.

1 6 The product of the slopes m1 5 26 and m2 5 16 is 21. The lines are perpendicular. m1 5 26; m2 5

EXERCISES Find the slope of the line PQ, if possible. 28. P 12,72 ; Q 125,272 27. P 13, 252 ; Q 11, 72 29. P 1b, a2 ; Q 1a, b2 30. P 1a 1 b, b2 ; Q 1b, b 2 a2 Find two points on the line and find the slope of the line. 32. y 5 5x 2 6 31. y 5 3x 1 6

Determine whether the slope of each line is 0 or undefined. y 33. 34. y

x

x

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Chapter Review

39. A line passes through 122, 52 and 16, 102 . A line parallel to it passes through 12, 22 and 110, y2 . Find y.

Determine whether the slope of each line is positive or negative. 35.

36.

y

y

x

x

Determine whether the lines with the given slopes are parallel, perpendicular, or neither. 1 2 7 38. m1 5 ; m2 5 37. m1 5 5; m2 5 2 5 7 2

SECTION 2.3

267

40. A line passes through 122, 52 and 16, 102 . A line perpendicular to it passes through 122, 52 and 1x, 232 . Find x. 41. Rate of descent If an airplane descends 3,000 feet in 15 minutes, what is the average rate of descent in feet per minute? 42. Rate of growth A small business predicts sales according to a straight-line method. If sales were $50,000 in the first year and $147,500 in the third year, find the rate of growth in dollars per year (the slope of the line).

Writing Equations of Lines

Definitions and Concepts

Examples

Point-slope form: An equation of the line passing through P 1x1, y12 and with slope m is

Write an equation of the line with slope 243 and passing through P 13, 222 . We will substitute 243 for m, 3 for x1, and 22 for y1 in the point-slope form y 2 y1 5 m 1x 2 x12 and simplify.

y 2 y1 5 m 1x 2 x12

y 2 y1 5 m 1x 2 x12

4 y 2 1222 5 2 1x 2 32 3 4 y1252 x14 3 4 y52 x12 3

4 Substitute 2 for m, 3 for 3 x1, and 22 for y1. Remove parentheses. Subtract 2 from both sides.

Slope-intercept form: An equation of the line with slope m and y-intercept 10, b2 is y 5 mx 1 b.

The equation of the line y 5 253x 1 4 is written in slope-intercept form.

Standard form of an equation of a line:

The above equation written in standard form Ax 1 By 5 C is 5x 1 3y 5 12.

Ax 1 By 5 C Slope-intercept form can be used to find the slope and the y-intercept from the equation of a line.

Find the slope and the y-intercept of the line with equation 2x 1 5y 5 210. We will write the equation in the form y 5 mx 1 b to find the slope m and the y-intercept 10, b2 . 2x 1 5y 5 210 5y 5 22x 2 10 2 y52 x22 5

Subtract 2x from both sides. Divide both sides by 5.

The slope of the graph is 225, and the y-intercept is 10, 222 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

268

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Definitions and Concepts

Examples

Horizontal line: y 5 b The slope is 0, and the y-intercept is 10, b2 .

The equation of the horizontal line with slope 0 and y-intercept 10, 72 is y 5 7

Vertical line: x 5 a There is no defined slope, and the x-intercept is 1a, 02 .

The equation of the vertical line with no defined slope and the x-intercept 126, 02 is x 5 26.

EXERCISES Find the slope and the y-intercept of the graph of each line. 51. 3x 2 2y 5 10 52. 2x 1 4y 5 28

Use point-slope form to write an equation of each line. Write each equation in standard form. 43. The line passes through the origin and the point 125, 72 . 44. The line passes through 122, 12 and has a slope of 24. 45. The line passes through 12, 212 and has a slope of 215. 46. The line passes through 17, 252 and 14, 12 . Use slope-intercept form to write an equation of each line. 47. The line has a slope of 23 and a y-intercept of 3. 48. The slope is 232 and the line passes through 10, 252 . Use slope-intercept form to graph each equation. 3 4 49. y 5 x 2 2 50. y 5 2 x 1 3 5 3 y

54. 2x 5 24y 2 8

55. 5x 1 2y 5 7

56. 3x 2 4y 5 14

Write an equation of each line. 57. The line has a slope of 0 and passes through 125, 172 . 58. The line has no defined slope and passes through 125, 172 . Write an equation of each line. Write the answer in slopeintercept form. 59. The line is parallel to 3x 2 4y 5 7 and passes through 12, 02 .

60. The line passes through 17, 222 and is parallel to the line segment joining 12, 42 and 14, 2102 .

y

x

53. 22y 5 23x 1 10

x

61. The line passes through 10, 52 and is perpendicular to the line x 1 3y 5 4. 62. The line passes through 17, 222 and is perpendicular to the line segment joining 12, 42 and 14, 2102 . Determine whether the graphs of each pair of equations are parallel, perpendicular, or neither. 63. y 5 3x 1 8, 2y 5 6x 2 19 64. 2x 1 3y 5 6, 3x 2 2y 5 15

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269

Chapter Review

SECTION 2.4

Graphs of Equations

Definitions and Concepts

Examples

To graph an equation:

Graph: y 5 x3 2 4x.

1. Find the x- and y-intercepts. 2. Find the symmetries of the graph. 3. Plot some additional points, if necessary, and draw the graph.

To graph y 5 x3 2 4x, we will find the x- and y-intercepts, test for symmetries, plot points, and join the points with a smooth curve.

Intercepts of a graph: To find the x-intercepts, let y 5 0 and solve for x. To find the y-intercepts, let x 5 0 and solve for y.

Step 1: Find the x- and y-intercepts. To find the x-intercepts, we let y 5 0 and solve for x. y 5 x3 2 4x 0 5 x3 2 4x

Substitute 0 for y.

0 5 x 1x2 2 42

Factor out x.

0 5 x 1x 1 22 1x 2 22

Factor x2 2 4.

x 5 0 or x 1 2 5 0

or x 2 2 5 0 x52

x 5 22

Set each factor equal to 0.

The x-intercepts are 10, 02 , 122, 02 , and 12, 02 . To find the y-intercepts, we let x 5 0 and solve for y. y 5 x3 2 4x y 5 03 2 4 102

Substitute 0 in for x.

y50

The y-intercept is 10, 02 . Test for x-axis symmetry: To test for x-axis symmetry, replace y with 2y. If the resulting equation is equivalent to the original one, the graph is symmetric about the x-axis.

Step 2: Test for symmetries. We test for symmetry about the x-axis by replacing y with 2y. (1)

y 5 x3 2 4x 2y 5 x3 2 4x

(2)

Test for y-axis symmetry: To test for y-axis symmetry, replace x with 2x. If the resulting equation is equivalent to the original one, the graph is symmetric about the y-axis.

y 5 2x3 1 4x

This is the original equation. Replace y with 2y. Multiply both sides by 21.

Since Equations 1 and 2 are different, the graph is not symmetric about the x-axis. To test for y-axis symmetry, we replace x with 2x. (1) y 5 x3 2 4x

(3)

This is the original equation.

y 5 12x2 2 4 12x2

Replace x with 2x.

y 5 2x3 1 4x

Simplify.

3

Since Equations 1 and 3 are different, the graph is not symmetric about the y-axis.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

270

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Definitions and Concepts

Examples

Test for origin symmetry: To test for symmetry about the origin, replace x with –x and y with 2y. If the resulting equation is equivalent to the original one, the graph is symmetric about the origin.

To test for symmetry about the origin, we replace x with –x and y with –y. (1)

y 5 x3 2 4x

This is the original equation.

2y 5 12x2 3 2 4 12x2

Replace x with 2x and y with 2y.

3

(4)

2y 5 2x 1 4x

Simplify.

y 5 x3 2 4x

Multiply both sides by 21.

Since Equations 1 and 4 are the same, the graph is symmetric about the origin. Plot some additional points and draw the graph.

Step 3: Graph the equation. To graph the equation, we plot the x- and y-intercepts and several other pairs 1x, y2 with positive values of x. We can use the property of symmetry about the origin to draw the graph for negative values of x. y

y 5 x3 2 4x

x

y = x3 – 4x

x

y

22

0

0

0

1

23

2

0

3

15

1x, y2

122, 02 10, 02

11, 232 12, 02

13, 152

Circles: A circle is the set of all points in a plane that are a fixed distance from a point called its center. The fixed distance is the radius of the circle. The standard equation of a circle with center 1h, k2 and radius r: The graph of any equation that can be written in the form 1x 2 h2 2 1 1y 2 k2 2 5 r2

is a circle with radius r and center at point 1h, k2 .

The standard equation of a circle with center 10, 02 and radius r: The graph of any equation that can be written in the form

Find the center and radius of the circle with the equation 1x 2 62 2 1 1y 1 52 2 5 4. Standard Form: Given Form:

1x 2 h2 2 1 1y 2 k2 2 5 r2

1x 2 62 2 1 3 y 2 1252 4 2 5 22

We see that h 5 6, k 5 25, and r 5 2. The center 1h, k2 of the circle is at (6, 25) and the radius is 2.

x2 1 y2 5 r2 is a circle with radius r and center at the origin.

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271

Chapter Review

Definitions and Concepts

Examples

The general form of an equation of a circle: The general form of an equation of a circle is

To convert the general form of the equation of the circle x2 1 y2 1 4x 2 2y 2 20 5 0 into standard form, we must find the coordinates of the center and the radius. To do so, we will complete the square on both x and y:

x2 1 y2 1 cd 1 dy 1 e 5 0 where c, d, and e are real numbers.

x2 1 y2 1 4x 2 2y 5 20 x2 1 4x 1 y2 2 2y 5 20 x2 1 4x 1 4 1 y2 2 2y 1 1 5 20 1 4 1 1

1x 1 22 2 1 1y 2 12 2 5 25

EXERCISES Find the x- and y-intercepts of each graph. Do not graph the equation. 65. y 5 4x 2 8x2

Add 4 and 1 to both sides to complete the square.

Factor x2 2 4x 1 4 and y2 2 2y 1 1.

76. y 5 0 x 1 1 0 1 2

75. y 5 "x 1 2

y

y

66. y 5 x2 2 10x 2 24 x x

Find the symmetries, if any, of the graph of each equation. Do not graph the equation. 67. y2 5 8x

68. y 5 3y4 1 6

69. y 5 22 0 x 0

Use a graphing calculator to graph each equation.

70. y 5 0 x 1 2 0

Graph each equation. Find all intercepts and symmetries. 71. y 5 x2 1 2

72. y 5 x3 2 2

y

y

77. y 5 0 x 2 4 0 1 2

78. y 5 2"x 1 2 1 3

79. y 5 x 1 2 0 x 0

80. y2 5 x 2 3

x

Identify the center and radius of each circle written in standard form. 82. x2 1 1y 2 62 2 5 100 81. x2 1 y2 5 64

x

73. y 5

1 0x0 2

83. 1x 1 72 2 1 y2 5

74. y 5 2"x 2 4 y

1 4

84. 1x 2 52 2 1 1y 1 12 2 5 9

y

Write an equation of each circle in standard form. x

x

85. Center at (0, 0); r 5 7

86. Center at 13,02 ; r 5

1 5

2 87. Center at 122, 122 , r 5 5 88. Center at a , 5b; r 5 9 7

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272

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Write an equation of each circle in standard form and general form. 89. Center at 123, 42 ; radius 12

95. x2 1 y2 2 2y 5 15

96. x2 1 y2 2 4x 1 2y 5 4

y

90. Ends of diameter at 126, 232 and 15, 82

y

x x

Convert the general form of each circle given into standard form. 91. x2 1 y2 1 6x 2 4y 1 4 5 0

Use a graphing calculator to solve each equation. If an answer is not exact, round to the nearest hundredth. 97. x2 2 11 5 0 98. x3 2 x 5 0 2 100. x2 2 3x 5 5 99. 0 x 2 2 0 2 1 5 0

92. 2x2 1 2y2 2 8x 2 16y 2 10 5 0

Graph each circle. 93. x2 1 y2 2 16 5 0

94. x2 1 y2 2 4x 5 5

y

y

x

SECTION 2.5

x

Proportion and Variation

Definitions and Concepts

Examples

Proportion: An equation indicating that two ratios are equal is called a proportion. In the proportion ab 5 dc , the numbers a and d are called the extremes, and the numbers b and c are called the means.

9 , 5 and 9 are called the means, In the proportion 35 5 15 and 3 and 15 are called the extremes.

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Chapter Review

273

Definitions and Concepts

Examples

Property of proportions: In any proportion, the product of the extremes is equal to the product of the means.

In the proportion on the previous page, the product of the means is equal to the product of the extremes:

Use the property of proportions to solve a proportion for a variable.

5 192 5 3 1152 5 45

x 2 5 for x. 6 x 1 11 We will use the property of proportions to solve the proportion. Solve the proportion:

x 2 5 6 x 1 11

x 1x 1 112 5 6 ? 2

The product of the extremes equals the product of the means.

2

x 1 11x 5 12

Remove parentheses and simplify.

x2 1 11x 2 12 5 0

Subtract 12 from both sides.

1x 1 122 1x 2 12 5 0 x 1 12 5 0

or

x 5 212

Factor the trinomial.

x2150

Set each factor equal to 0.

x51

Thus, x 5 212 or x 5 1. Direct variation: The words “y varies directly with x,” or “y is directly proportional to x,” mean that y 5 kx for some realnumber constant k. The number k is called the constant of proportionality. Inverse variation: The words “y varies inversely with x,” or “y is inversely proportional to x,” mean that y 5 xk for some real-number constant k.

In the direct variation formula, y 5 3x, find y when x 5 5. y 5 3x 5 3 152 5 15

In the inverse variation formula y 5 xk , find the constant of variation if y 5 5 when x 5 20. k x k 55 20 100 5 k y5

Substitute 5 for y and 20 for x.

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274

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Definitions and Concepts

Examples

Joint variation: The words “y varies jointly with w and x” mean that y 5 kwx for some real-number constant k.

Kinetic energy of an object varies jointly with its mass and the square of its velocity. A 50-gram mass moving at the rate of 20 centimeters per second has a kinetic energy of 40,000 dyne-centimeters. Find the kinetic energy of a 10-gram mass that is moving at 60 centimeters per second. We will use joint variation to solve the problem. If we let E, m, and v represent the kinetic energy, mass, and velocity, respectively, the phrase energy varies jointly with the mass and the square of its velocity translates into the formula E 5 kmv2. The constant k can be evaluated by substituting 40,000 for E, 50 for m, and 20 for v in the formula E 5 kmv2 40,000 5 k 1502 1202 2 40,000 5 20,000k k52 We can now substitute 2 for k, 10 for m, and 60 for v in the formula and evaluate E. E 5 kmv2 5 2 1102 1602 2 5 72,000 A 10-gram mass that is moving at 60 centimeters per second has a kinetic energy of 72,000 dyne-centimeters.

EXERCISES Solve each proportion. x13 x21 101. 5 10 x

102.

x21 12 5 2 x11

103. Hooke’s Law The force required to stretch a spring is directly proportional to the amount of stretch. If a 3-pound force stretches a spring 5 inches, what force would stretch the spring 3 inches? 104. Kinetic energy A moving body has a kinetic energy directly proportional to the square of its velocity. By what factor does the kinetic energy of an automobile increase if its speed increases from 30 mph to 50 mph?

105. Gas laws The volume of gas in a balloon varies directly as the temperature and inversely as the pressure. If the volume is 400 cubic centimeters when the temperature is 300 K and the pressure is 25 dynes per square centimeter, find the volume when the temperature is 200 K and the pressure is 20 dynes per square centimeter. 106. The area of a rectangle varies jointly with its length and width. Find the constant of proportionality. 107. Electrical resistance The resistance of a wire varies directly as the length of the wire and inversely as the square of its diameter. A 1,000-foot length of wire, 0.05 inch in diameter, has a resistance of 200 ohms. What would be the resistance of a 1,500-foot length of wire that is 0.08 inch in diameter?

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Chapter Test

108. Billing for services Angie’s Painting and Decorating Service charges a fixed amount for accepting a wallpapering job and adds a fixed dollar amount for each roll hung. If the company bills a customer $177 to hang 11 rolls and $294 to hang 20 rolls, find the cost to hang 27 rolls.

275

between x and y, and graph the equation. If Rolf tutors algebra for 180 hours, how long must he tutor Spanish? y

109. Paying for college Rolf must earn $5,040 for next semester’s tuition. Assume he works x hours tutoring algebra at $14 per hour and y hours tutoring Spanish at $18 per hour and makes his goal. Write an equation expressing the relationship

x

CHAPTER TEST Indicate the quadrant in which the point lies or the axis on which it lies. 2. (0, 28) 1. 123, p2

Find the distance between points P and Q. 10. P 10, p2 ; Q 12p, 02 9. P 11, 212 ; Q 123, 42

Find the x- and y-intercepts and use them to graph the equation. 3. x 1 3y 5 6 4. 2x 2 5y 5 10

Find the midpoint of the line segment PQ. 11. P 13, 272 ; Q 123, 72

12. PQ0, "2R; QQ"8, "18R

y

y

Find the slope of the line PQ. x

x

13. P 13, 292 ; Q 125, 12 14. PQ"3, 3R; QQ2"12, 0R

Graph each equation. 5. 2 1x 1 y2 5 3x 1 5

6. 3x 2 5y 5 3 1x 2 52

y

16. 2x 2 3y 5 5; 3x 1 2y 5 7

y

x

7.

Determine whether the two lines are parallel, perpendicular, or neither. 15. y 5 3x 2 2; y 5 2x 2 3

1 1x 2 2y2 5 y 2 1 2

x

8.

x1y25 5 3x 7

20. Perpendicular to 2x 2 y 5 3; b 5 5

y

y

Write an equation of the line with the given properties. Your answers should be written in slope-intercept form, if possible. 17. Passing through 13,252 ; m 5 2 1 18. m 5 3; b 5 2 19. Parallel to 2x 2 y 5 3; b 5 5

3 1 21. Passing through a2, 2 b and a3, b 2 2 22. Parallel to the y-axis and passing through 13, 242

x x

Find the x- and y-intercepts of each graph. 23. y 5 x3 2 16x 24. y 5 0 x 2 4 0

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276

Chapter 2

The Rectangular Coordinate System and Graphs of Equations

Find the symmetries of each graph. 2

Graph each equation. 33. x2 1 y2 5 9

4

25. y 5 x 2 1

26. y 5 x 1 1

34. x2 2 4x 1 y2 1 3 5 0

y

Graph each equation. Find all intercepts and symmetries. 28. x 5 0 y 0 27. y 5 x2 2 9 y

y 2

x

y

2

x

x

x

Write each statement as an equation. 35. y varies directly as the square of z. 36. w varies jointly with r and the square of s. 37. P varies directly with Q. P 5 7 when Q 5 2. Find P when Q 5 5.

30. x 5 y3

29. y 5 2"x

y

y

x x

Write an equation of each circle in standard form.

38. y is directly proportional to x and inversely proportional to the square of z, and y 5 16 when x 5 3 and z 5 2. Find x when y 5 2 and z 5 3. Use a graphing calculator to find the positive root of each equation. Round to two decimal places. 39. x2 2 7 5 0

40. x2 2 5x 2 5 5 0

31. Center at (5, 7); radius of 8 32. Center at (2, 4); passing through (6, 8)

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3

Functions

CAREERS AND MATHEMATICS:

Computer Scientist

Computer scientists are highly trained innovative workers who design and invent new technology. They solve complex business, scientific, and general computing problems. Computer scientists conduct research on a variety of topics, including computer hardware, virtual reality, and robotics.

Education and Mathematics Required

Benis Arapovic/Shutterstock.com

•

•

Most are required to possess a Ph.D. in computer science, computer engineering, or a closely related discipline. An aptitude for math is important. College Algebra, Trigonometry, Calculus, Linear Algebra, Ordinary Differential Equations, Theory of Analysis, Abstract Algebra, Graph Theory, Numerical Methods, and Combinatorics are math courses required.

3.1

Functions and Function Notation

3.2

Quadratic Functions

3.3

Polynomial and Other Functions

3.4

Transformations of the Graphs of Functions

3.5

Rational Functions

3.6

Operations on Functions

3.7

Inverse Functions Chapter Review Chapter Test Cumulative Review Exercises

How Computer Scientists Use Math and Who Employs Them •

•

Computer scientists use mathematics as they span a range of topics from theoretical studies of algorithms to the computation of implementing computing systems in hardware and software. Many computer scientists are employed by Internet service providers; Web search portals; and data processing, hosting, and related services firms. Others work for government, manufacturers of computer and electronic products, insurance companies, financial institutions, and universities.

Career Outlook and Earnings • •

Employment of computer scientists is expected to grow by 24 percent through 2018, which is much faster than the average for all occupations. The median annual wages of computer and information scientists is approximately $98,000. Some earn more than $150,000 a year.

For more information see: www.bls.gov/oco

In this chapter, we will discuss one of the most important concepts in mathematics—the concept of a function.

277 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

278

Chapter 3

Functions

3.1 Functions and Function Notation In this section, we will learn to 1. Understand the concept of a function. 2. Determine whether an equation represents a function. 3. Find the domain of a function. 4. 5. 6. 7.

Evaluate a function. Evaluate the difference quotient for a function. Graph a function by plotting points. Use the Vertical Line Test to identify functions.

8. Use linear functions to model applications. Correspondences between the elements of two sets is a common occurrence in everyday life. For example,

© Photos 12/Alamy

• • • •

To every Motorola cell phone, there corresponds exactly one phone number. To every Honda Civic car, there corresponds exactly one vehicle identification number. To every case on the television show “Deal or No Deal,” there corresponds exactly one amount of money. To every item’s barcode at the Target store, there corresponds exactly one price.

This table shows the four highest-grossing movies of all time and the year each movie was released.

Movie

Year

Avatar

2009

The Lord of the Rings: The Return of the King

2003

Pirates of the Caribbean: Dead Man’s Chest

2006

The Dark Knight

2008

The information shown in the table sets up a correspondence between a movie and the year it was released. Note that for each of the movies, there corresponds exactly one year in which it was released. The phase there corresponds exactly one is extremely important in mathematics, and we will use this idea to solve problems.

1. Understand the Concept of a Function Correspondences in which exactly one quantity corresponds to (or depends on) another quantity according to some specific rule are called functions. Equations are frequently used in mathematics to represent functions. For example, the equation y 5 x2 2 1 sets up a correspondence between two infinite sets of real numbers, x and y, according to the rule square x and subtract 1. Equation: y 5 x2 2 1 Correspondence: Each real number x determines exactly one real number y. Rule: Square x and subtract 1. Since the value of y depends on the number x, we call y the dependent variable and x the independent variable. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.1

279

Functions and Function Notation

The equation y 5 x2 2 1 determines what output value y will result from each input value x. This idea of inputs and outputs is shown in Figure 3-1(a). In the equation y 5 x2 2 1, if the input x is 2, the output y is y 5 22 2 1 5 3 This is illustrated in Figure 3-1(b). Input 2

y = x2 – 1 3

Rule of correspondence

Output (b)

(a) FIGURE 3-1

Comment Correspondences can be set up by a verbal expression, by a table, by a set of ordered pairs, by an equation, and by a graph.

The equation y 5 x2 2 1 also determines the table of ordered pairs, and the graph shown in Figure 3-2. To see how the table determines the correspondence, we find an input in the x-column and read across to find the corresponding output in the y-column. If we select x 5 2 as an input, we get y 5 3 for the output. To see how the graph of y 5 x2 2 1 determines the correspondence, we draw a vertical and horizontal line through any point (say, point P) on the graph shown in Figure 3-2. Because these lines intersect the x-axis at 2 and the y-axis at 3, the point P 12, 32 associates 3 on the y-axis with 2 on the x-axis. y 5 x2 2 1

y y = x2 – 1 P

3

2

x

1x, y2

x

y

22

3

(22, 3)

21

0

(21, 0)

0

21

(0, 21)

1

0

(1, 0)

2

3

(2, 3)

FIGURE 3-2

Any correspondence that assigns exactly one value of y to each number x is called a function. We refer to the set of inputs as the domain of the function and the set of outputs as the range of the function.

Function

A function f is a correspondence between a set of input values x and a set of output values y, where to each x-value there corresponds exactly one y-value.

Domain

The set of input values x is called the domain of the function.

Range

The set of output values y is called the range of the function.

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280

Chapter 3

Functions

2. Determine Whether an Equation Represents a Function EXAMPLE 1

SOLUTION

Determining Whether an Equation Represents a Function Determine whether the following equations define y to be a function of x. a. 7x 1 y 5 5 b. y = 0 x 0 1 1 c. y2 5 x 1 2

In each case, we will solve each equation for y, if necessary, and determine whether each input value x determines exactly one output value y. a. First, we solve 7x 1 y 5 5 for y to get y 5 5 2 7x From this equation we see that for each input value x there corresponds exactly one output value y. For example, if x 5 3, then y 5 5 2 7 132 5 216. The equation defines y to be a function of x. b. Since each input number x that we substitute for x determines exactly one output y, the equation y 5 0 x 0 1 1 defines y to be a function of x.

Caution Some equations represent functions and some do not. If to some input x in an equation there corresponds more than one output y, the equation will not represent a function.

Self Check 1

c. First, we solve y2 5 x 1 2 for y using the Square Root Property covered in Section 1.3 to get y 5 6"x 1 2. Note that each input value x for which the function is defined (except 22) gives two outputs y. For example, if x 5 7, then y 5 6"7 1 2 5 6"9 5 63. For this reason, the equation does not define y to be a function of x.

Determine whether each equation defines y to be a function of x. a. y 5 0 x 0 b. y 5 "x c. y2 5 2x Now Try Exercise 17.

3. Find the Domain of a Function The domain of a function is the set of all real numbers x for which the function is defined. Thus, to find the domain of a function we must find the set of numbers that are permissible inputs for x. It is often helpful to determine any restrictions on the input values for x. For example, we consider the information in the following table.

Equation 3

Restriction 2

y5x 2x 1x25 y 5 "x y5

1 x

There are no restrictions on x because any real number can be cubed, squared and combined with constants. The restriction that x cannot be a negative real number is placed on x because the square root of a negative number is an imaginary number. The restriction that x cannot be 0 is placed on x because division by 0 isn’t defined.

EXAMPLE 2

Domain

all real numbers, 12`, ` 2

x $ 0, the interval 3 0, ` 2 x 2 0, 12`, 02 c 10, ` 2

Finding the Domain of a Function Find the domain of the function defined by each equation: 3 a. y 5 2x 2 5 b. y 5 "3x 2 2 c. y 5 x12

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Section 3.1

SOLUTION

Functions and Function Notation

281

We must determine what numbers are permissible inputs for x. This set of numbers is the domain. a. Any real number that we input for x can be multiplied by 2 and then 5 can be subtracted from the result. Thus, the domain is the interval 12`, ` 2 .

b. Since the radicand must be nonnegative, we have 3x 2 2 $ 0 3x $ 2 x$

Add 2 to both sides.

2 3

Divide both sides by 3.

Thus, the domain is the interval S23, `R.

c. Since the fraction x 13 2 is undefined when x 5 22, 22 is not a permissible input value. Since all other values of x are permissible inputs, the domain is 12`, 222 c 122, ` 2 . Self Check 2

Find the domain of each function: a. y 5 0 x 0 1 2 Now Try Exercise 27.

EXAMPLE 3

3 b. y 5 " x12

Finding the Domain of a Function Find the domain of the function defined by the equation y 5

SOLUTION

1 . x 2 5x 2 6 2

We can factor the denominator to see what values of x will give 0’s in the denominator. These values are not in the domain. x2 2 5x 2 6 5 0

1x 2 62 1x 1 12 5 0

x 2 6 5 0 or x 1 1 5 0 x56

x 5 21

The domain is 12`, 212 c 121, 62 c 16, ` 2 . Self Check 3

Find the domain of the function defined by the equation y 5

2 . x 2 16 2

Now Try Exercise 41.

4. Evaluate a Function To indicate that y is a function of x, we often use function notation and write y 5 f 1x2

Read as “y is a function of x.”

The notation y 5 f 1x2 provides a way of denoting the value of y (the dependent variable) that corresponds to some input number x (the independent variable). For example, if y 5 f 1x2 , the value of y that is determined when x 5 2 is denoted by f 122 , read as “f of 2.” If f 1x2 5 5 2 7x, we can evaluate f 122 by substituting 2 for x. f 1x2 5 5 2 7x

f 122 5 5 2 7 122

Substitute the input 2 for x.

5 29

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282

Chapter 3

Functions

If x 5 2, then y 5 f 122 5 29. To evaluate f 1252 , we substitute 25 for x. f 1x2 5 5 2 7x

f 1252 5 5 2 7 1252

Substitute the input 25 for x.

5 40

If x 5 25, then y 5 f 1252 5 40.

Comment

To see why function notation is helpful, consider the following sentences. Note that the second sentence is much more concise. 1. In the function y 5 3x2 1 x 2 4, find the value of y when x 5 23.

2. In the function f 1x2 5 3x2 1 x 2 4, find f 1232 .

In this context, the notations y and f 1x2 both represent the output of a function and can be used interchangeably, but function notation is more concise.

Sometimes functions are denoted by letters other than f. The notations y 5 g 1x2 and y 5 h 1x2 also denote functions involving the independent variable x.

EXAMPLE 4 SOLUTION

Evaluating a Function

Let g 1x2 5 3x2 1 x 2 4. Find a. g 1232

b. g 1k2

c. g 12t32

d. g 1k 1 12

In each case, we will substitute the input value into the function and simplify. a.

g 1x2 5 3x2 1 x 2 4

g 1232 5 3 1232 2 1 1232 2 4 5 3 192 2 3 2 4

b. g 1x2 5 3x2 1 x 2 4

g 1k2 5 3k2 1 k 2 4

5 20

c.

g 1x2 5 3x2 1 x 2 4

g 12t32 5 3 12t32 2 1 12t32 2 4 5 3t6 2 t3 2 4

d.

g 1x2 5 3x2 1 x 2 4

g 1k 1 12 5 3 1k 1 12 2 1 1k 1 12 2 4

5 3 1k2 1 2k 1 12 1 k 1 1 2 4 5 3k2 1 6k 1 3 1 k 1 1 2 4 5 3k2 1 7k

Self Check 4

Evaluate: a. g 102

Now Try Exercise 49.

ACCENT ON TECHNOLOGy

b. g 122

c. g 1k 2 12

Evaluating a Function Functions can be easily evaluated on a graphing calculator. There are several ways to do this, but one of the easiest is to use the graph editor and the table. Press WINDOW and use Y= and input 3x2 1 x 2 4 in the graph editor. Next, press GRAPH and enter values the table setup shown on the next page. Then press for x. The function values will appear in the table.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.1

(a)

Functions and Function Notation

(b)

283

(c)

FIGURE 3-3

5. Evaluate the Difference Quotient for a Function The fraction

f 1x 1 h2 2 f 1x2 h

is called the difference quotient and is important in cal-

culus. The difference quotient can be used to find quantities such as the velocity of a guided missile or the rate of change of a company’s profit.

EXAMPLE 5

SOLUTION

Finding the Difference Quotient

If f 1x2 5 x2 2 2x 2 5, evaluate

f 1x 1 h2 2 f 1x2 . h

We will evaluate the difference quotient in three steps. Find f 1x 1 h2 . Then subtract f 1x2 . Then divide by h.

Step 1: Find f 1x 1 h2 .

f 1x2 5 x2 2 2x 2 5

f 1x 1 h2 5 1x 1 h2 2 2 2 1x 1 h2 2 5 2

Substitute x 1 h for x.

2

5 x 1 2xh 1 h 2 2x 2 2h 2 5

Step 2: Find f 1x 1 h2 2 f 1x2 . We can use the result from Step 1.

f 1x 1 h2 2 f 1x2 5 x2 1 2xh 1 h2 2 2x 2 2h 2 5 2 1x2 2 2x 2 52 5 x2 1 2xh 1 h2 2 2x 2 2h 2 5 2 x2 1 2x 1 5

Remove parentheses.

5 2xh 1 h2 2 2h

Combine like terms.

Step 3: Find the difference quotient

Comment After the completion of Step 2, in a polynomial function, when f 1x 1 h2 2 f 1x2 is simplified, each term will always include an h. Use that fact to check your work as you progress through the problem.

Self Check 5

Subtract f 1x2 .

f 1x 1 h2 2 f 1x2 . We can use the result from Step 2. h

f 1x 1 h2 2 f 1x2 2xh 1 h2 2 2h 5 h h h 12x 1 h 2 22 5 h 5 2x 1 h 2 2

If f 1x2 5 x2 1 2, evaluate

f 1x 1 h2 2 f 1x2 . h

Divide both sides by h.

In the numerator, factor out h. Divide out h:

h 5 1. h

Now Try Exercise 65. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

284

Chapter 3

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6. Graph a Function by Plotting Points If f is a function whose domain and range are sets of real numbers, its graph is the set of all points 1x, f 1x2 2 in the xy-plane that satisfy the equation y 5 f 1x2 . For example, the graph of the function y 5 f 1x2 5 27x 1 5 is a line with slope 27 and y-intercept (0, 5). (See Figure 3-4.) If the graph of a function is a nonvertical line, the function is called a linear function. y 5

y = f(x) = – 7x + 5 x FIGURE 3-4

Graph of a Function

EXAMPLE 6 SOLUTION

The graph of a function f in the xy-plane is the set of all points 1x, y2 where x is in the domain of f, y is in the range of f, and y 5 f 1x2 .

Graphing a Function by Plotting Points

Graph the functions. a. f 1x2 5 22 0 x 0 1 3

b. f 1x2 5 "x 2 2

In each case, we will make a table of solutions and plot the points given by the table. Then we will connect the points by drawing a smooth curve through them and obtain the graph of the function. a. f 1x2 5 22 0 x 0 1 3 y

x f(x) = – 2|x| + 3

x

22

21

21

1

0

3

1

1

2

21

FIGURE 3-6

122, 212 121, 12 10, 32 11, 12

12, 212

f 1x2 5 "x 2 2

1x, f 1x22

6

2

11

3

18

4

111, 32

2 x

1x, f 1x22

f 1x2

x

y = f(x) = x – 2

Self Check 6

f 1x2

FIGURE 3-5

b. f 1x2 5 "x 2 2

y

f 1x2 5 22 0 x 0 1 3

0

(2, 0)

(6, 2)

118, 42

Graph: f 1x2 5 0 x 1 3 0 . Now Try Exercise 83.

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Section 3.1

285

Functions and Function Notation

Both the domain and the range of a function can be identified by viewing the graph of the function. The inputs or x-values that correspond to points on the graph of the function can be identified on the x-axis and used to state the domain of the function. The outputs or f 1x2 values that correspond to points on the graph of the function can be identified on the y-axis and used to state the range of the function. (See Figure 3-7).

y

y

Range of the function x

x

Domain of the function (a)

(b) FIGURE 3-7

From Figure 3-6, we can see that the domain of the function f 1x2 5 22 0 x 0 1 3 is the set of all real numbers, and that the range is the set of real numbers that are less than or equal to 3. From Figure 3-6, we can see that the domain of the function f 1x2 5 "x 2 2 is the set of all real numbers greater than or equal to 2, and that the range is the set of all real numbers greater than or equal to 0. In Figure 3-8, we see the graphs of several basic functions.

y

y

y

f(x) = x x

x f (x) =

x2

Domain: (– ∞, ∞) Range: [0, ∞) Squaring function (b)

Domain: (– ∞, ∞) Range: (– ∞, ∞) Identity function (a)

Domain: (– ∞, ∞) Range: (–∞, ∞) Cubing function (c)

y

y

f (x) = x 3

x

y

f(x) = |x| f(x) = x x 3

f(x) = x x Domain: (–∞, ∞) Range: [0, ∞) Absolute value function (d)

x Domain: [0, ∞) Range: [0, ∞) Square root function (e)

Domain: (–∞, ∞) Range: (–∞, ∞) Cube root function (f )

FIGURE 3-8 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

286

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ACCENT ON TECHNOLOGy

Graphing the Absolute Value Function To graph the absolute value function on a graphing calculator, we must call up the function abs( into the graphing window. We press MATH , then scroll right to NUM as shown below in Figure 3-9. Press

Y=

ZOOM

. Go to MATH NUM and press as the window.

. Input

and

GRAPH

. Use

FIGURE 3-9

ACCENT ON TECHNOLOGy

Graphing the Cube Root Function We can graph this function on a graphing calculator using the cube root key by selecting MATH from the graph window as shown in Figure 3-10. Input and GRAPH . Use ZOOM as the window.

FIGURE 3-10

7. Use the Vertical Line Test to Identify Functions We can use a Vertical Line Test to determine whether a graph represents a function. Vertical Line Test

• If every vertical line that can be drawn intersects the graph in no more than one point, the graph represents a function. See Figure 3-11(a). • If a vertical line can be drawn that intersects the graph at more than one point, the graph does not represent a function. See Figure 3-11(b). y

y

One y

One intersection

Three y's

Three intersections

x

x One x

One x A function

Not a function

(a)

(b) FIGURE 3-11

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Section 3.1

EXAMPLE 7

Functions and Function Notation

287

Using the Vertical Line Test to Identify Functions Determine which of the following graphs represent functions.

x

x

(a)

SOLUTION

y

y

y

(b)

x (c)

We will use the Vertical Line Test by drawing several vertical lines through each graph. If every vertical line that intersects the graph does so exactly once, the graph represents a function. Otherwise, the graph does not represent a function. a. This graph fails the Vertical Line Test, so it does not represent a function. y Two points x

b. This graph passes the Vertical Line Test, so it does represent a function. y One point x

c. This graph fails the Vertical Line Test, so it does not represent a function. y

Two points x

Self Check 7

Does the graph shown in Figure 3-2 represent a function? Now Try Exercise 97.

Comment Some graphs represent functions and some do not. Graphs that pass the Vertical Line Test are functions.

Not all equations define functions. For example, the equation x 5 0 y 0 does not define a function, because two values of y can correspond to one number x. For example, if x 5 2, then y can be either 2 or 22. The graph of the equation is shown in Figure 3-12. Since the graph does not pass the Vertical Line Test, it does not represent a function.

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288

Chapter 3

Functions

y

x = |y|

x 5 0y0

x

x

y

1x, y2

2

22

(2, 22)

1

21

(1, 21)

0

0

(0, 0)

1

1

(1, 1)

2

2

(2, 2)

FIGURE 3-12

Correspondences between a set of input values x (called the domain) and a set of output values y (called the range), where to each x-value in the domain there corresponds one or more y-values in the range, are called relations. Although the graph in Figure 3-12 does not represent a function, it does represent a relation. We note that all functions are relations, but not all relations are functions. Another way to visualize the definition of function is to consider the diagram shown in Figure 3-13(a). The function f that assigns the element y to the element x is represented by an arrow leaving x and pointing to y. The set of elements in X from which arrows originate is the domain of the function. The set of elements in Y to which arrows point is the range. To constitute a function, each element of the domain must determine exactly one y-value in the range. However, the same value of y could correspond to several numbers x. In the function shown in Figure 3-13(b), the single value y corresponds to the three numbers x1, x2, and x3 in the domain. The correspondence shown in Figure 3-13(c) is not a function, because two values of y correspond to the same number x. However, it is still a relation. X

Y ƒ

x1

X y1

x2

x1 x2

y2

x3

y

x3

y3

Domain of ƒ

Y

Range of ƒ

Domain of

Range of

(a)

(b) X

Y y1

x y2

(c) FIGURE 3-13

8. Use Linear Functions to Model Applications We have seen that the equation of a nonvertical line defines a linear function—an important function in mathematics and its applications. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.1

Linear Functions

EXAMPLE 8

Functions and Function Notation

289

A linear function is a function determined by an equation of the form f 1x2 5 mx 1 b or

y 5 mx 1 b

Using a Linear Function to Model an Application Cost of a Fraternity Dance The cost associated with a fraternity dance is $300 for Country Club rental and $24 for each couple that attends. a. Write the cost C of the dance in terms of the number of couples x attending. b. Find the cost if 45 couples attend the dance.

SOLUTION

Use the information stated in the problem to write a linear function that models the application. a. The cost C for the dance is $24 per couple plus the $300 rental fee. If x couples attend, the cost is $24x plus 300. Therefore, the linear cost function is C 1x2 5 24x 1 300

b. To find the cost when 45 couples attend, we find C(45). C 1x2 5 24x 1 300

C 1452 5 24 1452 1 300

Substitute 45 for x.

5 1,080 1 300 5 1,380

If 45 couples attend, the cost of the dance is $1,380. Self Check 8

Find the cost when 60 couples attend. Now Try Exercise 105.

Nicholas Piccillo/Shutterstock.com

EXAMPLE 9

Writing a Linear Function that Models an Application Heart Rates The target heart rate at which a person should train to get an effective workout is a linear function of his age. For a 20-year-old starting an exercise program, the target heart rate should be 120 beats per minute. For a 40-year-old, it should be 108 beats per minute. Express the target heart rate R as a function of age A.

SOLUTION

We will use the information stated in the problem to find the slope of the linear function and write a linear function that models the application. Since the target heart rate (R) is given to be a linear function of age (A), there are constants m and b such that R 5 mA 1 b

Since R 5 120 when A 5 20, the point 1A1, R12 5 120, 1202 lies on the straight-line graph of this function. Since R 5 108 when A 5 40, the point 1A2, R22 5 140, 1082 also lies on that line. The slope of the line is R2 2 R1 A2 2 A1 108 2 120 5 40 2 20 12 52 20 3 52 5 5 20.6

m5

Substitute 108 for R2, 120 for R1, 40 for A2, and 20 for A1.

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290

Chapter 3

Functions

Thus, m 5 20.6. To determine b, we can substitute 20.6 for m and the coordinates of one point, say P(20, 120), in the equation R 5 mA 1 b and solve for b. R 5 mA 1 b

120 5 20.6 1202 1 b

Substitute 120 for R, 20 for b, and 20.6 for m.

132 5 b

Add 12 to both sides.

120 5 212 1 b

If we substitute 20.6 for m and 132 for b in R 5 mA 1 b, we obtain R 5 20.6A 1 132. Self Check 9

Find the target heart rate for a 30-year-old person. Now Try Exercise 111.

Self Check Answers

1. b. c. 8.

a. a function b. a function c. not a function 2. a. 12`, ` 2 12`, ` 2 3. 12`, 242 c 124, 42 c 14,` 2 4. a. 24 b. 10 y 3k2 2 5k 2 2 5. 2x 1 h 6. 7. yes $1,740 9. 114 beats per minute x f(x) = |x + 3|

Exercises 3.1 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. A correspondence that assigns exactly one value of y to any number x is called a . 2. A correspondence that assigns one or more values of y to any number x is called a . 3. The set of input numbers x in a function is called the of the function. 4. The set of all output values y in a function is called the of the function. 5. The statement “y is a function of x” can be written . as the equation 6. The graph of a function y 5 f 1x2 in the xy-plane is the set of all points that satisfy the equation, where x is in the of f and y is in the of f. 7. In the function of Exercise 5, is called the independent variable. 8. In the function of Exercise 5, y is called the variable. 9. If every line that intersects a graph does so , the graph represents a function.

10. A function that can be written in the form y 5 mx 1 b is called a function.

Practice Assume that all variables represent real numbers. Determine whether each equation determines y to be a function of x. 11. y 5 x 12. y 2 2x 5 0 2 14. 0 y 0 5 x 13. y 5 x 15. y 5 x2 2

17. y 2 4x 5 1 19. 0 x 0 5 0 y 0 21. y 5 7

16. y 2 7 5 7 18. 0 x 2 2 0 5 y 20. x 5 7 22. 0 x 1 y 0 5 7

Let the function f be defined by the equation y 5 f 1x2 , where x and f 1x2 are real numbers. Find the domain of each function. 24. f 1x2 5 25x 1 2 23. f 1x2 5 3x 1 5 25. f 1x2 5 x2 2 x 1 1

26. f 1x2 5 x3 2 3x 1 2

27. f 1x2 5 "x 2 2

28. f 1x2 5 "2x 1 3

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Section 3.1

29. f 1x2 5 "4 2 x

30. f 1x2 5 3"2 2 x

3 33. f 1x2 5 " x11

3 34. f 1x2 5 " 52x

31. f 1x2 5 "x2 2 1 3 35. f 1x2 5 x11 x 37. f 1x2 5 x23 39. f 1x2 5 40. f 1x2 5 41. f 1x2 5

42. f 1x2 5

32. f 1x2 5 "x2 2 2x 2 3

27 36. f 1x2 5 x13

x12 38. f 1x2 5 x21

44. f 1x2 5 5x 1 7 1 45. f 1x2 5 x 1 3 2 2 46. f 1x2 5 x 1 5 3 47. f 1x2 5 x2 48. f 1x2 5 3 2 x2

49. f 1x2 5 x2 1 3x 2 1 50. f 1x2 5 2x2 2 2x 1 1 51. f 1x2 5 0 x2 1 1 0

52. f 1x2 5 0 x2 1 x 1 4 0 2 53. f 1x2 5 x14 3 54. f 1x2 5 x25 1 55. f 1x2 5 2 x 21 3 56. f 1x2 5 2 x 13 57. f 1x2 5 "x2 1 1 58. f 1x2 5 "x2 2 1

Evaluate the difference quotient for each function f 1x2 . 60. f 1x2 5 5x 2 1 59. f 1x2 5 3x 1 1 2 61. f 1x2 5 x 1 1 62. f 1x2 5 x2 2 3

291

63. f 1x2 5 4x2 2 6

64. f 1x2 5 5x2 1 3

67. f 1x2 5 2x2 2 4x 1 2

68. f 1x2 5 3x2 1 2x 2 3

65. f 1x2 5 x2 1 3x 2 7

69. f 1x2 5 x3

66. f 1x2 5 x2 2 5x 1 1 70. f 1x2 5

1 x

Graph each function. Use the graph to identify the domain and range of each function. 72. f 1x2 5 3x 1 2 71. f 1x2 5 2x 1 3 y

x 2 x 24 2x x2 2 9 1 2 x 2 4x 2 5 x 2x2 2 16x 1 30

Let the function f be defined by y 5 f 1x2 , where x and f 1x2 are real numbers. Find f 122 , f 1232 , f 1k2 , and f 1k2 2 12 . 43. f 1x2 5 3x 2 2

Functions and Function Notation

y

x

x

3 73. f 1x2 5 2 x 1 4 4

1 74. f 1x2 5 x 2 3 2 y

y

x

x

76. 3x 5 2 1y 1 12

75. 2x 5 3y 2 3

y

y

x

x

77. f 1x2 5 x2 2 4

78. f 1x2 5 2x2 1 3

y

y

x

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x

292

Chapter 3

Functions

79. f 1x2 5 2x3 1 2

80. f 1x2 5 2x3 1 1

y

89. f 1x2 5 "2x 2 4

y

x

90. f 1x2 5 2"2x 2 4 y

y

x

x x

81. f 1x2 5 2 0 x 0

82. f 1x2 5 2 0 x 0 2 3

3 x12 91. f 1x2 5 "

y

y

3 x11 92. f 1x2 5 2"

y

y

x x

x x

83. f 1x2 5 0 x 2 2 0

84. f 1x2 5 2 0 x 2 2 0

y

Draw lines to indicate the domain and range of each function as intervals on the x- and y-axes. y y 93. 94.

y

x

x

x

1 85. f 1x2 5 ` x 1 3` 2

Use the Vertical Line Test to determine whether each graph represents a function. y y 95. 96.

1 86. f 1x2 5 2 ` x 1 3` 2

y

x

y

x

x x

x

97. 87. f 1x2 5 2"x 1 1

98.

y

88. f 1x2 5 "x 1 2

y

x

x

y

y

99.

100.

y

y

x x x

x

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Section 3.1

Use a graphing calculator to graph each function. Then determine the domain and range of the function. 101. f 1x2 5 0 3x 1 2 0

3 103. f 1x2 5 " 5x 2 1

102. f 1x2 5 "2x 2 5

3 104. f 1x2 5 2" 3x 1 2

Applications 105. Cost of t-shirts A chapter of Phi Theta Kappa, an honors society for two-year college students, is purchasing t-shirts for each of its members. A local company has agreed to make the shirts for $8 each plus a graphic arts fee of $75. a. Write a linear function that describes the cost C for the shirts in terms of x, the number of t-shirts ordered. b. Find the total cost of 85 t-shirts. 106. Service projects The Circle “K” Club is planning a service project for children at a local children’s home. They plan to rent a “Dora the Explorer Moonwalk” for the event. The cost of the moonwalk will include a $60 delivery fee and $45 for each hour it is used. Express the total bill b in terms of the hours used h. 107. Cell phone plans A grandmother agrees to purchase a cell phone for emergency use only. AT&T now offers such a plan for $9.99 per month and $0.07 for each minute t the phone is used. a. Write a linear function that describes the monthly cost C in terms of the time in minutes t the phone is used. b. If the grandmother uses her phone for 20 minutes during the first month, what was her bill? 108. Concessions A concessionaire at a football game pays a vendor $40 per game for selling hot dogs at $2.50 each. a. Write a linear function that describes the income I the vendor earns for the concessionaire during the game if the vendor sells h hot dogs.

Functions and Function Notation

293

109. Home construction In a proposal to prospective clients, a contractor listed the following costs: 1. Fees, permits, site preparation 2. Construction, per square foot

$14,000 $95

a. Write a linear function the clients can use to determine the cost C of building a house having f square feet. b. Find the cost to build a 2,600-square-foot house. 110. Temperature conversion The Fahrenheit temperature reading (F) is a linear function of the Celsius reading (C). If C 5 0 when F 5 32 and the readings are the same at 240°, express F as a function of C. 111. Cost of electricity The cost C of electricity in Eagle River is a linear function of x, the number of kilowatt-hours (kwh) used. If the cost of 100 kwh is $17 and the cost of 500 kwh is $57, find an equation that expresses C in terms of x. 112. Water billing The cost C of water is a linear function of n, the number of gallons used. If 1,000 gallons cost $4.70 and 9,000 gallons cost $14.30, express C as a function of n. 113. Coffee locations Suppose that in 2008 there were approximately 6,400 of your coffee company locations. Suppose that in 2012 this number had grown to approximately 13,168. Write a linear function that represents the number of coffee locations n as a function of time t. Let t 5 0 represent 2008. 114. Cliff divers The cliff divers of Acapulco amaze tourists with their diving skills. The velocity v of a diver is a function of the time t the diver has fallen. If the initial velocity of the diver is 2 feet per second and v 5 266 feet per second when t 5 2 seconds, express v as a function t. 115. Exchange rates If fifty U.S. dollars can be exchanged for 69.5550 Euros and 125 U.S. dollars can be exchanged for 173.8875 Euros, write a linear function that represents the number of Euros E in terms of U.S. dollars D. 116. Exchange rates If fifty U.S. dollars can be exchanged for 600.1100 Mexican pesos and 125 U.S. dollars can be exchanged for 1500.275 Mexican pesos, write a linear function that represents the number of Mexican pesos P in terms of U.S. dollars D.

b. Find the income if the vendor sells 175 hot dogs.

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294

Chapter 3

Functions

Discovery and Writing

124. Which numbers are rational numbers?

Find all values of x that will make f 1x2 5 0. 117. f 1x2 5 3x 1 2

118. f 1x2 5 22x 2 5

119. Write a paragraph explaining how to find the domain of a function. 120. Write a paragraph explaining how to find the range of a function. 121. Explain why all functions are relations, but not all relations are functions. 122. Use a graphing calculator to graph the function f 1x2 5 "x, and use TRACE and ZOOM to find "5 to three decimal places.

125. Which numbers are prime numbers? 126. Which numbers are even numbers? Write each in interval notation. 127.

(

]

−4

7

128.

)

[

–3

5

Graph each union of two intervals. 130. 12`, 02 c 10, ` 2 129. 123, 52 c 3 6, ` 2

Review

Consider this set: 5 23, 21, 0, 0.5, 34, 1, p, 7, 8 6 123. Which numbers are natural numbers?

3.2 Quadratic Functions In this section, we will learn to 1. 2. 3. 4. 5.

Recognize the characteristics of a quadratic function. Find the vertex of a parabola whose equation is in standard form. Graph a quadratic function. Find the vertex of a parabola whose equation is in general form. Use a quadratic function to solve maximum and minimum problems.

Mayskyphoto/Shutterstock.com

Quadratic functions are important because we can use them to model many real-life problems. For example, the path of a basketball jump shot by Shaquille O’Neal and the path of a guided missile can be modeled with quadratic functions. Businesses like Coca Cola and Best Buy can use quadratic functions to help maximize the profit and revenue for the products they produce and sell.

1. Recognize the Characteristics of a Quadratic Function The linear function f 1x2 5 mx 1 b 1m 2 02 is a first-degree polynomial function, because its right side is a first-degree polynomial in the variable x. A function defined by a polynomial of second degree is called a quadratic function.

Quadratic Function

A quadratic function is a second-degree polynomial function in one variable of the form f 1x2 5 ax2 1 bx 1 c or

y 5 ax2 1 bx 1 c,

where a, b, and c are real numbers and a 2 0.

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Section 3.2

Quadratic Functions

295

Some examples of quadratic functions are f 1x2 5 x2 2 2x 2 3 and

f 1x2 5 22x2 2 8x 2 3.

Quadratic functions can be graphed by plotting points. For example, to graph the function f 1x2 5 x2 2 2x 2 3, we plot several points with coordinates that satisfy the equation. We then join them with a smooth curve to obtain the graph shown in Figure 3-14.

x=1

y

f 1x2 5 x2 2 2x 2 3

22

f 1x2

1x, f 1x22

21

0

(21, 0)

0

23

(0, 23)

1

24

(1, 24)

2

23

(2, 23)

3

0

(3, 0)

4

5

(4, 5)

x

x f(x) = x2 – 2x – 3 (1, –4)

Domain: 12`, ` 2 , Range: 3 24, ` 2

5

(22, 5)

FIGURE 3-14

A table of values and the graph of f 1x2 5 22x2 2 8x 2 3 are shown in Figure 3-15. y (–2, 5) f(x) = –2x 2 – 8x – 3

x

x = –2

x

f 1x2 5 22x2 2 8x 2 3 f 1x2

1x, f 1x22

(24, 23)

24

23

23

3

(23, 3)

22

5

(22, 5)

21

3

(21, 3)

0

23

(0, 23)

Domain: 12`, ` 2 , Range: 12`, 5 4

FIGURE 3-15

The graph of a quadratic function is called a parabola, a cup-shaped curve that either opens upward c or downward d . The graphs in Figures 3-14 and 3-15 suggest that the graph of a quadratic function has the following characteristics.

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296

Chapter 3

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Characteristics of Quadratic Functions Characteristics

Examples

Equation of a quadratic function f 1x2 5 ax2 1 bx 1 c

If a . 0, the parabola opens up. If a , 0, the parabola opens down. The vertex is the turning point of the parabola. The minimum or maximum point occurs at the vertex. The axis of symmetry is the vertical line that intersects the parabola at the vertex. The parabola is symmetric about this vertical line.

ACCENT ON TECHNOLOGy

f 1x2 5 x2 2 2x 2 3

f 1x2 5 22x2 2 8x 2 3

a 5 1, opens up

a 5 22, opens down

Vertex is 11, 242

Vertex is 122, 52

The graph of x 5 1 is the axis of symmetry.

The graph of x 5 22 is the axis of symmetry.

11,242 is the minimum or lowest point on the graph.

122, 52 is the maximum or highest point on the graph.

Graphing Quadratic Functions We can use a graphing calculator to graph quadratic functions. If we use window settings of [210, 10] for x and [210, 10] for y, the graph of f 1x2 5 x2 2 2x 2 3 will look like Figure 3-16(a). The graph of f 1x2 5 22x2 2 8x 2 3 will look like Figure 3-16(b).

(a)

(b) FIGURE 3-16

2. Find the Vertex of a Parabola Whose Equation Is in Standard Form

In Figure 3-14, we considered the function f 1x2 5 x2 2 2x 2 3 with vertex 11, 242 . If we complete the square on the right side of the equation we will obtain f 1x2 5 1x2 2 2x 1 12 2 3 2 1

f 1x2 5 1x 2 12 1x 2 12 2 4 f 1x2 5 1x 2 12 2 2 4

One-half of 22 is 21, and 1212 2 is 1. Add 1 and subtract 1 on the right side of the equation. Factor x2 2 2x 1 1.

In this factored form, the coordinates of the vertex of the parabola can be read from the equation. The vertex of f 1x2 5 1x 2 12 2 2 4 is 11, 242 . We call this factored form the standard form of an equation of a quadratic function. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.2

Standard Form of an Equation of a Quadratic Function

EXAMPLE 1

SOLUTION

Quadratic Functions

297

The graph of a quadratic function

y 5 f 1x2 5 a 1x 2 h2 2 1 k 1a 2 02

is a parabola with vertex at 1h, k2 . The parabola opens upward when a . 0 and downward when a , 0. The axis of symmetry of the parabola is the vertical line graph of the equation x 5 h.

Finding the Vertex of a Parabola in Standard Form Find the vertex of the graph of each quadratic function: a. f 1x2 5 2 1x 2 32 2 1 5 b. f 1x2 5 23 1x 1 22 2 2 4

In each case, the equation of the quadratic function is given in standard form. From the equations, we can identify h and k, because the vertex is the point with coordinates 1h, k2 . a. We identify the values of h and k.

Standard Form: f 1x2 5 a 1x 2 h2 2 1 k

Given Function: f 1x2 5 2 1x 2 32 2 1 5 ▲

h 5 3 and k 5 5.

▲

Since h 5 3 and k 5 5, the vertex is the point with coordinates of 13, 52 .

b. We identify the values of h and k.

Standard Form: f 1x2 5 a 1x 2 h2 2 1 k

Given Function: f 1x2 5 23 1x 1 22 2 2 4

f 1x2 5 23 3 x 2 1222 4 2 1 1242

h 5 22 and k 5 24.

Since h 5 22 and k 5 24, the vertex is the point with coordinates of 122, 242 .

Self Check 1

Find the vertex of the graph of the quadratic function f 1x2 5 2 1x 1 52 2 2 4. Now Try Exercise 19.

3. Graph a Quadratic Function The easiest way to graph a quadratic function is to follow these steps.

Strategy for Graphing a Quadratic Function

To graph a quadratic function 1. Determine whether the parabola opens upward or downward. 2. Find the vertex of the parabola. 3. Find the x-intercept(s). 4. Find the y-intercept. 5. Identity one additional point on the graph. 6. Draw a smooth curve through the points found in Steps 225.

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Chapter 3

Functions

EXAMPLE 2

SOLUTION

Graphing a Quadratic Function Written in Standard Form

Graph the quadratic function f 1x2 5 2 1x 1 12 2 2 8.

We first determine whether the parabola opens upward or downward. Then we will find the vertex and the x- and y-intercepts. Finally, we will find one additional point and draw a smooth curve through the plotted points. Step 1: Determine whether the parabola opens upward or downward. Standard Form: f 1x2 5 a 1x 2 h2 2 1 k f 1x2 5 2 1x 1 12 2 2 8 ▲

298

Given Form:

Since a 5 2 and 2 is positive, the parabola opens upward. Step 2: Find the vertex of the parabola.

Standard Form: f 1x2 5 a 1x 2 h2 2 1 k f 1x2 5 2 1x 1 12 2 2 8

Given Form:

f 1x2 5 2 3 x 2 1212 4 2 1 1282

Since h 5 21 and k 5 28 the vertex is the point with coordinates of (21, 282 . Step 3: Find the x-intercept(s). To find the x-intercepts, we substitute 0 for f 1x2 and solve for x. f 1x2 5 2 1x 1 12 2 2 8 0 5 2 1x 1 12 2 2 8 8 5 2 1x 1 12 2 4 5 1x 1 12

2

x 1 1 5 62

x 5 21 6 2 x 5 1 or

Substitute 0 for f 1x2 . Add 8 to both sides of the equation. Divide both sides by 2.

Write 1x 1 12 2 on the left side and use the Square Root Property. Subtract 1 from both sides.

x 5 23

The x-intercepts are the points with coordinates of 11, 02 and 123, 02 . Step 4: Find the y-intercept. To find the y-intercept, we substitute 0 in for x and solve for y. f 1x2 5 2 1x 1 12 2 2 8 y 5 2 1x 1 12 2 2 8 y 5 2 10 1 12 2 2 8 y 5 2 112 2 8

Substitute y for f 1x2 . Substitute 0 in for x.

2

y5228 y 5 26

The y-intercept is the point with coordinates of 10, 262 .

Step 5: Identify one additional point on the graph. Because of symmetry, the point 122, 262 is on the graph.

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Section 3.2

Quadratic Functions

299

Step 6: Draw a smooth curve through the points found in Steps 2–5. We can now draw the graph of the function as shown in Figure 3-17. y

Comment x

Sometimes the vertex of the graph of a quadratic function occurs at the origin, 10, 02 . In this case, both h 5 0 and k 5 0, and the equation of the parabola is of the form f 1x2 5 a 1x 2 02 2 1 0 or f 1x2 5 ax2 and a 2 0.

f (x) = 2(x + 1) 2 – 8

FIGURE 3-17

Self Check 2

Graph the function: f 1x2 5 2 1x 2 22 2 1 4.

Now Try Exercise 41.

Comment The graph of a quadratic function can have one, two, or no x-intercepts. However, it will always have one y-intercept.

EXAMPLE 3 SOLUTION

4. Find the Vertex of a Parabola Whose Equation Is in General Form

To graph a quadratic function given in the form f 1x2 5 ax2 1 bx 1 c (a 2 0), called general form, we must find the coordinates of the vertex of the parabola. To do so, we can complete the square on ax2 1 bx to change the equation into standard form y 5 a 1x 2 h2 2 1 k. As we have seen, we can read the coordinates 1h, k2 of the vertex from this form. Once the quadratic function is in standard form, we can graph the function following the steps stated earlier.

Finding the Vertex of a Parabola Written in General Form

Find the vertex of the parabola whose equation is f 1x2 5 22x2 1 12x 2 16.

We will complete the square on x, write the equation in standard form, and identify h and k, the coordinates of the vertex. We begin by completing the square on 22x2 1 12x. f 1x2 5 22x2 1 12x 2 16

f 1x2 5 22 1x2 2 6x2 2 16

Identify a: a 5 22.

f 1x2 5 22 1x2 2 6x 1 92 2 2 1292 2 16

Add and subtract 9 within the parentheses.

f 1x2 5 22 1x2 2 6x 1 9 2 92 2 16 f 1x2 5 22 1x 2 32 1 18 2 16 2

f 1x2 5 22 1x 2 32 2 1 2

Factor a 5 22 from 22x2 1 12x.

One-half of 26 is 23 and 1232 2 5 9. Distribute the multiplication by 22. Factor x2 2 6x 1 9 and multiply. Simplify.

The equation is now in standard form with h 5 3 and k 5 2. Therefore, the vertex is the point with coordinates (h, k) 5 13, 22 . Self Check 3

Find the vertex of the graph of y 5 4x2 2 16x 1 19. Now Try Exercise 27.

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300

Chapter 3

Functions

To find formulas for the coordinates of the vertex of a parabola defined by y 5 ax2 1 bx 1 c 1a 2 02 , we can complete the square on x to write the equation in standard form (y 5 a 1x 2 h2 2 1 k): y 5 ax2 1 bx 1 c

b y 5 aax2 1 xb 1 c a

Factor a from ax2 1 bx.

b b2 b2 y 5 aax2 1 x 1 2 2 2 b 1 c a 4a 4a

b2 b b2 y 5 aax2 1 x 1 2 b 2 aa 2 b 1 c a 4a 4a y 5 aax 1

Comment You don’t need to memorize the formula for the y-coordinate of the vertex of a parabola. It is usually convenient to find b the y-coordinate by substituting 22a for x in the function and solving for y.

Vertex of a Parabola

b 2 b2 b 1c2 2a 4a

y 5 a c x 2 a2

b 2 b2 bd 1 c 2 2a 4a

Add and subtract

b2 within the parentheses. 4a2

Distribute the multiplication of a. b2 b b2 Factor x2 1 x 1 2 and simplify aa 2 b. a 4a 4a 2a2

b b b5 2a 2a

If we compare the last equation to the form y 5 a 1x 2 h2 2 1 k, we see that 2

b b h 5 22a and k 5 c 2 4a . This result gives the following fact.

The graph of the function y 5 f 1x2 5 ax2 1 bx 1 c 1a 2 02 is a parabola with b b vertex at Q22a , c 2 4a R. 2

EXAMPLE 4 SOLUTION

Graphing a Quadratic Function Written in General Form

Graph the function: y 5 f 1x2 5 22x2 2 5x 1 3.

We begin by determining whether the parabola opens upward or downward. Then b . Next we find the x- and y-intercepts we find the vertex by using the formula h 5 22a and one additional point and then draw a smooth curve through the plotted points. Step 1: Determine whether the parabola opens up or downward. The equation has the form y 5 ax2 1 bx 1 c, where a 5 22, b 5 25, and c 5 3. Since a , 0, the parabola opens downward. Step 2: Find the vertex. To find the x-coordinate of the vertex, we substitute the values of a and b into the b . formula x 5 22a x52

b 25 5 52 52 2a 2 1222 4

The x-coordinate of the vertex is 254. To find the y-coordinate, we substitute 254 for x in the equation and solve for y. y 5 22x2 2 5x 1 3

5 2 5 y 5 22a2 b 2 5a2 b 1 3 4 4

5 Substitute 2 for x. 4

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Section 3.2

301

Quadratic Functions

25 25 13 b1 16 4 25 50 24 1 52 1 8 8 8 49 5 8 5 22a

Since the vertex is the point Q254, 49 8 R, we can plot it on the coordinate system in

Figure 3-18(a) and draw the axis of symmetry.

Step 3: Find the x-intercept(s). To find the x-intercepts, we substitute 0 for y and solve for x. y 5 22x2 2 5x 1 3 0 5 22x2 2 5x 1 3

Substitute 0 for y.

0 5 2x2 1 5x 2 3

Divide both sides by 21 to make the leading coefficient positive.

0 5 12x 2 12 1x 1 32

Factor the trinomial.

2x 2 1 5 0 or x 1 3 5 0 Set each factor equal to 0. 1 x5 x 5 23 Solve each linear equation. 2 The x-intercepts are Q12, 0R and 123, 02 . We plot these intercepts as shown in Figure 3-18(a). Step 4: Find the y-intercept. To find the y-intercept, we let x 5 0 and solve for y. y 5 22x2 2 5x 1 3

5 22 102 2 2 5 102 1 3

Substitute 0 for x.

502013 53

The y-intercept is 10, 32 . We plot the intercept as shown in Figure 3-18(a). Step 5: Plot one additional point.

Because of symmetry, we know that the point Q2212, 3R is on the graph. We plot this point on the coordinate system in Figure 3-18(a).

Step 6: We can now draw the graph of the function, as shown in Figure 3-18(b). y

y

(– 54– , 49––8 )

f(x) = –2x 2 – 5x + 3

Comment The y-intercept of a parabola written in the general form f 1x2 5 ax2 1 bx 1 c (a 2 0) is the point (0, c). This is because when we substitute 0 for x, y is always c.

(–2 1–2 , 3)

(0, 3)

( 1–2 , 0)

x

x

(–3, 0)

(b)

(a) FIGURE 3-18

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302

Chapter 3

Functions

Self Check 4

Graph the function: f 1x2 5 23x2 1 7x 2 2.

Now Try Exercise 53.

ACCENT ON TECHNOLOGy

Finding the Maximum Point or Minimum Point (Vertex) of a Parabola We can use a graphing calculator to find the maximum point or minimum point (vertex) of a parabola. Consider the f 1x2 5 22x2 2 5x 1 3 given in Example 4. We will apply the following steps to determine the maximum point or vertex of the parabola. 1. Enter the function.

2. Set a window that shows the function.

3. Graph the function.

4. Go to the CALC menu by TRACE and select pressing maximum.

5. Use the

TRACE

key to get a left bound.

6. Use the bound.

7. Use the

TRACE

key to make a guess.

8. Press ENTER and find the maximum point.

TRACE

key to get a right

FIGURE 3-19

We see that the maximum point or the vertex is 121.250001, 6.1252 .

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Section 3.2

Quadratic Functions

303

5. Use a Quadratic Function to Solve Maximum and Minimum Problems EXAMPLE 5

Using a Quadratic Function to Solve a Maximum Area Problem The Montana Dude Rancher’s Association has 400 feet of fencing to enclose a rectangular corral. To save money and fencing, the association intends to use the bank of a river as one boundary of the corral, as in Figure 3-20. Find the dimensions that will enclose the largest area.

SOLUTION

We will represent the fenced area with a quadratic function. Since its parabolic graph opens downward, the largest or maximum area will occur at the vertex. We can use the vertex formula to find the vertex. Step 1: Represent the area with a quadratic function. Let x represent the width of the fenced area. Then 400 2 2x represents the length. Because the area A of a rectangle is the product of the length and the width, we have A 5 1400 2 2x2 x or A 1x2 5 22x2 1 400x

40

0–

2x

x

x FIGURE 3-20

The graph of this area function is a parabola. Since the coefficient of x2 is negative, the parabola opens downward and its vertex is its highest point. The A-coordinate of the vertex 1x, A2 represents the maximum area, and the x-coordinate represents the width of the corral that will give the maximum area. Step 2: Find the vertex of the parabola. We compare the equations A 1x2 5 22x2 1 400x and

y 5 ax2 1 bx 1 c

to see that a 5 22, b 5 400, and c 5 0. Using the vertex formula, the vertex of the parabola is the point with coordinates a2

Comment

b b2 400 4002 , c 2 b 5 a2 ,02 b 5 1100, 20,0002 2a 4a 2 1222 4 1222

Note that we could have determined the y-coordinate of 20,000 by finding A 11002 . A 1x2 5 22x2 1 400x

A 11002 5 22 11002 2 1 400 11002

5 22 110,0002 1 40,000

Substitute 100 for x.

5 220,000 1 40,000 5 20,000

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Chapter 3

Functions

If the fence runs 100 feet out from the river, 200 feet parallel to the river, and 100 feet back to the river, it will enclose the largest possible area, which is 20,000 square feet. Self Check 5

Find the largest area possible if the association has 1,200 feet of fencing available. Now Try Exercise 55.

EXAMPLE 6

Using a Quadratic Function to Solve a Minimum Cost Problem A company that makes and sells water skis has found that the total weekly cost C of producing x water skis is given by the function C 1x2 5 0.5x2 2 210x 1 26,250. Find the production level that minimizes the weekly cost and find that weekly minimum cost.

SOLUTION

manzrussali/Shutterstock.com

304

Comment

The weekly cost function C 1x2 is a quadratic function whose graph is a parabola that opens upward. The minumum value of C 1x2 occurs at the vertex of the parabola. We will use the vertex formula to find the vertex of the parabola. Since the coefficient of x2 is 0.5 (a positive real number), the x-coordinate of the vertex is the production level that will minimize the cost, and the y-coordinate is that minimum cost. We compare the equations C 1x2 5 0.5x2 2 210x 1 26,250

and

y 5 ax2 1 bx 1 c

to see that a 5 0.5, b 5 2210, and c 5 26,250. Using the vertex formula, we see that the vertex of the parabola is the point with coordinates a2

122102 2 b b2 2210 , c 2 b 5 a2 , 26,250 2 b 5 1210, 4,2002 2a 4a 2 10.52 4 10.52

We can also determine the y-coordinate of 4,200 by finding C 12102 . C 1x2 5 0.5x2 2 210x 1 26,250

C 12102 5 0.5 12102 2 2 210 12102 1 26,250

Substitute 210 for x.

5 0.5 144,1002 2 44,100 1 26,250 5 22,050 2 17,850 5 4,200

If the company makes 210 water skis each week, it will minimize its production cost. The minimum weekly cost will be $4,200. Self Check 6

A company that makes and sells baseball caps has found that the total monthly cost C of producing x caps is given by the function C 1x2 5 0.2x2 2 80x 1 9,000. Find the production level that will minimize the monthly cost and find the minimum cost. Now Try Exercise 67.

Self Check Answers

1. 125, 242 5. 180,000 square feet 6. 200, $1,000

2.

3. 12, 32

y f (x) = – (x – 2)2 + 4

4.

y

x x

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Section 3.2

305

Quadratic Functions

Exercises 3.2 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

27. y 5 22x2 1 12x 2 17

28. y 5 2x2 1 16x 1 33

29. y 5 3x2 2 4x 1 5

30. y 5 24x2 1 3x 1 4

1 31. y 5 x2 1 4x 2 3 2

2 32. y 5 2 x2 1 3x 2 5 3

Fill in the blanks. 1. A quadratic function is defined by the equation 1a 2 02 .

2. The standard form for the equation of a parabola is 1a 2 02 .

3. The vertex of the parabolic graph of the equation . y 5 2 1x 2 32 2 1 5 will be at 4. The vertical line that intersects the parabola at its . vertex is the 5. If the parabola opens the vertex will be a minimum point. 6. If the parabola opens the vertex will be a maximum point. 7. The x-coordinate of the vertex of the parabolic graph of f 1x2 5 ax2 1 bx 1 c is

.

8. The y-coordinate of the vertex of the parabolic graph of f 1x2 5 ax2 1 bx 1 c is

Graph each quadratic function given in standard form. 33. f 1x2 5 x2 2 4 34. f 1x2 5 x2 1 1 y

y

x x

35. f 1x2 5 23x2 1 6

36. f 1x2 5 24x2 1 4

y

y

.

Practice Determine whether the graph of each quadratic function opens upward or downward. State whether a maximum or minimum point occurs at the vertex of the parabola. 1 9. f 1x2 5 x2 1 3 10. f 1x2 5 2x2 2 3x 2 11. f 1x2 5 23 1x 1 12 2 1 2

13. f 1x2 5 22x 1 5x 2 1 2

12. f 1x2 5 25 1x 2 12 2 2 1

x

1 37. f 1x2 5 2 x2 1 8 2

14. f 1x2 5 2x 2 3x 1 1

x

1 38. f 1x2 5 x2 2 2 2 y

y

2

x

Find the vertex of each parabola. 15. y 5 x2 2 1 17. f 1x2 5 1x 2 32 2 1 5

19. f 1x2 5 22 1x 1 62 2 2 4

2 21. f 1x2 5 1x 2 32 2 3

16. y 5 2x2 1 2 18. f 1x2 5 22 1x 2 32 2 1 4 1 20. f 1x2 5 1x 1 12 2 2 5 3

22. f 1x2 5 7 1x 1 22 2 1 8

23. f 1x2 5 x2 2 4x 1 4

24. y 5 x2 2 10x 1 25

25. y 5 x2 1 6x 2 3

26. y 5 2x2 1 9x 2 2

x

39. f 1x2 5 1x 2 32 2 2 1

40. f 1x2 5 1x 1 32 2 2 1

x

x

y

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y

306

Chapter 3

Functions

41. f 1x2 5 2 1x 1 12 2 2 2

3 42. f 1x2 5 2 1x 2 22 2 4

y

y

49. f 1x2 5 x2 2 4x 1 1

50. f 1x2 5 x2 2 6x 2 7 y

y

x x x x

43. f 1x2 5 2 1x 1 42 2 1 1 y

44. f 1x2 5 23 1x 2 42 2 1 3

51. f 1x2 5 2x2 2 12x 1 10

52. f 1x2 5 2x2 2 4x 1 1

y

y

y

x x x

45. f 1x2 5 23 1x 2 22 2 1 6 y

46. f 1x2 5 2 1x 2 32 2 2 4

x

53. f 1x2 5 23x2 2 6x 2 9 y

54. f 1x2 5 23x2 2 3x 1 18 y

y

x

x x x

Applications

Graph each quadratic function given in general form. 47. f 1x2 5 x2 1 2x 48. f 1x2 5 x2 2 6x y

y

x

x

55. Police investigations A police officer seals off the scene of an accident using a roll of yellow tape that is 300 feet long. What dimensions should be used to seal off the maximum rectangular area around the collision? Find the maximum area. 56. Maximizing area The rectangular garden shown has a width of x and a perimeter of 100 feet. Find x such that the area of the rectangle is maximum.

100 ft

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.2

57. Maximizing storage area A farmer wants to partition a rectangular feed storage area in a corner of his barn, as shown in the illustration. The barn walls form two sides of the stall, and the farmer has 50 feet of partition for the remaining two sides. What dimensions will maximize the area?

307

Quadratic Functions

60. Landscape design A gardener will use D feet of edging to border a rectangular plot of ground. Show that the maximum area will be enclosed if the rectangle is a square. 61. Architecture A parabolic arch has an equation of x2 1 20y 2 400 5 0, where x is measured in feet. Find the maximum height of the arch. 62. Path of a guided missile A guided missile is propelled from the origin of a coordinate system with the x-axis along the ground and the y-axis vertical. Its path, or trajectory, is given by the equation y 5 400x 2 16x2. Find the object’s maximum height.

50 ft

58. Maximizing grazing area A rancher wishes to enclose a rectangular partitioned corral with 1,800 feet of fencing. (See the illustration.) What dimensions of the corral would enclose the largest possible area? Find the maximum area.

63. Height of a basketball The path of a basketball thrown from the free throw line can be modeled by the quadratic function f 1x2 5 20.06x2 1 1.5x 1 6, where x is the horizontal distance (in feet) from the free throw line and f 1x2 is the height (in feet) of the ball. Find the maximum height of the basketball. 64. Ballistics A child throws a ball up a hill that makes an angle of 45° with the horizontal. The ball lands 100 feet up the hill. Its trajectory is a parabola with equation y 5 2x2 1 ax for some number a. Find a.

10

0

ft

y

59. Sheet metal fabrication A 24-inch-wide sheet of metal is to be bent into a rectangular trough with the cross section shown in the illustration. Find the dimensions that will maximize the amount of water the trough can hold. That is, find the dimensions that will maximize the cross-sectional area.

Depth

24 in. Width

45° x

65. Maximizing height A ball is thrown straight up from the top of a building 144 ft. tall with an initial velocity of 64 ft per second. The distance s 1t2 (in feet) of the ball from the ground is given by s 1t2 5 144 1 64t 2 16t2. Find the maximum height attained by the ball. 66. Flat-panel television sets A wholesaler of appliances finds that she can sell 12,400 2 p2 flat-panel television sets each week when the price is p dollars. What price will maximize revenue?

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

308

Chapter 3

Functions

67. Digital camera A company that produces and sells digital cameras has determined that the total weekly cost C of producing x digital cameras is given by the function C 1x2 5 1.5x2 2 144x 1 5,856. Determine the production level that minimizes the weekly cost for producing the digital cameras and find that weekly minimum cost.

77. y 5 1x 2 72 15x 1 22

78. y 5 2x 10.2 1 0.1x2

Find all values of x that will make f 1x2 5 0. 79. f 1x2 5 x2 2 5x 1 6 80. f 1x2 5 6x2 1 x 2 2

Discovery and Writing

81. Find the dimensions of the largest rectangle that can be inscribed in the right triangle ABC shown in the illustration. AVAVA/Shutterstock.com

y A(0, 9)

(x, y)

68. Finding mass transit fares The Municipal Transit Authority serves 150,000 commuters daily when the fare is $1.80. Market research has determined that every penny decrease in the fare will result in 1,000 new riders. What fare will maximize revenue? 69. Finding hotel rates A 300-room hotel is two-thirds filled when the nightly room rate is $90. Experience has shown that each $5 increase in cost results in 10 fewer occupied rooms. Find the nightly rate that will maximize income. 70. Selling concert tickets Tickets for a concert are cheaper when purchased in quantity. The first 100 tickets are priced at $10 each, but each additional block of 100 tickets purchased decreases the cost of each ticket by 50¢. How many blocks of tickets should be sold to maximize the revenue? Use this information: At a time t seconds after an object is tossed vertically upward, it reaches a height s in feet given by the equation s 5 80t 2 16t2. 71. In how many seconds does the object reach its maximum height? 72. In how many seconds does the object return to the point from which it was thrown? 73. What is the maximum height reached by the object? 74. Show that it takes the same amount of time for the object to reach its maximum height as it does to return from that height to the point from which it was thrown. Use a graphing calculator to determine the coordinates of the vertex of each parabola. You will have to select appropriate viewing windows. 75. y 5 2x2 1 9x 2 56

76. y 5 14x 2

C

x

O

B(12, 0)

82. Point P lies in the first quadrant and on the line x 1 y 5 1 in such a position that the area of triangle OPA is maximum. Find the coordinates of P. (See the illustration.) y

x+y=1 P(x, y)

O

A

x

83. The sum of two numbers is 6, and the sum of the squares of those two numbers is as small as possible. What are the numbers? 84. What number most exceeds its square? The maximum or minimum value of a quadratic function can be found automatically by using a computer program, such as Excel. 85. Find the minimum value of the function f 1x2 5 2x2 2 3x 2 4 by using the Solver in Excel. Give the value of x that minimizes the function as well as the minimum value of the function. See Problem 119 in Section 1.3. 86. Find the maximum value of the function f 1x2 5 22x2 1 3x 1 4 by using the Solver in Excel. Give the value of x that maximizes the function as well as the maximum value of the function. See Problem 119 in Section 1.3.

x2 5

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Section 3.3

Find f 1a2 and f 12a2 . 87. f 1x2 5 x2 2 3x

Review

89. f 1x2 5 15 2 x2 2

88. f 1x2 5 x3 2 3x 90. f 1x2 5

Polynomial and Other Functions

91. f 1x2 5 7

309

92. f 1x2 5 2 0 x 0

1 x 24 2

3.3 Polynomial and Other Functions In this section, we will learn to 1. 2. 3. 4.

Understand the characteristics of polynomial functions. Graph polynomial functions. Determine whether a function is even, odd, or neither. Identify the open intervals on which a function is increasing, decreasing, or constant. 5. Graph piecewise-defined functions. 6. Evaluate and graph the greatest-integer function.

© Istockphoto.com/Steve Maehl

So far, we have discussed two types of polynomial functions—first-degree (or linear) functions, and second-degree (or quadratic) functions. In this section, we will discuss polynomial functions of higher degree. Polynomial functions can be used to model the path of a roller coaster or to model the fluctuation of gasoline prices over the past few months. Goliath, a hypercoaster, is located in Atlanta at Six Flags over Georgia. It climbs to a height of 200 feet and reaches speeds of nearly 70 mph. It has over 4,400 feet of steel track. Portions of Goliath’s tracks can be modeled with a polynomial function.

Polynomial Functions

A polynomial function in one variable (say, x) is a function of the form f 1x2 5 anxn 1 an21xn21 1 c 1 a1x 1 a0

where an, an21, …, a1, and a0 are real numbers and n is a whole number. The degree of a polynomial function is the largest power of x that appears in the polynomial.

The table shows three basic polynomial functions that we have already covered:

Name

Function

Constant function

f 1x2 5 2x 2 7

Linear function Quadratic function

f 1x 2 5 5

f 1x2 5 22x2 1 4x 2 5

Degree

Graph

0

Horizontal line

1

Nonvertical line

2

Parabola

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310

Chapter 3

Functions

Here are two examples of higher-degree polynomial functions, along with their graphs. y

y

30 10

20 10 x –4

–2

2

4 –4

–10

–2

2

4

x

1 f(x) = – x4 – 3x 2 + 2 2

f(x) = 2x3 – 2x2 – 18x + 18 FIGURE 3-21

FIGURE 3-22

1. Understand the Characteristics of Polynomial Functions There are several basic characteristics common to all polynomial functions. We will list several of them. 1: The graphs of polynomial functions are smooth and continuous curves. Like the graphs of linear and quadratic functions, the graphs of higher-degree polynomial functions are smooth continuous curves. Because their graphs are smooth, they have no cusps or corners. Because they are continuous, their graphs have no breaks or holes. They can always be drawn without lifting the pencil from the paper. •

The graph of a polynomial function y Smooth and continuous

x

Graph of a polynomial function FIGURE 3-23

•

The graphs of two functions that are not polynomial functions y

y

Cusp Hole Break x

x

Not the graph of a polynomial function

Not the graph of a polynomial function

Corner

FIGURE 3-24 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.3

Polynomial and Other Functions

311

2: Many polynomial functions have graphs similar to the graphs of f 1x2 5 x, f 1x2 5 x2, and f 1x2 5 x3.

Many polynomial functions are of the form f 1x2 5 xn, and several of their graphs are shown in Figure 3-25. Note that when n is even, the graph has the same general shape as y 5 x2. When n is odd and greater than 1, the graph has the same general shape as y 5 x3. However, the graphs are flatter at the origin and steeper as n becomes large. y

y

y

1

1 –1

0

1

x

1

–1

0 –1

–1

1

x

–1 –1

y = x2

y=x

0

(b)

(c)

y

y

y

1

–1

0 –1

(d)

1 y = x4

x

x

y = x3

(a)

1

1

1

–1

0 –1

1

x

–1

0 –1

y = x5

(e)

1

x

y = x6

(f)

FIGURE 3-25

3: The end behavior of the graph of a polynomial function is similar to the graph of its term with highest degree. The ends of the graph of any polynomial function will be similar to the graph of its term with the highest power of x, because when n becomes large, the other terms become relatively insignificant. Consider the polynomial function f 1x2 5 x3 1 x2 2 2x. The end behavior of its graph will be similar to the ends of the graph of its leading term x3. In Figure 3-25 (c), we see that the graph of y 5 x3 falls on the far left and rises on the far right. Therefore, the graph of f 1x2 will also fall on the far left and rise on the far right. See Figure 3-26. y f(x) = x 3 + x 2 – 2 x

x

FIGURE 3-26

4: Polynomial functions can be symmetric about the y-axis or the origin. In Section 2.4, we learned that a graph is symmetric about the y-axis if the graph of y 5 f 1x2 has the same y-coordinate when the function is evaluated at x or at 2x. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

312

Chapter 3

Functions

Thus, a function is symmetric about the y-axis if f 1x2 5 f 12x2 for all values of x that are in the domain of the function. See Figure 3-27(a). Also recall that a graph is symmetric about the origin if the point 12x, 2f 1x2 2 lies on the graph whenever 1x, f 1x2 2 does. In this case, f 12x2 5 2f 1x2 . See Figure 3-27(b). y

y

f(–x) = f(x)

f(–x) = –f(x)

(–x, f(–x))

(x, f(x)) x

(x, f(x)) (–x, –f(x))

x

(b)

(a) FIGURE 3-27

2. Graph Polynomial Functions To graph polynomial functions, we can use the following steps. Strategy for Graphing Polynomial Functions

EXAMPLE 1 SOLUTION

1. 2. 3. 4.

Find any symmetries of the graph. Find the x- and y-intercepts of the graph. Determine where the graph is above and below the x-axis. Plot a few points, if necessary, and draw the graph as a smooth continuous curve.

Graphing a Polynomial Function of Degree 3

Graph the function: f 1x2 5 x3 2 4x.

We will use the four steps stated above to graph the polynomial function. Step 1: Find any symmetries of the graph. To test for symmetry about the y-axis, we check to see whether f 1x2 5 f 12x2 . To test for symmetry about the origin, we check to see whether f 12x2 5 2f 1x2 . f 1x2 5 x3 2 4x

f 12x2 5 12x2 3 2 4 12x2

Substitute 2x for x.

f 12x2 5 2x3 1 4x

Simplify.

Since f 1x2 2 f 12x2 , there is no symmetry about the y-axis. However, since 1 f 2x2 5 2f 1x2 , there is symmetry about the origin.

Step 2: Find the x- and y-intercepts of the graph. To find the x-intercepts, we let f 1x2 5 0 and solve for x. x3 2 4x 5 0

x 1x2 2 42 5 0

x50

Factor out x.

x 1x 1 22 1x 2 22 5 0

or x 1 2 5 0

x 5 22

Factor x2 2 4.

or x 2 2 5 0

Set each factor equal to 0.

x52

The x-intercepts are (0, 0), (22, 0), and (2, 0). If we let x 5 0 and solve for f 1x2 , we see that the y-intercept is also (0, 0). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.3

Polynomial and Other Functions

313

Step 3: Determine where the graph is above or below the x-axis. To determine where the graph is above or below the x-axis, we plot the solutions of x3 2 4x 5 0 (the x-coordinates of the x-intercepts) on a number line and establish the four intervals shown in Figure 3-28. We then test a number from each interval to determine the sign of f 1x2 . (For a review of this process, see Example 7 in Section 1.7.) Sign of f(x) = x3 – 4x

Test point Graph of f(x)

–

+

–

+

( –∞, –2)

(–2, 0)

(0, 2)

(2, ∞)

f(–3) = –15 –2 f(–1) = 3 0 f(1) = –3 2 below the x-axis

above the x-axis

below the x-axis

f (3) = 15 above the x-axis

FIGURE 3-28

Comment Note that the far right and far left ends of the graph are similar to the ends of the graph of f 1x2 5 x3, which is the leading term of the function f 1x2 5 x3 2 4x or the term with highest degree. On the far right the graph rises and on the far left the graph falls.

Step 4: Plot a few points and draw the graph as a smooth continuous curve. We now plot the intercepts and one additional point. In the previous step we found that f 112 5 23. This will be the additional point we plot 11, 232 . Making use of our knowledge of symmetry and where the graph is above and below the x-axis we now draw the graph as shown in Figure 3-29(a). A calculator graph, using the standard viewing window, is shown in Figure 3-29(b). y

f 1x2 5 x3 2 4x f 1x2

1x, f 1x22

0

0

(0, 0)

1

23

2

0

x 22

0

(22, 0)

x

(1, 23) (2, 0)

f(x) = x 3 – 4x

(b)

(a) FIGURE 3-29

Self Check 1

Graph: f 1x2 5 x3 2 9x.

Now Try Exercise 13.

The peak and valley of the graph shown in Figure 3-29 are called turning points. In calculus, such points are called local minima and local maxima. Although we cannot find these points without using calculus, we can approximate them by plotting points or by using the TRACE feature on a graphing calculator. We can also use the CALC Minimum and CALC Maximum features on a graphing calculator to find these local extrema. Note that Figure 3-29 shows the graph of a third-degree polynomial and the graph has two turning points. This suggests the following result from calculus that helps us understand the shape of many polynomial graphs.

Number of Turning Points

If f 1x2 is a polynomial function of nth degree, then the graph of f 1x2 will have n 2 1, or fewer, turning points.

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314

Chapter 3

Functions

EXAMPLE 2 SOLUTION

Graphing a Polynomial Function of Degree 4

Graph the function: f 1x2 5 x4 2 5x2 1 4.

We will use the four steps for graphing a polynomial function.

Step 1: Find any symmetries of the graph. Because x appears with only even exponents, f 1x2 5 f 12x2 , and the graph is symmetric about the y-axis. The graph is not symmetric about the origin. Step 2: Find the x- and y-intercepts of the graph. To find the x-intercepts, we let f 1x2 5 0 and solve for x. x4 2 5x2 1 4 5 0

1x2 2 42 1x2 2 12 5 0

1x 1 22 1x 2 22 1x 1 12 1x 2 12 5 0

x1250

or x 2 2 5 0

or x 1 1 5 0

x52

x 5 22

or x 2 1 5 0 x51

x 5 21

The x-intercepts are (22, 0), (2, 0), (21, 0), and (1, 0). To find the y-intercept, we let x 5 0 and see that the y-intercept is (0, 4). Step 3: Determine where the graph is above or below the x-axis. To determine where the graph is above or below the x-axis, we plot the x-coordinates of the x-intercepts on a number line and establish the five intervals shown in Figure 3-30. We then test a number from each interval to determine the sign of f 1x2 .

+

–

+

–

+

(–∞, –2)

( –2, –1)

(–1, 1)

(1, 2)

(2, ∞)

Sign of f(x) = x4 – 5x2 + 4

Test point Graph of f(x)

f (–3) = 40 –2 f

(– 3–2 ) ~ –2.2 –1

above the x-axis

below the x-axis

f(0) = 4

1 f

above the x-axis

(3–2) ~ –2.2

2 f

(5–2) ~ 11.8

below the x-axis

above the x-axis

FIGURE 3-30

Comment Note that the far right and far left ends of the graph are similar to the ends of the graph of f 1x2 5 x4, which is the leading term of the function f 1x2 5 x4 2 5x2 1 4 or the term with highest degree. On the far right and far left the graph rises. Also note that the graph has three turning points.

Step 4: Plot a few points and draw the graph as a smooth continuous curve. We can now plot the intercepts and use our knowledge of symmetry and where the graph is above and below the x-axis to draw the graph, as shown in Figure 3-31(a). A calculator graph, using the standard viewing window, is shown in Figure 3-31(b). f 1x2 5 x4 2 5x2 1 4

22

f 1x2

1x, f 1x22

21

0

(21, 0)

0

4

(0, 4)

1

0

(1, 0)

3 2

35 2 16

x

2

0

0

y

(22, 0)

3 35 a ,2 b 2 16 (2, 0)

x (b) f(x) = x4 – 5x2 + 4 (a) FIGURE 3-31

Self Check 2

Graph: f 1x2 5 x4 2 10x2 1 9.

Now Try Exercise 19.

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Section 3.3

315

Polynomial and Other Functions

3. Determine Whether a Function Is Even, Odd, or Neither Functions can be classified as being even, odd, or neither even nor odd. These definitions are given next.

Even Function

A function is an even function if f 12x2 5 f 1x2 for all x in the domain of f. Even functions are symmetric about the y-axis.

Odd Function

A function is an odd function if f 12x2 5 f 1x2 for all x in the domain of f. Odd functions are symmetric about the origin.

Since the graph of the function f 1x2 5 x3 2 4x in Example 1 is symmetric about the origin, it represents an odd function. Since the graph of the function f 1x2 5 x4 2 5x 1 4 in Example 2 is symmetric about the y-axis, it represents an even function. If f 1x2 does not have either of these symmetries, it is neither even nor odd.

EXAMPLE 3

SOLUTION

Determining Whether a Function Is Even, Odd, or Neither Determine whether each function is even, odd, or neither: a. f 1x2 5 x2 1 1 b. f 1x2 5 x3

To check whether the function is an even function, we find f 12x2 and see whether f 12x2 5 f 1x2 . To check whether the function is an odd function, we find f 12x2 and see whether f 12x2 5 2f 1x2 .

a. f 12x2 5 12x2 2 1 1 5 x2 1 1 5 f 1x2 Since f 12x2 5 f 1x2 , the function is an even function.

The graph of f 1x2 5 x2 1 1 is shown in Figure 3-32. We see that the graph is symmetric about the y-axis.

b. f 12x2 5 12x2 3 5 2x3 5 2f 1x2

Since f 12x2 2 f 1x2 , the function is not an even function. However, the function is an odd function, because f 12x2 5 2f 1x2 .

The graph of f 1x2 5 x3 is shown in Figure 3-33. Note that the graph is symmetric about the origin. y

y

x x

FIGURE 3-32

f (x) = x 3

FIGURE 3-33

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316

Chapter 3

Functions

Self Check 3

Classify each function as even, odd, or neither: a. f 1x2 5 x3 1 x b. f 1x2 5 x2 1 4 Now Try Exercise 23.

4. Identify the Open Intervals on Which a Function Is Increasing, Decreasing, or Constant

If we trace the graph of a function from left to right and the values f 1x2 increase as shown in Figure 3-34(a), we say that the function is an increasing function. If the values f 1x2 decrease as in Figure 3-34(b), we say that the function is a decreasing function. If the values f 1x2 remain unchanged as x increases, we say that the function is a constant function. y

a

y

b

x

y

a

x

b

a

b

Increasing function

Decreasing function

Constant function

(a)

(b)

(c)

x

FIGURE 3-34

As we would expect, when we trace the graphs of some functions from left to right, there will be intervals where the function increases, intervals where the function decreases, and intervals where the function is constant. We will always state these intervals as open intervals because of the following definition.

Increasing on an Open Interval Decreasing on an Open Interval Constant on an Open Interval

EXAMPLE 4

A function f is increasing on an open interval 1a, b2 if for any x1 and x2 in 1a, b2 , where x1 , x2, then f 1x12 , f 1x22 .

A function f is decreasing on an open interval 1a, b2 if for any x1 and x2 in 1a, b2 , where x1 , x2, then f 1x12 . f 1x22 . A function f is constant on an open interval 1a, b2 if for any x1 and x2 in 1a, b2 , f 1x12 5 f 1x22 .

Identify the Open Intervals Where a Function Increases or Decreases Use the graph of the polynomial function shown to determine the intervals on which the function is increasing or decreasing. y

–5

5

x

–10

FIGURE 3-35 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.3

Polynomial and Other Functions

317

SOLUTION

We trace the graph of the function from left to right and identify the open intervals on which f 1x2 values increase and where they decrease. We see from the graph shown in Figure 3-35 that the values of f 1x2 increase on the open intervals 12`, 222 and 10, 22 . They decrease on the open intervals 122, 02 and 12, ` 2 . Hence, the function is increasing on 12`, 222 c 10, 22 and is decreasing on 122, 02 c 12, ` 2 .

Self Check 4

Use the graph shown in Figure 3-35 to identify the open intervals, if any, on which the function is constant. Now Try Exercise 41.

Comment Intervals on which a function is increasing, decreasing, or constant are open intervals. We use interval notation and x-values to write the open intervals. Note that y-values are not used.

5. Graph Piecewise-Defined Functions Some functions, called piecewise-defined functions, are defined by using different equations for different intervals in their domains. To illustrate, we will graph the piecewise-defined function f given by f 1x2 5 e

22 if x # 0 x 1 1 if x . 0

To evaluate this piecewise-defined function, we must determine which part of the function’s definition to use. •

If x # 0, we use the top part of the definition and the corresponding value of f 1x2 is 22. Some examples are: f 1222 5 22,

f 1212 5 22, and

f 102 5 22

In the interval 12`, 0 4 , the function value is always 22 and the graph is the horizontal line y 5 22. •

If x > 0, we use the bottom part of the definition and the corresponding value of f 1x2 is x 1 1. Some examples are: f 112 5 1 1 1 5 2 and f 122 5 2 1 1 5 3

In the interval 10, ` 2 , the graph of the function is the line f 1x2 5 x 1 1. This is a linear function with slope m 5 1 and y-intercept 10, b2 5 10, 12 . The graph of this piecewise-defined function appears in Figure 3-36. y

Comment Note that when x 5 0, we have f 1x2 5 22. For this reason, the point 10, 222 is shown as a closed point and the point 10, 12 is shown as an open point.

x f(x) =

–2 if x ≤ 0 x + 1 if x > 0

FIGURE 3-36

EXAMPLE 5

Graphing a Piecewise-Defined Function 2x if x , 0 Graph the function: f 1x2 5 • x2 if 0 # x # 1. 1 if x . 1

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318

Chapter 3

Functions

SOLUTION

This piecewise-defined function is defined in three parts. We will graph each part of the function. That is, we will graph f 1x2 5 2x in the interval 12`, 02 , f 1x2 5 x2 in the interval 3 0, 1 4 , and f 1x2 5 1 in the interval 11, ` 2 . If x , 0, the value of f 1x2 is determined by the equation f 1x2 5 2x. We graph the line with slope m 5 21 and y-intercept 10, b2 5 10, 02 in the interval 12`, 02 . In the interval 12`, 02 , the function is decreasing. If 0 # x # 1, the value of f 1x2 is x2. We graph the parabola in the interval [0, 1]. In the open interval (0,1) the function is increasing. If x . 1, the value of f 1x2 is 1. In the open interval 11, ` 2 , the function is constant and its graph is the same as the graph of y 5 1. The graph of the piecewise function appears in Figure 3-37. y

1

–1

0

1 f(x) =

x

–x if x < 0 x 2 if 0 ≤ x ≤ 1 1 if x > 1

FIGURE 3-37

Self Check 5

Graph: f 1x2 5 e

2x if x # 0 . x 2 1 if x . 0

Now Try Exercise 57.

ACCENT ON TECHNOLOGy

Piecewise-Defined Functions Piecewise-defined functions can be graphed on a calculator. We will graph the 2x if x , 0 piecewise-defined function f 1x2 5 • x2 if 0 # x # 1 given in Example 5. 1 if x . 1 • • • •

MATH The inequality symbols are found by pressing (TEST) shown in Figure 3-38(a). The “and” command is found in the LOGIC menu which is accessed via the TEST menu. See Figure 3-38(b). When graphing these functions, place the calculator in DOT mode. See Figure 3-38(c). Enter the function as shown in Figure 3-38(d). Note that the use of parentheses is essential. A viewing window of 3 25, 5 4 for x and 3 23, 3 4 for y is used here.

The graph of the function is shown in Figure 3-38(e).

(a)

(b)

(c)

FIGURE 3-38 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.3

Polynomial and Other Functions

(d)

319

(e) FIGURE 3-38 (Continued)

6. Evaluate and Graph the Greatest-Integer Function The greatest-integer function is important in many business applications and in the field of computer science. This function is determined by the equation f 1x2 5 Œxœ, where the value of f 1x2 that corresponds to x is the greatest integer that is less than or equal to x.

Comment Recall that the set of integers is 5 % , 22,21, 0, 1, 2, % 6 .

For example,

f 12.712 5 Œ2.71œ 5 2

f 123.52 5 Œ23.5œ 5 23 f 1102 5 Œ10œ 5 10 f 1p2 5 Œpœ 5 3

f 122.52 5 Œ22.5œ 5 23

EXAMPLE 6 SOLUTION

Graphing the Greatest-Integer Function

Graph: f 1x2 5 Œxœ.

We will list several intervals and determine the corresponding values of the greatest-integer function. Then we will use these values to graph the function. 3 0, 12 3 1, 22 3 2, 32

f 1x2 5 Œxœ 5 0

f 1x2 5 Œxœ 5 1 f 1x2 5 Œxœ 5 2

For numbers from 0 to 1 (not including 1), the greatest integer in the interval is 0. For numbers from 1 to 2 (not including 2), the greatest integer in the interval is 1. For numbers from 2 to 3 (not including 3), the greatest integer in the interval is 2.

Within each interval, the values of y are constant, but they jump by 1 at integer values of x. The graph is shown in Figure 3-39. From the graph, we can see that the domain of the greatest integer function is the interval 12`, ` 2 . The range is the set of integers. y

x y = [[x]]

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Chapter 3

Functions

Self Check 6

Find a. Œ7.61œ and b. Œ23.75œ. Now Try Exercise 63.

Since the greatest-integer function is made up of a series of horizontal line segments, it is an example of a group of functions called step functions.

EXAMPLE 7

Graphing a Step Function Occurring in an Application To print business forms, a printing company charges customers $10 for the order, plus $20 for each box containing 200 forms. The printing company counts any portion of a box as a full box. Graph this step function.

SOLUTION

To graph the step function, we will determine the cost for printing various amounts of boxes of forms. Then we will graph our result. If we order the forms and then change our minds before the forms are printed, the cost will be $10. Thus, the ordered pair (0, 10) will be on the graph. If we purchase up to one full box, the cost will be $10 for the order and $20 for the printing, for a total of $30. Thus, the ordered pair (1, 30) will be on the graph. The cost for 112 boxes will be the same as the cost for 2 full boxes, or $50. Thus, the ordered pairs (1.5, 50) and (2, 50) are on the graph. The complete graph is shown in Figure 3-40.

Cost

320

100 90 80 70 60 50 40 300 20 10 0 1

2

3 4 Boxes

5

FIGURE 3-40

Self Check 7

Find the cost of 412 boxes. Now Try Exercise 69.

Self Check Answers

y

1.

y

2.

10 8 6 4 2 –4

–2 –2 –4 –6 –8 –10

3. a. odd

b. even

12 8 4 –4 1 2 3 4 5

x

–2 –4 –8 –12 –16

1 2 3 4

x

f(x) = x4 – 10x2 + 9

f(x) = x 3 – 9x

4. no intervals

y

5.

6. a. 7

b. 24

7. $110

x

f(x) =

2x if x ≤ 0 x – 1 if x > 0

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Section 3.3

Exercises 3.3 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. The degree of the function y 5 f 1x2 5 x4 2 3 is . 2. Peaks and valleys on a polynomial graph are called points. 3. The graph of a nth degree polynomial function can have at most turning points. 4. If the graph of a function is symmetric about the , it is called an even function. 5. If the graph of a function is symmetric about the function. origin, it is called an 6. If the values of f 1x2 get larger as x increases on an interval, we say that the function is on the open interval. 7. functions are defined by different equations for different intervals in their domains. 8. If the values of f 1x2 get smaller as x increases on an interval, we say that the function is on the open interval. 9. Œ3.69œ 5 . 10. If the values of f 1x2 do not change as x increases on an interval, we say that the function is on the open interval.

15. f 1x2 5 x3 1 x2

Graph each polynomial function. 11. f 1x2 5 x3 2 9x 12. f 1x2 5 x3 2 16x

321

16. f 1x2 5 2x3 1 1

y

y

x x

17. f 1x2 5 x3 2 x2 2 4x 1 4 y

x

18. f 1x2 5 4x3 2 4x2 2 x 1 1 y

x

19. f 1x2 5 x4 2 2x2 1 1

Practice

y

Polynomial and Other Functions

y

20. f 1x2 5 x4 2 5x2 1 4 y

y

40 20

10 –2 –10

x

2

x

–5

5

x

21. f 1x2 5 2x4 1 5x2 2 4

–20 –40

13. f 1x2 5 2x3 2 4x2

14. f 1x2 5 x3 2 x

y

y

y

40 x 20 –5

5

x

x

–20 –40

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x

322

Chapter 3

Functions

22. f 1x2 5 x 1x 2 32 1x 2 22 1x 1 12

State the open intervals where each function is increasing, decreasing, or constant.

y

37.

38.

y

y

x x x

Determine whether each function is even, odd, or neither. 23. f 1x2 5 x4 1 x2 24. f 1x2 5 x3 2 2x 25. f 1x2 5 x3 1 x2 27. f 1x2 5 x5 1 x3 29. f 1x2 5 2x3 2 3x

39.

40.

y

26. f 1x2 5 x6 2 x2 28. f 1x2 5 x3 2 x2 30. f 1x2 5 4x2 2 5

y

x x

Determine whether each function is even, odd, or neither. 31. 32. y y

41.

42.

y

y

20 5 10 –3 –2 –1 –10

1

2

3

x

–3

–2 –1

1

2

3

x x

x

–5 –10

33.

34.

y

y

43. f 1x2 5 x2 2 4x 1 4

10

20 5 10 –3

–2

–1 –10

1

2

3

x

–3

–2 –1

1

2

3

x

–5

Evaluate each piecewise-defined function.

–10

35.

–6

36.

y

–4

40

20

20

–2 –20

2

4

6

x

–6

–4

–2 –20

45. f 1x2 5 e

2x 1 2 if x , 0 3 if x $ 0

46. f 1x2 5 e

x 2 2 if x , 1 x2 if x $ 1

a. f 1222

y

40

44. f 1x2 5 4 2 x2

2

4

6

x

a. f 112

b. f 102 b. f 152

2 if x , 0 47. f 1x2 5 • 2 2 x if 0 # x , 2 x 1 1 if x $ 2 a. f 1212

b. f 112

2x if x , 0 48. f 1x2 5 • 3 2 x if 0 # x , 2 0x0 if x $ 2 1 b. f 102 a. f 20.52

c. f 122 c. f 122

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Section 3.3

Graph each piecewise-defined function. x 1 2 if x , 0 49. f 1x2 5 e 2 if x $ 0

53. f 1x2 5 e

Polynomial and Other Functions

24 2 x if x , 1 3 if x $ 1 y

y

x x

50. f 1x2 5 e

2x if x , 0 22x if x $ 0

54. f 1x2 5 e

y

25 2 x if x , 1 23 if x $ 1 y x

x

51. f 1x2 5 e

x if x # 0 2 if x . 0

55. f 1x2 5 e

y

2x if x , 0 x2 if x $ 0 y

x x

2x if x , 0 52. f 1x2 5 • 1 x if x . 0 2

56. f 1x2 5 e

0 x 0 if x , 0 "x if x $ 0

y

y

x x

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323

324

Chapter 3

Functions

1 64. y 5 fi x 1 3fl 3

0 if x , 0 57. f 1x2 5 • x2 if 0 # x # 2 4 2 2x if x . 2

y

y x

x

65. y 5 Œxœ 2 1 y

2 if x , 0 1 2 58. f x 5 • 2 2 x if 0 # x , 2 x if x $ 2

x

66. y 5 Œx 1 2œ

y

y

x x

Evalute each function at the indicated x-values. a. f 132 b. f 1242 c. f 122.32 59. f 1x2 5 Œxœ 60. f 1x2 5 Œ3xœ

61. f 1x2 5 Œx 1 3œ

a. f 142

a. f 1212

62. f 1x2 5 Œ4xœ 2 1 a. f 1232

b. f 1222 c. f 121.22

2 b. f a b c. f 11.32 3 b. f 102

c. f 1p2

Applications 67. Grading scales A mathematics instructor assigns letter grades according to the following scale. From

Up to but less than

Grade

60% 70% 80% 90%

70% 80% 90% 100% (including 100%)

D C B A

Graph the ordered pairs 1p, g2 , where p represents the percent and g represents the grade. Find the final semester grade of a student who has test scores of 67%, 73%, 84%, 87%, and 93%.

Graph each function. 63. y 5 Œ2xœ y

g x

A B C D 50% 60% 70% 80% 90% 100%

p

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Section 3.3

68. Calculating grades See Exercise 67 and find the final semester grade of a student who has test scores of 53%, 65%, 64%, 73%, 89%, and 82%. 69. Renting a jeep A rental company charges $20 to rent a Jeep Wrangler for one day, plus $4 for every 100 miles (or portion of 100 miles) that it is driven. Graph the ordered pairs 1m, C2 , where m represents the miles driven and C represents the cost. Find the cost if the Jeep is driven 275 miles in one day. C

Polynomial and Other Functions

325

72. iPad repair There is a charge of $30, plus $40 per hour (or fraction of an hour), to repair an iPad. Graph the points 1t, C2 , where t is the time it takes to do the job and C is the cost. If it takes 4 hours to repair the iPad, how much did it cost? C 150 110 70 30

36 32 28 24 20

t 1

2

3

4

73. Rounding numbers Measurements are rarely exact; they are often rounded to an appropriate precision. Graph the points 1x, y2 , where y is the result of rounding the number x to the nearest ten.

m 100 200 300 400

70. Riding in a taxi A taxicab company charges $3 for a trip up to 1 mile, and $2 for every extra mile (or portion of a mile). Graph the ordered pairs 1m, C2 , where m represents the miles traveled and C represents the cost. Find the cost to ride 1014 miles.

y 30 20 10 10

20

30

x

40

C 9 8 7 6 5 4 3 2 1

74. Signum function Computer programmers often use the following function, denoted by y 5 sgn x. Graph this function and find its domain and range.

m 2

1

3

4

5

71. Computer communications An on-line information service charges for connect time at a rate of $12 per hour, computed for every minute or fraction of a minute. Graph the points 1t, C2 , where C is the cost of t minutes of connect time. Find the cost of 712 minutes.

21 if x , 0 y 5 • 0 if x 5 0 1 if x . 0 y

x

C 0

1.00 .80 .60 .40 .20

0

75. Graph the function defined by y 5 xx and compare it to the graph in Exercise 74. Are the graphs the same? t 1

2

3

y

4

x

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326

Chapter 3

Functions

76. Graph: y 5 x 1 0 x 0 .

79. Graph the function y 5 1x 2 a2 1x 2 b2 for several values of a and b. What is the relationship between the x-intercepts and the equation?

y

80. Use the insight you gained in Exercise 79 to factor x3 2 3x2 2 4x 1 12. 81. If f 1x2 5 3x 1 2, find f 1x 1 12 and f 1x2 1 1.

x

Review

Discovery and Writing Use a graphing calculator to explore the properties of graphs of polynomial functions. Write a paragraph summarizing your observations. 77. Graph the function y 5 x2 1 ax for several values of a. How does the graph change? 78. Graph the function y 5 x3 1 ax for several values of a. How does the graph change?

82. If f 1x2 5 x2, find f 1x 2 22 and f 1x2 2 2. 83. If f 1x2 5

3x 1 1 , find f 1x 2 32 and f 1x2 2 3. 5

84. If f 1x2 5 8, find f 1x 1 82 and f 1x2 1 8. 85. Solve: 2x2 2 3 5 x. 86. Solve: 4x2 5 24x 2 37.

3.4 Transformations of the Graphs of Functions In this section, we will learn to 1. 2. 3. 4.

Use vertical translations to graph functions. Use horizontal translations to graph functions. Graph functions using two translations. Use reflections about the x- and y-axes to graph functions.

EVRON/Shutterstock.com

5. Use vertical stretching and shrinking to graph functions. 6. Use horizontal stretching and shrinking to graph functions. 7. Graph functions using a combination of transformations. We can often transform the graph of a function into the graph of another function by shifting the graph vertically or horizontally. Also, we can reflect a graph about the x- or y-axis, and stretch or shrink a graph horizontally or vertically to transform the graph of a function into the graph of another function. In this section, we will graph new functions from known ones using these methods. Consider a white water rafting trip on the Ocoee River in Tennessee. Suppose one company charges a group of students $20 for each hour on the river, plus $100 for a guide and equipment. The cost of the rafting trip can be represented by the function C1 1t2 5 20t 1 100

where C1 1t2 represents the cost in dollars to raft t hours on the river. If the company increases its charge for the guide and equipment to $150, the new cost function can be represented by C2 1t2 5 20t 1 150

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Section 3.4

327

Transformations of the Graphs of Functions

The graphs of the two cost functions are shown in Figure 3-41. Co s t C2(t) = 20t + 150

250 200 150

C1(t) = 20t + 100

100 50

Ti me 1

2

3

4

FIGURE 3-41

Note that if we shift the graph of C1 1t2 50 units vertically, we obtain the graph of C2 1t2 . This shift is called a translation. As we continue our study of translations, it will be helpful to review the graphs of the basic functions that are shown in Figure 3-8 in Section 3.1. In this section, 3 the graphs of f 1x2 5 x2, f 1x2 5 x3, f 1x2 5 0 x 0 , f 1x2 5 "x, and f 1x2 5 " x will be translated and stretched in various ways.

1. Use Vertical Translations to Graph Functions

Comment The graphs of the functions shown in Figure 3-42 are exactly the graphs we would expect based on our previous study of quadratic functions. y

The graphs of functions can be identical except for their position in the xy-plane. For example, Figure 3-42 shows the graph of y 5 x2 1 k for three values of k. If k 5 0, we have the graph of y 5 x2. The graph of y 5 x2 1 2 is identical to the graph of y 5 x2, except that it is shifted 2 units upward. The graph of y 5 x2 2 3 is identical to the graph of y 5 x2, except that it is shifted 3 units downward. These shifts are called vertical translations.

y = x2 + 2

y=x

y 5 x2

x

y

4

1x, y2

(22, 4)

22

21

1

(21, 1)

0

0

y = x2 – 3

1 2

y 5 x2 2 3 x

y

6

1x, y2

(22, 6)

22

1

21

3

(21, 3)

21

22

(21, 22)

(0, 0)

0

2

(0, 2)

0

23

(0, 23)

1

(1, 1)

1

3

(1, 3)

1

22

(1, 22)

4

(2, 4)

2

6

(2, 6)

2

1

x

y

22

2

x

y 5 x2 1 2

1x, y2

(22, 1)

(2, 1)

FIGURE 3-42

In general, we can make the following observations:

Vertical Translations

If f is a function and k is a positive number, then

• The graph of y 5 f 1x2 1 k is identical to the graph of y 5 f 1x2 except that it is translated k units upward.

• The graph of y 5 f 1x2 2 k is identical to the graph of y 5 f 1x2 except that it is translated k units downward. y y = f(x) + k

x y = f(x) y = f(x) – k

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328

Chapter 3

Functions

EXAMPLE 1 SOLUTION

Using Vertical Translations to Graph Functions

Graph each function: a. g 1x2 5 0 x 0 2 2

b. h 1x2 5 0 x 0 1 3

We will use vertical translations of f 1x2 5 0 x 0 to graph each function.

a. The graph of g 1x2 5 0 x 0 2 2 is identical to the graph of f 1x2 5 0 x 0 , except that it is translated 2 units downward. It is translated downward because 2 is subtracted from 0 x 0 . The graph of g 1x2 is shown in Figure 3-43(a). b. The graph of h 1x2 5 0 x 0 1 3 is identical to the graph of f 1x2 5 0 x 0 , except that it is translated 3 units upward. It is translated upward because 3 is added to 0 x 0 . The graph of h 1x2 is shown in Figure 3-43(b). y

y

g(x) = |x| − 2 h(x) = |x| + 3

x

x

(b)

(a) FIGURE 3-43

Self Check 1

Fill in the blanks: The graph of g 1x2 5 x2 1 3 is identical to the graph of f 1x2 5 x2 , except that it is translated units . The graph of h 1x2 5 x2 2 4 is identical to the graph of f 1x2 5 x2 , except that it is translated units .

Now Try Exercise 27.

2. Use Horizontal Translations to Graph Functions

Figure 3-44 shows the graph of y 5 1x 1 k2 2 for three values of k. If k 5 0, we have the graph of y 5 x2. The graph of y 5 1x 2 22 2 is identical to the graph of y 5 x2, except that it is shifted 2 units to the right. The graph of y 5 1x 1 32 2 is identical to the graph of y 5 x2, except that it is shifted 3 units to the left. These shifts are called horizontal translations.

Comment The graphs of the functions shown in Figure 3-44 are exactly the graphs we would expect based on our previous knowledge of quadratic functions.

y 5 x2

y

y = (x – 2)2

x

y 5 1x 1 32 2

4

1x, y2

(0, 4)

25

4

(25, 4)

1

1

(1, 1)

24

1

(24, 1)

(0, 0)

2

0

(2, 0)

23

0

(23, 0)

1

(1, 1)

3

1

(3, 1)

22

1

(22, 1)

4

(2, 4)

4

4

(4, 4)

21

4

(21, 4)

x

y

22

4

(22, 4)

0

21

1

(21, 1)

0

0

1 2

y = x2

y = (x + 3)2

1x, y2

y 5 1x 2 22 2

x

y

x

y

1x, y2

FIGURE 3-44

In general, we can make the following observations.

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Section 3.4

Horizontal Translations y y = f(x + k)

y = f(x)

y = f(x – k) x

EXAMPLE 2

SOLUTION

Transformations of the Graphs of Functions

329

If f is a function and k is a positive number, then

• The graph of y 5 f 1x 2 k2 is identical to the graph of y 5 f 1x2 except that it is translated k units to the right.

• The graph of y 5 f 1x 1 k2 is identical to the graph of y 5 f 1x2 except that it is translated k units to the left.

Using Horizontal Translations to Graph Functions

Graph each function: a. g 1x2 5 0 x 2 4 0

b. h 1x2 5 0 x 1 2 0

We will use horizontal translations of the graph of f 1x2 5 0 x 0 to graph each function. a. The graph of g 1x2 5 0 x 2 4 0 is identical to the graph of f 1x2 5 0 x 0 , except that it is translated 4 units to the right. It is translated 4 units to the right because within the absolute value symbols 4 is subtracted from x. The graph of g 1x2 is shown in Figure 3-45(a). b. The graph of h 1x2 5 0 x 1 2 0 is identical to the graph of f 1x2 5 0 x 0 , except that it is translated 2 units to the left. It is translated 2 units to the left because within the absolute value symbols 2 is added to x. The graph of h 1x2 is shown in Figure 3-45(b). y

y

g(x) = |x – 4|

h(x) = |x + 2|

x

x (b)

(a) FIGURE 3-45

Self Check 2

Fill in the blanks: The graph of g 1x2 5 1x 2 32 2 is identical to the graph of f 1x2 5 x2 except that it is translated units to the . The graph of h 1x2 5 1x 1 22 2 is identical to the graph of f 1x2 5 x2 except that it is translated units to the . Now Try Exercise 29.

Caution

When using horizontal translations to graph a function, it is easy to shift the function in the wrong direction. • If we see a positive constant subtracted from x, we have the tendency to shift the graph left. This is incorrect. • If we see a positive constant added to x, we have the tendency to shift the graph right. This too is incorrect. We should avoid making these common errors.

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330

Chapter 3

Functions

3. Graph Functions Using Two Translations Sometimes we can obtain a graph by using both a horizontal and a vertical translation.

EXAMPLE 3

SOLUTION

Using a Horizontal and a Vertical Translation to Graph a Function Graph each function: a. g 1x2 5 1x 2 52 3 1 4

b. h 1x2 5 1x 1 22 2 2 2

By inspection, we see that the function in part a involves two translations of f 1x2 5 x3 and the function in part b involves two translations of f 1x2 5 x2. We will perform the horizontal translation first, followed by a vertical translation to obtain the graph of each function.

a. The graph of g 1x2 5 1x 2 52 3 1 4 is identical to the graph of f 1x2 5 x3, except that it is translated 5 units to the right and 4 units upward, as shown in Figure 3-46(a).

b. The graph of h 1x2 5 1x 1 22 2 2 2 is identical to the graph of f 1x2 5 x2, except that it is translated 2 units to the left and 2 units downward, as shown in Figure 3-46(b). y

y f(x) = x 3 (5, 4)

f(x) = x2 –2

4

x

–2 5

x g(x) = (x – 5)3 + 4

(a)

h(x) = (x + 2)2 – 2

(b) FIGURE 3-46

Self Check 3

Fill in the blanks: The graph of g 1x2 5 0 x 2 4 0 1 5 is identical to the graph of f 1x2 5 0 x 0 , except that it is translated units to the and units . Now Try Exercise 31.

ACCENT ON TECHNOLOGy

Translations of Functions We can use a graphing calculator to help understand translations of functions. Compare the graph of each function listed to the graph of f 1x2 5 "x. a. g 1x2 5 "x 2 4

b. h 1x2 5 "x 1 2

c. k 1x2 5 "x 2 2 1 3

a. Graph the functions f 1x2 5 "x and g 1x2 5 "x 2 4 on the same screen using the window shown in Figure 3-47(a). From these graphs, we can see that the graph of g is obtained by shifting the graph of f downward 4 units.

b. Graph the functions f 1x2 5 "x and h 1x2 5 "x 1 2 on the same screen using the window shown in Figure 3-47(b). From these graphs, we can see that the graph of h is obtained by shifting the graph of f to the left 2 units.

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Section 3.4

331

Transformations of the Graphs of Functions

c. Graph the functions f 1x2 5 "x and k 1x2 5 "x 2 2 1 3 on the same screen using the window shown in Figure 3-47(c). From these graphs, we can see that the graph of k is obtained by shifting the graph of f right 2 units and upward 3 units.

Comment

To make the graph of f 1x2 5 "x appear

(a) Vertical shift downward 4 units. The graph of f 1x2 5 "x is the bold graph.

bold, scroll to the left of Y1 and press ENTER

.

(b) Horizontal shift left 2 units. The graph of f 1x2 5 "x is the bold graph.

(c) Horizontal shift 2 units right and vertical shift 3 units upward. The graph of f 1x2 5 "x is the bold graph.

FIGURE 3-47

4. Use Reflections about the x- and y- Axes to Graph Functions Figure 3-48(a) shows that the graph of y 5 2"x is identical to the graph of y 5 "x except that it is reflected about the x-axis. Figure 3-48(b) shows that the graph of y 5 "2x is identical to the graph of y 5 "x except that it is reflected about the y-axis.

y

y

y 5 2"x

y= x

x

y =– x

x

y

0

0

1

21

4

22

1x, y2

y = –x

y= x

x

y 5 "2x

1x, y2

y

0

0

(0, 0)

(1, 21)

21

1

(21, 1)

(4, 22)

24

2

(24, 2)

(0, 0)

x

(a)

(b) FIGURE 3-48

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332

Chapter 3

Functions

In general, we can make the following observations.

Reflections

If f is a function, then

• The graph of y 5 2f 1x2 is identical to the graph of f 1x2 except that it is reflected about the x-axis. y y = f(x) x y = –f(x)

• The graph of y 5 f 12x2 is identical to the graph of f 1x2 except that it is reflected about the y-axis. y y = f(x)

y = f(– x)

x

EXAMPLE 4 SOLUTION

Using Reflections to Graph Functions

Graph each function: a. g 1x2 5 2 0 x 1 1 0

b. h 1x2 5 0 2x 1 1 0

By inspection, we see that the function given in part a involves a reflection about the x-axis and the function given in part b involves a reflection about the y-axis. We will use reflections to draw the graph of each function.

a. The graph of g 1x2 5 2 0 x 1 1 0 is identical to the graph of f 1x2 5 0 x 1 1 0 , except that it is reflected about the x-axis. This is because g 1x2 5 2f 1x2 . The graphs of both functions are shown in Figure 3-49(a).

Comment It is often helpful to think of a reflection as a mirror image of the graph about the x- or y-axis.

b. The graph of h 1x2 5 0 2x 1 1 0 is identical to the graph of f 1x2 5 0 x 1 1 0 , except that it is reflected about the y-axis. This is because f 12x2 5 h 1x2 . The graphs of both functions are shown in Figure 3-49(b). y

y f(x) = |x + 1|

h(x) = |–x + 1| f(x) = |x + 1|

x x g(x) = –|x + 1| (b)

(a) FIGURE 3-49

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Section 3.4

Self Check 4

333

Transformations of the Graphs of Functions

3 3 x is identical to the graph of f 1x2 5 " x, Fill in the blanks: The graph of g 1x2 5 2"

except that it is reflected about the axis. The graph of h 1x2 5 "2x 2 4 is 1 2 identical to the graph of f x 5 "x 2 4, except that it is reflected about the axis. Now Try Exercise 51.

5. Use Vertical Stretching and Shrinking to Graph Functions Figure 3-50 shows the graphs of y 5 x2, y 5 3x2, and y 5 13x2 .

1 y 5 x2 3

y y = 3x2 y = x2

y 5 x2

1 2 y = –x 3 x

y 5 3x2 x

y

4

1x, y2

(22, 4)

22

12

(22, 12)

21

1

(21, 1)

21

3

(21, 3)

0

0

(0, 0)

0

0

(0, 0)

1

1

(1, 1)

1

3

(1, 3)

2

4

(2, 4)

2

12

(2, 12)

x

y

22

1x, y2

x

y

22

4 3

21

1 3

0

0

1

1 3

2

4 3

1x, y2

4 a22, b 3 1 a21, b 3 (0, 0)

1 a1, b 3

4 a2, b 3

FIGURE 3-50

Because each value of y 5 3x2 is 3 times greater than the corresponding value of y 5 x2, its graph is stretched vertically by a factor of 3. Because each value of y 5 13x2 is 3 times smaller than the corresponding value of y 5 x2, its graph shrinks vertically by a factor of 13. In general, we can make the following observations.

Vertical Stretching and Shrinking

If f is a function and k >1, then

• The graph of y 5 kf 1x2 can be obtained by stretching the graph of y 5 f 1x2 vertically by multiplying each value of f 1x2 by k. y y = kf(x), k > 1 y = f(x) x

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334

Chapter 3

Functions

If f is a function and 0 , k , 1, then

• The graph of y 5 kf 1x2 can be obtained by shrinking the graph of y 5 f 1x2 vertically by multiplying each value of f 1x2 by k. y y = f(x) y = kf(x), 0 < k < 1 x

EXAMPLE 5

SOLUTION

Graphing a Function by Vertical Stretching or Shrinking

Graph each function: a. g 1x2 5 2 0 x 0

b. h 1x2 5

1 0x0 2

We will vertically stretch or vertically shrink the basic function f 1x2 5 0 x 0 to graph each of the given functions.

a. The graph of g 1x2 5 2 0 x 0 is identical to the graph of f 1x2 5 0 x 0 except that it is vertically stretched by a factor of 2. This is because each value of 0 x 0 is multiplied by 2. The graphs of both functions are shown in Figure 3-51. y g(x) = 2 x

f(x) = x x

FIGURE 3-51

b. The graph of g 1x2 5 12 0 x 0 is identical to the graph of f 1x2 5 0 x 0 except that it is vertically shrunk by a factor of 12. This is because each value of 0 x 0 is multiplied by 12. The graphs of both functions are shown in Figure 3-52. y

f(x) = x

Comment Note that vertically stretching the graph of a function narrows the graph of the function. Vertically shrinking the graph of a function widens the graph.

1 g(x) = – x 2 x

FIGURE 3-52

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Section 3.4

Self Check 5

Transformations of the Graphs of Functions

335

Fill in the blanks: The graph of g 1x2 5 5x3 is identical to the graph of f 1x2 5 x3 , except that it is vertically by a factor of . The graph of h 1x2 5 15x3 is 3 identical to the graph f 1x2 5 x , except that it is vertically by a factor of . Now Try Exercise 57.

6. Use Horizontal Stretching and Shrinking to Graph Functions Functions can also be graphed by using horizontal stretchings and shrinkings.

Horizontal Stretching and Shrinking

If f is a function and k . 1, then

• The graph of y 5 f 1kx2 can be obtained by shrinking the graph of y 5 f 1x2 horizontally by multiplying each x-value of f 1x2 by k1 . y y = f(kx), k > 1 y = f(x) x

If f is a function and 0 , k , 1, then

• The graph of y 5 f 1kx2 can be obtained by stretching the graph of y 5 f 1x2 horizontally by multiplying each x-value of f 1x2 by k1 . y y = f(x)

x y = f(kx), 0 < k < 1

Because we multiply each x-value of f 1x2 by k to obtain the graph of y 5 f 1kx2 , the graphs are horizontally stretched or shrunk by a factor of k1 .

EXAMPLE 6 SOLUTION

Graphing a Function by Horizontally Shrinking or Stretching

Graph y 5 13x2 2 2 1 using the graph of the function y 5 x2 2 1.

Since 3 . 1, the graph of y 5 13x2 2 2 1 can be obtained by shrinking the graph of y 5 x2 2 1 horizontally by multiplying each x-coordinate of y 5 x2 2 1 by 13.

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336

Chapter 3

Functions

First, we complete the table of solutions for y 5 x2 2 1, shown in Figure 3-53. Next, we multiply each x-value in the table by 13 to obtain the table of solutions for y 5 13x2 2 2 1, also shown in Figure 3-53. y

(2–3 , 3)

(–2, 3)

(– 2–3 , 3) (– 1–3 , 0)

x

y 5 x2 2 1

(2, 3)

y = x2 – 1 1 –,0 3

( )

1x, y2

x

y

22

3

(22, 3)

21

0

(21, 0)

0

21

(0, 21)

1

0

(1, 0)

2

3

(2, 3)

x (–1, 0)

(1, 0) y = (3x)2 – 1

y 5 13x2 2 2 1 y

2

2 3

3

2

1 3

0

0

21

1 3

0

2 3

3

1x, y2

2 a2 , 3b 3 1 a2 , 0b 3 (0, 21) 1 a , 0b 3 2 a , 3b 3

FIGURE 3-53

We now draw each graph as shown in Figure 3-53. Self Check 6

Fill in the blank: The graph of g 1x2 5 Q13xR 2 1 is identical to the graph of f 1x2 5 x2 2 1, except that it is horizontally

2

by a factor of

.

Now Try Exercise 63.

We can summarize the ideas in this section as follows.

Summary of Transformations

If f is a function and k represents a positive number, then The graph of

y 5 f 1x2 1 k y 5 f 1x2 2 k y 5 f 1 x 1 k2

can be obtained by graphing y 5 f 1x2 and translating the graph k units upward.

translating the graph k units downward.

y 5 f 1x 2 k2

translating the graph k units to the left.

y 5 kf 1x2

reflecting the graph about the y-axis.

y 5 2f 1x2

y 5 f 12x2 y 5 kf 1x2

y 5 f 1kx2 y 5 f 1kx2

k.1 0,k,1 k.1 0,k,1

translating the graph k units to the right. reflecting the graph about the x-axis. stretching the graph vertically by multiplying each value of f 1x2 by k. shrinking the graph vertically by multiplying each value f 1x2 by k.

shrinking the graph horizontally by multiplying each x-value of f 1x2 by k1 . stretching the graph horizontally by multiplying each x-value of f 1x2 by k1 .

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Section 3.4

EXAMPLE 7

337

Transformations of the Graphs of Functions

Applying Transformations of Graphs

Figure 3-54 shows the graph of y 5 f 1x2 . Use this graph and a translation to sketch b. y 5 f 1x 2 22 c. y 5 2f 1x2 the graph of a. y 5 f 1x2 1 2 y y = f(x) x

FIGURE 3-54

SOLUTION

We will use the summary of transformations in the section given earlier to graph each function.

a. The graph of y 5 f 1x2 1 2 is identical to the graph of y 5 f 1x2 , except that it is translated 2 units upward. See Figure 3-55(a). b. The graph of y 5 f 1x 2 22 is identical to the graph of y 5 f 1x2 , except that it is translated 2 units to the right. See Figure 3-55(b).

c. The graph of y 5 2f 1x2 is identical to the graph of y 5 f 1x2 , except that it is stretched vertically by multiplying each y-value of f 1x2 by 2. See Figure 3-55(c). y

y

y

y = f(x – 2)

y = f(x) + 2

y = 2f(x) x

x

x

(a)

(b)

(c)

FIGURE 3-55

Self Check 7

Use Figure 3-50 and a reflection to sketch the graph of a. y 5 2f 1x2 b. y 5 f 12x2 Now Try Exercise 79.

7. Graph Functions Using a Combination of Transformations To graph functions involving a combination of transformations, we must apply each translation or stretching to the function.

Strategy for Graphing Using a Sequence of Transformations

To graph a function using a combination of transformations, perform the transformations in the following order: 1. horizontal translation 2. stretching or shrinking 3. reflection 4. vertical translation

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338

Chapter 3

Functions

EXAMPLE 8 SOLUTION

Graphing a Function by Using a Combination of Transformations

Use the function f 1x2 5 0 x 0 to graph g 1x2 5 3 0 x 2 2 0 1 4.

We will graph g 1x2 5 3 0 x 2 2 0 1 4 by applying three transformations to the basic function f 1x2 5 0 x 0 :

Step 1: Translate f 1x2 5 0 x 0 horizontally 2 units to the right. Step 2: Vertically stretch the graph by a factor of 3.

Step 3: Translate the graph vertically 4 units upward.

Step 1: Translate f 1x2 5 0 x 0 horizontally 2 units to the right to obtain the graph of y 5 0x 2 20. y

y

y = |x – 2| f(x) = |x|

x x

(b)

(a) FIGURE 3-56

Step 2: Vertically stretch the graph of y 5 0 x 2 2 0 by a factor of 3 to obtain the graph of y 5 3 0 x 2 2 0 . y

y = 3|x – 2| x FIGURE 3-57

Step 3: Translate the graph of y 5 3 0 x 2 2 0 vertically 4 units upward to obtain the graph of g 1x2 5 3 0 x 2 2 0 1 4. y

g(x) = 3|x – 2| + 4 x FIGURE 3-58

Self Check 8

Use the graph of the function f 1x2 5 x3 to graph g 1x2 5 13 1x 1 12 3 2 2.

Now Try Exercise 69.

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Section 3.4

Self Check Answers

Transformations of the Graphs of Functions

339

1. 3, upward; 4, downward 4. x; y 7. a.

2. 3, right; 2, left 3. 4, right; 5, upward 1 5. stretched, 5; shrunk, 6. stretched, 3 5 y y y b. 8. y = –f(x) x

x

y = f(–x)

x g(x) = 1 – (x + 1)3 – 2 3

Exercises 3.4 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. The graph of y 5 f 1x2 1 5 is identical to the graph of y 5 f 1x2 except that it is translated 5 units . is identical to the graph of 2. The graph of y 5 f 1x2 except that it is translated 7 units downward. 3. The graph of y 5 f 1x 2 32 is identical to the graph of y 5 f 1x2 except that it is translated 3 units . 4. The graph of y 5 f 1x 1 22 is identical to the graph of y 5 f 1x2 except that it is translated 2 units . 5. To draw the graph of y 5 1x 1 22 2 2 3, translate the graph of y 5 x2 units to the left and 3 units . 6. To draw the graph of y 5 1x 2 32 3 1 1, translate and 1 unit the graph of y 5 x3 3 units to the . 7. The graph of y 5 f 12x2 is a reflection of the graph of y 5 f 1x2 about the . is a reflection of the graph 8. The graph of of y 5 f 1x2 about the x-axis.

9. The graph of y 5 f 14x2 shrinks the graph of by multiplying each x-value y 5 f 1x2 of f 1x2 by 14 . 10. The graph of y 5 8f 1x2 stretches the graph of y 5 f 1x2 by a factor of 8.

Practice

The graph of each function is a translation of the graph of f 1x2 5 x2. Graph each function.

11. g 1x2 5 x2 2 2

12. g 1x2 5 1x 2 22 2

y

y

x x

13. g 1x2 5 1x 1 32 2

14. g 1x2 5 x2 1 3

y

y

x x

15. h 1x2 5 1x 1 12 2 1 2 y

16. h 1x2 5 1x 2 32 2 2 1 y

x x

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340

Chapter 3

Functions

1 2 1 17. h 1x2 5 ax 1 b 2 2 2 y

3 2 5 18. h 1x2 5 ax 2 b 1 2 2 y

The graph of each function is a translation of the graph of f 1x2 5 0 x 0 . Graph each function. 28. g 1x2 5 0 x 0 2 2 27. g 1x2 5 0 x 0 1 2 y

y

x x x x

The graph of each function is a translation of the graph of f 1x2 5 x3. Graph each function. 19. g 1x2 5 x3 1 1 20. g 1x2 5 x3 2 3 y

29. g 1x2 5 0 x 2 5 0

30. g 1x2 5 0 x 1 4 0

y

y

y

x

x

x

31. f 1x2 5 0 x 1 2 0 2 1

32. h 1x2 5 0 x 2 3 0 1 3

x

21. g 1x2 5 1x 2 22 3

22. g 1x2 5 1x 1 32 3

y

y

y

y

x

x x x

23. h 1x2 5 1x 2 22 3 2 3 y

24. h 1x2 5 1x 1 12 3 1 4 y

The graph of each function is a translation of the graph of f 1x2 5 "x. Graph each function.

33. g 1x2 5 "x 1 1

34. g 1x2 5 "x 2 3

y

y

x

x x

26. y 2 7 5 1x 2 52 3

25. y 1 2 5 x3

x

35. g 1x2 5 "x 1 2

y

y

36. g 1x2 5 "x 2 4

y

y

x

x

x

x

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Section 3.4

37. h 1x2 5 "x 2 2 2 1 y

38. h 1x2 5 "x 1 2 1 3 y

Transformations of the Graphs of Functions

47. h 1x2 5 2x3

341

48. f 1x2 5 2 0 x 0

y

y

x x

x x

The graph of each function is a translation of the graph of 3 f 1x2 5 " x. Graph each function. 3 39. g 1x2 5 "x 2 4

3 40. g 1x2 5 " x13

y

3 50. g 1x2 5 " 2x

49. f 1x2 5 2"x y

y

x

y

x

x

x

52. g 1x2 5 12x2 2

51. f 1x2 5 0 2x 0 y

3 41. g 1x2 5 "x 2 2

3 42. g 1x2 5 "x 1 5

y

y

y

x x

x

3 43. h 1x2 5 " x1121

y

x

3 44. h 1x2 5 " x2121

y

The graph of each function is a vertical stretching or shrinking of the graph of y 5 x2, y 5 x3, or y 5 0 x 0 . Graph each function. 1 54. g 1x2 5 x2 53. f 1x2 5 2x2 2 y

y

x x

x

The graph of each function is a reflection of the graph of 3 y 5 x2, y 5 x3, y 5 0 x 0 , y 5 "x, or y 5 " x. Graph each function. 45. f 1x2 5 2x2 46. g 1x2 5 12x2 3 y

55. h 1x2 5 23x2

x

1 56. f 1x2 5 2 x2 3

y

y

x

y

x x

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x

342

Chapter 3

Functions

67. h 1x2 5 22 0 x 0 1 3

58. g 1x2 5 2x3

1 57. f 1x2 5 x3 2 y

68. f 1x2 5 22 0 x 1 3 0 y

y

y x x x

x

69. f 1x2 5 2 0 x 2 2 0 1 1

1 60. f 1x2 5 0 x 0 3

59. h 1x2 5 23 0 x 0

y

70. f 1x2 5 23 0 x 1 5 0 2 2 y

x

y

y

x x

x

72. g 1x2 5 2"x 1 3 1x $ 232

71. f 1x2 5 2"x 1 3 1x $ 02

The graph of each function is a horizontal stretching or shrinking of the graph of y 5 x2 or y 5 x3. Graph each function. 1 3 61. f 1x2 5 a xb 62. f 1x2 5 12x2 3 2

y

y

y

y

x x x

x

63. f 1x2 5 12x2 2

73. h 1x2 5 2"x 2 2 1 1 1x $ 22

64. f 1x2 5 122x2 3

y

1 74. h 1x2 5 "x 1 5 2 2 2 1x $ 252 y

y

y

x x

x

75. g 1x2 5 22 1x 1 22 3 2 1

x

Graph each function using a combination of transformations applied to the graph of a basic function. 65. g 1x2 5 3 1x 1 22 2 2 1 y

1 66. g 1x2 5 2 1x 1 12 2 1 1 3

y

1 76. g 1x2 5 1x 1 12 3 2 1 3 y

x x

y

x

x

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Section 3.4 3 77. f 1x2 5 2" x14

85. y 5 2f 12x2

3 78. f 1x2 5 22" x11

343

86. y 5 f 1x 1 12 2 2

y

y

y

Transformations of the Graphs of Functions

y

x

x

x

x

Discovery and Writing

Use the following graph and a translation, stretching, or reflection to sketch the graph of each function.

Use a graphing calculator to perform each experiment. Write a brief paragraph describing your findings.

y

87. Investigate the translations of the graph of a function by graphing the parabola y 5 1x 2 k2 2 1 k for several values of k. What do you observe about successive positions of the vertex?

y = f(x) x

88. Investigate the translations of the graph of a function by graphing the parabola y 5 1x 2 k2 2 1 k2 for several values of k. What do you observe about successive positions of the vertex?

80. y 5 f 1x 1 12

79. y 5 f 1x2 1 1 y

89. Investigate the horizontal stretching of the graph of a function by graphing y 5 "ax for several values of a. What do you observe?

y

x

81. y 5 2f 1x2

x

x 82. y 5 f a b 2

y

y

x

83. y 5 f 1x 2 22 1 1

x

Simplify each function. x2 1 x 2 6 93. 2 x 1 5x 1 6

y

x

Write a paragraph using your own words. 91. Explain why the effect of vertically stretching a graph by a factor of 21 is to reflect the graph about the x-axis. 92. Explain why the effect of horizontally stretching a graph by a factor of 21 is to reflect the graph about the y-axis.

Review

84. y 5 2f 1x2

y

90. Investigate the vertical stretching of the graph of a function by graphing y 5 b"x for several values of b. What do you observe? Are these graphs different from the graphs in Exercise 89?

x

94.

2x2 1 3x 2x2 1 x 2 3

Find the domain of each function. Use interval notation to write each answer. 95. f 1x2 5

x17 x23

96. f 1x2 5

x2 1 1 x2 1 3x 1 2

Perform each division and write the answer in quotient 1 remainder divider form. 97.

x2 1 3x x11

98.

x2 1 3 x11

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3.5 Rational Functions In this section, we will learn to 1. Find the domain of a rational function. 2. Understand the characteristics of rational functions and their graphs. 3. Find vertical asymptotes of rational functions. 4. Find horizontal asymptotes of rational functions. 5. Identify slant asymptotes of rational functions. 6. Graph rational functions. 7. Understand when a graph has a missing point. 8. Solve problems modeled by rational functions.

Walter G Arce/Shutterstock.com

We have discussed polynomial functions and now focus on another class of functions called rational functions. Rational functions are defined by rational expressions that are quotients of polynomials. For example, consider the time t it takes a Nascar driver to drive the 500 miles in the Daytona 500 race. The time t can be defined as a function of the average rate of the driver’s speed r. That is t 5 f 1r2 5

500 r

If a driver averages a speed of 170 mph, we can evaluate f 11702 to determine the time in hours driven. f 11702 5

500 50 5 170 17

This is approximately 2.94 hours.

We will discuss this type of function in this section.

Rational Functions

A rational function is a function defined by an equation of the form y5

P 1x2 Q 1x2

where P 1x2 and Q 1x2 are polynomials and Q 1x2 2 0.

1. Find the Domain of a Rational Function

Because rational functions are quotients of polynomials and Q 1x2 is the denominator of a fraction, Q 1x2 cannot equal 0. Thus, the domain of a rational function must exclude all values of x for which Q 1x2 5 0. Here are some examples of rational functions and their domains:

Function f 1x2 5

f 1x2 5

3 x17

5x 1 2 x2 2 4 2x f 1x2 5 2 x 13

Domain

12`, 272 c 127, ` 2 , x cannot equal 27

12`, 222 c 122, 22 c 12, ` 2 , x cannot equal 2 or 22

12`, ` 2 , x can equal any real number

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Section 3.5

EXAMPLE 1

SOLUTION

Rational Functions

345

Finding the Domain of a Rational Function

Find the domain of f 1x2 5

3x 1 2 . x 2 7x 1 12 2

We can factor the denominator to see what values of x will give 0’s in the denominator. These values are not in the domain. To find the numbers x that make the denominator 0, we set x2 2 7x 2 12 equal to 0 and solve for x. x2 2 7x 1 12 5 0

Caution Do not exclude values of the variable from the domain that make the numerator equal to zero, unless the denominator has the same factor. We will discuss this situation later in this section.

Self Check 1

1x 2 42 1x 2 32 5 0

Factor x2 2 7x 1 12.

x 2 4 5 0 or x 2 3 5 0 x54

Set each factor equal to 0.

x53

Solve each linear equation.

Since 4 and 3 make the denominator 0, the domain is the set of all real numbers except x 5 4 and x 5 3. In interval notation, we have 12`, 32 c 13, 42 c 14, ` 2 . Find the domain of f 1x2 5

2x 2 3 . x 2x22 2

Now Try Exercise 23.

ACCENT ON TECHNOLOGy

FIGURE 3-59

Comment When using a graphing calculator to graph a rational function, make sure that the function is entered properly. To avoid making an error, place both the numerator and denominator in parentheses.

Domain and Range of a Rational Function We can use a graphing calculator to find domains and ranges of rational functions. If we use settings of [210, 10] for x and [210, 10] for y and graph 1 1 f 1x2 5 2x x 2 1 , we will obtain Figure 3-59. From the graph, we can see that every real number x except 1 gives a value of y. Thus, the domain of the function is 12`, 12 c 11, ` 2 . We can also see that y can be any value except 2. The range of the function is 12`, 22 c 12, ` 2 .

2. Understand the Characteristics of Rational Functions and Their Graphs

Consider the rational function t 5 f 1r2 5 500 r given at the beginning of the section. A graph of this rational function is shown in Figure 3-60. Time (in hours) 80 60 40 500 t = f(r) = –––– r

20

0

50

100

150

200

250

300

Rate (of speed)

FIGURE 3-60 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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We see from the graph that as the rate of speed increases, the time it takes to complete the race decreases. In fact, if we drive at rocket speed, we will arrive in almost no time at all. We can express this by saying that “as the rate increases without bound (or approaches `), the time it takes to complete the race approaches 0 hours.” When a graph approaches a line as shown in the figure, we call the line an asymptote. The horizontal line representing the rate axis shown in the graph is a horizontal asymptote. We also see from the graph that as the rate of speed decreases, the time it takes to complete the race increases. In fact, if the car goes at turtle speed, it will take almost forever to finish the race. We can express this by saying that “as the rate gets slower and slower (or approaches 0 mph), the time approaches `.” The vertical line representing the time axis shown on the graph is a vertical asymptote. A vertical asymptote is a vertical line that the graph approaches, but never touches. One important characteristic of the graph of a rational function is that asymptotes often occur. We will first consider vertical and horizontal asymptotes. Later in the section, we will discuss slant asymptotes.

Vertical Asymptote

Horizontal Asymptote

The line x 5 a is a vertical asymptote of the graph of a function y 5 f 1x2 if f 1x2 either increases or decreases without bound (approaches ` or 2`) as x approaches a.

The line y 5 b is a horizontal asymptote of the graph of a function y 5 f 1x2 if f 1x2 approaches b as x increases or decreases without bound (approaches ` or 2`).

Figure 3-61 shows a typical vertical and typical horizontal asymptote. Vertical asymptote at x = a f(x) approaches ∞ as x approaches a from left.

Horizontal asymptote at y = b

f(x) approaches –∞ as x approaches a from right. a

f(x) approaches b as x approaches ∞ or as x approaches –∞. y

x

b a

x

FIGURE 3-61

The graph of the rational function f 1x2 5 x1 , called the reciprocal function, is shown in Figure 3-62. The domain of the function is 12`, 02 c 10, ` 2 and the range is 12`, 02 c 10, ` 2 . y

1 f (x) = –x x

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Section 3.5

Rational Functions

347

The reciprocal function has a vertical asymptote of x 5 0 because • •

We see from the graph that as x approaches 0 from the right that y or f 1x2 approaches `. We see from the graph that as x approaches 0 from the left that y or f 1x2 approaches 2`.

The reciprocal function has a horizontal asymptote of y 5 0 because • •

We see from the graph that as x approaches ` that y or f 1x2 approaches 0. We see from the graph that as x approaches 2` that y or f 1x2 approaches 0.

3. Find Vertical Asymptotes of Rational Functions To find the vertical asymptotes of a rational function written in simplest form, we must find the values of x for which the denominator of the rational function is 0 and 2 1 the function is undefined. For example, since the denominator of f 1x2 5 2x x 1 2 is 0 when x 5 22, there are no corresponding values of y and the line x 5 22 is a vertical asymptote. We note that when x approaches 22 from the right or from the left, f 1x2 approaches 2` and ` respectively. A graph of the function appears in Figure 3-63. y f (x) = 2x – 1 x+2

x

FIGURE 3-63

Strategy for Locating Vertical Asymptotes

Px To locate the vertical asymptotes of the rational function f 1x2 5 Q 1x2 , we fol1 2

low these steps:

Step 1: Factor P 1x2 and Q 1x2 and remove any common factors. Step 2: Set the denominator equal to 0 and solve the equation.

If a is a solution of the equation found in Step 2, x 5 a is a vertical asymptote.

EXAMPLE 2

Finding Vertical Asymptotes of Rational Functions Find the vertical asymptotes, if any, of each function: 2x x24 5x a. f 1x2 5 2 b. g 1x2 5 2 c. h 1x2 5 2 x 2 16 x 2 16 x 1 16

SOLUTION

We will locate the vertical asymptotes of the graph of each function by factoring its numerator and/or denominator and removing any common factors. Then we will set the resulting denominator equal to zero, solve the equation, and identify the vertical asymptotes. a. f 1x2 5 f 1x2 5

2x x2 2 16 2x 1x 1 42 1x 2 42

Factor the denominator completely.

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348

Chapter 3

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We then set the denominator equal to 0 and solve for x. 1x 1 42 1x 2 42 5 0 x1450

or x 2 4 5 0

x 5 24

x54

The vertical asymptotes are x 5 4 and x 5 24. See Figure 3-64. y

x = –4

x=4

x

2x f (x) = –––––– x 2 – 16

b. g 1x2 5 g 1x2 5 g 1x2 5

FIGURE 3-64

x24 x2 2 16 x24 1x 1 42 1x 2 42

1 x14

Factor the denominator completely.

Simplify:

x24 5 1. x24

We set the denominator equal to 0 and solve for x. x1450 x 5 24

Comment

The vertical asymptote is x 5 24. See Figure 3-65.

A rational function can have zero, one, or several vertical asymptotes.

y x = –4 x x–4 g(x) = –––––– x 2 – 16

y 2 1 –20 –10 –1 –2

10

20

x

5x h(x) = –––––– x 2 + 16

FIGURE 3-66

Self Check 2

c. h 1x2 5

FIGURE 3-65

5x x 1 16 Because the denominator is the sum of two squared quantities, it cannot be factored over the set of real numbers. Thus, no values of x will make the denominator equal to 0. Thus, the rational function has no vertical asymptotes. See Figure 3-66. 2

Find the vertical asymptotes, if any, of each function. x15 5 a. g 1x2 5 2 b. h 1x2 5 2 x 2 25 x 1 25 Now Try Exercise 31.

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Section 3.5

Rational Functions

349

4. Find Horizontal Asymptotes of Rational Functions To find the horizontal asymptote of a rational function, we must find the value of y that the function approaches as x approaches ` or 2`. To illustrate, we will consider three functions and some given function values. First we consider a function where the degree of the numerator is less than the degree of the denominator. f 1x2 5

• • •

x12 x2 2 1

f 129992 < 20.001

f 19992 < 0.001

The degree of the numerator is 1 and the degree of the denominator is 2. Since 1 , 2, the degree of the numerator is less than the degree of the denominator. If f is evaluated at large x-values (in both the positive and negative directions) such as 999 and 2999, we obtain y-values that are close to 0. Conclusion: The line y 5 0 (the x-axis) is a horizontal asymptote. See Figure 3-67. y

x x+2 f (x) = ––––– x2 – 1 FIGURE 3-67

Second we consider a function where the degree of the numerator is equal to the degree of the denominator. g 1x2 5

• • • •

2x x21

g 19992 < 2.002

g 129992 < 1.998

The degree of the numerator is 1 and the degree of the denominator is 1. Thus, the degrees of the numerator and denominator are equal. The leading coefficient of the numerator is 2 and the leading coefficient of the denominator is 1. Note that 2 4 1 is 2. If g is evaluated at large x-values (in both the positive and negative directions) such as 999 and 2999, we obtain y-values that are close to 2. Conclusion: The line y 5 2 is a horizontal asymptote. See Figure 3-68. y 2x g(x) = –––– x –1 y=2 x

FIGURE 3-68

Finally we consider a function where the degree of the numerator is greater than the degree of the denominator. h 1x2 5

x3 x21

h 19992 < 999,001

h 129992 < 997,003

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• •

•

The degree of the numerator is 3 and the degree of the denominator is 1. Thus, the degree of the numerator is greater than the degree of the denominator. If h is evaluated at large x-values (in both the positive and negative directions) such as 999 and 2999, we obtain y-values that are extremely large and do not seem to approach a finite number. Conclusion: There is no horizontal asymptote. See Figure 3-69. y

x3 h(x) = –––– x –1

x FIGURE 3-69

We summarize the conclusions made in the previous examples.

Strategy for Locating Horizontal Asymptotes

EXAMPLE 3

Px To locate the horizontal asymptote of the rational function f 1x2 5 Q 1x2 , we consider three cases: 1 2

Case 1: If the degree of P 1x2 is less than the degree of Q 1x2 , the line y 5 0 is a horizontal asymptote. Case 2: If the degree of P 1x2 and Q 1x2 are equal, the line y 5 pq, where p and q are the leading coefficients of P 1x2 and Q 1x2 , is a horizontal asymptote. Case 3: If the degree of P 1x2 is greater than the degree of Q 1x2 , there is no horizontal asymptote.

Finding Horizontal Asymptotes of Rational Functions Find the horizontal asymptote, if any, of each function: a. f 1x2 5

SOLUTION

3x 2x 2 1 2

b. g 1x2 5

3x2 1 1 x 2 2x 1 1 2

c. h 1x2 5

x3 1 2 x25

In each part, we will locate the horizontal asymptote by comparing the degree of the numerator to the degree of the denominator. Then we will decide which case applies and identify the horizontal asymptote. a. Because the degree of the numerator is 1, the degree of the denominator is 2, and 1 , 2, Case 1 applies. The horizontal asymptote is the line y 5 0. b. The degree of the numerator is 2 and the degree of the denominator is 2. Since the degrees are the same, Case 2 applies. Because the leading coefficient of the numerator is 3 and the leading coefficient of the denominator is 1, we divide 3 by 1 to obtain 31 5 3. The horizontal asymptote is the line y 5 3. c. Because the degree of the numerator is 3, the degree of the denominator is 1, and 3 . 1, Case 3 applies. There is no horizontal asymptote.

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Section 3.5

Self Check 3

351

Rational Functions

Find the horizontal asymptotes, if any, of each function. 4x 2 5 3x2 a. g 1x2 5 b. h 1x2 5 3 52x x 25 Now Try Exercise 39.

5. Identify Slant Asymptotes of Rational Functions A third type of asymptote is called a slant asymptote. These asymptotes occur when the degree of the numerator of a rational function is one more than the degree of the denominator. As the name implies, it is a slanted line, neither vertical nor horizontal. 2 To illustrate a slant asymptote, we consider the graph of f 1x2 5 x x2 2 shown in Figure 3-70.

y y

x2 f (x) = –––– x –2 x

x2 f (x) = –––– x –2 x

FIGURE 3-70

FIGURE 3-71

When x increases without bound to the right and to the left, the graph of the rational function approaches the slant asymptote shown in Figure 3-71. The equation of the slant asymptote is y 5 x 1 2. To find this equation, we perform a long division, write the result in quotient 1 remainder divisor form, and ignore the remainder. x12 x 2 2qx 1 0x 1 0 x2 2 2x 2x 1 0 2x 2 4 4 2

2

Thus, y 5 x x2 2 5 x 1 2 1 x 24 2 . Because the last fraction approaches 0 as x approaches ` and 2`, the equation of the slant asymptote is y 5 x 1 2.

Strategy for Locating Slant Asymptotes

If the degree of P 1x2 is 1 greater than the degree of Q 1x2 for the rational funcP 1x2 tion f 1x2 5 Q 1x2 , there is a slant asymptote. To find it, divide P 1x2 by Q 1x2 and ignore the remainder.

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352

Chapter 3

Functions

EXAMPLE 4

SOLUTION

Finding a Slant Asymptote of a Rational Function

Find the slant asymptote of y 5 f 1x2 5

3x3 1 2x2 1 2 . x2 2 1

To find the slant asymptote, we divide the numerator by the denominator, write the result in quotient 1 remainder divisor form, and ignore the remainder. 3x 1 2 x2 2 1q3x3 1 2x2 12 3 3x 2 3x 2x2 1 3x 1 2 2x2 22 3x 1 4 Thus, y5

3x3 1 2x2 1 2 3x 1 4 5 3x 1 2 1 2 2 x 21 x 21

The last fraction approaches 0 as x approaches ` and 2`. Thus the graph of the rational function approaches the slant asymptote with equation of y 5 3x 1 2. Self Check 4

Find the slant asymptote of f 1x2 5

2x3 2 3x 1 1 . x2 2 4

Now Try Exercise 45.

Comment The graph of a rational function can cross horizontal and slant asymptotes but can never cross a vertical asymptote.

Strategy for Graphing Rational Functions of the Form P 1x2 f 1x2 5 Q 1x2

6. Graph Rational Functions

Px We will use the following strategy to graph the rational function f 1x2 5 Q 1x2 , where 1 2

Px P 1x2 and Q 1x2 are polynomials written in descending powers of x and Q 1x2 is in sim1 2

plest form (no common factors).

Step 1: Check for symmetry. • If f 12x2 5 f 1x2 , the graph is symmetric about the y-axis. • If f 12x2 5 2f 1x2 , the graph is symmetric about the origin.

Step 2: Find the vertical asymptotes. The real roots of Q 1x2 5 0, if any, determine the vertical asymptotes of the graph. Step 3: Find the y- and x-intercepts, if any. • f 102 is the y-coordinate of the y-intercept of the graph. • Set P 1x2 5 0 and solve to find the x-intercepts of the graph.

Step 4: Find the horizontal asymptotes, if any. • If the degree of P 1x2 is less than the degree of Q 1x2 , the line y 5 0 is a horizontal asymptote. • If the degree of P 1x2 is equal to the degree of Q 1x2 , the graph of y 5 pq is a horizontal asymptote, where p and q are the leading coefficients of P 1x2 and Q 1x2 . Step 5: Find the slant asymptotes, if any. • If the degree of P 1x2 is 1 greater than the degree of Q 1x2 , there is a slant asymptote. To find it, divide P 1x2 by Q 1x2 and ignore the remainder. Step 6: Draw the graph. Find additional points (if necessary) near the asymptotes.

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Section 3.5

EXAMPLE 5

SOLUTION

Rational Functions

353

Graphing a Rational Function

Graph: y 5 f 1x2 5

x2 2 4 . x2 2 1

We will use the steps outlined on the previous page to graph the rational function. Step 1: Symmetry Because x appears to even powers only, f 12x2 5 f 1x2 and there is symmetry about the y-axis. There is no symmetry about the origin. Step 2: Vertical asymptotes To find the vertical asymptotes, we factor the numerator and denominator of f 1x2 and simplify, if possible. f 1x2 5

1x 1 22 1x 2 22 x2 2 4 5 2 1x 1 12 1x 2 12 x 21

There are no common factors.

We then set the denominator equal to 0 and solve for x. 1x 1 12 1x 2 12 5 0 x1150

or x 2 1 5 0 x51

x 5 21

There will be vertical asymptotes at x 5 21 and x 5 1. Step 3: y- and x-intercepts We can find the y-intercept by finding f 102 . f 102 5

24 02 2 4 5 54 2 0 21 21

The y-intercept is (0, 4). We can find the x-intercepts by setting the numerator equal to 0 and solving for x: x2 2 4 5 0

1x 1 22 1x 2 22 5 0 x1250

or x 2 2 5 0 x52

x 5 22

The x-intercepts are (2, 0) and (22, 0). Step 4: Horizontal asymptotes Since the degrees of the numerator and denominator of the polynomials are the same, the line

y x=1 (0, 4)

x2 – 4 f(x) = _____ x2 – 1

y5

x = –1

(2, 0)

The leading coefficient of the numerator is 1. The leading coefficient of the denominator is 1.

is a horizontal asymptote.

y=1 (–2, 0)

1 51 1

x

FIGURE 3-72

Self Check 5

Step 5: Slant asymptotes Since the degree of the numerator is not 1 greater than the degree of the denominator, there are no slant asymptotes. Step 6: Graph We plot the intercepts, draw the asymptotes, and make use of symmetry to graph the rational function. The graph is shown in Figure 3-72. Graph: f 1x2 5

x2 2 9 . x2 2 1

Now Try Exercise 59. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

354

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Functions

ACCENT ON TECHNOLOGy

Graphing a Rational Function A graphing calculator can be very useful in helping us draw the graph of a rational function. However, certain precautions must be used or we will get a graph that is not going to give us a good idea of how the function looks. For example, in Figure 3-73 (a) the graph of the function from Example 5 is shown. Using the ZOOM Standard window, a good representation of the graph is shown. In Figure 3-73 (b) the ZOOM Decimal window is used, and part of the graph is not shown.

(a)

(b) FIGURE 3-73

EXAMPLE 6

SOLUTION

Graphing a Rational Function

Graph: y 5 f 1x2 5

3x . x22

We will use the steps outlined earlier to graph the rational function. Step 1: Symmetry We find f 12x2 . f 12x2 5

23x 3 12x2 3x 5 5 . 12x2 2 2 2x 2 2 x12

Because f 12x2 2 f 1x2 and f 12x2 2 2f 1x2 , there is no symmetry about the y-axis or the origin. Step 2: Vertical asymptotes We first note that f 1x2 is in simplest form. We then set the denominator equal to 0 and solve for x. Since the solution is 2, there will be a vertical asymptote at x 5 2. Step 3: y- and x-intercepts We can find the y-intercept by finding f 102 . f 102 5

3 102 0 5 50 022 22

The y-intercept is (0, 0). We can find the x-intercepts by setting the numerator equal to 0 and solving for x: 3x 5 0 x50 The x-intercept is (0, 0). Step 4: Horizontal asymptotes Since the degrees of the numerator and denominator of the polynomials are the same, the line y5

3 53 1

The leading coefficient of the numerator is 3. The leading coefficient of the denominator is 1.

is a horizontal asymptote. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.5

Rational Functions

355

Step 5: Slant asymptotes Since the degree of the numerator is not 1 greater than the degree of the denominator, there are no slant asymptotes. Step 6: Graph First, we plot the intercept (0, 0) and draw the asymptotes. We then find one additional point on our graph to see what happens when x is greater than 2. To do so, we choose 3, a value of x that is greater than 2, and evaluate f 132 . f 132 5

3 132 9 5 59 322 1

Since f 132 5 9, the point (3, 9) lies on our graph. We sketch the graph as shown in Figure 3-74. y 10 8

(3, 9) 3x f(x) = ____ x– 2

6 4 2 –6 –4 –2 –2 –4

y=3

2

4

6

8 10

x

x=2

FIGURE 3-74

Self Check 6

Graph: f 1x2 5

3x . x12

Now Try Exercise 53.

EXAMPLE 7

Graphing a Rational Function Graph the function: y 5

SOLUTION

1 . x 1x 2 12 2

We will use the steps outlined earlier to graph the rational function. Step 1: Symmetry We find f 12x2 . f 1x2 5

1 x 1x 2 12 2 1 f 12x2 5 12x2 12x 2 12 2 1 5 12x2 3 1212 1x 1 12 4 2 1 5 12x2 1212 2 1x 1 12 2 1 12x2 1x 1 12 2 21 5 x 1x 1 12 2

Replace x with 2x

.

Factor out 21.

Square 21 and x 1 1.

5

Simplify.

Because f 12x2 2 f 1x2 and f 12x2 2 2f 1x2 , there is no symmetry about the y-axis or origin. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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Step 2: Vertical asymptotes We set the denominator equal to 0 and solve for x. Since 0 and 1 make the denominator 0, the vertical asymptotes are x 5 0 and x 5 1. Step 3: y- and x-intercepts Since x cannot be 0, the graph has no y-intercept. Since the numerator cannot be 0, y cannot be 0, and the graph has no x-intercepts. Step 4: Horizontal asymptotes Since the degree of the numerator is 0 and the degree of the denominator is 3, and 0 , 3, we conclude that the horizontal asymptote is the line y 5 0. Step 5: Slant asymptotes There are no slant asymptotes because the degree of the numerator is not 1 greater than the degree of the denominator. Step 6: Graph Because there are no intercepts to plot, we draw the asymptotes and find a few additional points to plot. The table gives three points lying in different intervals separated by the asymptotes. We sketch the graph as shown in Figure 3-75. y

1 x 1x 2 12 2 1x, y2 y

y5 x

1 f (x) = _______ x(x – 1)2

21

x=0

x x=1

2

1 4

1 2

8

2

1 2

1 a21, 2 b 4 1 a , 8b 2

1 a2, b 2

FIGURE 3-75

Self Check 7

Graph the function: f 1x2 5 Now Try Exercise 69.

EXAMPLE 8

SOLUTION

1 . x 1x 2 22 2

Graphing a Rational Function

Graph: y 5 f 1x2 5

1 . x2 1 1

We will use the steps outlined earlier to graph the rational function. Step 1: Symmetry Because x appears to an even power and f 12x2 5 f 1x2 , there is symmetry about the y-axis. There is no symmetry about the origin. Step 2: Vertical asymptotes Since no number x makes the denominator 0, the graph has no vertical asymptotes. Step 3: y- and x-intercepts We can find the y-intercept by finding f 102 . f 102 5

1 1 5 51 02 1 1 1

The y-intercept is (0, 1). Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.5

Rational Functions

357

Because the denominator is always positive, the fraction is always positive and the graph lies entirely above the x-axis. There are no x-intercepts. We also see that if we set the numerator equal to 0 there are no solutions for x. Step 4: Horizontal asymptotes Since the degree of the numerator is less than the degree of the denominator, the line y 5 0 is a horizontal asymptote. Step 5: Slant asymptotes There are no slant asymptotes because the degree of the numerator is not 1 greater than the degree of the denominator. Step 6: Graph We first plot the y-intercept 10, 12 . Then, we make use of the fact that the graph has y-axis symmetry and a horizontal asymptote of y 5 0 and sketch the graph. The graph is shown in Figure 3-76. y 1 f (x) = _____ x2 + 1 x FIGURE 3-76

Self Check 8

Graph: f 1x2 5

4 . x 11 2

Now Try Exercise 73.

EXAMPLE 9

SOLUTION

Graphing a Rational Function with a Slant Asymptote

Graph: y 5 f 1x2 5

x2 1 x 2 2 . x23

We first factor the numerator of the expression y 5 outlined earlier to graph the rational function.

1x 2 12 1x 1 22 x 2 3

and use the steps

Step 1: Symmetry We find f 12x2 . f 1x2 5

x2 1 x 2 2 x23 12x2 2 1 12x2 2 2 f 12x2 5 12x2 2 3 x2 2 x 2 2 2x 2 3 x2 2 x 2 2 5 1212 1x 1 32 2x2 1 x 1 2 5 x13 5

Replace x with 2x.

Simplify.

Factor 21 out of the denominator.

Divide by 21.

Because f 12x2 2 f 1x2 and f 1x2 2 2f 1x2 , there is no symmetry about the y-axis or origin. Step 2: Vertical asymptotes We set the denominator equal to 0 and solve for x. Since 3 makes the denominator 0, the vertical asymptote is x 5 3. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

358

Chapter 3

Functions

Step 3: y- and x-intercepts

The y-intercept is Q0, 23R because f 102 5 23. The x-intercepts are 11, 02 and 122, 02

because x values of 1 and 22 make the numerator 0.

Step 4: Horizontal asymptotes Since the degree of the numerator is 2, the degree of the denominator is 1, and 2 . 1, we conclude that there is no horizontal asymptote. Step 5: Slant asymptotes Because the degree of the numerator is 1 greater than the degree of the denominator, this graph will have a slant asymptote. To find it, we perform a long division. x14 x 2 3qx 1 x 2 2 x2 2 3x 4x 2 2 4x 2 12 10 2

We write the function as y5

x2 1 x 2 2 10 5x141 x23 x23

The fraction x 10 2 3 approaches 0 as x approaches ` or 2`, and the graph approaches a slant asymptote: the line y 5 x 1 4. Step 6: Graph The graph appears in Figure 3-77. y 18 16 14 12 10 8

y=x+4

6 4 2

x2 + x – 2 f(x) = _________ x –3

–8 –6 –4 –2 –2

2

4

6

8 10 12

x

–4 –6 –8

x=3

FIGURE 3-77

Self Check 9

Graph: y 5 f 1x2 5

x2 1 x 1 2 . x13

Now Try Exercise 75.

7. Understand When a Graph Has a Missing Point We have discussed rational functions where the fraction is in simplified form. We P 1x2 now consider a rational function f 1x2 5 Q 1x2 , where P 1x2 and Q 1x2 have a common Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.5

Rational Functions

359

factor. Graphs of such functions have gaps or missing points that are not the result of vertical asymptotes.

EXAMPLE 10

SOLUTION

Graphing a Rational Function that has a Missing Point

Find the domain of the function f 1x2 5

x2 2 x 2 12 and graph it. x24

We find the values of x that make the denominator 0. These values of x will not be included in the domain. We will then write the rational function in simplest form and graph it using the methods of this section. Since the denominator cannot be 0, x 2 4. Therefore, the domain is the set of all real numbers except 4. We now write the rational function in simplest form by factoring the numerator of the expression. We see that the numerator and denominator have a common factor of x 2 4. f 1x2 5

x2 2 x 2 12 x24 1x 1 32 1x 2 42 5 x24

Factor x2 2 x 2 12.

If x 2 4, the common factor of x 2 4 can be divided out. The resulting function is equivalent to the original function only when we keep the restriction that x 2 4. Thus, f 1x2 5

1x 1 32 1x 2 42 5x13 x24

(provided that x 2 4)

When x 5 4, the function is not defined. The graph of the function appears in Figure 3-78. It is a line with the point with x-coordinate of 4 missing. The line has a hole in it.

y

Caution x2 – x – 12 f(x) = __________ x –4

A calculator graph of a rational function does not indicate a missing point or hole.

x

FIGURE 3-78

Self Check 10

Graph: f 1x2 5

x2 1 x 2 12 . x14

Now Try Exercise 83.

8. Solve Problems Modeled by Rational Functions Rational expressions often define functions that occur in the real world. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

360

Chapter 3

Functions

EXAMPLE 11

Solving an Application Using a Rational Function

Dmitrijs Dmitrijevs/Shutterstock.com

Suppose the cost C of a no contract cell phone is $20 per month plus $0.15 per minute. Write a rational function that represents the average (mean) cost per minute, C. Let n represent the number of minutes used per month. Find the mean cost per minute when the service is used for 180 minutes. SOLUTION

Because the mean cost is the total monthly cost divided by the number of minutes used, we can write a rational function that models the problem. We will then evaluate the mean cost function at 180 to find the mean cost for 180 minutes used. We know that the mean cost C is total cost C divided by the number of access minutes n. That is, C C5 n Since total cost C is $20 per month plus $0.15 for each minute n, we can write C 1n2 5 0.15n 1 20. Thus, the function 0.15n 1 20 1n . 02 C 1n2 5 n gives the mean cost per minute of using the service for n minutes per month. To find the mean cost for 180 minutes of usage, we substitute 180 for n and simplify. 0.15 11802 1 20 < 0.26 C 11802 5 180 The mean cost per minute for 180 minutes of usage is approximately $0.26 per minute.

Self Check 11

Find the mean hourly cost when the cell phone described above is used for 240 minutes. Now Try Exercise 87.

ACCENT ON TECHNOLOGy

Graphs of Rational Functions

We can use a graphing calculator to graph the function C 1n2 5 0.15nn1 20 in Example 11. We enter the function as shown in Figure 3-79(a) using the window settings as shown in Figure 3-79 (b). The graph of the rational function appears in Figure 3-79(c). Note the following characteristics: • The graph of the function passes the Vertical Line Test, as expected. • From the graph in Figure 3-79(c), we see that the mean cost per minute decreases as the number of minutes of usage increases. Since the cost of each extra minute of usage of time is $0.15, the mean minute cost can approach $0.15 but will never drop below it. The graph of the function approaches the line y 5 0.15 as n increases without bound. The line y 5 0.15 is a horizontal asymptote of the graph. • As n gets smaller and approaches 0, the graph approaches the y-axis. The y-axis is a vertical asymptote of the graph.

(a)

(b)

(c)

FIGURE 3-79

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Section 3.5

Self Check Answers

1. 12`, 212 c 121, 22 c 12, ` 2 b. y 5 0 4. y 5 2x 5.

2. a. x 5 5

361

Rational Functions

b. none

3. a. y 5 24

y 20 x2 – 9 f(x) = –––––––– x2 – 1 x 2 4

10 –4 –2 –10 –20

6.

7.

y

8.

y

y

6 1 f(x) = –––––––2 x(x – 2)

3x –––– 4 f(x) = x + 2

x

2 –10 –5

5

9.

10

x 4 f(x) = ––––– x2 + 1

x

10.

y

y

11. $0.23 per minute

20

x

10 –10 –5

5

10

–10

x

x2 + x + 2 f(x) = –––––––– x+3 –20

x 2 + x – 12 f(x) = ––––––––– x+4

Exercises 3.5 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

8. In a rational function, if the degree of the numerator is 1 greater than the degree of the denominator, the graph will have a .

Fill in the blanks.

9. A graph can cross a never cross a

1. When a graph approaches a vertical line but never touches it, we call the line an . 2. A rational function is a function with a polynomial numerator and a polynomial denominator. 3. To find a asymptote of a rational function in simplest form, set the denominator polynomial equal to 0 and solve the equation. of a rational function, let 4. To find the x 5 0 and solve for y or find f 102 . 5. To find the of a rational function, set the numerator equal to 0 and solve the equation. P 1x2 6. In the function Q 1x2 , if the degree of P 1x2 is less than the degree of Q 1x2 , the horizontal asymptote is . P 1x2 7. In the function f 1x2 5 Q 1x2 , if the degree of P 1x2 and Q 1x2 are the , the horizontal symptote is y5

asymptote but can asymptote.

10. The graph of f 1x2 5 xx 12 24 will have a point. 2

Find the equations of the vertical and horizontal asymptotes of each graph. Find the domain and range. y y 11. 12.

x

the leading coefficient of the numerator . the leading coefficient of the denominator

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

362

Chapter 3

Functions

Practice The time t it takes to travel 600 miles is a function of the mean rate of speed r: t 5 f 1r2 5

600 r

Find t for the given values of r. 13. 30 mph 14. 40 mph 15. 50 mph 16. 60 mph Suppose the cost (in dollars) of removing p% of the pollution in a river is given by the function C 5 f 1 p2 5

50,000p 100 2 p

10 # p , 1002

Find the cost of removing each percent of pollution. 17. 10% 18. 30% 19. 50% 20. 80%

Find the horizontal asymptotes, if any, of each rational function. Do not graph the function. 37. f 1x2 5

2x 2 1 x

38. f 1x2 5

x2 1 1 3x2 2 5

39. f 1x2 5

x2 1 x 2 2 2x2 2 4

40. f 1x2 5

5x2 1 1 5 2 x2

41. f 1x2 5

x11 x3 2 4x

42. f 1x2 5

x 2x2 2 x 1 11

43. f 1x2 5

x2 x22

44. f 1x2 5

x4 1 1 x23

Find the slant asymptote, if any, of each rational function. Do not graph the function.

Find the domain of each rational function. Do not graph the function.

45. f 1x2 5

x2 2 5x 2 6 x22

46. f 1x2 5

x2 2 2x 1 11 x13

48. f 1x2 5

5x3 1 1 x15

21. f 1x2 5

x2 x22

22. f 1x2 5

x3 2 3x2 1 1 x13

23. f 1x2 5

2x2 2 5x 1 1 x24

2x2 1 7x 2 2 x2 2 25

24. f 1x2 5

47. f 1x2 5

5x2 1 1 x2 1 5

49. f 1x2 5

x3 1 2x2 2 x 2 1 x2 2 1

25. f 1x2 5 27. f 1x2 5

x21 x3 2 x

3x2 1 5 x2 1 1

26. f 1x2 5 28. f 1x2 5

x12 2x 2 9x 1 9 2

7x2 2 x 1 2 x4 1 4

50. f 1x2 5

Find all vertical, horizontal, and slant asymptotes, x- and y-intercepts, and symmetries, and then graph each function. Check your work with a graphing calculator. 51. y 5

Find the vertical asymptotes, if any, of each rational function. Do not graph the function. x 2x 29. f 1x2 5 30. f 1x2 5 x23 2x 1 5 31. f 1x2 5

x12 x2 2 1

1 33. f 1x2 5 2 x 2x26 35. f 1x2 5

2

x x 15 2

32. f 1x2 5

1 x22

52. y 5

3 x13

y

y

x

x

x24 x2 2 16

x12 34. f 1x2 5 2 2x 2 6x 2 8

36. f 1x2 5

2x3 1 3x2 2 x 1 1 x2 1 1

3

53. y 5

x x21

54. y 5

y

x x12 y

2

x 2 3x 1 1 2x2 1 3 x x

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.5

55. f 1x2 5

56. f 1x2 5

x11 x12 y

x21 x22

63. y 5

x2 1 2x 2 3 x3 2 4x

y

58. f 1x2 5

3x 1 2 x2 2 4

y

65. y 5

x2 2 9 x2

y

60. g 1x2 5

x

3x2 2 12 x2

y

y

x x

67. f 1x2 5

x2 2 4 x2 2 9

1 1 – 3

––1 2

66. y 5

x

x2 2 9 x2 2 4

3x2 2 4x 1 1 2x3 1 3x2 1 x

–1 x

x

59. g 1x2 5

363

y

x

2x 2 1 x21

64. y 5

y

x

57. f 1x2 5

Rational Functions

68. f 1x2 5

x 1x 1 32 2 y

x 1x 2 12 2

y

y

y

x

x x

x

69. f 1x2 5 x 2x22 61. g 1x2 5 2 x 2 4x 1 3 2

70. f 1x2 5

x11 x2 1x 2 22

x21 x2 1x 1 22 2

y

x 1 7x 1 12 62. g 1x2 5 2 x 2 7x 1 12

y

2

x

y

x

y

x

–4

–3 –200

3

4

x

71. y 5

x x 11

72. y 5

2

y

x21 x2 1 2 y

x

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

364

Chapter 3

73. y 5

Functions

3x2 x 11

74. y 5

2

Graph each rational function. Note that the numerator and denominator of the fraction share a common factor.

x2 2 9 2x2 1 1 y

y

x

79. f 1x2 5

80. f 1x2 5

x2 x y

x2 2 1 x21 y

x

x

75. h 1x2 5

x2 2 2x 2 8 x21

76. h 1x2 5

x2 1 x 2 6 x12

81. f 1x2 5

y

y 8

x

82. f 1x2 5

x3 1 x x

x3 2 x2 x21 y

y

4 x –8 –4

4

8

x

–4

77. f 1x2 5

x x

x3 1 x2 1 6x x2 2 1 y

83. f 1x2 5

x2 2 2x 1 1 x21

84. f 1x2 5

2x2 1 3x 2 2 x12

y

y x

x x

78. f 1x2 5

x3 2 2x2 1 x x2 2 4 y

85. f 1x2 5

86. f 1x2 5

x3 2 1 x21 y

x2 2 x x2 y

x

x x

Applications A service club wants to publish a directory of its members. Some investigation shows that the cost of typesetting and photography will be $700, and the cost of printing each directory will be $3.25. 87. a. Find a function that gives the total cost C of printing x directories. b. Find the total cost of printing 500 directories.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.5

c. Find a function that gives the mean cost per directory C of printing x directories. d. Find the mean cost per directory if 500 directories are printed. e. Find the mean cost per directory if 1,000 directories are printed. f. Find the mean cost per directory if 2,000 directories are printed.

365

Rational Functions

Discovery and Writing 91. Can a rational function have two horizontal asymptotes? Explain. 92. Can a rational function have two slant asymptotes? Explain. In Exercises 93–96, a, b, c, and d are nonzero constants. 93. Show that y 5 0 is a horizontal asymptote of the ax 1 b graph of y 5 cx 2 1 d.

An electric company charges $10 per month plus 20¢ for each kilowatt-hour (kwh) of electricity used.

94. Show that y 5 acx is a slant asymptote of the

88. a. Find a function that gives the total cost C of n kwh of electricity. b. Find the total cost for using 775 kwh. c. Find a function that gives the mean cost per kwh, C, when using n kwh.

95. Show that y 5 ac is a horizontal asymptote of the

d. Find the mean cost per kwh when 775 kwh are used. Round to the nearest hundredth. e. Find the mean cost per kwh when 3,200 kwh are used. Round to the nearest hundredth. 89. Utility costs An overseas electric company charges $8.50 per month plus 9.5¢ for each kilowatt-hour (kwh) of electricity used. a. Find a linear function that gives the total cost C of n kwh of electricity. b. Find a rational function that gives the average cost per kwh when using n kwh.

3

graph of y 5 ax cx2

2

graph of y 5 ax cx2

1 b 1 d.

1 b 1 d. 3

96. Graph the rational function y 5 x

1 1 x

and explain

why the curve is said to have a parabolic asymptote. Use a graphing calculator to perform each experiment. Write a brief paragraph describing your findings. 97. Investigate the positioning of the vertical asymptotes of a rational function by graphing y 5 x 2x k for several values of k. What do you observe? 98. Investigate the positioning of the vertical asymptotes of a rational function by graphing y 5 x2 x2 k for k 5 4, 1, 21, and 0. What do you observe? 2

c. Find the average cost per kwh when 850 kwh are used. 90. Scheduling work crews The following rational function gives the number of days it would take two construction crews, working together, to frame a house that crew 1 (working alone) could complete in t days and crew 2 (working alone) could complete in 1t 1 32 days. f 1t2 5

2

t 1 3t 2t 1 3

a. If crew 1 could frame a certain house in 21 days, how long would it take both crews working together? b. If crew 2 could frame a certain house in 25 days, how long would it take both crews working together?

99. Find the range of the rational function y 5 x2kx1 1 for several values of k. What do you observe? 100. Investigate the positioning of the x-intercepts of a 2

rational function by graphing y 5 x

2 k x

for

k 5 1, 21, and 0. What do you observe?

Review Perform each operation.

101. 12x2 1 3x2 1 1x2 2 2x2 103. 15x 1 22 12x 1 52

102. 13x 1 22 2 1x2 1 22 104.

2x2 1 3x 1 1 x11

105. If f 1x2 5 3x 1 2, find f 1x 1 12 .

106. If f 1x2 5 x2 1 x, find f 12x 1 12 .

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

366

Chapter 3

Functions

3.6 Operations on Functions In this section, we will learn to 1. Add, subtract, multiply, and divide functions, specifying domains. 2. Write functions as sums, differences, products, or quotients of other functions. 3. Evaluate composite functions. 4. Determine domains for composite functions. 5. Write functions as compositions. 6. Use operations on functions to solve problems.

© Istockphoto.com/Ian McDonnell

Functions can be combined by addition, subtraction, multiplication, and division. In this section, we will explore these operations on functions and give careful attention to their domains and ranges. Suppose that the functions R 1x2 5 140x and C 1x2 5 120,000 1 40x model a company’s yearly revenue and cost for producing and selling surfboards. By subtracting the functions, R 1x2 2 C 1x2 , we would arrive at a new function represented by 1R 2 C2 1x2 5 140x 2 1120,000 1 40x2 5 140x 2 120,000 2 40x

Remove parentheses.

5 100x 2 120,000 This function represents the profit made by the company when it sells x surfboards. We will now discuss how to add, subtract, multiply, and divide functions.

1. Add, Subtract, Multiply, and Divide Functions, Specifying Domains With the following definitions, it is possible to perform arithmetic operations on algebraic functions. Adding, Subtracting, Multiplying, and Dividing Functions

If the ranges of functions f and g are subsets of the real numbers, then 1. The sum of f and g, denoted as f 1 g, is defined by 1 f 1 g2 1x2 5 f 1x2 1 g 1x2

2. The difference of f and g, denoted as f 2 g, is defined by 1 f 2 g2 1x2 5 f 1x2 2 g 1x2

3. The product of f and g, denoted as f ? g, is defined by 1 f ? g2 1x2 5 f 1x2 g 1x2

4. The quotient of f and g, denoted as f/g, is defined by f 1x2 1 f/g2 1x2 5 1g 1x2 2 02 g 1x2

The domain of each function, unless otherwise restricted, is the set of real numbers x that are in the domains of both f and g. In the case of the quotient f/g, there is the restriction that g 1x2 2 0. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.6

EXAMPLE 1

SOLUTION

Operations on Functions

367

Finding the Sum and Difference of Two Functions and Specifying Domains

Let f 1x2 5 3x 1 1 and g 1x2 5 2x 2 3. Find each function and its domain: a. f 1 g b. f 2 g We will find the sum of f and g by using the definition 1 f 1 g2 1x2 5 f 1x2 1 g 1x2

We will find the difference of f and g by using the definition 1 f 2 g2 1x2 5 f 1x2 2 g 1x2 .

To find the domain of each result, we will consider the domains of both f and g. a. 1 f 1 g2 1x2 5 f 1x2 1 g 1x2

5 13x 1 12 1 12x 2 32 5 5x 2 2

Since the domain of both f and g is the set of real numbers, the domain of f 1 g is the interval 12`, ` 2 .

b. 1 f 2 g2 1x2 5 f 1x2 2 g 1x2

5 13x 1 12 2 12x 2 32 5 3x 1 1 2 2x 1 3

Remove parentheses.

5x14 Since the domain of both f and g is the set of real numbers, the domain of f 2 g is the interval 12`, ` 2 . Self Check 1

Find g 2 f . Now Try Exercise 15.

EXAMPLE 2

SOLUTION

Finding the Product and Quotient of Two Functions and Specifying Domains

Let f 1x2 5 3x 1 1 and g 1x2 5 2x 2 3. Find each function and its domain: a. f ? g b. f/g We will multiply f and g by using the definition 1 f ? g2 1x2 5 f 1x2 g 1x2

We will divide f by g by using the definition f 1x2 1 f/g2 1x2 5 1g 1x2 2 02 g 1x2

We will find the domains of each result by considering the domains of both f and g. a. 1 f ? g2 1x2 5 f 1x2 ? g 1x2

5 13x 1 12 12x 2 32 5 6x2 2 7x 2 3

Since the domain of both f and g is the set of real numbers, the domain of f ? g is the interval 12`, ` 2 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

368

Chapter 3

Functions

f 1x2 g 1x2 3x 1 1 5 2x 2 3

b. 1f/g2 1x2 5

Since

3 2

1g 1x2 2 02

12x 2 3 2 02

will make 2x 2 3 equal to 0, the domain of f /g is the set of all real

numbers except 32 . This is Q2`, 32R c Q32, `R. Self Check 2

Find g/f and its domain. Now Try Exercise 17.

EXAMPLE 3

Using Operations on Functions and Specifying Domains

Let f 1x2 5 x2 2 4 and g 1x2 5 "x. Find each function and its domain: a. f 1 g b. f ? g c. f/g d. g/f

SOLUTION

To determine each result, we use the definitions of the sum, product, and quotient of two functions. We will determine the domain of each result by considering the domains of both f and g. First, we find the domains of f and g. Because 4 can be subtracted from any real number squared, the domain of f is the interval 12`, ` 2 . Because "x is to be a real number, the domain of g is the interval 3 0, ` 2 .

a. 1 f 1 g2 1x2 5 f 1x2 1 g 1x2

5 x2 2 4 1 "x

The domain of f 1 g consists of the numbers x that are in the domain of both f and g. This is 12`, ` 2 d 3 0, ` 2 , which is 3 0, ` 2 . The domain of f 1 g is 3 0, ` 2 .

b. 1 f ? g2 1x2 5 f 1x2 g 1x2

5 1x2 2 42 "x

5 x2"x 2 4"x

Distribute the multiplication of "x.

The domain of f ? g consists of the numbers x that are in the domain of both f and g. The domain of f ? g is 3 0, ` 2 .

c. 1 f/g2 1x2 5

5

f 1x2 g 1x2

1g 1x2 2 02

x2 2 4 "x

The domain of f/g consists of the numbers x that are in the domain of both f and g, except 0 (because division by 0 is undefined). The domain of f /g is 10, ` 2 . To write 1 f/g2 1x2 in a different form, we can rationalize the denominator of x2 2 4 and simplify. !x 1 f/g2 1x2 5 5

5

x2 2 4 "x

1x2 2 42 "x "x ? "x

x2"x 2 4"x x

Multiply numerator and denominator by "x.

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Section 3.6

d. 1g/f 2 1x2 5 5

g 1x2 f 1x2

"x x2 2 4

Operations on Functions

369

1 f 1x2 2 02

The domain of g/f consists of the numbers x that are in 3 0, ` 2 , the domain of both f and g, except 2 (because division by 0 is undefined). The domain of g/f is 3 0, 22 c 12, ` 2 . Self Check 3

Find g 2 f and its domain. Now Try Exercise 19. We can perform these operations using a graphing calculator.

ACCENT ON TECHNOLOGy

Operations on Functions A graphing calculator can be used to graph operations on function. Consider Example 3. •

We will use a bold setting to indicate the function f 1 g. In Figure 3-80, we see in the left graph that the graph of f 1 g appears to be the same as f. However, if we zoom in to a small part of the graph as shown in the right graph, we can see that the functions are different. To input Y1 and Y2, press VARS , scroll right to Y-VARS, press ENTER , and select the desired function.

FIGURE 3-80

•

We use the bold setting to indicate the function f ? g. Again, by zooming in to just part of the window, we can see the graphs are different. See Figure 3-81.

FIGURE 3-81

•

We demonstrate the graph of f/g using a graphing calculator. See Figure 3-82. f /g appears bold.

FIGURE 3-82

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370

Chapter 3

Functions

•

We demonstrate the graph of g/f using a graphing calculator. See Figure 3-83. g/f appears bold.

FIGURE 3-83

EXAMPLE 4 SOLUTION

Evaluating the Sum of Two Functions

Find 1 f 1 g2 132 when f 1x2 5 x2 1 1 and g 1x2 5 2x 1 1.

We will first find 1 f 1 g2 1x2 and then find 1 f 1 g2 132 . To find 1 f 1 g2 1x2 , we proceed as follows: 1 f 1 g2 1x2 5 f 1x2 1 g 1x2

5 x2 1 1 1 2x 1 1 5 x2 1 2x 1 2

Comment

We can also determine 1 f 1 g2 132 by finding f 132 1 g 132 . Note that f 132 5 10, g 132 5 7, and f 132 1 g 132 5 10 1 7 5 17.

To find 1 f 1 g2 132 , we proceed as follows: 1 f 1 g2 1x2 5 x2 1 2 1x2 1 2 1 f 1 g2 132 5 32 1 2 132 1 2

Substitute 3 for x.

591612 5 17

Self Check 4

Find 1 f ? g2 1222 .

Now Try Exercise 25.

2. Write Functions as Sums, Differences, Products, or Quotients of Other Functions EXAMPLE 5

SOLUTION

Writing Functions as Combinations of Other Functions

Let h 1x2 5 x2 1 3x 1 2. Find two functions f and g such that a. f 1 g 5 h b. f ? g 5 h For part a, we must find two functions f and g whose sum is h. For part b, we must find two functions f and g whose product is h. a. There are many possibilities. One is f 1x2 5 x2 and g 1x2 5 3x 1 2, then 1 f 1 g2 1x2 5 f 1x2 1 g 1x2

5 1x22 1 13x 1 22

5 x2 1 3x 1 2 5 h 1x2

Another possibility is f 1x2 5 x2 1 2x and g 1x2 5 x 1 2. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.6

Operations on Functions

371

b. Again, there are many possibilities. One is suggested by factoring x2 1 3x 1 2. x2 1 3x 1 2 5 1x 1 12 1x 1 22

If we let f 1x2 5 x 1 1 and g 1x2 5 x 1 2, then 1 f ? g2 1x2 5 f 1x2 ? g 1x2

5 1x 1 12 1x 1 22 5 x2 1 3x 1 2

5 h 1x2

Another possibility is f 1x2 5 3 and g 1x2 5 x3 1 x 1 23 . Self Check 5

2

Find two functions f and g such that f 2 g 5 h. Now Try Exercise 35.

3. Evaluate Composite Functions Often one quantity is a function of a second quantity that depends, in turn, on a third quantity. For example, the cost of a car trip is a function of the gasoline consumed. The amount of gasoline consumed, in turn, is a function of the number of miles driven. Such chains of dependence are analyzed mathematically as composition of functions. Suppose that y 5 f 1x2 and y 5 g 1x2 define two functions. Any number x in the domain of g will produce a corresponding value g 1x2 in the range of g. If g 1x2 is in the domain of function f, then g 1x2 can be substituted into f, and a corresponding value f 1g 1x2 2 will be determined. This two-step process defines a new function, called a composite function, denoted by f + g. (See Figure 3-84.) f°g

g

f

x

Domain of g

g(x)

f(g(x))

Range of g Domain of f

Range of f

FIGURE 3-84

Composite Function

The composite function f + g is defined by 1 f + g2 1x2 5 f 1 g 1x2 2

The domain of f + g consists of all those numbers in the domain of g for which g 1x2 is in the domain of f. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

372

Chapter 3

Functions

To illustrate the previous definition, we consider the functions f 1x2 5 5x 1 1 and g 1x2 5 4x 2 3 and find 1 f + g2 1x2 and 1g + f 2 1x2 . 1 f + g2 1x2 5 f 1 g 1x2 2

1g + f 2 1x2 5 g 1 f 1x2 2

5 g 15x 1 12

5 f 14x 2 32

5 5 14x 2 32 1 1

5 4 15x 1 12 2 3

5 20x 2 14

Caution Note that in the previous example 1 f + g2 1x2 2 1g + f 2 1x2 .

5 20x 1 1

Since we get different results, the composition of functions is not commutative. We have seen that a function can be represented by a machine. If we put a number from the domain into the machine (the input), a number from the range comes out (the output). For example, if we put 2 into the machine shown in Figure 3-85(a), the number f 122 5 5 122 2 2 5 8 comes out. In general, if we put x into the machine shown in Figure 3-85(b), the value f 1x2 comes out. x

2

f(x) = 5x –2

y = f(x)

f(x)

8 (a)

(b) FIGURE 3-85

The function machines shown in Figure 3-86 illustrate the composition f + g. When we put a number x into the function g, the value g 1x2 comes out. The value g 1x2 then goes into function f, and f 1g 1x2 2 comes out. x

y = g(x)

g(x)

y = f (x)

f(g(x)) FIGURE 3-86

To further illustrate these ideas, we let f 1x2 5 2x 1 1 and g 1x2 5 x 2 4. •

1 f + g2 192 means f 1 g 192 2 . In Figure 3-87(a), function g receives the number 9 and subtracts 4, and the number g 1x2 5 5 comes out. The 5 goes into the f function, which doubles it and adds 1. The final result, 11, is the output of the composite function f + g: 1 f + g2 192 5 f 1 g 192 2 5 f 152 5 2 152 1 1 5 11

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Section 3.6

•

•

373

Operations on Functions

1 f + g2 1x2 means f 1 g 1x2 2 . In Figure 3-87(a), function g receives the number x and subtracts 4, and the number x 2 4 comes out. The x 2 4 goes into the f function, which doubles it and adds 1. The final result, 2x 2 7, is the output of the composite function f + g. 1 f + g2 1x2 5 f 1g 1x2 2 5 f 1x 2 42 5 2 1x 2 42 1 1 5 2x 2 7

1g + f 2 1222 means g 1 f 1222 2 . In Figure 3-87(b), function f receives the number 22, doubles it and adds 1, and releases 23 into the g function. Function g subtracts 4 from 23 and releases a final output of 27. Thus, 1g + f 2 1222 5 g 1 f 1222 2 5 g 1232 5 23 2 4 5 27 –2 f(x) = 2x + 1

x 9 g(x) = x – 4

–3 5

x–4 g(x) = x – 4 f(x) = 2x + 1 –7 11

2x – 7

(a)

(b) FIGURE 3-87

EXAMPLE 6 SOLUTION

Finding the Composition of Two Functions

If f 1x2 5 2x 1 7 and g 1x2 5 x2 2 1, find a. 1 f + g2 1x2

b. 1g + f 2 1x2

In part a, because 1 f + g2 1x2 means f 1 g 1x2 2 , we will replace x in f 1x2 5 2x 1 7 with g 1x2 . In part b, because 1g + f 2 1x2 means g 1 f 1x2 2 we will replace x in g 1x2 5 x2 2 1 with f 1x2 .

a. 1 f + g2 1x2 5 f 1 g 1x2 2

5 f 1x2 2 12

Substitute x2 2 1 for g 1x2 .

5 2x2 2 2 1 7

Remove parentheses.

5 2 1x2 2 12 1 7

Evaluate f 1x2 2 12 .

2

5 2x 1 5

b. 1g + f 2 1x2 5 g 1 f 1x2 2

5 g 12x 1 72

Substitute 2x 1 7 for f 1x2 .

5 4x2 1 28x 1 49 2 1

Square the binomial.

5 12x 1 72 2 2 1

Evaluate g 12x 1 72 .

2

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374

Chapter 3

Functions

Self Check 6

If f 1x2 5 2x 1 7 and h 1x2 5 x 1 1, find 1 f + h2 1x2 . Now Try Exercise 51.

EXAMPLE 7 SOLUTION

Evaluating the Composition of Two Functions

If f 1x2 5 3x 2 2 and g 1x2 5 3x2 1 6x 2 5, find 1 f + g2 1222 .

Because 1 f + g2 1222 means f 1g 1222 2 , we first find g 1222 . We then find f 1g 1222 2 . g 1x2 5 3x2 1 6x 2 5

g 1222 5 3 1222 2 1 6 1222 2 5 5 3 142 2 12 2 5 5 25

1 f + g2 1222 5 f 1 g 1222 2 5 f 1252

5 3 1252 2 2 5 217

Self Check 7

Find 1g + f 2 1212 .

Now Try Exercise 43.

4. Determine Domains of Composite Functions To be in the domain of the composite function f + g, a number x has to be in the domain of g, and the output of g must be in the domain of f. Thus, the domain of f + g consists of those inputs x that are in the domain of g and for which g 1x2 is in the domain of f.

Strategy to Determine the Domain of f + g

To determine the domain of 1 f + g2 1x2 5 f 1g 1x2 2 , apply the following restrictions to the composition: 1. If x is not in the domain of g, it will not be in the domain of f + g. 2. Any x that has an output g 1x2 that is not in the domain of f will not be in the domain of f + g.

EXAMPLE 8 SOLUTION

Finding the Domain of Composite Functions

Let f 1x2 5 "x and g 1x2 5 x 2 3. Find the domain of a. f + g

b. g + f

We will first find the domains of f 1x2 and g 1x2 . Then, we will find the domain of f + g and g + f by applying the restrictions stated above. For "x to be a real number, x must be a nonnegative real number. Thus, the domain of f is the interval 3 0, ` 2 . Since any real number x can be an input into g, the domain of g is the interval 12`, ` 2 .

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Section 3.6

Operations on Functions

375

a. The domain of f + g is the set of real numbers x such that x is in the domain of g and g 1x2 is in the domain of f. We have seen that all values of x are in the domain of g. However, g 1x2 must be nonnegative, because g 1x2 must be in the domain of f. So we must find the values of x such that g 1x2 is greater than or equal to 0. g 1x2 $ 0

x23$0

x$3

g 1x2 must be nonnegative.

Substitute x 2 3 for g 1x2 . Add 3 to both sides.

Since x $ 3, the domain of f + g is the interval 3 3, ` 2 .

b. The domain of g + f is the set of real numbers x such that x is in the domain of f and f 1x2 is in the domain of g. We have seen that only nonnegative values of x are in the domain of f. Because all values of f 1x2 are in the domain of g, the domain of g + f is the domain of f, which is the interval 3 0, ` 2 . Self Check 8

Find the domain of f + f . Now Try Exercise 55.

ACCENT ON TECHNOLOGy

Composition of Functions and Domain Consider the two functions given in Example 8(a). We can use a graphing calculator to graph the composite function f + g. To enter the composition, we enter y1 1y22 using VARS . See Figure 3-88. From the graph, we can determine the domain of f + g. We see that the domain is 3 3, ` 2 .

FIGURE 3-88

EXAMPLE 9

SOLUTION

Finding the Domain of Composite Functions

Let f 1x2 5

x13 1 and g 1x2 5 . x22 x a. Find the domain of f + g. b. Find f + g.

We will first find the domains of f 1x2 and g 1x2 . Then, we will find the domain of f + g and g + f by applying the restrictions stated earlier. Since any real number except 2 can be an input into f, the domain of f is 12`, 22 c 12, ` 2 . For x1 to be a real number, x cannot be 0. Thus, the domain of g is 12`, 02 c 10, ` 2 . a. The domain of f + g is the set of real numbers x such that x is in the domain of g and g 1x2 is in the domain of f. We have seen that all values of x but 0 are in the domain of g and that all values of g 1x2 but 2 are in the domain of f. So we must

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376

Chapter 3

Functions

exclude 0 from the domain of f + g and all values of x where g 1x2 5 2. To find the excluded values, we proceed as follows: g 1x2 5 2 1 52 x 1 5 2x x5

1 2

Substitute

1 for g 1x2 . x

Since x 2 0, we can multiply both sides by x. Divide both sides by 2.

The domain of f + g is the set of all real numbers except 0 and 12 , which is

12`, 02 c Q0, 12R c Q12, `R.

b. To find f + g, we proceed as follows. 1 f + g2 1x2 5 f 1g 1x2 2 1 5 fa b x

1 13 x 5 1 22 x 1 1 3x 5 1 2 2x

Thus, 1 f + g2 1x2 5 11 Self Check 9

Let f 1x2 5

Substitute

1 for g 1x2 . x

Substitute

1 for x in f. x

Multiply numerator and denominator by x.

1 3x 2 2x .

1 x and g 1x2 5 . a. Find the domain of f + g x21 x

b. Find f + g

Now Try Exercise 63.

Caution

Example 9 illustrates that the domain of the composite function f + g cannot always be found by finding f + g and analyzing it. In the example, we found that the domain is all real numbers, except for 0 and 12 . If we only analyze the form f + g, we would state the domain incorrectly as all real numbers except 12 .

5. Write Functions as Compositions Comment The result of a decomposition is not unique. There are several possibilities. In the example to the right, another possibility is f 1x2 5 x2 and g 1x2 5 12x2 2 6x 1 32 2 .

When we form a composite function f + g, we obtain a function h. It is possible to reverse this composition process and begin with a function h and express it as a composition of two functions. This process is called decomposition. For example, consider h 1x2 5 12x2 2 6x 1 32 4 . The function h takes 2 2x 2 6x 1 3 and raises it to the fourth power. To write the function h as a composition of functions f and g, we can let f 1x2 5 x4

and

g 1x2 5 2x2 2 6x 1 3

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Section 3.6

Operations on Functions

377

Then

1 f + g2 1x2 5 f 1g 1x2 2

5 f 12x2 2 6x 1 32 5 12x2 2 6x 1 32 4 5 h 1x2

EXAMPLE 10 SOLUTION

Writing a Function as a Composition of Two Functions

Let h 1x2 5 "x 1 1. Find two functions f and g such that f + g 5 h.

Because h takes the square root of the algebraic function x 1 1, we let f 1x2 5 "x and g 1x2 5 x 1 1. f 1x2 5 "x

and

g 1x2 5 x 1 1

We can check the composition f + g to see that it gives the original function h. 1 f + g2 1x2 5 f 1g 1x2 2

5 f 1x 1 12

5 "x 1 1 5 h 1x2

Self Check 10

3 2 Let h 1x2 5 " x 2 5. Find functions f and g such that f + g 5 h.

Now Try Exercise 73.

6. Use Operations on Functions to Solve Problems EXAMPLE 11

Solving an Application Using Composition of Functions A laboratory sample is removed from a cooler at a temperature of 15 + F. Technicians then warm the sample at the rate of 3 + F per hour. Express the sample’s temperature in degrees Celsius as a function of the time t (in hours) since it was removed from the cooler.

SOLUTION

We first write a Fahrenheit temperature function that represents the warming of the sample. We then use composition of functions to write degrees Celsius as a function of degrees Fahrenheit. The temperature of the sample is 15° F when t 5 0. Because the sample warms at 3° F per hour, it warms 3t° after t hours. Thus, the Fahrenheit temperature after t hours is given by the function F 1t2 5 3t 1 15

F 1t2 is the Fahrenheit temperature and t represents the time in hours.

The Celsius temperature is a function of the Fahrenheit temperature F 1t2 , given by the formula 5 C 1F 1t2 2 5 1F 1t2 2 322 9

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378

Chapter 3

Functions

To express the sample’s Celsius temperature as a function of time, we find the composition function C + F . 1C + F2 1t2 5 C 1F 1t2 2

5 C 13t 1 152 5 5 3 13t 1 152 2 32 4 9 5 5 13t 2 172 9 15 85 5 t2 9 9 5 85 5 t2 3 9

Self Check 11

Substitute for F 1t2 .

Substitute 3t 1 15 for F 1t2 inC 1F 1t22 . Simplify.

Find the 1C + F2 1102 .

Now Try Exercise 93. Self Check Answers

1. 1g 2 f 2 1x2 5 2x 2 4

2. 1g/f 2 1x2 5

2x 2 3 1 1 , a2`, 2 b c a2 , `b 3x 1 1 3 3

3. 1g 2 f 2 1x2 5 "x 2 x2 1 4, 3 0, ` 2 4. 1 f ? g2 122 5 215 5. One possibility is f 1x2 5 2x2 and g 1x2 5 x2 2 3x 2 2. 7. 1g + f 2 1212 5 40 8. 3 0, ` 2 6. 1 f + h2 1x2 5 2x 1 9 1 9. 12`, 02 c 10, 12 c 11, ` 2 ; 1 f + g2 1x2 5 12x 3 x, g 1x2 5 x2 2 5 10. One possibility is f 1x2 5 " 11. approximately 7.2°C

Exercises 3.6 Getting Ready

Fill in the blanks. 1. 1 f 1 g2 1x2 5 2. 1 f 2 g2 1x2 5 3. 1 f ? g2 1x2 5 4. 1 f/g2 1x2 5

, where g 1x2 2 0 5. The domain of f 1 g is the domains of f and g. 6. 1 f + g2 1x2 5 7. 1g + f 2 1x2 5 8. To determine 1 f + g2 1252 , first find

Let f 1x2 5 2x 1 1 and g 1x2 5 3x 2 2. Find each function and its domain. 12. f 2 g 11. f 1 g

Practice

You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises.

13. f ? g

of the

14. f/g

Let f 1x2 5 x2 1 x and g 1x2 5 x2 2 1. Find each function and its domain. 15. f 2 g 16. f 1 g

.

9. Composition of functions is not . 10. To be in the domain of the composite function f + g, a number x has to be in the of g, and of f. the output of g must be in the Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.6

17. f/g

18. f ? g

Let f 1x2 5 x2 2 7 and g 1x2 5 "x. Find each function

49. f + f

Operations on Functions

379

50. g + g

Let f 1x2 5 x2 and g 1x2 5 2x. Determine the domain of each composite function and then find the composite function. 52. f + g 51. g + f

and its domain. 19. f 1 g

20. f 2 g

53. g + g

21. f/g

22. f ? g

Let f 1x2 5 "x and g 1x2 5 x 1 1. Determine the domain of each composite function and then find the composite function. 56. g + f 55. f + g

Let f 1x2 5 x2 2 1 and g 1x2 5 3x 2 2. Find each value, if possible. 24. 1 f 1 g2 1232 23. 1 f 1 g2 122 26. 1 f 2 g2 1252 25. 1 f 2 g2 102 27. 1 f ? g2 122 2 29. 1 f/g2 a b 3

28. 1 f ? g2 1212

30. 1 f/g2 1t2

Find two functions f and g such that h 1x2 can be expressed as the function indicated. Several answers are possible. 31. h 1x2 5 3x2 1 2x; f 1 g 32. h 1x2 5 3x2 ; f ? g 3x2 ; f/g 33. h 1x2 5 2 x 21 34. h 1x2 5 5x 1 x2 ; f 2 g 35. h 1x2 5 x 13x2 1 12 ; f 2 g

36. h 1x2 5 13x 2 22 13x 1 22 ; f 1 g 37. h 1x2 5 x2 1 7x 2 18; f ? g 38. h 1x2 5 5x5 ; f/g

Let f 1x2 5 2x 2 5 and g 1x2 5 5x 2 2. Find each value. 40. 1g + f 2 1232 39. 1 f + g2 122

57. f + f

54. f + f

58. g + g

Let f 1x2 5 "x 1 1 and g 1x2 5 x2 2 1. Determine the domain of each composite function and then find the composite function. 59. g + f 60. f + g

61. g + g

62. f + f

Let f 1x2 5 x 21 1 and g 1x2 5 x 21 2 . Determine the domain of each composite function and then find the composite function. 63. f + g

64. g + f

66. g + g

1 41. 1 f + f 2 a2 b 2

3 42. 1g + g2 a b 5

65. f + f

45. 1 f + f 2 Q"3R

46. 1g + g2 1242

Find two functions f and g such that the composition f + g 5 h expresses the given correspondence. Several answers are possible. 68. h 1x2 5 7x 2 5 67. h 1x2 5 3x 2 2

Let f 1x2 5 3x2 2 2 and g 1x2 5 4x 1 4. Find each value. 44. 1g + f 2 132 43. 1 f + g2 1232

Let f 1x2 5 3x and g 1x2 5 x 1 1. Determine the domain of each composite function and then find the composite function. 48. g + f 47. f + g

69. h 1x2 5 x2 2 2

70. h 1x2 5 x3 2 3

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Chapter 3

Functions

71. h 1x2 5 1x 2 22 2

72. h 1x2 5 1x 2 32 3

73. h 1x2 5 "x 1 2

1 74. h 1x2 5 x25

75. h 1x2 5 "x 1 2

76. h 1x2 5

77. h 1x2 5 x

78. f 1x2 5 3

1 25 x

Use the graphs of functions f and g to answer each problem. y

f(x) x

Applications

91. DVD camcorder Suppose that the functions R 1x2 5 300x and C 1x2 5 60,000 1 40x model a company’s monthly revenue and cost for producing and selling DVD camcorders. a. Find 1R 2 C2 1x2 , the function that models the monthly profit, P 1x2 . b. Find the company’s profit if 500 camcorders are produced and sold in one month. © Istockphoto.com/Wittelsbach bernd

380

92. TV screen The height of the television screen shown is 13 inches.

y 13 in.

d

w g(x) x

79. 1 f 1 g2 1242

80. 1 f 2 g2 112

82. 1 f/g2 1212 84. 1g + f 2 122

81. 1 f ? g2 152 83. 1 f + g2 132

85. 1 f + f 2 1222

86. 1g + g2 1252

Use the tables of values of f and g to answer each problem. f 1x2

x

4

0

4

9

2

4

6

13

3

9

13

17

4

16

x 2

87. 1 f 1 g2 122 89. 1 f + g2 122

g 1x2 0

88. 1 f/g2 142

90. 1g + f 2 122

a. Write a formula to find the area of the viewing screen. b. Use the Pythagorean Theorem to write a formula to find the width w of the screen. c. Write a formula to find the area of the screen as a function of the diagonal d. 93. Area of an oil spill Suppose an oil spill from a tanker is spreading in the shape of a circular ripple. If the function d 1t2 5 3t represents the diameter of the spill in inches at time t minutes, express the area, A, of the oil spill as a function of time. Find the area of the oil spill after 2 hours. Round to one decimal place. 94. Area of a square Write a formula for the area A of a square in terms of its perimeter P. 95. Perimeter of a square Write a formula for the perimeter P of a square in terms of its area A. 96. Ceramics When the temperature of a pot in a kiln is 1,200 + F, an artist turns off the heat and leaves the pot to cool at a controlled rate of 81 + F per hour. Express the temperature of the pot in degrees Celsius as a function of the time t (in hours) since the kiln was turned off.

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Section 3.7

97. Let f 1x2 5 3x. Show that 1 f 1 f 2 1x2 5 f 1x 1 x2 . 98. Let g 1x2 5 x2 . Show that 1g 1 g2 1x2 2 g 1x 1 x2 . 2 1 99. Let f 1x2 5 xx 1 1 . Find 1 f + f 2 1x2 .

Discovery and Writing

100. Let g 1x2 5 x 2x 1 . Find 1g + g2 1x2 .

381

Review Solve each equation for y. 105. x 5 3y 2 7 107. x 5

Let f 1x2 5 x2 2 x, g 1x2 5 x 2 3, and h 1x2 5 3x. Use a graphing calculator to graph both functions on the same axes. Write a brief paragraph summarizing your observations. 101. f and f + g 102. f and g + f 103. f and f + h

Inverse Functions

y y13

7 y y21 108. x 5 y 106. x 5

104. f and h + f

3.7 Inverse Functions In this section, we will learn to 1. 2. 3. 4.

Understand the definition of a one-to-one function. Determine whether a function is one-to-one. Verify inverse functions. Find the inverse of a one-to-one function.

contax66/Shutterstock.com

5. Understand the relationship between the graphs of f and f 21 . In this section, we will discuss inverse functions. A function and its inverse do opposite things. Suppose we climb the Great Wall of China on a summer day when the temperature reaches a high of 35°C. The linear function defined by F 5 95C 1 32 gives a formula to convert degrees Celsius to degrees Fahrenheit. If we substitute a Celsius reading into the formula, a Fahrenheit reading comes out. For example, if we substitute 35 for C, we obtain a Fahrenheit reading of 95°: 9 F 5 C 1 32 5 9 5 1352 1 32 5 5 63 1 32 5 95 If we want to find a Celsius reading from a Fahrenheit reading, we need a formula into which we can substitute a Fahrenheit reading and have a Celsius reading come out. Such a formula is C 5 59 1F 2 322 , which takes the Fahrenheit reading of 95° and turns it back into a Celsius reading of 35°. 5 C 5 1F 2 322 9 5 5 195 2 322 9 5 5 1632 9 5 35

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Chapter 3

Functions

The functions defined by these two formulas do opposite things. The first turns 35°C into 95° Fahrenheit, and the second turns 95° Fahrenheit back into 35°C. Such functions are called inverse functions. Some functions have inverses that are functions and some do not. To guarantee that the inverse of a function will also be a function, we must know that the function is one-to-one.

1. Understand the Definition of a One-to-One Function In this section, we will find inverses of functions that are one-to-one. One-to-one functions are functions whose inverses are also functions. We now examine what it means for a function to be one-to-one. Consider the following two functions: Function 1: To each student, there corresponds exactly one eye color Function 2: To each student, there corresponds exactly one college identification number Function 1 is not a one-to-one function because two different students can have the same eye color. Function 2 is a one-to-one function because two different students will always have two different ID numbers. Recall that each element x in the domain of a function has a single output y. For some functions, different numbers x in the domain can have the same output. See Figure 3-89(a). For other functions, called one-to-one functions, different numbers x have different outputs. See Figure 3-89(b). y

y

Same output

Different outputs

x x1

x2

x

x3

x1

Different x's

x2

Different x's

Not a one-to-one function

A one-to-one function

(a)

(b) FIGURE 3-89

One-to-One Functions

A function f from a set X to a set Y is called a one-to-one function if and only if different numbers in the domain of f have different outputs in the range of f. The previous definition implies that if x1 and x2 are two numbers in the domain of f and x1 2 x2, then f 1x12 2 f 1x22 .

2. Determine Whether a Function Is One-to-One EXAMPLE 1

Determining Whether a Function Is One-to-One Determine whether each function is one-to-one. a. f 1x2 5 x4 1 x2 b. f 1x2 5 x3

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Section 3.7

SOLUTION

383

Inverse Functions

We will examine the functions and determine whether the definition of a one-toone function applies. If different x-values always produce different y-values, the function is one-to-one. a. The function f 1x2 5 x4 1 x2 is not one-to-one, because different numbers in the domain have the same output. For example, 2 and –2 have the same output: f 122 5 f 1222 5 20.

b. The function f 1x2 5 x3 is one-to-one, because different numbers x produce different outputs f 1x2 . This is because different numbers have different cubes. Self Check 1

Determine whether f 1x2 5 "x is one-to-one. Now Try Exercise 11.

A Horizontal Line Test can be used to determine whether the graph of a function represents a one-to-one function. If every horizontal line that intersects the graph of a function does so exactly once, the function passes the Horizontal Line Test and is one-to-one. See Figure 3-90(a). If any horizontal line intersects the graph of a function more than once, the function fails the Horizontal Line Test and is not one-to-one. See Figure 3-90(b). y

y Three intersections One intersection

Comment A one-to-one function satisfies both the Horizontal and Vertical Line Tests.

x

x

One intersection A one-to-one function

Not a one-to-one function

(a)

(b) FIGURE 3-90

EXAMPLE 2

Using the Horizontal Line Test Use the Horizontal Line Test to determine whether each graph represents a one-toone function. a.

b.

y

y

x x

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384

Chapter 3

Functions

SOLUTION

We will use the Horizontal Line Test and draw many horizontal lines. If every horizontal line that intersects the graph does so exactly once, the function is one-to-one. If any horizontal line intersects the graph more than once, the function is not oneto-one. a. Because the horizontal line drawn in Figure 3-91 intersects the graph in two places, we know that the function fails the Horizontal Line Test and is not a oneto-one function. y

x FIGURE 3-91

b. Several horizontal lines are drawn in Figure 3-92, and each one intersects the graph exactly once. We conclude that the graph passes the Horizontal Line Test and represents a one-to-one function. y

x

FIGURE 3-92

Self Check 2

Determine whether the graph in the margin represents a one-to-one function. Now Try Exercise 19.

y

3. Verify Inverse Functions x

Figure 3-93(a) illustrates a function f from set X to set Y. Since three arrows point to a single y, the function f is not one-to-one. If the arrows in Figure 3-93(a) were reversed, the diagram would not represent a function. If the arrows of the one-to-one function f in Figure 3-93(b) were reversed, as in Figure 3-93(c), the diagram would represent a function. This function is called the inverse of function f and is denoted by the symbol f 21 .

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.7

X

Y

X x1

y = f (x1) = f(x2) = f(x3)

x2 x3 Domain of f

Y

f

f x1

385

Inverse Functions

f(x1) x2

f(x2)

x3

Range of f

f(x3)

Domain of f

(a)

Range of f (b)

X

f

Y

–1

x1

f(x1) x2

f(x2)

x3

f(x3)

Domain of f

Range of f (c) FIGURE 3-93

Consider the functions:

Caution The 21 in the notation for inverse function is not an exponent. Remember that f

21

1 1x2 2 f 1x2

f 1x2 5 4x

and

g 1x2 5

x 4

These functions are inverses of each other because f multiplies any input x by 4 and function g will take the result and divide it by 4. The final result will be the original input x. We can show that the composition of these functions (in either order) is the identity function.

x x 1 f + g2 1x2 5 f 1g 1x2 2 5 f a b 5 4a b 5 x 4 4 f

Inverse Functions

21

and

1g + f 2 1x2 5 g 1 f 1x2 2 5 g 14x2 5

4x 5x 4

Since g 1x2 is the inverse of f 1x2 ,we can write g 1x2 using inverse notation as 1x2 5 x4 . Thus, 1 f + f 212 5 x and 1 f 21 + f 2 1x2 5 x. We can now define inverse functions.

If f and g are two one-to-one functions such that 1 f + g2 1x2 5 x for every x in the domain of g and 1g + f 2 1x2 5 x for every x in the domain of f , then f and g are inverse functions. Function g can be denoted as f 21 and is called the inverse function of f.

We can also list two important properties of one-to-one functions.

Properties of a One-to-One Function

Property 1: If f is a one-to-one function, there is a one-to-one function f 21 1x2 such that 1 f 21 + f 2 1x2 5 x

and

1 f + f 212 1x2 5 x.

Property 2: The domain of f is the range of f 21 and the range of f is the domain of f 21 .

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386

Chapter 3

Functions

Figure 3-94 shows a one-to-one function f and its inverse f 21 . To the number x in the domain of f, there corresponds an output f 1x2 in the range of f. Since f 1x2 is in the domain of f 21 , the output for f 1x2 under the function f 21 is f 21 1 f 1x2 2 5 x. Thus, 1 f 21 + f 2 1x2 5 f 21 1 f 1x2 2 5 x. X

Comment

Y f

To show that one function is the inverse of another, we must show that their compositions are the identity function, x.

x = f –1( f(x))

y = f (x)

f –1

Domain of f Range of f –1

Range of f Domain of f –1 FIGURE 3-94

EXAMPLE 3 SOLUTION

Verifying that Two Functions are Inverses

3 x are inverse functions. Verify that f 1x2 5 x3 and g 1x2 5 "

To show that f and g are inverse functions, we must show that f + g and g + f are x, the identity function. 3 3 1 f + g2 1x2 5 f 1g 1x2 2 5 f 1" x2 5 1"x2 3 5 x 3 3 1g + f 2 1x2 5 g 1 f 1x2 2 5 g 1x32 5 " x 5x

Because g is the inverse of f , we can use inverse notation and write g 1x2 5 "x 3 as f 21 1x2 5 " x. Because f is the inverse of g, we can use inverse notation and write f 1x2 5 x3 as g21 1x2 5 x3. Self Check 3

3

If x $ 0, are f 1x2 5 x2 and g 1x2 5 "x inverse functions? Now Try Exercise 21.

4. Find the Inverse of a One-to-One Function

If f is the one-to-one function y 5 f 1x2 , then f 21 reverses the correspondence of f. That is, if f 1a2 5 b, then f 21 1b2 5 a. To determine f 21 , we follow these steps.

Strategy for Finding f 21 from a Given Function f 1x2

Step 1: Replace f 1x2 with y. Step 2: Interchange the variables x and y. Step 3: Solve the resulting equation for y. Step 4: Replace y with f 21 1x2 .

Once f 21 1x2 is determined, it should be verified by showing that 1 f + f 212 1x2 5 x and 1 f 21 + f 2 1x2 5 x.

EXAMPLE 4

SOLUTION

Finding the Inverse of a One-to-One Function

3 Find the inverse of f 1x2 5 x 1 2 and verify the result. 2

We will use the strategy given above to find f 21 . We then will verify the result by showing that 1 f + f 212 1x2 5 x and 1 f 21 + f 2 1x2 5 x.

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Section 3.7

Inverse Functions

387

To find f 21, we use the following steps.

Step 1: Replace f 1x2 with y. 3 f 1x2 5 x 1 2 2 3 y5 x12 2

Step 2: Interchange the variables x and y. 3 x5 y12 2

Caution

Step 3: Solve the resulting equation for y.

After completing Step 3, if y does not represent a function of x, the process ends and f does not have an inverse.

3 x5 y12 2 2x 5 3y 1 4 2x 2 4 5 3y y5

2x 2 4 3

Multiply both sides by 2. Subtract 4 from both sides. Divide both sides by 3.

Step 4: Replace y with f 21 1x2 . 2x 2 4 3 2x 24 f 21 1x2 5 3 y5

The inverse of f 1x2 5 32x 1 2 is f 21 1x2 5 2x 32 4 . To verify the result, we will use f 1x2 5 32x 1 2 and f 21 1x2 5 2x 32 4 and show that 1 f + f 212 1x2 5 x and 1 f 21 + f 2 1x2 5 x. 1 f + f 212 1x2 5 f 1 f 21 1x2 2 5fa

2x 2 4 b 3 3 2x 2 4 5 a b12 2 3 5x2212 5x

1 f 21 + f 2 1x2 5 f 21 1 f 1x2 2

3 5 f 21 a x 1 2b 2 3 2a x 1 2b 2 4 2 5 3 3x 1 4 2 4 5 3 5x

Self Check 4

Find f 122 . Then find f 21 152 . Explain the significance of the results. Now Try Exercise 27.

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388

Chapter 3

Functions

5. Understand the Relationship Between the Graphs of f and f 21 Because we interchange the positions of x and y to find the inverse of a function, the point 1b, a2 lies on the graph of y 5 f 21 1x2 whenever the point 1a, b2 lies on the graph of y 5 f 1x2 . Thus, the graph of a function and its inverse are reflections of each other about the line y 5 x.

EXAMPLE 5

SOLUTION

Finding f 21 and Graphing Both f and f 21

Find the inverse of f 1x2 5 x3 1 3. Graph the function and its inverse on the same set of coordinate axes.

We will find the inverse of the function f 1x2 using the strategy given in the section. We will use translations to graph both f and f 21 . We first find f 21 and proceed as follows: Step 1: Replace f 1x2 with y. f 1x2 5 x3 1 3

y

y 5 x3 1 3

3

f(x) = x + 3 (1, 4)

y=x

(0, 3)

Step 2: Interchange the variables x and y.

3

= x–3 (4, 1) (3, 0)

x 5 y3 1 3 Step 3: Solve the resulting equation for y.

x

x 2 3 5 y3

3 y5" x23

FIGURE 3-95

Comment We can also graph f and f 21 by completing a table of solutions. The x and y columns of f can be reversed to obtain the table of solutions for f 21 .

Self Check 5

Step 4: Replace y with f 21 1x2 . 3 f 21 1x2 5 "x 2 3

We now graph f and f 21 . To graph f 1x2 5 x3 1 3, we translate the graph 3 of y 5 x3 vertically upward 3 units. To graph f 21 1x2 5 " x 2 3 we translate the 3 graph of y 5 "x horizontally 3 units to the right. The graphs of f and f 21 are shown in Figure 3-95. In the graph that appears in Figure 3-95, the line y 5 x is the axis of symmetry. Find f 122 . Then find f 21 1112 . Explain the significance of the result.

Now Try Exercise 45.

In the next example, we will consider a function that is not one-to-one but becomes so when we restrict its domain. By restricting the domain of the function and making it one-to-one, we are able to find its inverse and examine the function and its inverse graphically.

EXAMPLE 6

Restricting the Domain of f to Make It One-to-One; Finding f 21 ; Graphing f and f 21 ; Stating Domain and Range

The function y 5 f 1x2 5 x2 1 3 is not one-to-one. However, it becomes one-to-one when we restrict its domain to the interval 12`, 0 4 . Under this restriction, a. Find the inverse of f . b. Graph each function and state each one’s domain and range. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 3.7

SOLUTION

Inverse Functions

389

We will find f 21 by using the four-step strategy given in the section. We will then graph f and f 21 by using translations and then identify the domain and range from the graphs of each. a. We first find f 21 and follow these steps: Step 1: Replace f 1x2 with y. f 1x2 5 x2 1 3 y 5 x2 1 3

1x # 02

Step 2: Interchange the variables x and y. x 5 y2 1 3

1y # 02

Interchange x and y.

Step 3: Solve the resulting equation for y. x 2 3 5 y2 To solve this equation for y, we take the square root of both sides. Because y # 0, we have 2"x 2 3 5 y 1y # 02

Step 4: Replace y with f 21 1x2 .

The inverse of f is defined by f 21 1x2 5 2"x 2 3.

b. We graph the function f 1x2 5 x2 1 3 with domain 12`, 0 4 by translating the graph of the parabola y 5 x2 with domain 12`, 0 4 vertically upward 3 units. From the graph, we see that the y coordinates are 3 and above and thus the range is the interval 3 3, ` 2 . (See Figure 3-96.) We graph the function f 21 1x2 5 2!x 2 3 by translating the graph of y 5 !x horizontally to the right 3 units and then reflecting the graph about the x-axis. It has domain 3 3, ` 2 and range 12`, 0 4 . (See Figure 3-96.) Note that the line of symmetry is shown and is y 5 x. y (–2, 7)

f(x) = x2+ 3 (x  0)

y=

x

x =– x–3 (y  0)

(7, –2)

FIGURE 3-96

Domain of f and Range of f 21 : 12`, 0 4 Range of f and Domain of f 21 : 3 3, ` 2

Self Check 6

Find the inverse of f when its domain is restricted to the interval 3 0, ` 2 . Now Try Exercise 53. If a function is defined by the equation y 5 f 1x2 , we can often find the domain of f by inspection. Finding the range can be more difficult. One way to find the range of f is to find the domain of f 21 .

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390

Chapter 3

Functions

EXAMPLE 7

SOLUTION

Using the Domain of f 21 1x2 to Find the Range of f 1x2

Find the domain and range of f 1x2 5 x2 1 3. Find its range by finding the domain of f 21 1x2 .

We will find the domain of f 1x2 5 x2 1 3 by identifying the values of x that make the function undefined. We will then find f 21 1x2 and find its domain. The domain of f 21 1x2 will be the range of f 1x2 . Because x cannot be 0, the domain of f is 12`, 02 c 10, ` 2 . Next, we find 21 1 f x2 . Step 1: Replace f 1x2 with y. f 1x2 5

2 13 x 2 y5 13 x

Step 2: Interchange the variables x and y. x5

2 13 y

Interchange x and y.

Step 3: Solve the resulting equation for y. xy 5 2 1 3y

Multiply both sides by y.

xy 2 3y 5 2

Subtract 3y from both sides.

y 1x 2 32 5 2

Factor out y.

2 y5 x23

Divide both sides by x 2 3.

Step 4: Replace y with f 21 1x2 . f 21 1x2 5

2 x23

The domain of f 21 1x2 5 x 22 3 is 12`, 32 c 13, ` 2 because x cannot be 3. Because the range of f is the domain of f 21 , the range of f is 12`, 32 c 13, ` 2 . Self Check 7

Find the range of y 5 f 1x2 5

3 2 1. x

Now Try Exercise 61.

Self Check Answers

1. yes 2. no 3. yes 7. 12`, 212 c 121, ` 2

4. 5, 2

5. 11, 2

6. f 21 1x2 5 "x 2 3

Exercises 3.7 Getting Ready You should be able to complete these vocabulary and concept statements before you proceed to the practice exercises. Fill in the blanks. 1. If different numbers in the domain of a function have different outputs, the function is called a function.

2. If every line intersects the graph of a function only once, the function is one-to-one. 3. Two functions f and g are inverses if their composition in either order is the function. 4. The graph of a function and its inverse are reflections of each other about the line .

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Section 3.7

Practice Determine whether each function is one-to-one. 1 5. f 1x2 5 3x 6. f 1x2 5 x 2 8. f 1x2 5 x4 2 x2 7. f 1x2 5 x2 1 3 9. f 1x2 5 x3 2 x

11. f 1x2 5 0 x 0

12. f 1x2 5 0 x 2 3 0

13. f 1x2 5 5

15. f 1x2 5 1x 2 22 2 ; x $ 2

14. f 1x2 5 "x 2 5

16. f 1x2 5

29. f 1x2 5 x3 1 2

30. f 1x2 5 1x 1 22 3

33. f 1x2 5

1 x13

34. f 1x2 5

1 x22

35. f 1x2 5

1 2x

36. f 1x2 5

1 x3

5 x 31. f 1x2 5 "

10. f 1x2 5 x2 2 x

1 x

Inverse Functions

5 x14 32. f 1x2 5 "

Find the inverse of each one-to-one function and graph both the function and its inverse on the same set of coordinate axes. 3 38. y 5 x 2

37. y 5 5x Use the Horizontal Line Test to determine whether each graph represents a one-to-one function. y 17. 18. y

y

y

x

x

x

x

19.

20.

y

3 40. y 5 x 2 2 2

39. y 5 2x 2 4 y

y

y

x

391

x

x x

Verify that the functions are inverses by showing that f + g and g + f are the identity function. 1 21. f 1x2 5 5x and g 1x2 5 x 5 x25 22. f 1x2 5 4x 1 5 and g 1x2 5 4 x11 1 23. f 1x2 5 and g 1x2 5 x x21 x11 x11 24. f 1x2 5 and g 1x2 5 x21 x21 Each equation defines a one-to-one function f. Determine f 21 and verify that f + f 21 and f 21 + f are both the identity function. 1 26. f 1x2 5 x 25. f 1x2 5 3x 3 28. f 1x2 5 2x 2 5 27. f 1x2 5 3x 1 2

41. x 2 y 5 2

42. x 1 y 5 0

y

y

x

43. 2x 1 y 5 4

x

44. 3x 1 2y 5 6 y

y

x

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x

392

Chapter 3

Functions

3 45. f 1x2 5 " x24

3 46. f 1x2 5 " x13

y

54. f 1x2 5

y

x

47. f 1x2 5 1x 2 62 3

48. f 1x2 5 x3 1 2

y

1x . 02

55. f 1x2 5 x4 2 8 56. f 1x2 5

x

1 x2

21 x4

1x , 02

57. f 1x2 5 "4 2 x2

y

1x $ 02

58. f 1x2 5 "x2 2 1

10 # x # 22 1x # 212

x

Find the domain and the range of f. Find the range by finding the domain of f 21 . x x22 59. f 1x2 5 60. f 1x2 5 x22 x13

x

49. f 1x2 5

1 2x y

61. f 1x2 5

x

3 1 2 x 2

Applications

1 x23

63. Buying pizza A pizzeria charges $8.50 plus 75¢ per topping for a medium pizza.

y

x

51. f 1x2 5

62. f 1x2 5

© Istockphoto.com/Christa Brunt

50. f 1x2 5

1 22 x

52. f 1x2 5

x11 x21 y

x21 x

a. Find a linear function that expresses the cost f 1x2 of a medium pizza in terms of the number of toppings x. b. Find the cost of a pizza that has four toppings.

y

x

x

c. Find the inverse of the function found in part (a) to find a formula that gives the number of toppings f 21 1x2 in terms of the cost x. The function f defined by the given equation is one-to-one on the given domain. Find f 21 1x2 . 53. f 1x2 5 x2 2 3 1x # 02

d. If Josh has $10, how many toppings can he afford?

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Chapter Review

64. Cell phone bills A phone company charges $11 per month plus a nickel per call. a. Find a rational function that expresses the average cost f 1x2 of a call in a month when x calls were made. b. To the nearest tenth of a cent, find the average cost of a call in a month when 68 calls were made. c. Find the inverse of the function found in part (a) to find a formula that gives the number of calls f 21 1x2 that can be made for an average cost x. d. How many calls need to be made for an average cost of 15¢ per call?

Discovery and Writing 65. Write a brief paragraph to explain why the range of f is the domain of f 21 . 66. Write a brief paragraph to explain why the graphs of a function and its inverse are reflections about the line y 5 x.

393

68. Let f 1x2 5 x5 1 x3 1 x 2 3. Find f 21 1232 . (Hint: Do not find f 21 1x2 . Use the fact that if f 1a2 5 b, then f 21 1b2 5 a.) Use a graphing calculator to graph each function for various values of a. 69. For what values of a is f 1x2 5 x3 1 ax a one-to-one function? 70. For what values of a is f 1x2 5 x3 1 ax2 a one-toone function?

Review Simplify each expression. 71. 163/4

72. 2521/2

73. 1282 2/3

74. 282/3

77. 4921/2

78. a

75. a

64 21/3 b 125

67. Let f 1x2 5 x5 1 x3 1 x 1 3. Find f 21 132 . (Hint: Do not find f 21 1x2 . Use observation and the fact that if f 1a2 5 b, then f 21 1b2 5 a.)

76. 493/2 9 23/2 b 25

CHAPTER REVIEW SECTION 3.1

Functions and Function Notation

Definitions and Concepts

Examples

A function f is a correspondence between a set of input values x and a set of output values y, where to each xvalue there corresponds exactly one y-value.

Determine whether the equation y2 5 4x defines y to be a function of x. First we solve for y. y 5 6"4x y 5 62"x Since for each real number input for x (except 0) there corresponds an output of two y-values, y is not a function of x.

The set of input values x is called the domain of a function. The set of output values y is called the range of the function.

Find the domain of g 1x2 5 "x 2 4. Since x 2 4 must be non-negative x24$0 x$4

The domain is 3 4, ` 2 . Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

394

Chapter 3

Functions

Definitions and Concepts

To evaluate a function f 1x2 at a given input value x, we substitute the input value for x.

The fraction

f 1x 1 h2 2 f 1x2 h

is called the difference quotient

and is important in calculus. The graph of a function f in the xy-plane is the set of all points 1x, y2 where x is in the domain of f, y is in the range of f, and y 5 f 1x2 .

Examples

Let f 1x2 5 x 26 6 . Find f 1232 . 6 6 2 f 1232 5 5 52 1232 2 6 29 3

See Example 5 in Section 3.1. Graph the function f 1x2 5 22 0 x 0 1 5 and determine the domain and range of the function. To graph the function, we make a table of values and plot the points by drawing a smooth curve through them. f 1x2 5 22 0 x 0 1 5

22

1x, f 1x22

21

3

0

5

1

3

10, 52

2

1

x

x f(x) = –2|x| + 5

The domain and the range of a function can be identified by viewing the graph of the function. The inputs or x-values that correspond to points on the graph of the function can be identified on the x-axis and used to state the domain of the function. The outputs or f 1x2 values that correspond to points on the graph of the function can be identified on the y-axis and used to state the range of the function.

f 1x2 5 22 0 x 0 1 5 f 1x2

y

1

122, 12 121, 32 11, 32 12, 12

The domain of f 1x2 5 22 0 x 0 1 5 is 12`, ` 2 . The range is 12`, 5 4 . Note that the graph of the function passes the Vertical Line Test.

Vertical line test: If every vertical line that intersects a graph does so exactly once, every number x determines exactly one value of y, and the graph represents a function. A linear function is a function determined by an equation of the form f 1x2 5 mx 1 b or y 5 mx 1 b.

f 1x2 5 3x 1 2

1 f 1x2 5 2 x 2 7 2

EXERCISES Determine whether each equation defines y to be a function of x. Assume that all variables represent real numbers. 1. y 5 3

2. y 1 5x2 5 2

2

4. y 5 0 x 0 1 x

3. y 2 x 5 5

Find the domain of each function. Write each answer using interval notation. 3x 5. f 1x2 5 3x2 2 5 6. f 1x2 5 x25 7. f 1x2 5 "x 2 1

8. f 1x2 5 "x2 1 1

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Chapter Review

Find f 122 , f 1232 , and f 1k2 . 9. f 1x2 5 5x 2 2

10. f 1x2 5

6 x25

11. f 1x2 5 0 x 2 2 0

12. f 1x2 5

x2 2 3 x2 1 3

Use the Vertical Line Test to determine whether each graph represents a function. y y 17. 18.

x

Evaluate the difference quotient for each function f 1x2 . 14. f 1x2 5 2x2 2 7x 1 3 13. f 1x2 5 5x 2 6 Graph each function. Use the graph to identify the domain and range of each function. 16. f 1x2 5 3 0 x 2 2 0 15. f 1x2 5 2x2 1 4

x x

Definitions and Concepts

or

•

y 5 ax2 1 bx 1 c

where a, b, and c are real numbers and a 2 0. The graph of a quadratic function of the form f 1x2 5 ax2 1 bx 1 c 1a 2 02 is a parabola with vertex b b at Q22a , c 2 4a R.

20. Cost of electricity The cost C of electricity in Boston is a linear function of x, the number of kilowatthours (kwh) used. If the cost of 100 kwh is $20 and the cost of 500 kwh is $80, find a linear function that expresses C in terms of x.

Examples

A quadratic function is a second-degree polynomial function in one variable of the form f 1x2 5 ax2 1 bx 1 c

b. Find the income if the vendor sells 200 hamburgers.

Quadratic Functions

SECTION 3.2

f 1x2 5 3x2 2 2x 1 1

a 5 3, the parabola opens upward. 1 y 5 2 x2 2 4 2

•

a 5 212 , the parabola opens downward.

2

• •

x

19. Concessions A concessionaire at a basketball game pays a vendor $50 per game for selling hamburgers at $3.50 each. a. Write a linear function that describes the income I the vendor earns for the concessionaire during the game if the vendor sells h hamburgers.

y

y

395

If a . 0, the parabola opens upward. If a , 0, the parabola opens downward.

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396

Chapter 3

Functions

Definitions and Concepts

Examples

The standard form of an equation of a quadratic function is: y 5 f 1x2 5 a 1x 2 h2 2 1 k

1a 2 02

The vertex is at 1h, k2 . • The parabola opens upward when a . 0 and downward when a , 0. • The axis of symmetry of the parabola is the vertical line graph of the equation x 5 h. Graphing a quadratic function: To graph a quadratic function: 1. Determine whether the parabola opens upward or downward. 2. Find the vertex of the parabola. 3. Find the x-intercept(s). 4. Find the y-intercept. 5. Identify one additional point on the graph. 6. Draw a smooth curve through the points found in Steps 2–5.

•

f 1x2 5 4 1x 2 22 2 2 8

a 5 4, the parabola opens upward. The vertex is 12, 282 . The axis of symmetry is x 5 2. 1 y 5 2 1x 1 22 2 1 5 3

•

a 5 213 , the parabola opens downward. The vertex is 122, 52 . The axis of symmetry is x 5 22.

Graph the quadratic function f 1x2 5 3 1x 1 22 2 2 3.

Step 1: Determine whether the parabola opens upward or downward. Standard Form: f 1x2 5 a 1x 2 h2 2 1 k Given Form:

f 1x2 5 3 1x 1 22 2 2 3

Since a 5 3 and 3 is positive, the parabola opens upward. Step 2: Find the vertex of the parabola.

Standard Form: f 1x2 5 a 1x 2 h2 2 1 k Given Form:

f 1x2 5 3 1x 1 22 2 2 3

f 1x2 5 3 3 x 2 1222 4 2 1 1232

Since h 5 22 and k 5 23 the vertex is the point 122, 232 .

Step 3: Find the x-intercept(s). To find the x-intercepts, we substitute 0 for f 1x2 and solve for x. f 1x2 5 3 1x 1 22 2 2 3 0 5 3 1x 1 22 2 2 3 3 5 3 1x 1 22 2 1 5 1x 1 22 2

x 1 2 5 61

x 5 22 6 1 x 5 21

or

x 5 23

The x-intercepts are the points 123, 02 and 121, 02 .

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Chapter Review

Definitions and Concepts

397

Examples Step 4: Find the y-intercept. To find the y-intercept, we substitute 0 in for x and solve for y. f 1x2 5 3 1x 1 22 2 2 3 y 5 3 1x 1 22 2 2 3 y 5 3 10 1 22 2 2 3

y 5 3 122 2 2 3 y 5 12 2 3 y59

The y-intercept is the point 10, 92 .

Step 5: Identify one additional point on the graph. Because of symmetry, the point 124, 292 is on the graph. Step 6: Draw a smooth curve through the points found in Steps 2–5. x = –2

y 8

4 –6 –4

2

x

–4

The axis of symmetry of the parabola is x 5 22 because h 5 22. If the graph of the parabola opens downward, then the vertex is the maximum point on the graph of the parabola. If the graph of the parabola opens upward, then the vetex is the minimum point on the graph of the parabola.

Minimum cost A company has found that the total monthly cost C of producing x air-hockey tables is given by C 1x2 5 1.5x2 2 270x 1 28,665. Find the production level that minimizes the monthly cost and find that monthly minimum cost.

The function C 1x2 is a quadratic function whose graph is a parabola that opens upward. The minumum value of C 1x2 occurs at the vertex of the parabola. We will use the vertex formula to find the vertex of the parabola. We compare the equations C 1x2 5 1.5x2 2 270x 1 28,665 and y 5 ax2 1 bx 1 c to see that a 5 1.5, b 5 2270, and c 5 28,665. Using the vertex formula, we see that the vertex of the parabola is the point with coordinates a2

122702 2 b b2 2270 , c 2 b 5 a2 , 28,665 2 b 2a 4a 2 11.52 4 11.52 5 190, 16,5152

If the company makes 90 air-hockey tables each month, it will minimize its production cost. The minimum monthly cost will be $16,515. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

398

Chapter 3

Functions

EXERCISES Determine whether the graph of each quadratic function opens upward or downward. State whether a maximum or minimum point occurs at the vertex of the parabola. 1 21. f 1x2 5 x2 1 4 22. f 1x2 5 24 1x 1 12 2 1 5 2

31. y 5 x2 2 3x 2 4

32. y 5 3x2 2 8x 2 3 y

y

x x

Find the vertex of each parabola. 23. f 1x2 5 2 1x 2 12 2 1 6 2

24. y 5 22 1x 1 42 2 2 5 2

25. y 5 x 1 6x 2 4

26. y 5 24x 1 4x 2 9

Graph each quadratic function and find its vertex. 27. f 1x2 5 1x 2 22 2 2 3

28. f 1x2 5 2 1x 2 42 2 1 4

y

y

x x

29.y 5 x2 2 x

30. y 5 x 2 x2 y

y

33. Architecture A parabolic arch has an equation of 3x2 1 y 2 300 5 0. Find the maximum height of the arch. 34. Puzzle problem The sum of two numbers is 1, and their product is as large as possible. Find the numbers. 35. Maximizing area A rancher wishes to enclose a rectangular corral with 1,400 feet of fencing. What dimensions of the corral will maximize the area? Find the maximum area. 36. Digital camera A company that produces and sells digital cameras has determined that the total weekly cost C of producing x digital cameras is given by the function C 1x2 5 1.5x2 2 150x 1 4,850. Determine the production level that minimizes the weekly cost for producing the digital cameras and find that weekly minimum cost.

x

x

SECTION 3.3

Polynomial and Other Functions

Definitions and Concepts

Examples

A polynomial function in one variable (say, x) is a function of the form f 1x2 5 anx 1 an21x n

n21

1 c 1 a1x 1 a0

f 1x2 5 3x2 1 4x 2 7, degree of 2

f 1x2 5 217x4 1 3x3 2 2x2 1 13, degree of 4

where an , an21 , …, a1 , and a0 are real numbers and n is a whole number. The degree of a polynomial function is the largest power of x that appears in the polynomial.

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399

Chapter Review

Definitions and Concepts Graphing polynomial functions: 1. Find any symmetries of the graph. 2. Find the x- and y-intercepts of the graph. 3. Determine where the graph is above and below the x-axis. 4. Plot a few points, if necessary, and draw the graph as a smooth continuous curve.

If f 12x2 5 f 1x2 for all x in the domain of f, the graph of the function is symmetric about the y-axis, and the function is called an even function.

If f 12x2 5 2f 1x2 for all x in the domain of f, the function is symmetric about the origin, and the function is called an odd function.

Examples

Graph the function f 1x2 5 2x3 1 9x and determine whether it is even, odd, or neither. Step 1: Find any symmetries of the graph. To test for symmetry about the y-axis, we check to see whether f 1x2 5 f 12x2 . To test for symmetry about the origin, we check to see whether f 1x2 5 2f 1x2 . f 1x2 5 2x3 1 9x

f 12x2 5 2 12x2 3 1 9 12x2 f 12x2 5 x3 2 9x

Since f 1x2 2 f 12x2 , there is no symmetry about the yaxis. However, since f 12x2 5 2f 1x2 , there is symmetry about the origin. Step 2: Find the x- and y-intercepts of the graph. To find the x-intercepts, we let f 1x2 5 0 and solve for x. 2x3 1 9x 5 0

2x 1x2 2 92 5 0

2x 5 0

2x 1x 1 32 1x 2 32 5 0

or

x1350

x50

or

x2350 x53

x 5 23

The x-intercepts are (0, 0), (–3, 0), and (3, 0).

If we let x 5 0 and solve for f 1x2 , we see that the y-intercept is also (0, 0). Step 3: Determine where the graph is above or below the x-axis. To determine where the graph is above or below the x-axis, we plot the solutions of 2x3 1 9x 5 0 on a number line and establish the four intervals shown in the figure. We then test a number from each interval to determine the sign of f 1x2 .

+

–

+

–

( –∞, –3)

(–3, 0)

(0, 3)

(3, ∞)

f(–4) = 28 –3 f (–1) = –8 0 above the x-axis

below the x-axis

f(1) = 8 above the x-axis

3 f(4) = –28 below the x-axis

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400

Chapter 3

Functions

Definitions and Concepts

Examples Step 4: Plot a few points and draw the graph as a smooth continuous curve. We now plot the intercepts and one additional point. In the previous step we found that f 112 5 8. This will be the additional point we plot 11, 82 . Making use of our knowledge of symmetry about the origin and where the graph is above and below the x-axis, we now draw the graph as shown. y

4 –2

2

x

–4

f (x) = – x 3 + 9x

The function f 1x2 5 2x3 1 9x is an odd function because the function is symmetric about the origin. f 1x2 5 e

Some functions, called piecewise-defined functions, are defined by using different equations for different intervals in their domains.

x 2 2 if x , 3 x2 if x $ 3

• f 122 5 2 2 2 5 0

The greatest-integer function is important in many business applications and in the field of computer science. This function is determined by the equation f 1x2 5 Œxœ , where the value of f 1x2 that corresponds to x is the greatest integer that is less than or equal to x.

• f 142 5 142 2 5 16

Let f 1x2 5 Œx 2 3œ . Find f 12.22 . f 1x2 5 Œx 2 3œ

f 12.22 5 Œ2.2 2 3œ 5 Œ20.8œ 5 21

21 is the greatest integer less than or equal to 20.8.

EXERCISES Graph each polynomial function and determine whether it is even, odd, or neither. 37. y 5 x3 2 x

39. y 5 x3 2 x2

40. y 5 1 2 x4 y

y

38. y 5 x2 2 4x y

y

x x

x x

Evaluate each piecewise-defined function. 41. f 1x2 5 e a. f 1222

x 2 2 if x , 3 x2 if x $ 3 b. f 132

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Chapter Review

Evaluate each function at the indicated x-values. 45. f 1x2 5 Œ2xœ Find f 11.72 .

2 if x , 0 42. f 1x2 5 • 2 2 x if 0 # x , 2 x 1 1 if x $ 2

46. f 1x2 5 Œx 2 5œ

b. f 122

3 a. f a b 2

Graph each piecewise-defined function and determine the open intervals on which it is increasing, decreasing, or constant. 43. y 5 f 1x2 5 e

401

Find f 14.992 .

Graph each function. 47. f 1x2 5 Œxœ 1 2

48. f 1x2 5 Œx 2 1œ

y

x 1 5 if x # 0 5 2 x if x . 0

y

x

x

y

x

49. Renting a Jeep A rental company charges $20 to rent a Jeep Wrangler for one day, plus $8 for every 100 miles (or portion of 100 miles) that it is driven. Find the cost if the Jeep is driven 295 miles in one day. 50. Riding in a taxi A taxicab company charges $4 for a trip up to 1 mile, and $2 for every extra mile (or portion of a mile). Find the cost to ride 1112 miles.

44. y 5 f 1x2 5 e

x 1 3 if x # 0 3 if x . 0 y

x

SECTION 3.4

Transformations of the Graphs of Functions

Definitions and Concepts y 5 f 1x2 1 k If k . 0, the graph of e y 5 f 1x2 2 k Vertical translations:

is identical to the graph of y 5 f 1x2 , except that it is translated k units e

Examples

The function g 1x2 5 "x 1 3 2 2 is a translation of the graph of f 1x2 5 "x. Graph both on one set of coordinate axes.

upward . downward

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402

Chapter 3

Functions

Definitions and Concepts

Examples

y 5 f 1x 2 k2 If k . 0, the graph of e y 5 f 1 x 1 k2

By inspection, we see that the function g 1x2 5 "x 1 3 2 2 involves two translations of f 1x2 5 "x. The graph of g 1x2 5 "x 1 3 2 2 is identical to the graph of f 1x2 5 "x except it is translated 3 units to the left and 2 units downward as shown in the figure.

Horizontal translations:

is identical to the graph of y 5 f 1x2 , except that it is translated k units to the e

right . left

Vertical stretchings: If f is a function and k >1, then • The graph of y 5 kf 1x2 can be obtained by stretching the graph of y 5 f 1x2 vertically by multiplying each value of f 1x2 by k.

If f is a function and 0 , k , 1, then • The graph of y 5 kf 1x2 can be obtained by shrinking the graph of y 5 f 1x2 vertically by multiplying each value of f 1x2 by k.

Horizontal stretchings: If f is a function and k . 1, then • The graph of y 5 f 1kx2 can be obtained by shrinking the graph of y 5 f 1x2 horizontally by multiplying each x-value of f 1x2 by k1 . If f is a function and 0 , k , 1, • The graph of y 5 f 1kx2 can be obtained by stretching the graph of y 5 f 1x2 horizontally by multiplying each x-value of f 1x2 by k1 . Reflections: If f is a function, then • The graph of y 5 2f 1x2 is identical to the graph of y 5 f 1x2 except that it is reflected about the x-axis. • The graph of y 5 f 12x2 is identical to the graph of y 5 f 1x2 except that it is reflected about the y-axis.

To graph functions involving a combination of transformations, we must apply each transformation to the basic function. We will apply these transformations in the following order: 1. Horizontal translation 2. Stretching or shrinking 3. Reflection 4. Vertical translation

y

f(x) = x x g(x) = x +3 – 2

1 Graph: g 1x2 5 2 0 x 0 . 3

The graph of g 1x2 5 213 0 x 0 is identical to the graph of f 1x2 5 0 x 0 except that it is vertically shrunk by a factor of 13 and reflected about the x-axis. This is because each value of 0 x 0 is multiplied by 213 . The graphs of both functions are shown in the figure. y f(x) = x

x g(x) = – 1– x 3

Graph g 1x2 5 2 1x 2 42 3 1 1.

We will graph g 1x2 5 2 1x 2 42 3 1 1 by applying three translations to the basic function f 1x2 5 x3 : translate f 1x2 5 x3 horizontally 4 units to the right, stretch the graph by a factor of 2, and translate the graph vertically 1 unit upward. The graphs of both functions are shown in the figure. y

f(x) = x 3 x

g(x) =2(x – 4)3 + 1

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403

Chapter Review

EXERCISES Graph each function using a combination of translations and stretchings. 1 57. g 1x2 5 2 0 x 2 4 0 1 3 58. g 1x2 5 0 x 2 4 0 1 1 4

Each function is a translation of a basic function. Graph both on one set of coordinate axes. 51. g 1x2 5 x2 1 5

52. g 1x2 5 1x 2 72 3 y

y

y

y

x

x x

x

53. g 1x2 5 "x 1 2 1 3 y

1 60. g 1x2 5 1x 1 32 3 1 2 3

59. g 1x2 5 3"x 1 3 1 2

54. g 1x2 5 0 x 2 4 0 1 2

y

y

y

x x x x

Each function is a stretching of f 1x2 5 x3 . Graph both on one set of coordinate axes. 1 55. g 1x2 5 x3 56. g 1x2 5 125x2 3 3 y

3 62. g 1x2 5 2" x25

y

y

x

y

x

SECTION 3.5

x

x

Rational Functions

Definitions and Concepts A rational function is a function defined by an equation P 1x2 of the form y 5 Q 1x2 , where P 1x2 and Q 1x2 are polynomials and Q 1x2 2 0.

61. f 1x2 5 "2x 1 3

Examples f 1x2 5

3 x12

f 1x2 5

3x 1 4 x 2 3x 1 4 2

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404

Chapter 3

Functions

Definitions and Concepts Locating vertical asymptotes: To locate the vertical asymptotes of rational function P 1x2 f 1x2 5 Q 1x2 , we follow these steps:

Step 1: Factor P 1x2 and Q 1x2 and remove any common factors.

Step 2: Set the denominator equal to 0 and solve the equation. If a is a solution of the equation found in Step 2, x 5 a is a vertical asymptote. Locating horizontal asymptotes: To locate the horizontal asymptote of the rational funcP 1x2 tion f 1x2 5 Q 1x2 , we consider three cases:

Case 1: If the degree of P 1x2 is less than the degree of Q 1x2 , the line y 5 0 is the horizontal asymptote.

Case 2: If the degree of P 1x2 and Q 1x2 are equal, the line y 5 pq , where p and q are the leading coefficients of P 1x2 and Q 1x2 , is the horizontal asymptote.

Case 3: If the degree of P 1x2 is greater than the degree of Q 1x2 , there is no horizontal asymptote.

A third type of asymptote is called a slant asymptote. These asymptotes occur when the degree of the numerator of a rational function is 1 more than the degree of the denominator. As the name implies, it is a slanted line, neither vertical nor horizontal. Locating slant asymptotes: If the degree of P 1x2 is 1 greater than the degree of Q 1x2 P 1x2 for the rational function f 1x2 5 Q 1x2 , there is a slant asymptote. To find it, divide P 1x2 by Q 1x2 and ignore the remainder.

Examples Graph: y 5 f 1x2 5

4x . x22 We will use the steps outlined to graph the function. Step 1: Symmetry We find f 12x2 . f 12x2 5

24x 4x 4 12x2 5 5 12x2 2 2 2x 2 2 x12

Because f 12x2 2 f 1x2 and f 12x2 2 2 f 1x2 , there is no symmetry about the y-axis or the origin. Step 2: Vertical asymptotes We first note that f 1x2 is in simplest form. We then set the denominator equal to 0 and solve for x. Since the solution is 2, there will be a vertical asymptote at x 5 2. Step 3: y- and x-intercepts We can find the y-intercept by finding f 102 . f 102 5

4 102 0 5 50 022 22

The y-intercept is (0, 0). We can find the x-intercepts by setting the numerator equal to 0 and solving for x: 4x 5 0 x50 The x-intercept is (0, 0). Step 4: Horizontal asymptotes Since the degrees of the numerator and denominator of the polynomials are the same, the line y5

4 54 1

The leading coefficient of the numerator is 4. The leading coefficient of the denominator is 1.

is a horizontal asymptote. Step 5: Slant asymptotes Since the degree of the numerator is not 1 greater than the degree of the denominator, there are no slant asymptotes.

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405

Chapter Review

Definitions and Concepts

Examples

Steps to graph a rational function:

Step 6: Graph First, we plot the intercept (0, 0) and draw the asymptotes. We then find one additional point on our graph to see what happens when x is greater than 2. To do so, we choose 3, a value of x that is greater than 2, and evaluate f 132 .

We will use the following steps to graph the rational P 1x2 P 1x2 function f 1x2 5 Q 1x2 , where Q 1x2 is in simplest form (no common factors).

1. 2. 3. 4. 5. 6.

Check symmetries. Look for vertical asymptotes. Look for the y- and x-intercepts. Look for horizontal asymptotes. Look for slant asymptotes. Graph the function.

f 132 5

4 132 12 5 5 12 322 1

Since f 132 5 12, the point (3, 12) lies on our graph. We sketch the graph as shown in the figure. y 12 4x f(x) = –––– x–2 8 4 –4

4

x

EXERCISES Find the domain of each rational function. 3x2 1 x 2 2 63. f 1x2 5 x2 2 25

2x2 1 1 64. f 1x2 5 2 x 17

Graph each rational function. 2x 75. f 1x2 5 x24 y

12

Find the vertical asymptotes, if any, of each rational function. x15 x27 65. f 1x2 5 2 66. f 1x2 5 2 x 21 x 2 49

67. f 1x2 5

x x 1x26 2

68. f 1x2 5

5x 1 2 2x 2 6x 2 8 2

76. f 1x2 5

24x x14 y x

–12 –8 –4

8

–4

4

–8 4

77. f 1x2 5

y

8

12

–12

x

78. f 1x2 5

x 1x 2 12 2

1x 2 12 2 x

y

Find the horizontal asymptotes, if any, of each rational function. 69. f 1x2 5

2x2 1 x 2 2 4x2 2 4

70. f 1x2 5

5x2 1 4 4 2 x2

71. f 1x2 5

x11 x3 2 4x

72. f 1x2 5

x3 2x 2 x 1 11 2

x x

79. f 1x2 5

80. f 1x2 5

x2 2 x 2 2 x2 1 x 2 2 y

x3 1 x x2 2 4 y

Find the slant asymptote, if any, for each rational function. 73. f 1x2 5

2x2 2 5x 1 1 x24

74. f 1x2 5

5x3 1 1 x15

4 x

–4

4 –4

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x

406

Chapter 3

Functions

SECTION 3.6

Operations on Functions

Definitions and Concepts Adding, subtracting, multiplying, and dividing functions: If the ranges of functions f and g are subsets of the real numbers, then 1.

The sum of f and g, denoted as f 1 g, is defined by 1 f 1 g2 1x2 5 f 1x2 1 g 1x2 .

Examples

Let f 1x2 5 3x 1 5 and g 1x2 5 4x 2 7. Find each function and its domain: a. f 1 g b. f 2 g c. f ? g d. f/g a. 1 f 1 g2 1x2 5 f 1x2 1 g 1x2

5 13x 1 52 1 14x 2 72

5 7x 2 2

2.

The difference of f and g, denoted as f 2 g, is defined by 1 f 2 g2 1x2 5 f 1x2 2 g 1x2 .

Since the domain of both f and g is the set of real numbers, the domain of f 1 g is the interval 12`, ` 2 .

b. 1 f 2 g2 1x2 5 f 1x2 2 g 1x2

5 13x 1 52 2 14x 2 72 5 3x 1 5 2 4x 1 7 5 2x 1 12

3. The product of f and g, denoted as f ? g, is defined by 1 f ? g2 1x2 5 f 1x2 ? g 1x2 .

Since the domain of both f and g is the set of real numbers, the domain of f 2 g is the interval 12`, ` 2 .

c. 1 f ? g2 1x2 5 f 1x2 ? g 1x2

5 13x 1 52 14x 2 72 5 12x2 2 x 2 35

Since the domain of both f and g is the set of real numbers, the domain of f ? g is the interval 12`, ` 2 .

4. The quotient of f and g, denoted as f/g, is defined by 1 f/g2 1x2 5 g 1x2 , g 1x2 2 0. f 1x2

The domain of each function, unless otherwise restricted, is the set of real numbers x that are in the domains of both f and g. In the case of the quotient f /g, there is the restriction that g 1x2 2 0. The composite function f + g is defined by 1 f + g2 1x2 5 f 1 g 1x2 2 .

d. 1 f/g2 1x2 5

5

f 1x2 g 1x2

3x 1 5 14x 2 7 2 02 4x 2 7

Since 74 will make 4x 2 7 equal to 0, the domain of f/g is the set of all real numbers except 74 . This is Q2`, 74R c Q74, `R.

If f 1x2 5 2x 1 7 and g 1x2 5 x2 1 1, find 1 f + g2 1x2 and its domain.

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Chapter Review

Definitions and Concepts

407

Examples

The domain of f + g consists of all those numbers in the domain of g for which g 1x2 is in the domain of f.

Because 1 f + g2 1x2 means f 1g 1x2 2 , we will replace x in f 1x2 5 2x 1 7 with g 1x2 . 1 f + g2 1x2 5 f 1g 1x2 2

5 f 1x2 1 12

5 2 1x2 1 12 1 7 5 2x2 1 9

The domain of 1 f + g2 1x2 is the interval 12`, ` 2 because the domain of both f and g consists of all real numbers.

EXERCISES

Let f 1x2 5 x2 2 1 and g 1x2 5 2x 1 1. Find each function and its domain. 82. f ? g 81. f 1 g

83. f 2 g

84. f/g

Let f 1x2 5 2x2 2 1 and g 1x2 5 2x 2 1. Find each value, if possible. 86. 1 f 2 g2 1252 85. 1 f 1 g2 1232 87. 1 f ? g2 122

SECTION 3.7

Let f 1x2 5 x2 2 1 and g 1x2 5 2x 1 1. Find each function and its domain. 90. g + f 89. f + g Let f 1x2 5 x2 2 5 and g 1x2 5 3x 1 1. Find each value. 92. 1g + f 2 1222 91. 1 f + g2 1222 Find two functions f and g such that the composition f + g 5 h expresses the given correspondence. Several answers are possible. 93. h 1x2 5 1x 2 52 2

94. h 1x2 5 1x 1 62 3

1 88. 1 f/g2 a b 2

Inverse Functions

Definitions and Concepts A function f from a set X to a set Y is called a one-to-one function if and only if different numbers in the domain of f have different outputs in the range of f. A Horizontal Line Test can be used to determine whether the graph of a function represents a one-to-one function. If every horizontal line that intersects the graph of a function does so exactly once, the function passes the Horizontal Line Test and is one-to-one.

Examples

Determine whether the function f 1x2 5 x4 2 2x2 is one-to-one. The function f 1x2 5 x4 2 2x2 is not one-to-one, because different numbers in the domain have the same output. For example, 2 and –2 have the same output: f 122 5 f 1222 5 8.

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408

Chapter 3

Functions

Definitions and Concepts

Examples

Inverse functions: If f and g are two one-to-one functions such that 1 f + g2 1x2 5 x for every x in the domain of g and 1g + f 2 1x2 5 x for every x in the domain of f , then f and g are inverse functions. Function g is denoted as f 21 and is called the inverse function of f.

5 Verify that f 1x2 5 x5 and g 1x2 5 " x are inverse

functions.

To show that f and g are inverse functions, we must show that f + g and g + f are x, the identity function. 5 5 1 f + g2 1x2 5 f 1 g 1x2 2 5 f Q" xR 5 Q" xR 5 x 5

5 5 1g + f 2 1x2 5 g 1 f 1x2 2 5 g 1x52 5 " x 5x

Properties of a one-to-one function: Property 1: If f is a one-to-one function, there is a oneto-one function f 21 1x2 such that 1 f 21 + f 2 1x2 5 x and 1 f + f 212 1x2 5 x.

Because g is the inverse of f , we can use inverse nota5 tion and write f 1x2 5 x5 and f 21 1x2 5 "x. Because f is the inverse of g, we can use inverse notation and 5 write g 1x2 5 "x and g21 1x2 5 x5 .

Property 2: The domain of f is the range of f 21 , and the range of f is the domain of f 21 .

Find the inverse of f 1x2 5 x3 1 5.

Strategy for finding f 21 : Step 1: Replace f 1x2 with y.

We will find the inverse of the function using the strategy given in the section.

Step 2: Interchange the variables x and y.

Step 1: Replace f 1x2 with y.

Step 3: Solve the resulting equation for y.

f 1x2 5 x3 1 5

Step 4: Replace y with f 21 1x2 .

y 5 x3 1 5

The graph of a function and its inverse are reflections of each other about the line y 5 x.

Step 2: Interchange the variables x and y. x 5 y3 1 5 Step 3: Solve the resulting equation for y. x 2 5 5 y3

3 x25 y5"

Step 4: Replace y with f 21 1x2 . 3 f 21 1x2 5 "x 2 5

EXERCISES Verify that the functions are inverses by showing that f + g and g + f are the identity function. 1 99. f 1x2 5 8x 2 3 100. f 1x2 5 22x

Determine whether each function is one-to-one. 95. f 1x2 5 x2 1 7

96. f 1x2 5 x3

21 Use the Horizontal Line Test to determine whether each graph Each equation defines a one-to-one function. Find f and verify that f + f 21 and f 21 + f are the identity function. represents a one-to-one function. 1 y y 97. 98. 102. y 5 101. y 5 7x 2 1 22x

12

x

8

103. y 5

4 –4

4

x 12x

104. y 5

3 x3

x

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Chapter Test

105. Find the inverse of the one-to-one function f 1x2 5 2x 2 5 and graph both the function and its inverse on the same set of coordinate axes.

409

2x 1 3 106. Find the range of y 5 5x 2 10 by finding the 21 domain of f .

y

x

CHAPTER TEST Find the domain of each function. Write each answer using interval notation. 3 2. f 1x2 5 "x 1 3 1. f 1x2 5 x25 Find f 1212 and f 122 . x 3. f 1x2 5 x21

13. Suspension bridges The cable of a suspension bridge is in the shape of the parabola x2 2 2,500y 1 25,000 5 0 in the coordinate system shown in the illustration. (Distances are in feet.) How far above the roadway is the cable’s lowest point? y

4. f 1x2 5 "x 1 7

Find the vertex of each parabola. 5. y 5 3 1x 2 72 2 2 3 6. y 5 x2 2 2x 2 3 7. f 1x2 5 3x2 2 24x 1 38

Graph each function. 9. f 1x2 5 x4 2 x2

x 500 ft

8. f 1x2 5 5 2 4x 2 x2

10. f 1x2 5 x 2 x 5

y

14. Refer to Question 13. How far above the roadway does the cable attach to the vertical pillars? Graph each function.

15. f 1x2 5 1x 2 32 2 1 1

3

y

x

500 ft

y

16. f 1x2 5 "x 2 1 1 5 y

x x x

Assume that an object tossed vertically upward reaches a height of h feet after t seconds, where h 5 100t 2 16t2. 11. In how many seconds does the object reach its maximum height?

Find all asymptotes of the graph of each rational function. Do not graph the function. x21 17. y 5 2 x 29

12. What is that maximum height?

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410

18. y 5

Chapter 3

Functions

Let f 1x2 5 3x and g 1x2 5 x2 1 2. Find each function.

x2 2 5x 2 14 x23

23. f 1 g 24. g + f

Graph each rational function. Check for asymptotes, intercepts, and symmetry. x x2 20. y 5 2 19. y 5 2 x 29 x 11

25. f/g 26. f + g

Assume that f 1x2 is one-to-one. Find f 21. x11 27. f 1x2 5 28. f 1x2 5 x3 2 3 x21

y

y

x

x

Graph each rational function. The numerator and denominator share a common factor. 2x2 2 3x 2 2 x 21. y 5 22. y 5 2 x22 x 2x

Find the range of f by finding the domain of f 21. 3 3x 2 1 30. y 5 29. y 5 2 2 x x23

y

y

x x

CUMULATIVE REVIEW EXERCISES Use the x- and y-intercepts to graph each equation. 1. 5x 2 3y 5 15 2. 3x 1 2y 5 12 y

Write the equation of the line with the given properties. Give the answer in slope-intercept form. 5. The line passes through (23, 5) and (3, 27).

y

6. The line passes through Q32, 52R and has a slope of 72.

x

x

Find the length, the midpoint, and the slope of the line segment PQ. 1 7 3. Pa22, b ; Qa3, 2 b 2 2 4. P 13, 72 ; Q 127, 32

7. The line is parallel to 3x 2 5y 5 7 and passes through (25, 3). 8. The line is perpendicular to x 2 4y 5 12 and passes through the origin.

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Cumulative Review Exercises

Graph each function. 27. f 1x2 5 x2 2 4

Graph each equation. Make use of intercepts and symmetries. 2

2

9. x 5 y 2 2

10. y 5 x 2 2

411

28. f 1x2 5 2x2 1 4

y

y

y

y

x

x

x

x

11. x2 1 y2 5 100

29. f 1x2 5 x3 1 x

12. x2 2 2x 1 y2 5 8

y

y

y

y

4 –4

30. f 1x2 5 2x4 1 2x2 1 1

x

x

4

x

x

–4

Graph each function. Show all asymptotes. Solve each proportion. 13.

x22 x26 5 x 5

14.

x12 3x 1 1 5 x26 2x 2 11

15. Dental billing The billing schedule for dental X-rays specifies a fixed amount for the office visit plus a fixed amount for each X-ray exposure. If 2 X-rays cost $37 and 4 cost $54, find the cost of 5 exposures. 16. Automobile collisions The energy dissipated in an automobile collision varies directly with the square of the speed. By what factor does the energy increase in a 50-mph collision compared with a 20-mph collision? Determine whether each equation defines a function. 17. y 5 3x 2 1 1 x22

19. y 5

2

18. y 5 x 1 3 20. y2 5 4x

Find the domain of each function.

21. f 1x2 5 x2 1 5

22. f 1x2 5

23. y 5 2"x 2 2

24. f 1x2 5 "x 1 4 26. f 1x2 5 2x2 1 5x 1 6

Find the vertex of each parabola. 25. y 5 x2 1 5x 2 6

7 x12

31. f 1x2 5

32. f 1x2 5

x x23

y

x2 2 1 x2 2 9 y

x

x

Let f 1x2 5 3x 2 4 and g 1x2 5 x2 1 1. Find each function and its domain. 33. 1 f 1 g2 1x2 34. 1 f 2 g2 1x2 35. 1 f ? g2 1x2 36. 1 f/g2 1x2

Let f 1x2 5 3x 2 4 and g 1x2 5 x2 1 1. Find each value. 38. 1g + f 2 122 37. 1 f + g2 122

39. 1 f + g2 1x2 40. 1g + f 2 1x2

Find the inverse of the function defined by each equation. 1 41. y 5 3x 1 2 42. y 5 x23

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412

Chapter 3

Functions

43. y 5 x2 1 5 1x $ 02

44. 3x 2 y 5 1

Write each sentence as an equation. 45. y varies directly with the product of w and z. 46. y varies directly with x and inversely with the square of t.

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4

Exponential and Logarithmic Functions

Kiselev Andrey Valerevich/Shutterstock.com

CAREERS AND MATHEMATICS:

Epidemiologists

Epidemiologists investigate and describe the determinants and distribution of disease, disability, and other health outcomes. They also develop means for prevention and control. Applied epidemiologists typically work for state health agencies and are responsible for responding to disease outbreaks and determining the cause and method of containment. Research epidemiologists work in laboratories studying ways to prevent future outbreaks. This career can be quite rewarding, both mentally and financially. Epidemiologists spend a lot of time saving lives and finding solutions for better health.

Education and Mathematics Required •

•

4.1

Exponential Functions and Their Graphs

4.2

Applications of Exponential Functions

4.3

Logarithmic Functions and Their Graphs

4.4

Applications of Logarithmic Functions

4.5

Properties of Logarithms

4.6

Exponential and Logarithmic Equations Chapter Review Chapter Test

Applied epidemiologists are generally required to have a master’s degree from a school of public health. Research epidemiologists may need a Ph.D. or medical degree, depending on their area of work. College Algebra, Trigonometry, Calculus, Applied Data Analysis, Survey and Research Methods, Mathematical Statistics, and Biostatistics are required math courses.

How Epidemiologists Use Math and Who Employs Them •

•

Epidemiologists use mathematical models when they are tracking the progress of an infectious disease. The SIR model consists of three variables: S (for susceptible), I (for infectious), and R (for recovered). It is used for infectious diseases such as measles, mumps, and rubella. Government agencies employ 57%; hospitals employ 12%; colleges and universities employ 11%, and 9% are employed in areas of scientific research and developmental services like the American Cancer Society.

Career Outlook and Earnings •

•

Employment growth is projected to be 15% over the 2008–2018 decade, which is faster than average. This is due to an increased threat of bioterrorism and rare but infectious diseases, such as West Nile Virus or Avian flu. The median annual income is $61,700, with the top 10% of salaries at $92,610.

For more information see: www.bls.gov/oco

In this chapter, we will discuss exponential functions, which are often used in banking, ecology, and science. We will also discuss logarithmic functions, which are applied in chemistry, geology, and environmental science.

413 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

414

Chapter 4

Exponential and Logarithmic Functions

4.1 Exponential Functions and Their Graphs In this section, we will learn to 1. Approximate and simplify exponential expressions. 2. Graph exponential functions. 3. Solve compound interest problems.

Extreme water slides, called “plunge” or “plummet” slides, are fearsome water slides because of their heights. With near vertical drops, the slides are designed to allow riders to reach the greatest possible speeds. Summit Plummet at Blizzard Beach, a part of Walt Disney World Resort in Florida, stands 120 ft tall. On this slide, riders can achieve speeds up to 55 mph. The shapes of extreme water slides can be modeled using exponential functions, the topic of this section. Exponential functions are also important in business. Consider the graph shown in Figure 4-1. It shows the balance in a bank account in which $5,000 was invested in 1990 at 8%, compounded monthly. The graph shows that in the year 2015, the value of the account will be approximately $38,000, and in the year 2030, the value will be approximately $121,000. The curve in Figure 4-1 is the graph of an exponential function. From the graph, we can see that the longer the money is kept on deposit, the more rapidly it will grow. Value of $5,000 invested at 8% compounded monthly

Value (in dollars)

bocky/Shutterstock.com

4. Define e and graph base-e exponential functions. 5. Use transformations to graph exponential functions.

121,000 120,000 110,000 100,000 90,000 80,000 70,000 60,000 50,000 40,000 38,000 30,000 20,000 10,000 1990

2000

2010 2015 2020

2030

Year FIGURE 4-1

Before we can discuss exponential functions, we must define irrational exponents.

1. Approximate and Simplify Exponential Expressions We have discussed expressions of the form bx, where x is a rational number. • • •

52 means “the square of 5.” 41/3 means “the cube root of 4.” 622/5 5 612/5 means “the reciprocal of the fifth root of 62.”

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Section 4.1

Exponential Functions and Their Graphs

415

To understand exponential functions and their graphs requires that we give meaning to bx when x is an irrational number. Consider the expression 3"2 where "2 is the irrational number 1.414213562 ...

We can use closer and closer approximations as shown below. Since "2 is an irrational number, we will use a calculator to find the approximations. 3"2 < 31.4 < 4.655536722

3"2 < 31.41 < 4.706965002

3"2 < 31.414 < 4.727695035 3"2 < 31.4142 < 4.72873393

Since the exponents of the expressions in the list are getting closer to "2, the values of the expressions are getting closer to the value of 3"2. On a scientific calculator, there is an exponential key, usually yx . On a graphing calculator, we can use the key.

EXAMPLE 1

Approximating Exponential Expressions Approximate each expression correct to 4 decimals. a. 42/3

SOLUTION

b. 52"3

4 p c. a b 7

We will use a calculator. a. 4^ 12/32 < 2.5198

b. 5^ Q2"3R < 0.0616 c. 14/72 ^p < 0.1724

Enter 2/3 in parentheses. Enter 2"3 in parentheses. Enter the base, 4/7, in parentheses.

Figure 4-2 shows the graphing calculator screen that was used to find the values.

FIGURE 4-2

Self Check 1

Approximate each expression correct to 4 decimals. b. 23"6

a. 53/5

c. 722.356

Now Try Exercise 15. If b is a positive number and x is a real number, the expression bx always represents a positive number. It is also true that the familiar properties of exponents hold for irrational exponents.

EXAMPLE 2

Simplifying Expressions with Irrational Exponents Simplify each expression: a. Q3"2 R

"2

b. a"8 ? a"2 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

416

Chapter 4

Exponential and Logarithmic Functions

SOLUTION

We will use properties of exponents to simplify each expression. a. Q3"2 R

"2

5 3"2"2

Keep the base and multiply the exponents. "2"2 5 "4 5 2

5 32 59

b. a"8 ? a"2 5 a"81"2

Keep the base and add the exponents. "8 5 "4"2 5 2"2

2"21"2

5a

2"2 1 "2 5 3"2

5 a3"2 Self Check 2

Simplify: a. Q2"3R

"12

b. x"20 ? x"5

Now Try Exercise 19.

2. Graph Exponential Functions If b . 0 and b 2 1, the function y 5 bx defines a function, because for each input x, there is exactly one output y. Since x can be any real number, the domain of the function is the set of real numbers. Since the base b of the expression bx is positive, y is always positive, and the range is the set of positive numbers. Since bx is an exponential expression, the function is called an exponential function. We make the restriction that b . 0 to exclude any imaginary numbers that might result from taking even roots of negative numbers. The restriction that b 2 1 excludes the constant function f 1x2 5 1x, in which f 1x2 5 1 for every real number x. Exponential Functions

An exponential function with base b is defined by the equation f 1x2 5 bx or

y 5 bx

(b . 0, b 2 1, and x is a real number)

The domain of any exponential function is the interval 12`, ` 2 . The range is the interval 10, ` 2 .

Since the domain and range of f 1x2 5 bx are sets of real numbers, we can graph exponential functions. For example, to graph f 1x2 5 2x

we find several points 1x, f 1x2 2 whose coordinates satisfy the equation, plot the points, and join them with a smooth curve, as in Figure 4-3(a). To graph the function 1 x f 1x2 5 a b 2

we find several points 1x, f 1x2 2 whose coordinates satisfy the equation, plot the points, and join them with a smooth curve, as shown in Figure 4-3(b) on the next page. y

x

(2, 4)

(−1, 1–2 ) f(x) = 2

(0, 1)

(a)

x

f 1x2

21

1 2

0

1

1

2

2

4

(1, 2)

x

f 1x2 5 2x

1x, f 1x22

1 a21, b 2 10, 12 11, 22 12, 42

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Section 4.1

Exponential Functions and Their Graphs

417

1 x f 1x2 5 a b 2

y

(−2, 4)

22

f 1x2

1x, f 1x22

21

2

0

1

1

1 2

10, 12

x 1 f(x) = – 2

x

()

(−1, 2) (0, 1)

x

(1, 1–2) (b)

4

122, 42 121, 22 1 a1, b 2

FIGURE 4-3

By looking at the graphs in Figure 4-3, we can see that the domain of each function is the interval 12`, ` 2 and that the range is the interval 10, ` 2 .

EXAMPLE 3 SOLUTION

Graphing an Exponential Function

Graph: f 1x2 5 4x .

We will find several points 1x, y2 that satisfy the equation, plot the points, and join them with a smooth curve, as in Figure 4-4. y

(1, 4) f(x) =

(

1 −1, – 4

)

(0, 1)

x

4x

x

f 1x2 5 4x f 1x2

21

1 4

0

1

1

4

1x, f 1x22

1 a21, b 4 10, 12 11, 42

FIGURE 4-4

Self Check 3

1 x Graph: f 1x2 5 a b . 4 Now Try Exercise 29. The graph of f 1x2 5 4x in Example 3 has the following properties: 1. 2. 3. 4.

It passes through the point 10, 12 . It passes through the point 11, 42 . It approaches the x-axis. The x-axis is a horizontal asymptote. The domain is the interval 12`, ` 2 , and the range is the interval 10, ` 2 .

This example illustrates the following properties of exponential functions.

Properties of Exponential Functions

• The domain of the exponential function f 1x2 5 bx is 12`, ` 2 , the set of real numbers.

• The range is 10, ` 2 , the set of positive real numbers.

• The graph has a y-intercept at (0, 1).

• The x-axis is a horizontal asymptote of the graph.

• The graph of f 1x2 5 bx passes through the point 11, b2 .

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

418

Chapter 4

Exponential and Logarithmic Functions

EXAMPLE 4

Determining the Base of an Exponential Function

The graph of an exponential function of the form f 1x2 5 bx is shown in Figure 4-5. Find the value of b. y (2, 25)

f(x) = bx

(0, 1)

x

We note that the graph passes through the point 10, 12 , a property of exponential functions of this form. Since the graph also passes through the point 12, 252 , we can find the base b by substituting 2 for x and 25 for f 122 in the equation f 1x2 5 bx and solving for b. FIGURE 4-5

SOLUTION

f 1x2 5 bx f 122 5 b2 25 5 b2 55b

The base b is 5. Note that the points 10, 12 and 12, 252 satisfy the equation f 1x2 5 5x . Self Check 4

b must be positive.

3 Can a graph passing through 10, 22 and a1, b be the graph of f 1x2 5 bx ? 2 Now Try Exercise 45.

In Figure 4-3(a) (where b 5 2 and 2 . 1), the values of y increase as the values of x increase. Since the graph rises as we move to the right, the function is an increasing function. Such a function is said to model exponential growth. In Figure 4-3(b) where Qb 5 12 and 0 , 21 , 1R, the values of y decrease as the

values of x increase. Since the graph drops as we move to the right, the function is a decreasing function. Such a function is said to model exponential decay. Two additional properties are stated here.

Additional Properties of Exponential Functions

If b . 1, then f 1x2 5 bx is an increasing function. This function models exponential growth. y f(x) = b x b>1

(1, b)

(0, 1) x 1

Increasing function

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 4.1

Exponential Functions and Their Graphs

419

If 0 , b , 1, then f 1x2 5 bx is a decreasing function. This function models exponential decay. y f(x) = b x 0, we draw the circle as a dashed circle, as in Figure 6-16(a). Step 3: Because the work is easy, we select the origin as the test point. Step 4: We substitute the coordinates of the origin into the inequality. x2 1 y2 . 25 02 1 02 . 25 0 . 25

False

Step 5: Since the coordinates of the origin do not satisfy the inequality, we shade the area outside the circle, as shown in Figure 6-16(b). y

y

x 2 + y 2 > 25

x 2 + y 2 = 25

x

x

(b)

(a) FIGURE 6-16

Self Check 3

Graph the inequality: x2 1 y2 # 36. Now Try Exercise 19.

2. Graph Systems of Inequalities We now consider systems of inequalities. To graph the solution set of the system b y

x=6

y=5

Solution

x FIGURE 6-17

EXAMPLE 4

y,5 x#6

we graph each inequality on the same set of coordinate axes, as in Figure 6-17. The graph of the inequality y , 5 is the half-plane that lies below the line y 5 5. The graph of the inequality x # 6 includes the half-plane that lies to the left of the line x 5 6 together with the line x 5 6. The portion of the xy-plane where the two graphs intersect is the graph of the system. Any point that lies in the doubly shaded region has coordinates that satisfy both inequalities in the system.

Graphing a System of Linear Inequalities Graph the solution set of b

x1y#1 . 2x 2 y . 2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

626

Chapter 6

Linear Systems

SOLUTION

On the same set of coordinate axes, we graph each inequality, as in Figure 6-18. The graph of x 1 y # 1 includes the graph of x 1 y 5 1 and all points below it. Because the boundary line is included, we draw it as a solid line.

x

y

1

1x, y2

(0, 1)

0

22

0

(1, 0)

1

0

x1y51

−y

1

2x

y=

=2

y x+

2x − y > 2

x

y

0 1

x

x+y≤1

2x 2 y 5 2

1x, y2

(0, 22) (1, 0)

Solution FIGURE 6-18

The graph of 2x 2 y . 2 contains only those points below the line graph of 2x 2 y 5 2. Because the boundary line is not included, we draw it as a dashed line. The area that is shaded twice represents the solution of the system of inequalities. Any point in the doubly-shaded region has coordinates that satisfy both inequalities.

Comment To be sure you graph the correct side of the boundary line, always use a test point.

Graph the solution set of b

Self Check 4

x1y,2 . x 2 2y $ 2

Now Try Exercise 29.

EXAMPLE 5

Graphing a System of Inequalities y , x2 Graph the solution set of c . x2 y. 22 4 The graph of y 5 x2 is a parabola opening upward with vertex at the origin, as shown in Figure 6-19. The points with coordinates that satisfy the inequality are the points below the parabola.

SOLUTION

2

The graph of y 5 x4 2 2 is also a parabola opening upward. The points that satisfy the inequality are the points above the parabola. The graph of the solution set is the shaded area between the two parabolas. x2

y > –– − 2 4

y

y 5 x2

y = x2

n

utio Sol

x x2 y = –– − 2 4

y
∞

@

>

² 

67.



(– 11, 11)

(

)

– 11

11

71. (–∞, –1) ∪ (–1, 0) ∪ (1, ∞)

(−3, 2) ( −3

() –1

) 2

) 0

( 1

73. (–∞, –3) ∪ (–3, –2] ∪ (2, ∞) () –3

( 2

] –2

75. (–∞, –2) ∪ (–2, –1/3) ∪ (1/2, ∞) 77.

79.

) –1/3

( 1/2

81.

(–∞, 0) ∪ (3/2, ∞) ( ( 0 3/2

83. (–∞, – 7) ∪ (–1, 1) ∪ ( 7, ∞) ( – 7

( –1

) 1

(0, 3/2) ( ) 0 3/2 (– ∞, 2) ∪ [13/5, ∞) ( [ 2 13/5

85. 19 min

87. 12

( 7

89. p # $1,124.12

5. b 2 c

7. >

9. linear

15.

[1, ∞) [ 1

(−∞, 1) ) 1

19.

[1, ∞) [ 1

21.

(−∞, 3] ] 3

23.

(−10/3, ∞) ( −10/3

25.

(5, ∞) ( 5

27.

[14, ∞) [ 14

29.

(−∞, 15/4] ] 15/4

31.

(−44/41, ∞) ( −44/41

) 1

17.

[5, 21]

53.

37. 2

Section 1.7

] 22

39.

] 4

( −4

65.

(–∞, 1) ∪ [2,∞) 2

49.

1 , −8 8 1 27. 2 3

[–2, 1)

)

1

45.

5

[ [

41.

(8, 22] ( 8

[−11, 4]

() –2

[ )

] 9

[ −11

17.

1 2 41. 3, 4 43. , 21 45. 3, 5 47. 22, 1 5 5 2, 2 51. 0, 4 53. 22 55. no solution 2 3 59. 1 61. 400 ft 63. 8 ft about $3,109 −5 −4 −3 −2 −1

71.

4 1 7. 0, , 2 3 2

5. 0, 25, 24

13. 3"2, 23"2, "5, 2"5 9. 5, 25, 1, 21

137. 4x2"2

141. "5 1 1

139. x 2 4"x 1 1 1 5

1. equal

117.

129. 6.2 2 0.7i

35.

(6, 9] ( 6

1 2"2 101. 2 6 i 5 5

99. 21 6 2"3i

103. 25 6 2"3i

33.

"7 95. 6 i 3

87. 1

91. a . $50,000 2 93. anything over $1,800 95. 16 cm , s , 20 cm 3 97. 40 1 2w , P , 60 1 2w 101. 0, 22, 2 103. 2 105. All are real.

Section 1.8 1. x 3. x 5 k or x 5 2k 7. x # 2k or x $ k 9. 7 17. x 2 5 19. 15. p 2 2 4 21. 0, 24 23. 2, 2 25. 3

5. 2k , x , k 11. 0 13. 2 x3 if x $ 0; 2x3 if x , 0 14 , 22 27. 7, 23 3

2 35. x $ 0 29. no solution 31. 5 33. , 2 7 3 3 37. 2 39. 0, 26 41. 0 43. , 3 2 5

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A6

Answers to Selected Exercises

45. 2

3 9 , 13 5

49.

57.

( −3

(–∞, –9) ∪ (3, ∞) ) ( –9 3 [−13/3, 1]

53.

[ ] −13/3 1 (−1, 1/5) ( −1 ( −5

46. two different rational numbers ) 9

(–∞, –7] ∪ [3, ∞) ] [ –7 3 55. (–∞, –3) ∪ (–3, ∞) () –3

59.

63. ) 7

(–∞, –7/9) ∪ (13/9, ∞) ) ( –7/9 13/9 (–2, –1/2) ∪ (–1/2, 1) ) ( )( –2 –1/2 1

65.

(–7/3, –2/3) ∪ (4/3, 3) ( ) ( ) –7/3 –2/3 4/3 3

67.

(–7, –1) ∪ (11, 17) ( ) ( ) –7 –1 11 17

69.

(–18, –6) ∪ (10, 22) ( ) ( ) –18 –6 10 22

71.

(–10, –7] ∪ [5, 8) [ ) ( ] –10 –7 5 8

1 73. c 2 , ` b 2

79. 3 0, ` 2

75. 12`, 02

1 77. a2`, 2 b 2

83. 0 c 2 0.6° 0 # 0.5°

85. 0 h 2 55 0 , 17 87. a. 26.45%, 24.76% b. It is less than or equal to 1%. 95. 3.725 3 104 97. 523,000 99. −4xy 81. 70° # t # 86°

5 2 190 yd

34. 6"5 38. 24, 22 42. 9, 2

2 3

5 6 4"2 4 1 6 "21 39. 10 21 6 "21 43. 10

35.

7 6 3"5 37. 3, 5 5 1 40. 0, 41. 2, −7 5

48. 10, 2

69. 72. 76. 80.

67. "10 + 0i

"2 1 "11 1 6 i 70. 6 i 71. 2, −3 3 3 3 3 4, −2 73. 1, 1, −1, −1 74. 1, −1, 6, −6 8, −27 77. 5 78. 0 79. 4, −4 no solution 81. (−∞, 7) 66. 0 2 i

68. 1 + 0i

75. 9

) 7

82.

[–1/5, ∞) [ –1/5

84.

(−∞, −12/7) ) −12/7

85.

[−3, 5) [ −3

Chapter Review 2. x 2 0 3. x 2 1 16 4. x 2 2, x 2 3 5. ; conditional equation 27 16 6. 214; conditional equation 7. ; conditional equation 5 8. no solution; contradiction 9. 7; conditional equation 10. all real numbers; identity 11. 7; conditional equation 1 12. ; conditional equation 13. 22; conditional equation 3 9 14. 0; conditional equation 15. F 5 C 1 32 5 ff2 is a 2 S 1 Sr 16. f 5 17. f1 5 18. l 5 Pn 2 l f2 2 f r 19. 60% 20. 22.5 ft by 27.5 ft 21. 3 hr 22. 0.5 hr 1 1 23. 1.5 liters 24. about 3.9 hr 25. 5 hr 26. 3 oz 7 3 3 27. $4,500 at 11%; $5,500 at 14% 28. 10 29. 2, 2 2 1 4 8 2 4 30. , 2 31. 0, 32. , 33. 62"2 4 3 5 3 9

1 3

50.

65. 0 1 i

87.

1. no restrictions

47.

8 ,5 51. either 95 by 110 yd or 55 by 5 52. 320 mph for prop plane; 440 mph for jet plane 1 53. 1 sec 54. 1 ft 55. 22 2 i 56. 22 2 5i 2 57. 5 2 2i 58. 21 2 9i 59. 0 2 3i 60. 0 2 2i 3 3 2 4 4 3 1 5 61. 2 i 62. 2 1 i 63. 1 i 64. 2 i 2 2 5 5 5 5 2 2 49. 2

51.

) 1/5 (−5, 7)

61.

(−3, 9)

47.

93.

90. ) 3

96. 0 (−6, 0)

( −6

101.

94.

4 97. 2 , 26 3 100.

(−5, 1)

(−4, 1] ] 1

(– ∞, 0) ∪ (5/2, ∞) ) ( 0 5/2

3 3 98. 2 , 8 10 (–∞, 2] ∪ [8/3, ∞) ] [ 2 8/3

102.

(– ∞, –29) ∪ (35, ∞) ) ( –29 35

104.

(–1, 4/3) ∪ (4/3, 11/3) ( )( ) –1 4/3 11/3

) 1

(–7/2, –2) ∪ (–1, 1/2) ( ) ( ) –7/2 –2 –1 1/2

) 1

(–∞, –3/2) ∪ (1, ∞) ) ( –3/2 1

( −4

) 0

( −5

(−4, 1)

92.

[–2, 1] ∪ (3, ∞) [ ] ( 1 3 –2

95. 5, −7

103.

(2, ∞) ( 2

( −4

(–∞, –2] ∪ (3, ∞) ] ( –2 3

99.

86.

88.

(−1, 3) ( −1

91.

(−∞, 5/3) ) 5/3

) 5

(–∞, –2) ∪ (4, ∞) ) ( –2 4

89.

83.

36.

44. 21 6 "6

45. 1

Chapter Test 1. x 2 0, x 2 1 5. x 5 m 1 zs

2 5 3. 4. 37 3 2 bc 6. a 5 7. 87.5 8. $14,000 c1b 2. x 2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A7

Answers to Selected Exercises

1 3 , 2 2

3 , −4 2

5 6 "133 12. 6 9.

10.

11. x 5

13. 5, −3

2b 6 "b2 2 4ac 2a

14. 8 sec

15. 7 2 12i

4 2 1 i 18. 0 1 i 19. i 5 5 "10 20. 1 21. 13 22. 23. 2, −2, 3, −3 10 1 24. 1, 2 25. 139 26. −1 32 27. 28. (−∞, 2] (−∞, 5) 16. 223 2 43i

17.

] 2

29.

31.

) 5

[

)

3

4

(2, ∞) ( 2

12`, 21 4 c 3 8, ` 2

]

[

–1

8

32.

33. 2, 2

[−2, 1) [ −2

35.

30.

[3, 4)

) 1

(–∞, 3/2) ∪ (7/2, ∞) ) ( 3/2 7/2

10 3

] 11/5

55. 3 23, 21 4 c 12, ` 2 [ –3

] –1

[ −4

17. 20. 24. 27. 30. 32. 34. 36. 39. 42. 43. 45.

56. 12`, 232 c 10, 32 ( 0

( 1/3

) 3

) 3

y = –5x + 5

y = 2x + 7

(– 27– , 0)

x x

4.

(–∞, 0) ∪ [2, ∞) ) [ 0 2

6. Transitive Prop. 7. 9a2 10 4 4y x 11. 12. 16y2 9x6 3 "2x2 a 3b 3 14. abc2 15. "3 16. x 3 1y 1 "32 3x 1"x 1 12 18. 19. 5"3 2 3"5 y2 2 3 x21 3"2 21. 5 2 2"6 22. 4 23. 28x 1 8 9x4 2 4x3 25. 6x2 1 11x 2 35 26. z3 1 z2 1 4 x 2x2 2 x 1 1 28. 3x2 1 1 2 2 29. 3t 1t 2 22 x 12 2 2 2 13x 1 22 1x 2 42 31. 1x 1 12 1x 2 12 1x 1 12 2 x25 1x 1 12 1x2 2 x 1 12 1x 2 12 1x2 1 x 1 12 33. x15 5x2 1 17x 2 6 12x 1 12 1x 1 22 35. 1x 1 32 1x 2 32 2 2x 1 x 1 7 1 37. b 1 a 38. 2 1x 1 22 1x 1 32 xy R1R2 0, 10 40. 34 41. R 5 R1 1 R2 a2S S2a r5 or r 5 l2S S2l either 8 ft by 24 ft or 12 ft by 16 ft 44. $8,000 3 4 3 1 1 i 46. 2 i 47. 5 48. 0 1 10i 5 5 2 2

35.

37.

y 10 y = 2x – –– 3

( )

5– , 0 3

( ) (0, – 10––3 ) 39.

y

1 0, – 6

x

5. Comm. Prop. of Addition 9y2 9. a6b4 10. 4 8. 81a2 x 13.

( 5

1. quadrants 3. to the right 5. first 7. linear 9. x-intercept 11. horizontal 13. midpoint 15. A 12, 32 17. C 122, 232 19. E 10, 02 23. QI 25. QIII 27. QI 21. G 125, 252 29. x-axis y 31. 33. y

] 6

2. 2, 5, 11 ) 7

) 3

1 58. a , 3b 3

[ 4

Cumulative Review 1. 22, 0, 2, 6 [−4, 7) 3.

52. 64, 729

) –3

( 2

57. 12`, 21 4 c 3 4, ` 2

[−9, 6] [ −9

] –1

54. 12`, 32 c 15, ` 2

51. 2, 7

Section 2.1

34. 0

36.

50. 2, −2, 3, −3 11 53. a2`, d 5 49. 3

1 1 y= –x+ – 2 6 x

(–2, – 5–6) 41.

y

(

y

x

x+y=5

1 y=––x 2

43.

)

–1 –,0 3

x

45.

y

y 3x + 2y = 6

x 2x – y = 4

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

A8

Answers to Selected Exercises

47.

49.

y

85.

y

DT Dt

is the hourly rate of change in temperature.

T

y=3 x

x 4x – 5y = 20

75 70 65 60 55 50 45 t 1

51.

53.

y

y

2

3 4

87. The slope is the speed of the plane. d (mi)

x

1,200

d = 590t

x

3(y + 2) = y 3x + 5 = –1

55.

77. 7 85. ¢

59. 4.67 79. 14, 62

93. "2 units

63. "13

61. 5 81. 10, 12

71. 13

"5 "5 , ≤ 2 2

87. 15, 62

97. $312,500

103. approx. 171 mi

1. y 2 y1 5 m 1x 2 x12

Section 2.3

x

69. 5

t (hr)

2

3 2 91. y 5 2 x 1 3 93. y 5 2 x 1 2 7 5 97. 1m 1 n2 1p 1 q2 95. 12p 1 32 13p 2 42

3(y + 2x) = 6x + y

67. 2

(1, 590)

1

y

57. 1.22

600

109.

65. "2

75. 8"2 1 83. a21, b 2

73. 2"13

89. 15, 152 99. 200

[ −3

101. 100 rpm ] 3

111. no graph; the intersection is the empty set

A B 7. 2x 2 y 5 0 9. 4x 2 2y 5 27 11. 2x 2 5y 5 27 15. x 5 26 17. px 2 y 5 p2 13. y 5 23 7 19. 2x 2 3y 5 211 21. y 5 x 23. y 5 x 2 3 3 9 2 1 25. y 5 2 x 1 27. y 5 3x 2 2 29. y 5 5x 2 5 5 5 1 31. y 5 ax 1 33. y 5 ax 1 a 35. 3x 2 2y 5 0 a

41. 1, 10, 212

39. "2x 2 y 5 2"2 2 43. , 10, 22 3

37. 3x 1 y 5 24

y

113. 212

3. slope-intercept

y=x−1

3. run

5. the change in 7. vertical 5 7 9. perpendicular 11. 1 13. 2 15. 2 12 4 5 19. undefined 21. 23. 2 1 25. 3 17. 2 2 3 1 2 27. 29. 31. 0 33. 0 35. undefined 2 3 37. negative 39. positive 41. undefined 43. perpendicular 45. parallel 47. perpendicular 49. perpendicular 51. perpendicular 53. parallel 55. perpendicular 57. neither 59. 5 61. 6 63. not on same line 65. on same line 67. No two are perpendicular. 69. PQ and PR are perpendicular. 71. PQ and PR are perpendicular. 81. 3.5 students per yr 83. $642.86 per year

y

2 y = –x + 2 3

Section 2.2 1. divided

5. 2

+2 +3

x

1

x

1

2 45. 2 , 10, 62 3

47.

y

3 , 10, 242 2

+3 2 y = −–x + 6 3

1 5 49. 2 , a0, 2 b 3 6 55. perpendicular 61. perpendicular

−2

x

7 , 10, 22 53. parallel 2 57. parallel 59. perpendicular 63. perpendicular 65. y 5 4x 51.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A9

Answers to Selected Exercises

4 26 1 69. y 5 x 2 71. y 5 2 x 5 5 4 1 11 5 73. y 5 2 x 1 75. y 5 2 x 1 3 4 2 4 4 2 79. m 5 2 ; 10, 42 81. x 5 22 77. m 5 2 ; 10, 42 5 3 67. y 5 4x 2 3

83. x 5 5

31. 35. 37. 43. 47.

about the y-axis 33. about the x-axis about the x-axis, the y-axis, and the origin about the y-axis 39. none 41. about the x-axis about the y-axis 45. about the x-axis y 49. y

85. y 5 23,200x 1 24,300

87. y 5 47,500x 1 475,000 91. $90

95. $37,200 5 101. C 5 1F 2 322 9

93. $890

99. about 838

9 103. y 5 2 x 1 47; 11% 10 107. 1,655 barrels per day 109. a. Chirps/min

y = x3

710 x 1 1,900 3 97. $230

89. y 5 2

x

x

1 105. y 5 3.75x 1 37.5; $52 2

y = x2+ 4x

51.

53.

y

y y = 3 – |x|

y = |x – 2|

250 225 200 175

x

x

150 125 100

55.

57.

y

y

2

y = –x

75

y2 = 9x

50

x

25 50 60 70 80 90 100

Temp (°F)

23 x 2 210 (answers may vary) 5 c. 204 (answers may vary) 111. y 5 4.44x 2 196.62

b. y 5

123. x5

125.

59.

61.

y

127. 2"3

129.

x24

x y = √x – 1

3. axis of symmetry 9. x2 1 y2 5 r2

11. 122, 02 , 12, 02 ; 10, 242

5. x-axis

1 13. 10, 02 , a , 0b ; 10, 02 2

15. 121, 02 , 15, 02 ; 10, 252 17. 11, 02 , 122, 02 ; 10, 222 19. 123, 02 , 10, 02 , 13, 02 ; 10, 02 21. 121, 02 , 11, 02 ; 10, 212 23. 25. y y y = –x2 + 2 x y = x2 x

27.

y

5Q"x 2 2R

125 729

29.

y

x xy = 4

Section 2.4 1. x-intercept 7. circle; center

x

63. 10, 02 ; 10

69. 75.

79. 83. 85. 87. 89. 93. 97.

1 65. 10, 52 ; 7 67. 126, 02 ; 2 1 14, 12 ; 3 71. Q , 22R; 3"5 73. x2 1 y2 5 25 4 1 2 2 x 1 1y 1 62 5 36 77. 1x 2 82 2 1 y2 5 25 1x 1 22 2 1 1y 2 122 2 5 169 81. x2 1 y2 2 1 5 0 2 2 x 1 y 2 12x 2 16y 1 84 5 0 x2 1 y2 2 6x 1 8y 1 23 5 0 x2 1 y2 2 6x 2 6y 2 7 5 0 x2 1 y2 1 6x 2 8y 5 0 91. 1x 2 32 2 1 1y 1 22 2 5 9 2 2 1x 2 52 1 1y 2 62 5 4 95. 1x 2 22 2 1 1y 2 42 2 5 9 y 99. y

y x x

x x (1, –2) y = x 2– 4 x

1 y = – x2– 2x 2

x2 + y2 – 25 = 0

(x – 1)2 + (y + 2)2 = 4

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A10

Answers to Selected Exercises

101.

103.

y

528.

y

y (–3, 5) QII

(–1, 0)

x x

x – 1– , 0 2 negative x-axis

(–2, –1)

x2 + y2 + 2x – 24 = 0

105.

( )

(5, –3) QIV

x2 + y2 + 4x + 2y –11 = 0

107.

y

(0, –7) negative y-axis

y

(0, 2–3)

(1–2 , –1)

9.

x

10.

y

x

y

x 2x + 5y = –10 x

4x2 + 4y2 – 4x + 8y + 1 = 0

9x2 + 9y2 – 12y = 5

109. (0.25, 0.88)

2x – y = 6

111. (0.50, 7.25) 11.

12.

y

x+y=7

115. 1.44

113. 62.65

y

x 3x – 5y = 15

x

117. 4 sec

119.

D ( 60 , 3 4 2 )

13.

14.

y

y

x x

V

121. x2 1 y2 5 36 123. x2 1 1y 2 352 2 5 900 2 2 125. x 1 y 2 14x 2 8y 1 40 5 0 127. (23, 2) 129. a single point 131. 6 133. 21 135. 20 oz

x – 5y = 5

x + y = –7

15.

16.

y y=4

x = −2

Section 2.5

9 47. The force is multiplied by . 4 3x 1 5 1x 1 22 1x 1 12

1. 12, 02

2. 122, 12

Chapter Review

x

x

1. quotient 3. means 5. extremes; means 7. inverse 9. joint 11. 14 13. 2, 23 15. 18 1 21 19. 1,000 21. 1 23. 25. 6 27. 2 8 17. 2 4 29. direct variation 31. neither 33. 202, 172, 136 6 3 35. about 2 gal 37. 14 ft 39. 3 sec 41. "2 m 11 43. 15 lumens 45. It is multiplied by 18.

57.

y

59.

"3 49. 4

1x 1 42 1x 1 12 x21

3. 10, 212

61. 2

4. 13, 212

1 x

22. 2"2 0 a 0

23. 10, 32

17. $12,150

18. $332,500

21. 2

26. 10, 02

25. Q"3, 8R

19. 10

27. 26

20. 4"2 15 24. a26, b 2 28. 2

29. 21

1 31. 3 32. 5 33. 0 34. undefined negative 36. positive 37. perpendicular neither 39. y 5 7 40. x 5 3 41. 200 ft/min $48,750 per year 43. 7x 1 5y 5 0 4x 1 y 5 27 45. x 1 5y 5 23 46. 2x 1 y 5 9 2 3 47. y 5 x 1 3 48. y 5 2 x 2 5 3 2

30. 35. 38. 42. 44.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Answers to Selected Exercises

49.

50.

y

84. 15, 212 ; 3

y 4 y = ––x + 3 3 x

x 3 y = –x – 2 5

3 , 10, 252 2 1 54. 2 , 10, 222 2 51.

57. y 5 17

1 52. 2 , 10, 222 2 5 7 55. 2 , Q0, R 2 2

3 , 10, 252 2 3 7 56. , Q0, 2 R 4 2

53.

60. y 5 27x 1 47 1 62. y 5 x 2 3 63. parallel 7

1 65. 10, 02 , a , 0b ; 10, 02 2

67. about the x-axis 69. about the y-axis 71.

86. 1x 2 32 2 1 y2 5

85. x2 1 y2 5 49

1 25

2 2 88. ax 2 b 1 1y 2 52 2 5 81 7

87. 1x 1 22 2 1 1y 2 122 2 5 25

89. 1x 1 32 2 1 1y 2 42 2 5 144 or x2 1 y2 1 6x 2 8y 2 119 5 0

1 2 5 2 121 or x2 1 y2 1 x 2 5y 2 54 5 0 90. ax 1 b 1 ay 2 b 5 2 2 2 91. 1x 1 32 2 1 1y 2 22 2 5 9 92. 1x 2 22 2 1 1y 2 42 2 5 25 93. 94. y y

3 3 59. y 5 x 2 4 2 61. y 5 3x 1 5

58. x 5 25

A11

(2, 0)

(0, 0)

x

x

64. perpendicular

66. 112, 02 , 122, 02 ; 10, 2242

95.

68. about the y-axis 70. none 72. y 5 x3 2 2

y

96.

y

y

x

(0, 1) x

y

(2, –1)

x x2 + y2 – 2y = 15

y = x2+ 2

y = x3 – 2

97. 3.32, 23.32 98. 21, 0, 1 99. 1.73, 21.73, 1, 21 100. 4.19, 21.19 101. x 5 2, x 5 5 102. x 5 25, x 5 5 9 25 1 103. lb 104. 105. 333 cc 106. 1 5 9 3 107. about 117 ohms 108. $385 y 109. 140 hr

x

73.

74.

y

y

1 y = – |x| 2 x

x

x2 + y2 – 4x + 2y = 4

280 140

y = –√x – 4

14x + 18y = 5,040

x 180

75.

76.

y

y

Chapter Test 1. QII 3. y

y = √x + 2 y = |x + 1| + 2

x

360

2. y-axis 4.

y

x + 3y = 6

x

x

x 2x – 5y = 10

77.

78.

79.

80.

5.

6.

y

y 3x – 5y = 3(x – 5)

81. 10, 02 ; 8

82. 10, 62 ; 10

83. 127, 02 ;

1 2

x 2(x + y) = 3x + 5

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

A12

Answers to Selected Exercises

7.

y

8.

37. domain: 12`, 32 c 13, ` 2 39. domain: 12`, 222 c 122, 22 c 41. domain: 12`, 212 c 121, 52 c

y

x+y–5 –––––––– = 3x 7

x 1 – (x – 2y) = y – 1 2

9. "41

x

10. approximately 4.44

12. Q"2, 2"2R

13. 2

22. 24. 26. 27.

14.

"3 3

15. neither

1 2 1 11 y 5 2x 1 5 20. y 5 2 x 1 5 21. y 5 2x 2 2 2 x53 23. ( 2 4, 0), (0, 0), (4, 0); (0, 0) (4, 0); (0, 4) 25. about the x-axis about the y-axis y 28. y

16. perpendicular 19.

5 4

11. (0, 0)

17. y 5 2x 2 11

18. y 5 3x 1

x

x = |y| x

12, ` 2 15, ` 2 3 1 1 5 43. 4; 211; 3k 2 2; 3k2 2 5 45. 4; ; k 1 3; k2 1 2 2 2 2 47. 4; 9; k2; k4 2 2k2 1 1 49. 9; 21; k2 1 3k 2 1; k4 1 k2 2 3 51. 5; 10; k2 1 1; k4 2 2k2 1 2 1 2 2 1 1 1 1 53. ; 2; ; 55. ; ; 2 ; 3 k 1 4 k2 1 3 3 8 k 2 1 k4 2 2k2 57. "5; "10; "k2 1 1; "k4 2 2k2 1 2 59. 3 61. 2x 1 h 63. 8x 1 4h 65. 2x 1 h 1 3 67. 4x 1 2h 2 4 69. 3x2 1 3xh 1 h2 y y 71. 73. f (x) = – 3 –x + 4 4 x x

f(x) = 2x + 3

domain: 12`, ` 2 range: 12`, ` 2 y 75.

domain: 12`, ` 2 range: 12`, ` 2 y 77.

2x = 3y – 3 y = x2 – 9

29.

x

x

30.

y

y

domain: 12`, ` 2 range: 12`, ` 2 79. y

x = y3

y = 2√x

x

domain: 12`, ` 2 range: 3 24, ` 4 81. y

f (x) = x 2 – 4

x

31. 1x 2 52 2 1 1y 2 72 2 5 64 33. y

32. 1x 2 22 2 1 1y 2 42 2 5 32 34. y 2

x x

2

x2 + y2 = 9

35. y 5 kz2 39. x 5 2.65

36. w 5 krs2

domain: 12`, ` 2 range: 12`, ` 2 83. y

domain: 12`, ` 2 range: 12`, 0 4 85.

1 f(x) = – x + 3 2

|

35 2

38. x 5

27 32

40. x 5 5.85

1. function 3. domain 5. y 5 f 1x2 7. x 9. vertical; once 11. function 13. not a function 15. function 17. not a function 19. not a function 21. function 23. domain: 12`, ` 2 25. domain: 12`, ` 2 27. domain: 3 2, ` 2 29. domain: 12`, 4 4 31. domain: 12`, 21 4 c 3 1, ` 2 33. domain: 12`, 2` 2 35. domain: 12`, 212 c 121, ` 2

Section 3.1

f (x) = –x 3 + 2

x2 – 4x + y2 + 3 = 0

37. P 5

f(x) = –|x|

x

f(x) = |x – 2|

domain: 12`, ` 2 range: 3 0, ` 2

x

y

| x

x

domain: 12`, ` 2 range: 3 0, ` 2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A13

Answers to Selected Exercises

87.

89.

y

y

37.

39.

y

y

f(x) = 2x – 4 x x

domain: 3 21, ` 2 range: 12`, 0 4 y 91.

domain: 3 2, ` 2 range: 3 0, ` 2 y 93.

f(x) = – x + 1

3

f(x) = x + 2

x

x 1 f (x) = – – x 2 + 8 2

41.

f (x) = (x – 3)2 – 1 y

43.

y

Ran g e

x f

x

x

Do mai n

x

domain: 12`, ` 2 range: 12`, ` 2 95. function 97. function 101.

45.

99. function 103.

domain: 12`, ` 2 ; domain: 12`, ` 2 ; range: 12`, ` 2 range: 3 0, ` 2 105. a. C 1x2 5 8x 1 75 b. $755 107. a. C 1t2 5 0.07t 1 9.99 b. $11.39 109. a. C 1 f 2 5 95f 1 14,000 b. $261,000 111. C 1x2 5 0.10x 1 7 113. n 1t2 5 1,692t 1 6,400 2 115. E 1D2 5 1.3911D 117. 2 123. 1, 7, 8 125. 7 3 ( ) [ 127. 124, 7 4 129. –3

f (x) = –(x + 4)2 + 1

f (x) = 2(x + 1)2 – 2

x x f ( x) = x 2 + 2x f (x) = –3(x – 2)2 + 6

49.

51.

y

y x

x

5 6

1. f 1x2 5 ax2 1 bx 1 c 3. (3, 5) 5. upward b 7. 2 9. upward; minimum 2a 11. downward; maximum 13. downward; maximum 15. (0, 21) 17. (3, 5) 19. (26, 24) 21. (3, 0) 23. (2, 0) 25. (23, 212) 27. (3, 1)

y

47.

y

Section 3.2

2 11 29. a , b 31. (24, 211) 3 3 y 33. 35.

f(x) = x2– 4x + 1 f (x) = 2x 2 – 12x + 10

53.

55. 75 ft by 75 ft, 5,625 ft 2

y x

y f (x) = –3x 2 – 6x – 9

x x f (x) = x 2 – 4 f (x) = –3x 2 + 6

57. 25 ft by 25 ft 59. w 5 12 in.; d 5 6 in. 61. 20 ft 63. 15.4 ft 65. 208 ft 67. 48 digital cameras; minimum cost $2,400 69. $95 5 71. sec 73. 100 ft 75. (22.25, 266.13) 2 1 77. (3.3, 268.5) 79. 2, 3 81. 6 by 4 units 2 83. Both numbers are 3. 87. a2 2 3a; a2 1 3a 91. 7; 7 89. 15 2 a2 2 ; 15 1 a2 2

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A14

Answers to Selected Exercises

Section 3.3 1. 4 11.

3. n 2 1

5. odd

y

7. piecewise-defined y 13.

59. a. 3 c. 23 c. 4

y

57. 9. 3

b. 24 61. a. 2

b. 3

40 20 f (x) = –x 3 – 4x 2

10 –2 –10

2

x

–5

x

–20 f (x) = x 3 – 9x y

15.

5

x

–40

0 if x < 0 x2 if 0 ≤ x ≤ 2 4 – 2x if x > 2

f(x) = y

17.

63.

65.

y

y

x x

x

x

f(x) = x 3+ x 2

y = [[x]] – 1

y = [[2x]] f (x) = x 3 – x 2 – 4x + 4

19.

y

21.

y

67. B g x x

A B C

f(x) = x4 – 2x 2 + 1

D f (x) = –x 4 + 5x 2 – 4 50% 60% 70% 80% 90% 100%

23. even 25. neither 27. odd 29. odd 31. even 33. odd 35. neither 37. decreasing on 12`, 02 ; increasing on 10, ` 2 39. increasing on 12`, 02 ; decreasing for 10, ` 2 41. decreasing on 12`, 222 ; constant on 122, 22 ; increasing on 12, ` 2 43. decreasing on 12`, 22 ; increasing on 12, ` 2 45. a. 22 b. 3 47. a. 2 b. 1 c. 3 49. 51. y y

p

69. $32 C 36 32 28 24 20 m 100 200 300 400

x

x

71. $1.60 C

f(x) =

53.

x + 2 if x < 0 if x ≥ 0 2

x if x ≤ 0 f(x) = 2 if x > 0

55.

y

1.00 .80 .60

y

.40 .20

x x f(x) = f(x) = –4 – x if x < 1 if x ≥ 1 3

–x if x < 0 x 2 if x ≥ 0

t 1

73.

2

3

4

y 30 20 10 10

20

30

40

x

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A15

Answers to Selected Exercises

75. no; this is not defined at x 5 0

31.

y

33.

y

y

f(x) = |x + 2| – 1 x |x| y = ––– x

81. 3x 1 5; 3x 1 3 3 85. 21, 2

g(x) = x + 1 x

3x 2 8 3x 1 1 3x 2 14 83. ; 2 3 or 5 5 5

35.

x

37. y

y g(x) = x + 2

h(x) = x – 2 – 1

Section 3.4 1. upward 3. to the right 9. horizontally 11. y

5. 2; downward 13.

x

x

7. y-axis y

39.

41.

y

y

x 3

x

g(x) = x – 2

x g(x) = (x + 3)2

3

g( x) = x 2 – 2

15.

x

g(x) = x – 4

17.

y

y

43.

45.

y

y

x h(x) = (x + 1)2 + 2

1 h(x) = x + – 2

2

( )

x

1 – – 2

f (x) = –x 2

x

x

19.

21.

y

y 3

h(x) = x + 1 – 1

47.

49.

y

y

x x

g(x) = (x – 2)3

g(x) = x 3 + 1

x h(x) = –x3 x

23.

25.

y

f(x) = – x

y

x

x

51.

53.

y

y

y + 2 = x3 h(x) = (x – 2)3 – 3 x

27.

29.

y

f(x) = |–x|

y

f(x) = 2x 2

g(x) = |x – 5| g(x) = |x| + 2 x

x

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

A16

Answers to Selected Exercises

55.

y

57.

y

y

83.

y

85.

x h(x) = –3x 2 x

x 1 f (x) = – x 3 2

59.

61.

y

y = 2 f(–x)

y

93.

x x 3

f(x) =

(1–2 x)

65.

y

y

f(x) = (2x)2

x

x g(x) = 3(x + 2)2 – 1

67.

69.

y

95. 12`, 32 c 13, ` 2

x22 x12

97. x 1 2 1

22 x11

Section 3.5

h(x) = –3|x|

63.

x

y = f(x – 2) + 1

1. asymptote 3. vertical 5. x-intercept 7. same 9. horizontal; vertical 11. vertical asymptote: x 5 2; horizontal asymptote: y 5 1; domain: 12`, 22 c 12, ` 2 ; range: 12`, 12 c 11, ` 2 13. 20 hr 15. 12 hr 17. $5,555.56 19. $50,000 21. 12`, 22 c 12, ` 2 23. 12`, 252 c 125, 52 c 15, ` 2 25. 12`, 212 c 121, 02 c 10, 12 c 11, ` 2 27. 12`, ` 2 29. x 5 3 31. x 5 1, x 5 21 33. x 5 22, x 5 3 1 35. none 37. y 5 2 39. y 5 41. y 5 0 2 43. none 45. y 5 x 2 3 47. y 5 2x 1 3 49. y 5 x 1 2 51. 53. y y 1 y = ____ x–2

y

x y = ____ x–1

h(x) = –2|x| + 3 x x x x f(x) = 2|x – 2| + 1

55. 71.

73.

y

57.

y

y

y h(x) = 2 x – 2 +1 x+1 f(x) = ____ x+2 x

x

75.

x 2x – 1 f(x) = _____ x–1

x

f(x) = 2 x + 3

77.

y

59.

61.

y

y

y

3

f(x) = 2 x + 4

x

x

x

x x2 – 9 g(x) = _____ x2 – 4

g(x) = –2(x + 2) 3 – 1

79.

81.

y y = f(x) + 1

y

x2 – x – 2 g(x) = ________ x2 – 4x + 3

y = 2f (x) x

x

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A17

Answers to Selected Exercises

63.

65.

y

x

x x2

+ 2x – 3 y = _________ x 3 – 4x

67.

x2 – 9 y = _____ x2 y

69.

y

x

x

x+1 f(x) = _______ x 2(x – 2)

x f(x) = ______2 (x + 3)

71.

73.

y

y

x x y = _____ x2 + 1

x 3x2 y = _____ 2+ 1 x

(2,8) y=

x–

1

8 4

1

x x

4

x+

–8 –4

8 x2 – 2x – 8 h(x) = _________ x–1

–4

y

79.

y

77.

y

3

x + x2 + 6x f(x) = __________ x2 – 1

y=

75.

81.

y

x

f(x) = –– x

3

x +x f(x) = ––––– x

85.

y

y

x x2 – 2x + 1 f (x) = ––––––––– x–1

73. f 1x2 5 "x; g 1x2 5 x 1 2 75. f 1x2 5 x 1 2; g 1x2 5 "x 77. f 1x2 5 x; g 1x2 5 x 79. 0 81. 0 83. 1 85. 1 87. 8 89. 9 91. a. 1R 2 C2 1x2 5 260x 2 60,000 b. $70,000 9 2 93. A 1t2 5 pt ; 101,787.6 square inches 95. P 5 4"A 4 x17 3x 105. y 5 107. y 5 3 12x

Section 3.7

x x2

83.

3. f 1x2 g 1x2 5. intersection 1. f 1x2 1 g 1x2 7. g 1 f 1x22 9. commutative 11. 1 f 1 g2 1x2 5 5x 2 1; 12`, ` 2 13. 1 f ? g2 1x2 5 6x2 2 x 2 2; 12`, ` 2 15. 1 f 2 g2 1x2 5 x 1 1; 12`, ` 2 x2 1 x x 17. 1 f /g2 1x2 5 2 5 ; 12`, 212 c 121, 12 c 11, ` 2 x 21 x21 19. 1 f 1 g2 1x2 5 x2 1 "x 2 7; 3 0, ` 2 x2 2 7 21. 1 f /g2 1x2 5 ; 10, ` 2 23. 7 25. 1 27. 12 "x 29. undefined 31. f 1x2 5 3x2 ; g 1x2 5 2x 33. f 1x2 5 3x2 ; g 1x2 5 x2 2 1 35. f 1x2 5 3x3 ; g 1x2 5 2x 37. f 1x2 5 x 1 9; g 1x2 5 x 2 2 39. 11 41. 217 43. 190 45. 145 47. 12`, ` 2 ; 1 f + g2 1x2 5 3x 1 3 51. 12`, ` 2 ; 1g + f 2 1x2 5 2x2 49. 12`, ` 2 ; 1 f + f 2 1x2 5 9x 53. 12`, ` 2 ; 1g + g2 1x2 5 4x 55. 3 21, ` 2 ; 1 f + g2 1x2 5 "x 1 1 4 57. 3 0, ` 2 ; 1 f + f 2 1x2 5 " x 59. 3 21, ` 2 ; 1g + f 2 1x2 5 x 61. 12`, ` 2 ; 1g + g2 1x2 5 x4 2 2x2 x22 63. 12`, 22 c 12, 32 c 13, ` 2 ; 1 f + g2 1x2 5 32x x21 65. 12`, 12 c 11, 22 c 12, ` 2 ; 1 f + f 2 1x2 5 22x 67. f 1x2 5 x 2 2; g 1x2 5 3x 69. f 1x2 5 x 2 2; g 1x2 5 x2 71. f 1x2 5 x2 ; g 1x2 5 x 2 2

Section 3.6

y

x x3 – 1 f (x) = ––––– x–1

87. a. C 1x2 5 3.25x 1 700 b. $2,325 3.25x 1 700 c. C 1x2 5 d. $4.65 e. $3.95 f. $3.60 x 0.095n 1 8.50 89. a. C 1n2 5 0.095n 1 8.50 b. C 1n2 5 n c. 10.5¢ 101. 3x2 1 x 103. 10x2 1 29x 1 10 105. 3x 1 5

1. one-to-one 3. identity 5. one-to-one 7. not one-to-one 9. not one-to-one 11. not one-to-one 13. not one-to-one 15. one-to-one 17. one-to-one 19. not a function 1 x22 25. f 21 1x2 5 x 27. f 21 1x2 5 3 3 3 29. f 21 1x2 5 " 31. f 21 1x2 5 x5 x22 1 1 33. f 21 1x2 5 2 3 35. f 21 1x2 5 x 2x y y 37. 39. f(x) = 5x f –1(x) = 1 –x 5 x

x+4 f –1(x) = ––––– 2 x

f(x) = 2x – 4

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A18

Answers to Selected Exercises

y

41.

43.

y 2x + y = 4 f(x) = 4 – 2x

y–x=2 f –1(x) = x + 2

2y + x = 4 4–x f –1(x) = ––––– 2

x

x

x–y=2 f(x) = x – 2

27.

y

45.

17. function 18. not a function 19. a. I 1h2 5 3.5h 2 50 1 2 b. $650 20. C x 5 0.15x 1 5 21. upward; minimum 22. downward; maximum 23. (1, 6) 24. (24, 25) 1 25. (23, 213) 26. a , 28b 2 28.

y

y f (x) = – (x – 4)2 + 4 (4, 4)

y

47.

x 3

f –1(x) = x 3 + 4

f –1(x) = x + 6

x f (x) = (x – 6)3

3

f(x) = x – 4

49.

x

(2, – 3) f (x) = (x – 2)2 – 3

29.

30.

y

y

x y

51.

y

(1–2 , 1–4) x

y = x – x2 y = x2 – x x

(

x

x 1 f(x) = f –1(x) = –– 2x

31.

1– 1 ,–– 2 4

) 32.

y

x

x+1 f(x) = f –1(x) = ––––– x–1

x

53. f 21 1x2 5 2"x 1 3 1x $ 232 4 55. f 21 1x2 5 " x 1 8 1x $ 282 21 1 2 57. f x 5 "4 2 x2 10 # x # 22 59. domain: 12`, 22 c 12, ` 2 ; range: 12`, 12 c 11, ` 2 61. domain: 12`, 02 c 10, ` 2 ; range: 12`, 222 c 122, ` 2 63. a. f 1x2 5 0.75x 1 8.50 b. $11.50 c. f 21 1x2 5

71. 8

x 2 8.50 0.75

73. 4

75.

d. 2

5 4

67. 0

77.

y

y = 3x2 – 8x – 3 y = x 2 – 3x – 4 25 ( 3–2 , – ––– 4)

25 (4–3 , – ––– 3)

1 34. Both numbers are . 2 35. 350 ft by 350 ft; 122,500 ft 2 36. 50 digital cameras; minimum cost $1,100 y 37. 38. y 33. 300 units

69. a $ 0

1 7

Chapter Review

1. function 2. function 3. not a function 4. function 5. domain: 12`, ` 2 6. domain: 12`, 5 4 c 3 5, ` 2 7. domain: 3 1, ` 2 8. domain: 12`, ` 2 9. 8; 217; 5k 2 2 3 6 1 1 k2 2 3 11. 0; 5; 0 k 2 2 0 12. ; ; 2 10. 22; 2 ; 4 k25 7 2 k 13 13. 5 14. 4x 1 2h 2 7 y y 15. 16.

x x y = x3– x

y = x2– 4x

odd function 39.

neither even nor odd 40.

y

y y = 1 – x4 x

x

x x

y = x 3– x 2

f (x) = 3|x – 2|

domain: 12`, ` 2 range: 12`, 4 4

f (x) = –x 2 + 4

domain: 12`, ` 2 range: 3 0, ` 2

neither even nor odd 41. a. 2

b. 9

1 42. a. 2

even function b. 3

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A19

Answers to Selected Exercises

43.

44.

y

59.

y

60.

y

y

x x

y = f(x) = x + 5 5–x

if x ≤ 0 if x > 0

increasing on 12`, 02 ; decreasing on 10, ` 2 45. 3 46. 21 47. y

x

y = f(x) = x + 3 if x ≤ 0 if x > 0 3

g ( x) = 3 x + 3 + 2

increasing on 12`, 02 ; constant on 10, ` 2

48.

61.

x –x + 3

63. 12`, 252 c 125, 52 c 15, ` 2 64. 12`, ` 2 65. x 5 1, x 5 21 66. x 5 27 67. x 5 2, x 5 23 1 68. x 5 4, x 5 21 69. y 5 70. y 5 25 2 71. y 5 0 72. none 73. y 5 2x 1 3 74. none y y 75. 76.

f(x) = [[x]] + 2

50. $26 y x3 x f(x) = x 2

g(x) = 2 x – 5

2x f(x) = –––– x–4

12 f(x) =

3

x

x

52.

8

–4

4

–8 4

g(x) = (x – 7)3

8

12

y

–12

–4x f(x) = –––– x+4

x

78.

y

y

54.

y

77.

x

– 12 –8 –4

x

53.

y

y

x

y g(x) = x 2 + 5

62.

y f( x) =

f(x) = [[x – 1]]

49. $44 51.

g(x) = 1 – (x + 3)3 + 2 3

x

f(x) = |x| x

g( x) = x + 2 + 3 x x f(x) = ______ (x –1)2

( – 2, 3 ) f ( x) = x

(4, 2)

x

(x – 1)2 f(x) = ______ x

x g(x) = |x – 4| + 2

79. 55.

f(x) = x 3

1 g(x) = – x 3 3

g(x) = (–5x) 3

f(x) = x 3 x

x

57.

58.

y

y

g(x) = – |x – 4| + 3

x

80.

y

y

56.

y

1 g(x) = – |x – 4| + 1 4 x

x2 – x – 2 f(x) = ________ x2 + x – 2 x

y x3 + x f(x) = _____ x2 – 4 4 –4

4

x

–4

81. 1 f 1 g2 1x2 5 x2 1 2x; domain: 12`, ` 2 82. 1 f ? g2 1x2 5 2x3 1 x2 2 2x 2 1; domain: 12`, ` 2 83. 1 f 2 g2 1x2 5 x2 2 2x 2 2; domain: 12`, ` 2 f 1x2 x2 2 1 1 1 84. 1 f /g2 1x2 5 5 ; domain: a2`, 2 b c a2 , ` b g 1x2 2x 1 1 2 2 85. 10 86. 60 87. 21 88. undefined 89. 1 f + g2 1x2 5 f 1g 1x22 5 4x2 1 4x; domain: 12`, ` 2 90. 1g + f 2 1x2 5 g 1 f 1x22 5 2x2 2 1; domain: 12`, ` 2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A20

Answers to Selected Exercises

91. 20 92. 22 93. f 1x2 5 x2 ; g 1x2 5 x 2 5 3 1 2 1 2 94. f x 5 x ; g x 5 x 1 6 95. not one-to-one 96. one-to-one 97. one-to-one 98. not one-to-one x 1 1 1 101. f 21 1x2 5 102. f 21 1x2 5 2 2 7 x "3x 3 3 5 Åx x 2 2 106. a2`, b c a , ` b 5 5

103. f 21 1x2 5

104. f 21 1x2 5

x x11 x15 105. f 21 1x2 5 2 y

3

21.

22.

y

y

2x2 – 3x – 2 y = –––––––––– x–2

2

x x y = ––––– x2 – x

x

23. 1 f 1 g2 1x2 5 f 1x2 1 g 1x2 5 x2 1 3x 1 2 24. 1g + f 2 1x2 5 g 1 f 1x22 5 9x2 1 2 f 1x2 3x 5 2 25. 1 f /g2 1x2 5 g 1x2 x 12 26. 1 f + g2 1x2 5 f 1g 1x22 5 3x2 1 6

x+5 f –1(x) = ––––– 2

27. f 21 1x2 5

x11 3 28. f 21 1x2 5 " x13 x21 29. range: 12`, 222 c 122, ` 2 30. range: 12`, 32 c 13, ` 2

x f (x) = 2x – 5

2. domain: 3 23, ` 2 1. domain: 12`, 52 c 15, ` 2 1 3. , 2 4. "6, 3 5. (7, 23) 6. (1, 24) 2 7. (4, 210) 8. (22, 9) y 9. 10. y

Cumulative Review

Chapter Test

1.

2.

y

y

x

3x + 2y = 12

5x – 3y = 15 x

x

x

f (x) = x 4 – x 2

11.

25 sec 8

15.

y

12.

f (x) =

625 ft 4

13. 10 ft 16.

x5 –

x3

1 3 4 3. "41; a , b ; 2 2 2 5 5. y 5 22x 2 1 8. y 5 24x y 9.

14. 110 ft

4. 2"29; (22, 5); 7 11 6. y 5 x 2 2 4 10.

y

2 5

3 7. y 5 x 1 6 5 y y2 = x – 2

x2 = y – 2 f ( x) = x – 1 + 5 (1, 5)

(3, 1) f(x) = (x – 3)2 + 1

x x

x x

11.

12.

y

y

10

17. vertical asymptotes: x 5 23 and x 5 3; horizontal asymptote: y 5 0 18. vertical asymptote: x 5 3; horizontal asymptote: none; slant asymptote: y 5 x 2 2 19. 20. y y

4 –4

4

10

–4

x

x (1, 0) x2 – 2x + y2 = 8

x2 y = _____ x2 – 9

x2 x

x x y = _____ x2 + 1

+

y2

= 100

13. x 5 1, x 5 10 14. x 5 2, x 5 8 15. $62.50 25 16. 17. function 18. function 19. function 4 20. not a function 21. domain: 12`, ` 2 22. domain: 12`, 222 c 122, ` 2 23. domain: 3 2, ` 2

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A21

Answers to Selected Exercises 24. domain: 3 24, ` 2 27.

5 49 25. a2 , 2 b 2 4

5 49 26. a , b 2 4

28.

y

27.

y

f(x) = 3x

y

x

()

1 f(x) = – 5

(1, 3) (0, 1)

f(x) = –x2 + 4

x

29.

y

(0, 1)

x

x

( ) 1, 1 – 5

x

31.

33.

y

y

f(x) = x2 – 4

29.

30.

y

3 f(x) = – 4

x

()

y

( )

(0, 1)

f(x) = –x4 + 2x2 + 1 x

x

f (x) = (1.5) x

1, 3– 4

f(x) = x3+ x

\

35.

y

39. yes

x

47. b 5 2 51.

41. no

²²

x

[



 [ ² I[  ² 

²

y

x

²



x2 – 1 f(x) = ––––– x2 – 9

(1, 1.5)

\

37.

I[  ²[

x2 2 1 32. f 1x2 5 2 x 29

31.

(0, 1)

x

[

43. b 5

1 2

45. no value of b

49. b 5 e 53.

y

y

x f(x) = –––– x–3

33. 1 f 1 g2 1x2 5 f 1x2 1 g 1x2 5 x2 1 3x 2 3 domain: 12`, ` 2 34. 1 f 2 g2 1x2 5 f 1x2 2 g 1x2 5 2x2 1 3x 2 5 domain: 12`, ` 2 35. 1 f ? g2 1x2 5 f 1x2 ? g 1x2 5 3x3 2 4x2 1 3x 2 4 domain: 12`, ` 2 f 1x2 3x 2 4 36. 1 f /g2 1x2 5 5 2 domain: 12`, ` 2 g 1x2 x 11 37. 1 f + g2 122 5 11 38. 1g + f 2 122 5 5 40. 1g + f 2 1x2 5 9x2 2 24x 1 17 39. 1 f + g2 1x2 5 3x2 2 1 x 2 2 1 41. f 21 1x2 5 42. f 21 1x2 5 1 3 3 x x11 43. f 21 1x2 5 "x 2 5 44. f 21 1x2 5 3 kx 45. y 5 kwz 46. y 5 2 t

Section 4.1 1. exponential 9. increasing 17. 451.8079 25. 1, 9

5. 10, ` 2 3. 12`, ` 2 11. 2.72 13. increasing

19. 52"2 5 25"2

21. a4

f(x) = 2x + 1

x f(x) =

55.

3x –

x

1

57.

y

y

f(x) = 3x – 1

f(x) = 3x + 1 x x

59.

61.

y

y

x x f ( x) = e x – 2

f ( x) = e x – 4

7. asymptote 15. 11.0357

63.

y

65.

y

23. 1, 25

x

y = 3x – 2 + 1

f(x) = 2x + 1 – 2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

A22

Answers to Selected Exercises

67.

69.

y

103. b 5 2 y 105.

y

107.

y

x

f (x) = –3 + 1 f(x) = log3 x

x x

x x f (x) = 2–x – 3 f(x) = log1/3 x

71.

73.

y

\

109.

f (x) = –e x + 2

111. I[  ²ORJ [

x

y

[

x f(x) = 2 + log2 x

75.

77. 113.

\

115.

y f(x) = log3 (x + 2)

79.

I[  ORJ [ [

x

81. $22,080.40 83. $15.79 85. $2,273,996.13 87. $1,263.77 89. $13,375.68 91. $7,647.95 from continuous compounding, $7,518.28 from annual compounding 93. $291.27 95. $12,155.61 97. 13,228 101. 125 103. x2 11 1 9x22 105. 1x 1 42 1x 2 32

117.

119.

y

y

x f(x) = ln(x – 4) x

f(x) = –3 + ln x

Section 4.2

1. birth; death 3. 0.1868 g 5. about 10 kg 7. about 35.4% 9. 0.1575 unit 11. 2 lumens 13. 8 lumens 15. about 56,570 17. 61.9 + C 19. 315 21. 13 23. 10.6 billion 25. 2.6 27. about 0.07% 29. 0 31. 18,394 33. about 24,060 35. about 492 37. 19.0 mm 39. 49 mps 41. about 7 million 43. about 72.2 years 49. 8 51. 3 53. 2 55. 3

121.

123.

y

x f(x) = 1– ln x

Section 4.3

3. range 5. inverse 7. exponent 1. x 5 by 9. 1b, 12 ; 11, 02 11. logex 13. 12`, ` 2 15. 10 1 17. log8 64 5 2 19. log4 5 22 21. log1/2 32 5 25 16 1 3 1 23. logx z 5 y 25. 34 5 81 27. a b 5 2 8 29. 423 5

39. 51. 63. 71. 79. 85. 93.

1 64

31. p1 5 p

33. 3

35. 23

37. 3

1 1 41. 23 43. 64 45. 7 47. 5 49. 2 25 1 3 2 53. 5 55. 57. 4 59. 61. 5 6 2 3 4 65. 0.5119 67. 22.3307 69. 3.8221 20.4055 73. 3.5596 75. 2.0664 77. 20.2752 undefined 81. 25.2522 83. 1.9498 3 1024 4.0645 87. 69.4079 89. 0.0245 91. 120.0719 4 95. 23 97. 7 99. 4 101. b 5 2

125.

127.

129. 7 37 135. a2 , 2 b 2 4 137. 425 ft by 850 ft 139. y 5 5x

131. b is larger.

Section 4.4 1. 20 log

E0 EI

5. E 5 RT lna

1 C 3. t 5 2 lna1 2 b k M

Vf Vi

b

7. 55 dB

9. 29 dB

11. 49.5 dB

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A23

Answers to Selected Exercises

4.4 15. 4 17. no 19. 19.8 min about 5.8 yr 23. about 9.2 yr about 3,654 joules 27. about 99% per year 3 yr old 31. about 10.8 yr 33. about 208,000 V 1 mi 39. y 5 7x 1 3 41. x 5 2 1 x11 43. 45. 2x 2 3 3 1x 2 22 13. 21. 25. 29. 35.

Chapter Review 1. 52"2 3.

2. 2"10

4.

y

f(x) = 3x

5. p 5 1, q 5 7 y 7. x

f (x) =

3. A022t/h

11. 24 19. 22, 21 27. 1.2702

15 2 21. 3

13. 2

5. 2

7. 22

15. 3, 21

17. 63

23. 65

25. 1.1610

29. 1.7095

31. 0

1 g(x) = –2

()

9.

6. domain: 12`, ` 2 ; range: 10, ` 2 y 8. 1 f(x) = – 2

()

x

x

x 1 g(x) = – 2

()

x

–2

10.

y

x+2

y

x f (x) = –5x x f (x) = –5x + 4

11.

12.

y

y

f(x) = ex + 1 x

5 9. 2 6

33. 61.0878 1 35. 0, 1.0566 37. ln 10 39. ln 6 41. 0 2 43. 0.2789 45. 1, 3 47. 0 49. 10, 210 51. 4 1 4 6 53. e 55. 1e 1 72 57. 7 59. 50 61. 20 2 63. 10 65. 7 67. 6 69. 5 71. 4 73. 3, 4 75. 1010 77. no solution 79. 6 81. 9 83. 4 85. 7, 1 87. 20 89. 1.81 91. about 5.1 yr 93. about 42.7 days 95. about 2,900 yr 97. about 5.6 yr 99. about 5.4 yr 101. because ln 2 < 0.70 103. about 3.2 days 2 lna b ln 0.75 3 109. 105. about 12 min 107. k 5 3 5 x22 21 2 117. f 1x2 5 119. 19 121. 5x 2 1 3

(1–2)

x

()

x

Section 4.6

1. exponential

1 f(x) = – 3

x

Section 4.5

5. x; y 7. x 9. 2 11. 0 1. 0 3. M; N 13. 7 15. 10 17. 1 25. logb 2 1 logb x 1 logb y 27. logb 2 1 logb x 2 logb y 29. 2 logb x 1 3 logb y 1 1 31. 1logb x 1 logb y2 33. logb x 1 logb z 3 2 1 1 1 35. logb x 2 logb y 2 logb z 37. 7 ln x 1 8 ln y 3 3 3 x11 39. ln x 2 4 ln y 2 ln z 41. logb x x 1x "z z x 2 3 43. logb x "y 45. logb 3 2 47. logb 5 logb y xy y 1y z z x 1x 1 52 51. ln 6 2 53. true 55. false 49. ln 9 xy 57. true 59. false 61. true 63. false 65. false 67. true 69. true 71. true 73. false 75. true 77. 1.4472 79. 0.3521 81. 1.1972 83. 2.4014 85. 2.0493 87. 0.4682 89. 1.7712 91. 0.9597 93. 1.8928 95. 2.3219 97. 7.20 99. 4.77 101. from 5.01 3 1024 to 1.26 3 1023 103. 19 dB 105. The original intensity must be raised to the 4th power. 107. The volume V is squared. 121. yes 123. no 125. 12`, ` 2 127. 12`, ` 2

y

22. 28. 34. 39.

x

2 3 0.19 lumen 17. about 635,000,000 18. about 2,708 10, ` 2 ; 12`, ` 2 20. 10, ` 2 ; 12`, ` 2 21. 2 1 1 1 2 23. 0 24. 22 25. 26. 27. 32 2 2 3 1 9 29. 8 30. 21 31. 32. 2 33. 4 8 1 2 35. 10 36. 37. 5 38. 3 25 y 40. y

13. $2,189,703.45 16. 19.

f(x) = ex – 3

14. $2,324,767.37

15.

f(x) = 3 + log x

f(x) = log (x – 2) x

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

x

A24

Answers to Selected Exercises

41.

y

42.

11.

y

12.

y

y

y = 4x 1 y= – 3

f(x) = 2 + ln x

f(x) = log (x – 1)

x

()

y = log4 x

x

x

x

x

y = log1/3 x

43. 6.1137 y 47.

45. 10.3398 48. y

44. 20.1111

46. 2.5715

x x

49. 12

50. 14x

54. 23 yr

51. 53 dB

55. 2,017 joules

1 53. 9 min 2 57. 1 58. 3

52. 4.4 56. 0

x3z 7 y5

y3"x

3 " z 77. 26.72

73. ln

71. logb 74. 3.36

"xy3 z7

72. ln

75. 1.56

x4 y 5z 6

76. 2.64

78. 1.7604

Chapter Test 2.

y

y

f(x) = 2x + 1 x x f(x) = ex – 2

3 g 64 7. 23 3.

18. 0.4259

8. 17

9. 2

5. $4,451.08

6. 3

20.

log e 1 or log p ln p 25. 21, 3 28. 1

5. factor 2 7. 4x2 1 2x 1 1 1 x21 23 9. 2x3 2 3x2 1 8x 2 1 1 x12 11. 21 13. 9 15. 3 17. 23 19. 69 21. 70,249 23. 128.3085938 25. true 27. true 29. false 31. true 33. 1x 2 12 13x2 1 x 2 52 2 9 35. 1x 2 32 13x2 1 7x 1 152 1 41 37. 1x 1 12 13x2 2 5x 2 12 2 3 39. 1x 1 32 13x2 2 11x 1 272 2 85 24 41. x2 1 2x 1 3 43. 7x2 2 10x 1 5 1 x11

3. any

245 x23 3x4 1 12x3 1 48x2 1 192x 49. 47 51. 2569 15 21 2 6i 55. 57. 5 59. 0 61. 384 8 228 2 16i 65. 16 1 2i 67. 40 2 40i 1 5 21, 25, 3 6 71. e 2 , 3, 23 f 2

45. 4x3 1 9x2 1 27x 1 80 1 47. 53. 63. 69.

73. U2, 2 1 "5, 2 2 "5V

75. U23, 3 1 "10i, 3 2 "10iV

77. U1, 1, "3, 2"3V 79. 5 2, 3, i, 2i 6 81. P 1x2 5 x2 2 9x 1 20 83. P 1x2 5 x3 2 3x2 1 3x 2 1 3 2 85. P 1x2 5 x 2 11x 1 38x 2 40 87. P 1x2 5 x4 2 3x2 1 2 89. P 1x2 5 x3 2 "2x2 1 x 2 "2 91. P 1x2 5 x3 2 2x2 1 2x

101. 10 4. $1,060.90

17. 1.3801

21. true 22. false 23. 6.4 24. 46 dB log 3 26. < 20.3133 27. ln 9 < 2.1972 log 3 2 2 29. 10 30. 9

1. whole

79. about 7.94 3 1024 gram-ions per liter 80. k ln 2 less 5 81. 2 82. 21, 23 83. 2 84. 3, 23 4 log 7 log 3 85. < 1.7712 86. < 2.7095 log 3 log 3 2 log 2 87. ln 8 < 2.0794 88. ln 7 < 1.9459 89. 23 90. 8 91. 25, 4 92. 4 93. 2 94. 4, 3 95. 6 ln 9 96. 31 97. < 3.1699 98. no solution ln 2 e 7 99. e < 1,096.6332 100. < 1.5820 101. 1 e21 102. about 3,300 yr

1.

3 a Å b2 16. ln c log 3 ln 3 19. or log 7 ln 7

1 1ln a 2 2 ln b 2 ln c2 2

Section 5.1

59. 4 60. 4 61. 0 62. 7 63. 3 64. 4 65. 9 66. 2 logb x 1 3 logb y 2 4 logb z 1 67. 1log8 x 2 log8 y 2 2 log8 z2 2 1 68. 4 ln x 2 5 ln y 2 6 ln z 69. 1ln x 1 ln y 1 ln z2 3 70. logb

14.

b"a 1 2 15. log c2

f(x) = ln(x + 1)

f(x) = 1 + ln x

13. 2 log a 1 log b 1 3 log c

93. 0

97. QIV

99. QI

103. 1

10. 23

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Answers to Selected Exercises

Section 5.2 1. zero

3. conjugate

9. lower bound 19. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 51. 63. 67.

11. 10

5. 0 13. 4

7. 2 15. 4

17. 4, 4

5, 5 21. x2 1 4 5 0 23. x2 2 6x 1 10 5 0 x3 2 3x2 1 x 2 3 5 0 x3 2 6x2 1 13x 2 10 5 0 x4 2 5x3 1 7x2 2 5x 1 6 5 0 x4 2 2x3 1 3x2 2 2x 1 2 5 0 x2 2 4xi 2 4 5 0 0 or 2 positive; 1 negative; 0 or 2 nonreal 0 positive; 1 or 3 negative; 0 or 2 nonreal 0 positive; 0 negative; 4 nonreal 1 positive; 1 negative; 2 nonreal 0 positive; 0 negative; 10 nonreal 0 positive; 0 negative; 8 nonreal; yes 1 positive; 1 negative; 2 nonreal 49. (22, 4) (21, 1) 53. (24, 6) 55. (24, 3) 57. (22, 4) k units to the right 65. reflected about the y-axis stretched vertically by a factor of k

1. 27

3. root

5. 61, 62, 63, 64, 66, 612 1 3 7. 61, 62, 63, 66, 6 , 6 2 2 1 5 1 5 9. 61, 62, 65, 610, 6 , 6 , 6 , 6 2 2 4 4 11. 1, 21, 5 13. 1, 2, 21 15. 1, 2, 22 17. 3, 23, 2 1 1 3 2 3 19. 1, 21, 21. 21, 21, 23. , , 2 2 3 2 3 5 2 3 3 2 5 25. , 2 , 4 27. , , 29. 1, 2, 3, 4 31. 2, 25 3 5 2 3 4 1 3 5 1 1 33. 1, 21, , 35. 22, 37. , 2 2 2 2 3 2 1 1 39. 1, 21, 2, 22, 23 41. 3, , 2 2 2

43. 0, 2, 2, 2, 22, 22, 22 45. 3, "2, 2"2 1 47. , i, 2i 49. 2, 22, 1 1 "5, 1 2 "5 2 1 51. , 21, 3i, 23i 53. 1, 1, 1, 5i, 25i 2 3 57. 3, 1 2 i 59. 3, 23, 1 2 i 55. , 1, 21, 2i, 22i 2 2 1 1 1 61. 2 , 3, 21 63. , , , 1, 1 65. 10, 20, 60 ohms 3 2 2 2 67. 13 in. 69. 1, 5, and 9 miles 79. "3 1 1

73. (3, 7) or approx. (1.54, 13.63) 77. 8a"2b

75. 6ab2"2abc

3. xl and xr 1. P 1a2 and P 1b2 9. P 1222 5 3; P 1212 5 22 The signs of the results are opposites. 11. P 142 5 240; P 152 5 30 The signs of the results are opposites. 13. P 112 5 8; P 122 5 21 The signs of the results are opposites.

Section 5.4

15. P 122 5 272; P 132 5 154 The signs of the results are opposites. 17. P 102 5 10; P 112 5 260 The signs of the results are opposites. 19. 1.7 21. 22.2 23. 1.7 25. 21.2 27. 22.2, 2.2 29. 1, 2; The Bisection Method fails to find the solution 2. 31. yes 33. (0.83, 0.56), (20.83, 20.56) 37. vertical: x 5 2, x 5 22; horizontal: y 5 0; slant: none 39. vertical: x 5 2; horizontal: none; slant: y 5 x 1 2

Chapter Review 1. 1

2. 66

7. true 10. 2x3 11. 5x4 12. 4x4 15.

Section 5.3

A25

13 8

3. 241

4. 34

6. false 277 8. true 9. 3x 1 9x 1 29x 1 90 1 x23 25 1 4x2 1 5x 1 13 1 x22 2259 2 14x3 1 31x2 2 64x 1 129 1 x12 1 2 2x3 1 x2 1 2x 1 13. 151 14. 2113 x11 1 16. 21 2 6i 17. e 3, , 22 f 2 3

5. false

2

18. U22, 22, "5, 2"5V

19. 20. 21. 22. 25. 30. 33. 35. 36. 37. 38. 39. 40. 41. 44. 46. 49. 52. 54. 57. 59.

P 1x2 5 2x3 2 5x2 2 x 1 6 P 1x2 5 2x3 1 3x2 2 8x 1 3 P 1x2 5 x4 1 3x3 2 9x2 1 3x 2 10 23. 6 24. 6 P 1x2 5 x4 1 x3 2 5x2 1 x 2 6 65 26. 1,984 27. 4, 4 28. 40, 40 29. 5, 5 32. i 3, 3 31. 2 2 i x3 2 4x2 1 x 2 4 5 0 34. x3 1 5x2 1 x 1 5 5 0 0 or 2 positive; 0 or 2 negative; 0, 2, or 4 nonreal 1 or 3 positive; 1 negative; 0 or 2 nonreal 1 positive; 0, 2, or 4 negative; 0, 2, or 4 nonreal 1 or 3 positive; 0 or 2 negative; 2, 4, or 6 nonreal 0 positive; 0 negative; 4 nonreal 0 positive; 1 negative; 6 nonreal 1 3 (21, 2) 42. (25, 2) 43. 61, 62, 63, 66, 6 , 6 2 2 1 5 1 5 61, 62, 65, 610, 6 , 6 , 6 , 6 45. 1, 4, 5 2 2 4 4 3 1 1, 21, 8 47. 25, 2 , 22 48. 2 3 3 3 1 1 2, 22, , 2 50. 4, 4, 22, 2 51. , 4i, 24i 2 2 2 3 2, 22, 1 1 "6, 1 2 "6 53. P 1212 5 29; P 102 5 18 55. 4.2 56. 2.5 P 112 5 28; P 122 5 21 1.67 58. 0.67 10 m by 6 m; 12 m by 5 m; 15 m by 4 m 60. 17.27 ft

Chapter Test

yes 2. 1 3. 230 4. no 1x 2 22 12x2 1 x 2 22 2 5 1x 1 12 12x2 2 5x 1 12 2 2 7. 2x 1 3 8. 3x2 1 x 11 9. 5 10. 228 11. 12. 6 2 3i 3 14. P 1x2 5 x4 2 2x2 2 3 13. P 1x2 5 x3 2 4x2 2 5x 16. x3 2 5x2 1 9x 2 5 5 0 15. x3 2 2x2 1 x 2 2 5 0 17. 3, 3 18. 3 1 2i

1. 5. 6.

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A26

Answers to Selected Exercises

19. 1 or 3 positive; 0 or 2 negative; 0, 2, or 4 nonreal 20. 1 positive; 0 or 2 negative; 0 or 2 nonreal 21. (23, 3) 22. (21, 6) 1 2 23. 61, 62, 64, 68 24. 61, 62, 6 , 6 5 5 1 25. 2, 23, 2 26. 2, i, 2i 27. 2, 23, 3i, 23i 2 28. 1, 21, 22, 2i, 22i 29. no 30. 3.3

19.

Cumulative Review

21. (2.2, 24.7)

1.

2.

y

y

x

1 41. a1, 2 b 2 is 1x, 5 2 2x2

x

4.

{

23. (1.7, 0.3) 25. (21, 22) 1 1 31. no solution; inconsistent 29. a , b 2 3

system 33. dependent equations; a general solution is 1x, 3x 2 62 35. (3, 1) 37. (3, 2) 39. (23, 0)

x f(x) = 3x – 2

y

dependent equations; infinite number of solutions y = –x + 6 5x + 5y = 30

27. (3, 22)

f(x) = 2e x

3.

y

y

f(x) = log3 x f(x) = ln(x – 2) x x

5. 6 6. 23 7. 3 8. 2 9. log a 1 log b 1 log c 10. 2 log a 1 log b 2 log c 1 1 11. 1log a 1 log b 2 3 log c2 12. ln a 1 ln b 2 ln c 2 2 log 8 a3 "ab3 13. ln 3 14. log 3 15. x 5 21 2 b log 3 "c

16. x 5 21 17. x 5 500 18. x 5 "11 19. 9 23. a factor 20. 236 21. 4 22. 2 2 i 24. not a factor 25. a factor 26. a factor 27. 12 28. 2,000 29. 2 or 0 positive; 2 or 0 negative; 4, 2, or 0 nonreal 30. 1 positive; 3 or 1 negative; 2 or 0 nonreal 31. 21, 23, 3 32. 21, 1, 2

43. dependent equations; a general solution 45. no solution; inconsistent system

47. (4, 27) 49. (2, 23) 51. (9, 21) 53. (1, 2, 0) 1 1 55. a0, 2 , 2 b 57. (1, 2, 21) 59. (1, 0, 5) 3 3 61. no solution; inconsistent system 63. (0, 1, 0) 2 1 1 1 1 1 65. a , , b 67. a , , b 3 4 2 2 2 2 69. dependent equations; a general solution is 1x, 2 2 x, 12 71. hamburger $2; fries $1 73. 225 acres of corn; 125 acres of soybeans 75. 8 kph 77. 40 g and 20 g 79. 10 ft 81. E 1x2 5 43.53x 1 742.72, R 1x2 5 89.95x; 16 pairs/day 83. 15 hr cooking hamburgers, 10 hr pumping gas, 5 hr janitorial 85. 1.05 million in 0214 group, 1.56 million in 15249 group, 0.39 million in 50-and-older group 87. 30°, 50°, 100° 95. y

(0, 1)

(1, 3) f(x) = 3x x

Section 6.1 1. system 3. consistent 5. independent 7. consistent 9. dependent 11. is y 13. 15.

y

(–2, 4) (1, 2)

3x + 2y = 2 x

x –2x + 3y = 16

x – 2y = –3

17.

y no solution; inconsistent system y = –x + 5 3x + 3y = 30

{

x

99. log x 2 2 log y 2 log z

Section 6.2

x = –2 y=4

x=1 y=2

y = –3x + 5

97. 8

101. log

"z xy3

1. matrix 3. coefficient 5. equation 7. row equivalent 9. interchanged 11. adding, multiple 13. (2, 3) 15. (7, 6) 17. (1, 2, 3) 19. (1, 1, 3) 21. row echelon form 23. reduced row echelon form 25. 12, 212 27. 122, 02 29. (3, 1) 31. no solution; inconsistent system 33. 10, 272 35. (1, 0, 2) 37. 12, 22, 12 39. (1, 1, 2)

41. dependent equations; a general solution is 7 3 43. no solution; inconsistent system ax, 2x 1 , 2 b 2 2 45. (13, 3) 47. 1213, 7, 222 49. (10, 3) 1 1 51. (0, 0) 53. 13, 1, 222 55. a , 1, b 4 4

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Answers to Selected Exercises

57. (1, 2, 1, 1)

9 3 61. a , 23, b 4 4

59. (1, 2, 0, 1)

63. a0,

20 1 65. dependent equations; a general ,2 b 3 3 1 67. (1, 23) solution is ax, x 2 3, 0b 2

69. dependent equations; a general solution is 8 1 10 4 71. dependent equations; a general 2 z, zb a 1 z, 7 7 7 7 solution is 11 1 z, 2z, 21 2 z, z2 73. 77. 79. 85. 89.

no solution; inconsistent system 75. 1,300 mi Dictionaries 4.5 in; atlases 3.5 in; thesauruses 4 in 2 niacin, 4 zinc, 6 vitamin C x 5 62; y 5 61; z 5 63 87. y 5 mx 1 b equal 91. y 5 2x 1 7 93. x 5 2

Section 6.3 1. i, j 3. corresponding 5. columns, rows 7. additive identity 9. x 5 2, y 5 5 11. x 5 1, y 5 2 13. c 17. c 21. c 25. c

21 26

2 0

26 1

5 21

25 210 2 3

4 29. £ 5 2

1 d 0

75 225

22 d 10 2 22 22

47. c

51. c

4 27

5 d 21

210 d 5

27. c

10 4§ 1

33. not possible 236.29 39. £ 16.2 § 219.26

0 d 0

25 2 27 15. £ 5 0 23 § 2 23 5 15 215 19. c d 0 210

222 2105

23. c

18 235

222 d 126

21 0

24 d 21

4 25 26 31. £ 28 10 12 § 212 15 18 16 35. £ 12 § 37. not possible 12

216.11 41. £ 219.6 2100.6

4.71 20.35 72.82

33.64 6.4 § 62.71

49. not possible

16 47 53. c d d 26 81 2,000 The cost to Supplier 1 is $2,000. 55. QC 5 c d 1,700 The cost to Supplier 2 is $1,700. 584.50 Adult males spent $584.50. 57. QP 5 £ 709.25 § Adult females spent $709.25. 1,036.75 Children spent $1,036.75. 1 1 0 59. £ 0 1 1 § 1 0 0 24 39

5 1 61. A2 5 ≥ 2 2

1 5 2 2

63. No; if A 5 c

65. LetA 5 c

2 2 6 0

1 1

2 2 ¥ 0 4

indicates the number of ways two cities can be linked with exactly one intermediate city to relay messages.

1 1 d and B 5 c 1 0

1 2 2 d and B 5 c 1 2 21 matrix, yet AB 5 0. x11 x21

67. 6x2 2 4x 2 8

69.

Section 6.4

3. 3 I 0 A21 4

1. AB 5 BA 5 I 5 7. c 22

27 d 3

1 15. £ 0 0

22 1 0

4 11. £ 25 21

19.

23. 25. 29. 33. 37. 45. 53.

1 21 21

0 d , then 1AB2 2 2 A2B2 . 0

2 d . Neither is the zero 21 71. a 5 5. c

22 3 23 9. £ 25 7 26 § 1 21 1 23 4§ 13. no inverse 1

1 22 § 1

A27

3 2

2s 2l n

4 d 3

17. no inverse

1 22 1 0 8 22 26 0 1 22 1 ≥ ¥ 21. £ 25 2 4§ 0 0 1 22 2 0 22 0 0 0 1 22.5 5 3 5.5 5.5 28 26 29.5 ≥ ¥ 21 3 1 3 25.5 9 6 10.5 x 5 23, y 5 17 27. x 5 0, y 5 0 x 5 1, y 5 2, z 5 2 31. x 5 54, y 5 237, z 5 249 x 5 2, y 5 1 35. x 5 1, y 5 2, z 5 1 2 of model A, 3 of model B 39. hi 41. no 0 X 5 £0§ 51. 12`, 222 c 122, 22 c 12, ` 2 0 12`, ` 2 55. 3 0, ` 2 57. 12`, ` 2

3. 0 5. 0 7. 8 9. 1 1. 0 A 0 , det 1A2 11. 242 13. 13 15. 42 17. 13 19. 254 21. 27 23. 86 25. 22 27. 2 29. 120 31. true 33. false 35. 3 37. 3 39. (1, 2) 41. (3, 0) 43. (1, 0, 1) 45. (1, 21, 2) 47. (6, 6, 12) 5 2 1 5 49. a , , , b 51. 3x 2 2y 5 0 53. 6x 1 7y 5 9 6 3 2 2 55. 30 sq. units 57. 73 sq. units 63. 8 65. 21 67. $5,000 in HiTech, $8,000 in SaveTel, $7,000 in OilCo 69. 10 71. 24 73. 8 77. domain: n 3 n matrices; range: all real numbers 79. yes 81. 21.468 83. 1x 2 12 1x 1 42

Section 6.5

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A28

Answers to Selected Exercises

85. x 13x 1 12 13x 2 12 x2 1 2x 1 1 89. x 1x2 1 12

87.

4x 2 5 1x 2 22 12x 2 12

13.

1 2 1 x x21 5 3 1 2 2 2 5. 2 7. 1 9. 2 x x23 x11 x21 x x22 1 3 8 5 11. 2 13. 2 x23 x12 x13 x21 5 2 2 3 1 15. 1 17. 1 2 2x 2 3 x25 x x21 x11 1 1 3 2 19. 1 2 21. 1 2 x x 13 x11 x 1 2x 1 3 3 2 1 1 2 3 1 1 23. 25. 1 22 1x 1 12 2 x x11 x x x21 2 1 2 1 4 27. 1 1 29. 2 1x 2 32 2 1x 2 12 3 x x23 x21 1 1 2 3 4 x11 31. 1 21 2 33. 1 21 2 x x x 1x11 x x x 11 1 3 x11 2 35. 2 2 2 37. 2 1 2 x11 x 12 x 12 x 1x12 1 x x12 1 2 1 2 39. 1x 1 2x 1 52 2 x x 1 2x 1 5 1 8 2 1 41. x 2 3 2 1 43. 1 1 1 2 x11 x12 3x 1 1 x 11 1 x 1 3 2 45. 1 1 1 2 47. 2 1 1 1 2 x x 1x11 x x21 x 11 49. No, it’s the sum of two cubes. 51. 2 0 a 0 "2ab 53. 3x2"2x

y 2y ≥ 3x – 2

1 y = –x + 1 2

Section 6.6

x

x 2y = 3x – 2

1 y ≤ –x + 1 2

3.

17.

19.

y

y x2 + y2 ≤ 4 x

x y < x2

21.

23.

y

y

y=3 y=1 x

x x=2

1. first-degree; second-degree

15.

y

25.

x=2

27.

y

y

x y = 2x + 1

x + y =2

y=x–2

x x + y =1

55. x 5 9

29.

31.

y

y

Section 6.7 1. half-plane; boundary 5. y

3. is not 7.

3x + 2y = 6

x + 2y = 3 y x

x

2x + 3y = 12

2x – 3y = 6 x 2x x x 4x – y > 4

y = x2 – 4

1 y = –x 2

y = 2x

37.

y = 4 – x2

39.

y

y

2x − y = 0

x – 2y = 0 x=0 x

x + 2y = 10 x

x–y=2

y=0

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Answers to Selected Exercises

41.

43.

y

6. x 5 2, y 5 21 7. x 5 0, y 5 23 8. x 5 1, y 5 1 9. dependent equations; a general solution is 1x, 3x 2 42 10. no solution; inconsistent system 11. x 5 23, y 5 2 12. x 5 22, y 5 5 13. x 5 2, y 5 21 14. no solution; inconsistent system 15. dependent equations; a general solution is 1x, 3x 2 42 16. x 5 1, y 5 0, z 5 1 17. x 5 1, y 5 1, z 5 21 18. x 5 0, y 5 1, z 5 2 19. $10,400 20. 900 adult tickets, 450 senior tickets, 450 children’s tickets 21. x 5 1, y 5 1 22. dependent equations; a general solution is 1x, 3x 1 42 23. x 5 3, y 5 1, z 5 22 24. x 5 210, y 5 1, z 5 10 25. no solution; inconsistent system 1 3 4 27. c 26. x 5 24, y 5 3 d 4 0 2 2 5 4 4 21 28. £ 22 26 29. c 6§ d 27 27 24 5 23 2 21 1 3 217 19 4 22 2 6 30. c 31. 3 5 4 32. ≥ ¥ d 2 21 1 3 10 212 10 25 5 15

y x + 2y = 10 x=0

x=0

x+y=4

x

y=0

y=0

x

3x – 2y = 6

x–y=4

6s 1 4l # 60 45. cs $ 0 l$0

53. one; one

55. 0

Section 6.8

1. constraints 3. objective 5. P 5 12 at (0, 4) 13 5 4 18 3 12 at a , b 9. P 5 at a , b 7. P 5 6 3 3 7 7 7 11. P 5 3 at (1, 0) 13. P 5 0 at (0, 0) 15. P 5 0 at (0, 0) 17. P 5 212 at (22, 0) 19. P 5 22 at the edge joining (1, 2) and (21, 0) 21. 3 tables, 12 chairs; $1,260 23. 30 IBMs, 30 Apple; $2,700 25. 15 DVRs, 30 TVs; $1,560 27. $150,000 in stocks, $50,000 in bonds; $17,000 29. 2 buses, 2 trucks; $1,100 1 0 0 1 1 0 1 0 33. ≥ ¥ 35. a 2 3y, y, 2 b 0 0 1 3 3 0 0 0

33. not possible 5 36. c 23

23 d 2

34. 3 224 4

1 3 37. 2 E 2 1

Chapter Review 1.

2.

y

y

2x – y = –1

2x – y = –5

5x + 2y = 1 (2, 5) (–1, 3)

x+y=7

x x

3.

4.

y

x=y–7

y = 5x + 7

5.

x

x

0 1 2 1 2 2

35. c

0 1 2U

0 d 26

0

9 16 256 38. £ 23 25 39. No inverse exists. 18 § 21 22 7 40. No inverse exists. 41. x 5 1, y 5 2, z 5 21 42. w 5 1, x 5 1, y 5 0, z 5 21 43. 27 44. 26 45. 3 46. 225 47. x 5 1, y 5 22 48. x 5 1, y 5 0, z 5 22 49. x 5 1, y 5 21, z 5 3 50. w 5 1, x 5 0, y 5 21, z 5 2 51. 21 52. 7 3 4 3 2 x21 53. 1 54. 1 21 2 x x11 x x x 11 1 1 1 2 2 55. 2 2 56. 2 1 1x 1 12 2 1x 1 12 3 x x 1x15 x11 y 57. 58. y

3x + 2y = 6 3 y = –– x + 3 2

a general solution is 3 ax, 2 x 1 3b 2

y no solution; inconsistent system

y dependent equations; infinite number of solutions

(0, 7)

A29

x2 + y2 > 4

y ≥ –2 x – 1

x

x

59.

60.

y

y

x–y=3

x 3x + 2y = 6

{

4x – y = 4 y = 4(x – 2)

x

x 2

y=x +1

y = x2 – 1

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A30

Answers to Selected Exercises

62. P 5 12 at 10, 242 2 5 63. P 5 2 at (1,1) 64. P 5 3 at a2 , b 3 3 65. 1,000 bags of x, 1,400 bags of y

10, 02 , 10, 32 , y 5 23 27. 10, 32 , 15, 32 , x 5 25 122, 12 , 122, 252 , y 5 7 31. x2 5 12y 2 2 1 2 35. x 2 3 5 212 1y 2 52 y 5 212x 1x 2 32 2 5 228 1y 2 52 39. x2 5 24 1y 2 22 1y 1 52 2 5 8 1x 2 12 1y 2 22 2 5 22 1x 2 22 or 1x 2 22 2 5 22 1y 2 22 16 9 45. 1x 1 42 2 5 2 1y 2 62 or 1y 2 62 2 5 1x 1 42 3 4 1 47. 1y 2 82 2 5 24 1x 2 62 49. 1x 2 32 2 5 1y 2 12 2 y 51. 53. y

61. P 5 6 at (3,0)

25. 29. 33. 37. 41. 43.

Chapter Test 1.

2.

y

y x = 2y + 5

x – 3y = –5

x

(1, 2) x

(1, –2)

2x – y = 0

y = 2x – 4

x 5 1, y 5 23 4. x 5 3, y 5 5 6 liters of 20% solution, 4 liters of 45% solution CD World 100 units, Ace 25 units, Hi-Fi 50 units x 5 2, y 5 1 8. x 5 1, y 5 2, z 5 1 x 5 1, y 5 0, z 5 22 2 7 8 7 10. x 5 2 y 1 , z 5 2 y 2 , y 5 any number 5 5 5 5 7 2 16 214 20 3 11. c 12. 3 21 4 13. ≥ d 0 26 213 2 3

(2, 3)

3. 5. 6. 7. 9.

14. 16. 19. 22. 23.

(–2, 1)

y = x2 + 4x + 5 or y – 1 = (x + 2)2

19 3 ¥ 5 2 3

213 23 14 17 4 £ 4 1 24 § 15. x 5 , y 5 2 3 3 12 3 213 x 5 236, y 5 11, z 5 34 17. 212 18. 224 5 3 1 2 20. 1 21. 1 4 2x 2 3 x11 1 2x 1 1 1 2 x x 12 y 24. y x – 3y = 3 x + 3y = 3

x

x

55.

y2 + 4x – 6y = –1 or (y – 3)2 = –4(x – 2)

57.

y

y

(3, 2)

(1, 2)

x

x

y2 – 4y = –8x + 20 or (y – 2)2 = –8(x – 3)

y2 – 4y = 4x – 8 or (y – 2)2 = 4(x – 1)

59.

61.

y

y

(1–2 , 3–2) (–2, 3)

3x + 4y = 12

x x

x x

25. P 5 7 at 11, 22

3x + 4y = 6

x2 – 6y + 22 = –4x or (x + 2)2 = 6(y – 3)

26. P 5 28 at (8, 0)

1. 2, 25, 3 3. 0, 0, "5 5. to the left 7. downward 9. directrix, focus 11. circle 13. parabola 15. x2 1 y2 5 49, x2 1 y2 2 49 5 0 17. 1x 2 22 2 1 1y 1 22 2 5 17, x2 1 y2 2 4x 1 4y 2 9 5 0 19. 1x 2 12 2 1 1y 1 22 2 5 36, x2 1 y2 2 2x 1 4y 2 31 5 0 y y 21. 23.

Section 7.1

x 2 + y2 = 4 (2, 1) x

4x2 – 4x + 32y = 47 or 3 1 2 x – – = –8 y – – 2 2

( )

( )

63.

71. 1x 2 72 2 1 y2 5 9 65. yes

67. 60 mi

69. 1x 2 42 2 1 y2 5 16 73. 2 ft

75. x2 5 2

45 y 2

77. 1 ft 79. 300 ft 81. about 12.6 cm 83. about 520 ft 85. 0x2 1 0xy 1 y2 2 8x 2 4y 1 12 5 0 89. y 5 x2 1 4x 1 3 93. 4 87. x2 1 1y 2 32 2 5 25 49 97. x 5 1, 25 99. x 5 22, 9 95. 4

x 3x2 + 3y2 – 12x – 6y = 12

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A31

Answers to Selected Exercises

Section 7.2

Section 7.3

1. sum, constant 3. vertices 5. a, 0; 2a, 0 7. 26-in string; thumbtacks 24 in. apart 9. circle y2 x2 1 51 11. parabola 13. ellipse 15. 16 9 y2 9y2 9x2 x2 1 51 19. 1 51 17. 25 16 16 25

1. difference, constant 3. a, 0; 2a, 0 7. circle 9. parabola 11. ellipse

21. 25.

y2 x2 1 51 7 16

15.

1y 2 42 2 1x 2 32 2 1 51 4 9

23.

1y 2 42 2 1x 2 32 2 1 51 9 4

1y 2 42 2 1x 2 32 2 1 51 41 16 2 2 y x 1 51 31. 100 64 y 33. 2 2 27.

29.

19. 23.

1y 2 42 2 1x 2 22 2 2 51 4 9

17.

1y 2 32 2 1x 2 52 2 2 51 9 9

1y 2 42 2 1x 2 12 2 2 51 4 32

21.

y2 x2 2 51 9 16

25.

3y2 x2 2 51 10 20

1y 1 42 2 1x 1 22 2 4 1y 1 42 2 4 1x 1 22 2 2 5 1 or 2 51 31. 4 81 4 81 16y2 x2 2 51 33. 36 25 y y 35. 37. 27. 24 sq. units

1y 2 42 2 x2 1 51 36 20

35.

y2 x2 2 51 25 24

5. transverse axis 13. hyperbola

29. 12 sq. units

y

y x –– + –– = 1 25 9

y2 x2 ––– – ––– = 1 9 4

x2 y2 –– + –– =1 25 49

37.

39.

y

4x2 – 3y2 = 36

y

x

x

x

x

x

39.

41.

y

(x + 2)2 y2 –––––– – –– = 1 9 4

y

(4, 2) x

(0, –2)

x (x – 4)2 (y – 2)2 –––––– + –––––– = 1 49 9

x2 (y + 2)2 –– + –––––– = 1 16 36

1y 2 12 2 x2 41. 1 51 4 16 45. y

1y 1 22 2 1x 1 12 2 43. 1 51 4 9 y 47.

x

y2 – x2 = 1

43.

45.

y

y

x x 2 + 4y 2 – 4x + 8y + 4 = 0 x

(5, 4)

x

x (2, –1)

49.

16x 2 + 25y 2 – 160x – 200y + 400 = 0

y2 x2 51. 1 51 900 400

y2 x2 1 5 1; 36 m 55. about 20.8 in. 2,500 900 y2 x2 57. 199,395 mi 63. 1 51 9 8 67. 4 of the 75 planets shown have eccentricity 0. 1 3 22 4 71. £ 2 2 § 69. £ 1 1 § 23 1 24 5

4(y – 2)2 – 9(x + 1)2 = 36

47.

4x2 – 2y2 + 8x – 8y = 8

49.

y

y –xy = 6

x x

53.

1y 2 12 2 1x 2 32 2 2 51 9 16 55. 57. 24 y2 – 4x2 + 6y + 32x = 59

51.

53. 4x2 2 5y2 2 60y 5 0 59. 3 units

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

A32

Answers to Selected Exercises

Chapter Review

y2 x2 2 51 144 25 y2 x2 2 51 hyperbola; 36 64 x12 2x f 21 1x2 5 71. f 21 1x2 5 3 52x f 1g 1x22 5 1x 1 12 4 1 1 f 1 f 1x22 5 1x2 1 12 2 1 1

61. hyperbola; 63. 69. 73. 75.

1. 3. 5. 6.

x2 1 y2 5 16 2. x2 1 y2 5 100 2 1x 2 32 1 1y 1 22 2 5 25 4. 1x 1 22 2 1 1y 2 42 2 5 25 2 2 1x 2 52 1 1y 2 102 5 85 1x 2 22 2 1 1y 2 22 2 5 89 7. 1x 2 32 2 1 1y 1 22 2 5 16 y

x

Section 7.4 1. graphs 3.

C(3, –2)

5.

y

y (–3, 9)

(3, 9)

x = 2y (4, 2) y=

x

2

y

C(–2, 5)

y

y = 2x – 4

(4, 3)

x

(3, 2) x

x

(4, –3)

11.

y

+

64y2

(1–5 , –18–5– )

= 768

11. 1x 1 22 2 5 2

(x + 2)2 + (y – 5)2 = 16

10. x2 5 16y 12x2

9. y2 5 22x

x2 – 13 = –y2

x2 + y2 = 25

(–4, –3)

(x – 3)2 + (y + 2)2 = 16

x2 + y2 = 90

9.

y

(–4, 3)

x2 x

(–4, –2) 8x2 + 32y2 = 256

7.

8. 1x 1 22 1 1y 2 52 5 16 2

12.

y

13.

4 1y 2 32 11

y

x2 – 6x – y = –5

(1, 3)

(1, 2) x2 – 4y – 2x + 9 = 0 (1, 0)

(5, 0)

x x 2

y – 6y = 4x – 13

x

y2 y2 x2 x2 1 51 15. 1 51 36 16 4 25 1y 2 32 2 1x 1 22 2 1 51 16. 16 9 1y 1 12 2 y 51 17. 1x 2 22 2 1 4 14.

15. (1, 0.67), 121, 20.672 19. (1, 1) 21. (1, 2), (2, 1)

x2 – 6x + y = –5

13. (1, 2), (21, 0) 17. (3, 0), (0, 5)

25. Q"5, 5R, Q2"5, 5R

27. Q"3, 0R, Q2"3, 0R

23. (22, 3), (2, 3)

x

31. Q2"15, 5R, Q"15, 5R, (22, 26), (2, 26)

(2, –1)

29. (2, 4), (2, 24), (22, 4), (22, 24)

33. (0, 24), (23, 5), (3, 5) 35. (22, 3), (2, 3), (22, 23), (2, 23)

37. (3, 3) 39. (6, 2), (26, 22), Q"42, 0R, Q2"42, 0R 1 1 1 1 41. a , b, a , b 43. 7 cm by 9 cm 2 3 3 2 45. 80 ft by 100 ft or 50 ft by 160 ft 47. either $750 at 9% or $900 at 7.5% 49. (30, 3) 51. yes at 122, 42 and (1, 1) 53. about 23 mi 61. vertical: x 5 1; horizontal: y 5 3 63. vertical: x 5 1, x 5 21; horizontal: y 5 0 65. y-axis 67. origin

4x2 + y2 – 16x + 2y = –13

y2 y2 x2 x2 2 51 19. 2 51 4 12 9 16 1x 2 32 2 y2 4 2 51 22. y 5 6 x 21. 9 16 5

18.

20.

1y 2 32 2 x2 2 51 9 16

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Answers to Selected Exercises

23.

1y 1 22 2 1x 2 12 2 2 51 4 9

y

11.

1y 2 32 2 1x 2 22 2 1 51 4 36

12.

A33

y

x (1, 2)

(3, –2)

(–1, –2)

x

(x – 1)2 (y – 2)2 –––––––––– + –––––––––– = 1 4 9

9x2 – 4y2 – 16y – 18x = 43

24.

25.

y

y (0, 4)

4 xy = 1 y=

x

26.

x+

4

x

(–4, 0)

x2 + y2 = 16 y

27.

y

y2 4y2 x2 x2 2 51 14. 2 51 25 144 36 25 2 2 1y 1 12 1x 2 22 2 51 15. 64 36 y 16. 13.

(–2, 3)

(2, 3) x

(4, 2)

(– 4, 2)

x

x (4, –2)

(– 4, –2)

(–2, –3) 2

x –

y2

= 12

3x2

+

y2

= 52

y2 x2 – –– 3

=1

(2, –3) y2 x2 –– + –– 16 12

=1

28. (24, 2), (24, 22), (4, 2), (4, 22) 29. (0, 4), Q2"3, 22R

30. (22, 3), (22, 23), (2, 3), (2, 23)

Chapter Test

1. 1x 2 22 2 1 1y 2 32 2 5 9 3. 1x 2 22 2 1 1y 1 52 2 5 169 y 4.

(y – 2)2 x2 –––––––––– – ––– = 1 2 8

2. 1x 2 22 2 1 1y 2 32 2 5 41

17. Q"7, 4R, Q2"7, 4R

18. Q3"2, 3R, Q23"2, 3R, Q3"2, 23R, Q23"2, 23R

19. 1y 2 22 2 5 6 1x 1 32 ; parabola 1y 1 22 2 1x 2 12 2 1 5 1; ellipse 20. 3 2

Cumulative Review 1. 16 6.

x

2.

1 2

1 121x x

10. x 1 3

(x – 2)2 + (y + 3)2 = 9

5. 1x 2 32 5 16 1y 2 22 6. 1y 1 62 5 24 1x 2 42 4 9 2 7. 1x 2 22 5 1y 1 32 or 1y 1 32 2 5 2 1x 2 22 3 2 y 8. 2

2

16. "xy 20. 23. 26. 30.

x (3, –2)

33. 37.

(x – 3)2 = 8(y + 2)

41. 2

9.

2

y x 1 51 100 64

2

10.

2

y x 1 51 169 144

45. 49.

8. 4t"3t

7. 23x

14.

3 2 5" x x

17. 2, 7

5. x4/3 2 x2/3 9. 4x

4 4 12. 212" 2 1 10" 3

11. 7"2

(2, –3)

13. 218"6

4. x17/12

3. y2

18.

15. 1 4

x 1 3"x 1 2 x21 19. 1, 2

3 2

22 6 "7 21. 7 1 2i 22. 25 2 7i 3 13 1 0i 24. 12 2 6i 25. 212 2 10i 3 1 1 i 27. "13 28. "61 29. 22 2 2 1 31. 12`, 222 c 13, ` 2 32. [22, 3] a1, b 2 36. 6x2 1 3 5 34. 27 35. 12x2 2 12x 1 5 y 2 5x 38. log3 a 5 b 39. 5 40. 3 1 x 42. 1 43. y 5 2 44. x 27 1.9912 46. 0.301 47. 1.6902 48. 0.1461 2 log 2 log 3 2 log 2

50. 16

51. $2,848.31

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A34

Answers to Selected Exercises

52. 1.16056 53. (2, 1) 54. (1, 1) 55. (2, 22) 56. (3, 1) 57. 21 58. 21 59. (21, 21, 3) 60. (1, 1, 1) 1. power 3. first 5. 7 ? 6 ? 5 ? 4 ? 3 ? 2 ? 1 7. 1n 2 12 ! 9. 120 11. 4,320 13. 1,440 1 5 15. 19. 18,564 17. 1,320 3 21. a5 1 5a4b 1 10a3b2 1 10a2b3 1 5ab4 1 b5 23. x3 2 3x2y 1 3xy2 2 y3 25. a3 1 3a2b 1 3ab2 1 b3 27. a5 2 5a4b 1 10a3b2 2 10a2b3 1 5ab4 2 b5 29. 8x3 1 12x2y 1 6xy2 1 y3 31. x3 2 6x2y 1 12xy2 2 8y3 33. 16x4 1 96x3y 1 216x2y2 1 216xy3 1 81y4 35. x4 2 8x3y 1 24x2y2 2 32xy3 1 16y4 37. x5 2 15x4y 1 90x3y2 2 270x2y3 1 405xy4 2 243y5 x3y 3x2y2 x4 39. 1 1 1 2xy3 1 y4 41. 6a2b2 16 2 2 51. 1,134a5b4

1. r n21 3. ar n21 5. infinite 7. Geometric means 9. 10, 20, 40, 80 11. 22, 26, 218, 254 13. 3, 3"2, 6, 6"2

Section 8.1

43. 35a3b4

Section 8.4

45. 2b5 47. 2,380a13b4 2 2 3x y 55r2s9 53. 55. 2 2 2,048

49. 24"2a3

n! n! 59. an23b3 an2r11br21 1r 2 12 ! 1n 2 r 1 12 ! 3! 1n 2 32 ! 63. 2252 71. 3xyz2 1x2yz2 2 2z3 1 5x2 31x 73. 1a2 1 b22 1a 1 b2 1a 2 b2 75. 32x 57.

29. 35. 39. 45. 59.

1. 1n 2 12 3. infinite 5. an 5 a 1 1n 2 12 d 7. Arithmetic means 9. 1, 3, 5, 7, 9, 11 7 1 5 23 33 43 11. 5, , 2, , 21, 2 13. 9, , 14, , 19, 2 2 2 2 2 2 25 35 15. 318 17. 6 19. 370 21. 44 23. , 15, 2 2 82 59 12 13 25. 2 , 2 , 2 , 2 27. 285 29. 555 15 15 5 15 1 33. 20,100 35. 1080°; 1800° 37. $460 31. 157 2 39. $1,587,500 41. 80 ft 43. 210 49. 6 51. 21 53. 1, 21, i, 2i

Section 8.3

4

23. 8, 32, 128, 512 25. 124 1,995 29. 31. 18 33. 8 32

27. 229,524 5 25 35. 37. 9 99

39. 23 45. 51. 57. 63. 71.

41. 12.96 ft, 400 ft 43. 5.13 m 1 49. about 1.03 3 1020 $69.82 47. about C 3 $180,176.87 53. $2,001.60 55. $2,013.62 $264,094.58 59. 5,000 61. 1.8447 3 1019 grains 1 8 no 65. 5 2 3i 67. 10 1 0i 69. 1 i 5 5 02i

Section 8.5 1. two 3. n 5 k 1 1 5. 5 5 5; 15 5 15; 30 5 30; 50 5 50 7. 7 5 7; 17 5 17; 30 5 30; 46 5 46 29. no 37. a. 1 b. 3 c. 7 d. 15 39. 41. y

y 2x + y > 5

x x

3x + 4y ≤ 12

1. domain 3. series 7. Summation notation 13. 0, 10, 30, 60, 100, 150 19.

17. 256

4 21. 10"2, 10"4 or 10"2, 10" 8

19. 2162

Section 8.2

5. infinite 9. 6 11. 5c 15. 21 17. a 1 4d 242 15 21. 15 23. 15 25. 27. 35 243 1 33. k, k2, k4, k8 3, 7, 15, 31 31. 24, 22, 21, 2 2 16 32 64 8, , 2 , 3 37. an alternating infinite series k k k not an alternating infinite series 41. 30 43. 250 7 40 47. 500 49. 51. 160 53. 3,725 12 6 cm 61. 26 ft

15. 2, 6, 18, 54 4

Section 8.6 1. 24 11. 21. 31. 39. 47. 57. 65. 75.

n! 1n 2 r2 ! 13. 840

5. 1

n! r! 1n 2 r2 ! 17. 120

9. 1 n! 15. 35 19. 5 a!b!... 1 23. 1,200 25. 40 27. 2,278 29. 144 8,000,000 33. 240 35. 6 37. 40,320 14,400 41. 24,360 43. 5,040 45. 48 96 49. 210 51. 24 53. 5,040 55. 60 2,721,600 59. 1,120 61. 59,400 63. 272 28 67. 252 69. 142,506 71. 66 73. 256 56 83. 2 85. 4 87. true 89. true 3.

7.

n 1E2 n 1S2 5. 5 11, H2 , 12, H2 , 13, H2 , 14, H2 , 15, H2 , 16, H2 , 11, T2 , 12, T2 , 13, T2 , 14, T2 , 15, T2 , 16, T2 6 7. {a, b, c, d, e, f, g, h, i, j, k, l, 1 2 m, n, o, p, q, r, s, t, u, v, w, x, y, z} 9. 11. 6 3 19 13 3 1 13. 15. 17. 19. 0 21. 42 42 8 12 5 1 23. 25. 27. about 6.3 3 10212 29. 0 12 169 3 1 1 5 31. 33. 35. 37. 13 6 8 16

Section 8.7

1. experiment

3.

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A35

Answers to Selected Exercises

39. {SSSS, SSSF, SSFS, SFSS, FSSS, SSFF, SFSF, FSSF, SFFS, FSFS, FFSS, SFFF, FSFF, FFSF, FFFS, FFFF} 1 1 32 1 41. 43. 45. 1 47. 49. 4 4 119 3 51. 0.18 53. 0.14 55. about 33% 57. no 59. 210, 4 61. 124, 102

Chapter Review

280 1. 720 2. 30,240 3. 8 4. 3 3 2 2 3 5. x 1 3x y 1 3xy 1 y 4 3 2 2 3 4 6. p 1 4p q 1 6p q 1 4pq 1 q 7. a5 2 5a4b 1 10a3b2 2 10a2b3 1 5ab4 2 b5 8. 8a3 2 12a2b 1 6ab2 2 b3 9. 56a5b3 10. 80x3y2 3 6 3 11. 84x y 12. 439,040x 13. 63 14. 9 15. 5, 17, 53, 161 16. 22, 8, 128, 32,768 17. 90 18. 60 19. 1,718 20. 2360 21. 117 22. 281 135 7 13 23. 292 24. 2 25. , 5, 2 2 2 26. 25, 40, 55, 70, 85 27. 3,320 28. 5,780 1 29. 25,220 30. 21,540 31. 32. 13,122 729 9 8 33. 34. 35. 2"2, 4, 4"2 16,384 15,625 3,280 36. 4, 28, 16, 232 37. 16 38. 39. 6,560 27 2,295 520,832 3,280 41. 42. 43. 16"2 40. 128 78,125 3 2 3 1 44. 45. 46. no sum 47. 1 48. 3 25 3

49. 1

50.

17 99

51.

5 11

52. $4,775.81

53. 6,516; 3,134 54. $3,486.78 55. 1 5 1; 9 5 9; 36 5 36; 100 5 100 57. 6,720 58. 35 59. 1 60. 4,050 62. 840 1 67. 6

63. 21 33 68. 66,640 1 73. 108,290

61. 564,480 56 64. 66 65. 120 66. 1,287 69. 20,160 70. 90,720

108,289 108,290 60 1 76. 77. 75. about 6.3 3 10212 143 2 1 33 15 79. 80. 81. 2,598,960 16,660 16

72. 24

74.

78.

7 13

Chapter Test 1. 144 2. 384 3. 10x4y 4. 112a2b6 5. 27 6. 236 7. 155 8. 2130 9. 9, 14, 19 10. 26, 218 11. 255.75 12. about 9 13. about $0.42C 14. about $1.46C 16. 800,000 17. 42 18. 24 19. 28 20. 1 21. 576 22. 120 23. 60 24. {(H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)} 1 2 33 1 87 25. 26. 27. 28. 29. 6 13 66,640 9 245 12 30. 19

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A36

Index

Index Absolute value of complex numbers, 136, 180 definition of, 10–11, 75, 183 equations of definition of, 184 graphing with a calculator, 238 graph of function of, 285 graphing, 286 inequalities properties of, 184 properties of, 171, 184 simplifying an expression with, 11 solving equations containing, 166–168 using definition of, 165–166 Accent on technology. See Calculator. Addition Associative Property of, 5 Commutative Property of, 5 Equality Property of, 87 property of fractions, 64 property of inequality, 151 Addition method of solving systems of equations, 554–555, 642, 718 Additive identity, 579 Additive identity matrix, 579 Additive inverse of a matrix, 580 Algebra, Fundamental Theorem of, 511–512, 539 Algebraic expression, 21 Algebraic fraction, 63 Alternating series, 733 Annual growth rate, 487 Annual population growth rate, 431 Applied problem, 95 Area problem, 119–120 Arithmetic means definition of, 741, 781 inserting between real numbers, 741–742 Arithmetic sequence definition of, 740, 780 determining terms of, 741 finding sums of terms in, 781 writing terms of, 741 Arithmetic series definition of, 742 finding sums of terms in, 742–743 Associative Property of addition, 5, 74 matrices and, 587 of multiplication, 5, 74 Asymptote definition of, 346 finding horizontal, 349–351, 404 finding slant, 351–352, 404 finding vertical, 347–348 Augmented matrix. See Systems matrix. Axis hyperbolic conjugate, 694 transverse, 688 parabolic, 658, 673 Axis of symmetry definition of, 235 quadratic functions and, 296 Back substitution definition of, 564 solving systems of equations using, 567–569 Base of exponential expressions, 15 Binomial definition of, 41, 78 expanding with Pascal’s Triangle, 724–725 using factorial notation, 725–726 using the Binomial Theorem, 727–728 multiplying, 44–45, 78

Binomial expansion, 728–729 Binomial Theorem definition of, 727, 768, 779 expanding binomials with, 727–728 writing with combination notation, 768 Bisection Method, 533–534, 544–545 Bounded interval, 9 Bounds on roots, 516–518, 541 Break point, 98 Break-point analysis problem, 98 Calculator adding and multiplying matrice, 585 applying the logarithmic Power Rule, 462 approximating base-10 logarithms, 441 approximating real roots, 534–535, 545 approximating solutions to a logarithmic equation, 479 approximating solutions to an exponential equation, 475 checking linear equation solutions, 89 combinations and, 767 complex numbers and, 135 confirming roots of a polynomial equation, 523, 526 evaluating a function, 282–283 evaluating determinants, 604 evaluating factorials, 726 exponential regression and, 433–434 finding domain and range of a rational function, 345 finding intercepts, 197–198 finding parabolic vertexes, 302 finding roots of a polynomial equation, 527 finding the intercepts of a graph, 239–240 finding the inverse of a matrix, 595 generating table values, 191 graphing a circle, 248 graphing a logarithmic function, 465 graphing a radical equation, 241 graphing a rational function, 354 graphing absolute value, 286 graphing an absolute value equation, 238 graphing circles, 656 graphing composite functions and domains, 375 graphing ellipses, 681–682 graphing equations, 195–196, 235, 243 graphing exponential functions, 419, 424 graphing hyperbolas, 696–697 graphing inequalities, 624 graphing logarithmic functions, 447 graphing operations on functions, 369–370 graphing parabolas, 663 graphing piecewise-defined functions, 318–319 graphing quadratic functions, 296 graphing rational functions, 360 graphing the cube root function, 286 linear regression and, 228 natural logarithms and, 442–443 permutations and, 764 reducing a matrix to row echelon form, 568–569 reducing a matrix to row reduced echelon form, 573

scientific notation and, 23 sequences and summation and, 737–738 simplifying imaginary numbers, 130 solving a system of equations using matrix inversion, 596 solving a system of linear equations, 552–553 solving equations, 248–250 solving population problems, 432 solving systems of equations, 703 translating functions, 330–331 using to find powers, 16 verifying logarithmic Product and Quotient Rules, 460 verifying solutions of an exponential equation, 472 Carbon-14 dating formula for, 496 problems in, 479–480 Carrying capacity, 613 Center of circle definition of, 243, 270 finding, 244 of ellipse, 673 of hyperbola, 688 Change-of-base formula, 464–465, 493 Charging battery, 453, 490 Cipher text, 591 Circle definition of, 243, 270 general form of equation of definition of, 245, 271 finding, given the center and radius, 245 finding, given the endpoints of its diameter, 246 graphing, 247–248, 656 standard form of equation of, 244, 270, 654–655, 710 writing equations of, 245, 654–656 Circular arrangement, 766, 784 Closed interval, 9, 74 Coefficient, 41 Coefficient matrix, 564, 643 Column operations, 604–606 Combination, computing using Pascal’s Triangle, 768 Combination formulas, 766, 784 Combination notation, 768 Combination problems, 766–768 Combined variation, 259–260 Common difference, 740 Common ratio, 747 Commutative Property for addition, 5, 74 matrices and, 585, 587 for multiplication, 5, 74 Completing the square, solving quadratic equations by, 108–111, 177 Complex conjugate, 133, 179, 513 Complex fraction, 70–71, 81 Complex number absolute value of, 136, 180 adding, 131, 179 definition of, 130, 179 dividing, 133–134, 179 equality of, 131, 179 multiplying, 132, 179 subtracting, 131, 179 Composite number, 3, 74 Compound inequality with absolute value, 170 definition of, 7 solving, 154–155, 182

Compound interest continuous, 421–422, 485–486 formulas for, 420, 421, 485–486 Compounding period, number of, 420 Computation, simplifying with scientific notation, 23 Conditional equation, 87 Conditional probability, 778 Conic section, 654. See also specific types. Conjugate binomial, 46, 79 Conjugate Pairs Theorem, 513–514, 540 Constant, in linear equation, 193 Constant function definition of, 309 identifying, 316 Constant k definition of, 257, 273 finding with discriminant, 115 Constant of proportionality. See Constant k. Constraint (in linear programming), 632 Contradiction (equation), 87, 174 Coordinate definition of, 6 plotting, 192 Correspondences definition of, 278 as functions, 278–279 relations as, 288 Cramer’s Rule definition of, 607, 647 solving systems of equations using, 607–609 Cube root definition of, 31 solving radical equations with, 143, 145 Cube root function, 285, 286 Cubing function, 285 Decibel, 451 Decibel voltage gain, 452, 490 Decimal changing repeating to a fraction, 750–751 rational numbers as, 4 Decomposition of fractions definition of, 613–614, 648 with denominator with distinct linear factors, 615 with denominator with distinct quadratic factors, 615–616 with denominator with repeated linear factors, 617 with denominator with repeated quadratic factors, 618–619 with numerator degree equal to or greater than denominator degree, 619 strategy for, 614 Decomposition of functions, 376 Decreasing function definition of, 419 exponential decay and, 429 identifying, 316–317 Degree of monomial, 41, 78 of polynomial, 41, 78 of polynomial functions, 309, 398 Denominator definition of, 63 rationalizing, 35–38, 46–47, 79 Dependent variable, 278 Depressed equation, 522–523, 524–525

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Index

Descartes’ Rule of Signs, 514–516, 540 Determinant. See Determinant function. Determinant function. See also Matrix. definition of, 600–601, 646–647 expanding by cofactors, 602 finding areas of triangles using, 609–610 of a higher-order matrix, 602 properties of, 604–606, 647 row and column operations and, 604–606 solving systems of equations using, 606–609 writing equations of lines using, 609 Difference of functions definition of, 366 process for finding, 406 of matrices, 580, 645 of two cubes, 58–59, 80 of two squares factoring, 54–55, 80 trinomial and, 59 Difference quotient definition of, 394 evaluating for a function, 283 Direct proportion definition of, 273 solving problems using, 257 Direct variation definition of, 273 solving problems using, 257 Directrix, 658, 711 Discriminant, 114 definition of, 177 determining roots with, 114–115 finding constant k with, 115 Distance formula definition of, 200, 264 finding, 199–200 Distributive Property formulas for, 5, 74 matrices and, 587 polynomials and, 41–44 verifying for matrix multiplication, 587–588 Dividend (polynomial), 48 Division Equality Property of, 87 property of fractions, 64 property of inequality, 151–152 Divisor (polynomial), 48 Domain of a composite function definition of, 371–372 finding, 374–376 definition of, 280 of an exponential function, 416–417 of a function definition of, 279, 366, 393 identifying, 285, 394 stating, 388–389 of an inverse function, 390 of a logarithmic function, 439 of a rational function, 344–345 specifying for functions, 366–369 Double Negative Rule, 5, 74 Eccentricity, elliptical, 682–683, 687 Elementary row operations, 566, 644 Ellipse circles and, 681 definition of, 671, 712 determining axis of, 680 eccentricity of, 682–683, 687 equations of with center at origin and horizontal on x-axis, 673–675, 712 with center at origin and vertical on y-axis, 675, 713 with center not at origin and horizontal on x-axis, 677, 713 with center not at origin and vertical on y-axis, 678, 714

standard form of, deriving, 673–679 writing, 673–679 graphing, 672, 679–682 properties of, 673 solving application problems with, 682–683 Encryption, 591 Epidemiology problems, 432–433 Equality properties of, 87, 174 property of fractions, 64 Equation. See also specific types. absolute value in, 166–168 of a circle general form of, 245, 270–271 standard form of, 270–271, 654–655 writing, 654–656 definition of, 86, 174 of ellipses with center at origin and horizontal on x-axis, 673–675 with center at origin and vertical on y-axis, 675 with center not at origin and horizontal on x-axis, 677 with center not at origin and vertical on y-axis, 678 graphing, 679–682 standard form of, deriving, 673–679 writing, 673–679 factoring, 141–143 as function, 278, 280 graphing with a calculator, 195–196, 243 to solve systems of equations, 551–553 strategy for, 236, 269 using intercepts and symmetry, 236–238, 240–242 using symmetry to, 235–236 of hyperbolas with center at origin and foci on x-axis, 690 with center at origin and foci on y-axis, 690 with center not at origin and foci parallel to x-axis, 691 with center not at origin and foci parallel to y-axis, 692 deriving standard form of, 688–689 non-horizontal or non-vertical, 697 of lines, 609 of a parabola finding, 659–661 in standard form, 657–662 of quadratic function, 297 second-degree, 654 solving using a graphing calculator, 248–250 strategy for modeling, 95 system of. See System of equations. types of, 87 Equivalent equation, 87 Equivalent inequality, 152 Even function, 315–316, 399–400 Event definition of, 762 experimental, 772 finding probability of, 773–776 probability of, 784–785 Excel spreadsheet solving linear regression problems with, 233 solving quadratic equations with, 118 Experiment definition of, 772 probability of, 773 showing sample space of, 773 Exponent. See also Rational exponents. definition of, 15 negative, 18

One-to-One Property of, 471, 494 Power Rules of, 17 Product Rule for, 17 Quotient Rule for, 19 rules of, 16–18, 76 zero, 18 Exponential decay, 419, 429 Exponential equation definition of, 471 finding approximate solutions to, 475 solving using like bases, 471–472 using logarithms, 473–474, 495 using natural logarithms, 474–475, 495 verifying solutions of using a calculator, 472 Exponential expression. See also Expression. approximating, 414–415 definition of, 15 simplifying, 414–416 Exponential form, logarithmic equivalent of, 439–440 Exponential function defining and graphing e and graph base-e, 421–424 definition of, 416, 484 determining the base of, 418 domain of, 416–417 graphing, 416–417 graphing using transformations, 422–424 graphing with a calculator, 419 properties of, 417, 418–419 range of, 416–417 Exponential growth, 418–419, 429 Exponential regression, 433–434 Expression. See also Algebraic expression; Exponential expression. evaluating using order of operations, 20 multiplying, 46 simplifying exponent rules and, 30 with integer exponents, 18–20 with natural-number exponents, 15 radical form of, 32–33 with rational exponents, 27–29, 30–31 Extraneous solution, 90, 142, 174 Extreme, in a proportion, 255, 272 Factor, 15, 53 Factor Theorem, 503–504, 537 Factorial notation definition of, 725, 779 expanding binomials with, 726 Factorial properties, 726 Factoring common monomial removal and, 53–54, 55, 80 definition of, 53 difference of two cubes, 58–59, 80 difference of two squares, 54–55, 80 grouping and, 54, 58 quadratic equations, 106–107 strategy for, 60 sum of two cubes, 58–59, 80 sum of two squares, 137–138 trinomial squares, 56 trinomials and, 80 Falling body problem, 122 Feasibility region, 633 Fibonacci sequence, 732 Finite arithmetic sequence, 740 Finite arithmetic series, 742 Finite geometric sequence, 747 Finite geometric series, 749 Finite sequence, 731, 779 First-degree equation, 88 First-degree polynomial equation, 88 First-degree polynomial function. See Linear function.

I1

Focal width, elliptical, 686 Focus elliptical, 673 of hyperbola, 688 parabolic, 658, 711 FOIL binomial multiplication and, 44, 78 polynomial multiplication and, 44 Formula definition of, 91 proving by induction, 755–759 solving for a specific variable, 91–92, 175 for a squared variable, 114 Fractal, 128, 140 Fractions. See also Complex fraction. changing repeating decimal to, 750–751 as decimals, 4 definition of, 63 linear equations and, 89 to negative powers, 20 partial decomposition of definition of, 613–614, 648 with denominator with distinct linear factors, 615 with denominator with distinct quadratic factors, 615–616 with denominator with repeated linear factors, 617 with denominator with repeated quadratic factors, 618–619 with numerator degree equal to or greater than denominator degree, 619 strategy for, 614 properties of, 64–65, 66, 80 as rational numbers, 3 Function. See also specific types. adding evaluating, 370 process for, 406 specifying domains and, 366–367 composite definition of, 371–372, 406–407 domain of, 407 evaluating, 371–374 finding the domain of, 374–376 composition of, 377–378 composition of two evaluating, 374 finding, 373–374 definition of, 278, 279, 393 dividing process for, 406 specifying domains and, 366–368 equation as representation of, 280 evaluating, 281–283, 394 finding difference quotient of, 283 finding domain of, 280–281 graph of, 394 graphing, 284–286 with inverse function, 388–389 using a combination of transformations, 337–339 using horizontal stretching and shrinking, 335–336 using horizontal translations, 328–329 using reflections about the x- and y- axes, 330–331 using two translations, 330 using vertical stretching and shrinking, 333–334 using vertical translations, 327–328 graphs of basic, 285 identifying, 286–288 multiplying process for, 406 specifying domains and, 366–368 restriction domain of, 388–389 subtracting process for, 406 specifying domains and, 366–367

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I2

Index

Function (continued) transforming graphs of applying, 337 overview, 326–327 summary of, 336 using a calculator, 330–331 using operations on, 368–370 writing as combinations of other functions, 370–371 writing as compositions, 376–377 Function notation, 281–282 Fundamental property of fractions, 64 Fundamental rectangle (hyperbolic), 694 Fundamental Theorem of Algebra, 511–512, 539 Future value, 420 Gaussian elimination definition of, 564 solving systems of equations using, 565–566 Gauss–Jordan elimination definition of, 571 matrices and, 593 solving systems of equations using, 572–574 General form of equation of a circle, 245, 710 of equation of a quadratic function, 299 Geology problems, 452–453 Geometric means definition of, 748, 782 finding, 748 inserting between integers, 748 Geometric problem, 96–97, 119–121 Geometric sequence definition of, 746–747, 781 finding sums of terms in, 750 finding terms of, 747–748 nth term of, 747 solving problems involving, 751–753 writing terms of, 747 Geometric series finding sums of terms in, 748–751, 782 solving problems involving, 751–753 Graph of absolute value, 286 of circles, 245–248, 656 of the cube root function, 286 finding x- and y-intercepts of, 234 of a function, 284, 394 of hyperbola, 693–697, 716 of inequalities, 621–625 of intervals on a number line, 7–10 of linear equations, 263 missing point on, 358–359 of nonlinear systems of equations, 702–703, 717 of numbers on a number line, 6–7 of parabolas, 662–663 of polynomial functions, 310–315 of quadratic function, 295–296 of rational functions, 345–347, 360 of second-degree equations, 654 of a slope, 206 symmetry and, 235–242 of systems of inequalities, 625–629, 649 vertical line test and, 286–287 Graphing calculator. See Calculator. Graphing method for solving systems of equations, 551–553, 641 Greatest-integer function determining, 400 evaluating and graphing, 319–320 Grouping polynomial factoring and, 54 trinomial factoring and, 58 Half-life, 429 Half-open interval, 9, 74 Harmonic series, 733

Horizon distance, 141 Horizontal asymptote definition of, 346 finding, 349–351, 404 Horizontal line test definition of, 383, 407 using, 383–384 Horizontal stretching, 333, 402 Horizontal translation, 328–329, 402 Hyperbola definition of, 687–688, 715 equations of with center at origin and foci on x-axis, 690, 715 with center at origin and foci on y-axis, 690, 715 with center not at origin and foci parallel to x-axis, 691, 716 with center not at origin and foci parallel to y-axis, 692, 716 non-horizontal or non-vertical, 697 writing, 688–693 graphing, 693–697, 716 strategy for graphing, 695 Hypotenuse, 106, 120 Identical Row or Column Theorem, 606 Identity (equation), 87, 174 Identity function graph of, 285 inverse functions and, 386 Identity matrix, 586–587 Imaginary number, simplifying, 129–130 Imaginary part of complex number, 130 Imaginary unit i definition of, 129 powers of, 134–135, 180 Increasing function definition of, 418–419 exponential growth and, 429 identifying, 316–317 Increment value, 191 Independent variable, 278 Index (in radical expressions), 31 Induction hypothesis, 757 Inequality. See also Linear inequality. graphing, 621–625 graphing on the number line, 7–10 graphing systems of, 625–629, 649 properties of, 150–152 real numbers and, 181 solving, 168–169, 170–171 types of, 182 Inequality symbols, 7, 150 Infinite arithmetic sequence, 740 Infinite arithmetic series, 742 Infinite geometric sequence, 747 Infinite geometric series, 748 Infinite sequence, 731, 779 Input value definition of, 190 equations and, 279 restrictions on for x, 280 Integer coefficient, factoring trinomials with, 56–57 Integers definition of, 3, 74 inserting geometric means between, 748 Intensity of light formula, 430, 487 Interest, calculating, 420 Intermediate Value Theorem, 531–533, 544 Intersection (interval), 2, 8 Interval notation, 8–10 Intervals. See also specific types. definition of, 8 graphing on the number line, 7–8 Inverse function definition of, 381–382, 385, 408 finding and graphing with function, 388–389

graph of, and one-to-one function graph, 388 of a one-to-one function, 386–387 strategy for finding, 408 verifying, 384–386 Inverse function of f, 384–385 Inverse of matrices, 592–597 Inverse proportion definition of, 273 solving problems using, 257–258 Inverse variation definition of, 273 solving problems using, 257–258 Investment problem, 97–98 Irrational numbers, 4, 74 Isothermal expansion formula for, 491 problems in, 455 Joint variation definition of, 274 solving problems using, 259 Key number, 58 Lead entry of a matrix, 566 Leading coefficient, factoring trinomials with, 56–57 Least common denominator (LCD) finding, 68–69 linear equations and, 89 Like radical, 34 Line determining type of defined by linear equations, 221–222 forms of equation of, 225 graphing. See also Linear equation. slope of horizontal, 210–211 slope of parallel, 211–212 slope of perpendicular, 212–214 slope of vertical, 210–211 horizontal equation of, 263, 268 graphing, 198–199 slope of, 266 parallel, slope of, 211–212, 266 perpendicular, slope of, 212–214, 266 slope of nonvertical, 206, 266 standard form of equation of, 223–224, 267 vertical equation of, 263, 268 graphing, 198–199 slope of, 266 writing equations of to describe parallel, 222–223 to describe perpendicular, 223 using point-slope form, 218–219 using slope-intercept form, 219–220 using the determinant, 609 Line segment, finding midpoint of, 200–201 Linear curve fitting, 226–228 Linear equation definition of, 174, 193 determining line type of, 221–222 graphing by finding intercepts, 196–197 standard form of, 192–195 using the slope and y-intercept, 220–221 in one variable, 88 solving, 88–90 solving application problems with, 175, 199, 225–226 standard form of, 193, 225, 267 Linear function definition of, 284, 289, 309, 394 modeling applications with, 288–290 Linear inequality definition of, 621–622 graphing, 621–624

solving, 152–153, 182 solving applications of, 153–154 Linear programming definition of, 632 solving applications of, 635–639 solving problems of, 632–635, 649 Linear regression, 228 Local maxima, 313 Local minima, 313 Location Theorem, 531–533, 544 Logarithm base-10, 441 common, 441, 488 evaluating, 438–442 examples of values of, 440 Napierian, 442 natural definition of, 488 evaluating, 442–443 properties of, 492 solving exponential equations using, 474–475 One-to-One Property of definition of, 461 solving logarithmic equations using, 476, 495 Power Rule of, 461 Product Rule for, 459–460 properties of approximating with, 463 Power Rule as, 461 Product and Quotient Rules as, 459–460 simplifying expressions using, 458–459 solving logarithmic equations using, 476–478 summary of, 463, 492 Quotient Rule for, 459–460 solving exponential equations using, 473–474 undefined, 439 Logarithmic equation approximating solutions to using a calculator, 479 definition of, 471 finding unknown terms in, 440 solving summary of strategies used, 478–479 using logarithmic properties, 476, 496 using the One-to-One Property, 476, 495 Logarithmic expression combining, 462 condensing, 493 expanding, 461, 493 Logarithmic form, exponential equivalent of, 439–440 Logarithmic function definition of, 439, 488 domain of, 439, 488 graphing, 443–444 graphing common, 444 graphing natural, 445 graphing with a calculator, 465 graphing with transformations, 445–447 introduction to, 437–438 range of, 439, 488 Logistic epidemiology model, 432 Logistic function, 432 Lowest terms, rational expressions in, 65 Lumen, 430 Malthusian model of population growth, 431, 487 Malthusian population growth problems, 431–432 Mathematical model, 95 Matrix. See also specific types. addition of, 577–579, 585, 645 additive inverse of, 580

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Index

cofactor of, 601–602 definition of, 564, 578 equality of, 578, 644 examples of, 567, 572 higher order, 601–604 inverse of definition of, 592, 645 finding using row operations, 592–595, 645 solving application problems using, 597 solving systems of equations using, 595–596 strategy for finding, 593 invertible, 592, 646 minor of, 601–602 multiplication of, 578, 645 multiplying, 581–585 multiplying by a constant, 580 nonsingular, 592 non-square, inverse of, 595 operations of, 566, 644 producing in reduced row echelon form, 571–574 real number properties and, 587 reducing to row echelon form, 568–569 row and column operations and determinants and, 604–606 singular, 592 solving application problems with, 586 solving inconsistent systems of equations using, 570 solving systems of dependent equations using, 570–571 square finding determinants of, 601–604 inverse of, 595 multiplicative inverses and, 593 subtraction of, 580, 645 Matrix equation, 581 Matrix of constants, 564, 643 Mean, in a proportion, 255, 272 Midpoint formula, 200, 264 Midpoint of a line segment, 200–201 Mixture problem, 99–100 Model, mathematical description as, 429 Monomial definition of, 41, 78 dividing, 48 factoring out common, 53–54, 55, 80 multiplying, 43, 78 Multiplication Associative Property of, 5 Commutative Property of, 5 Equality Property of, 87 property of fractions, 64 property of inequality, 151–152 Multiplication Principle for Events, 761–762, 783 Multiplication Property of Probabilities, 774–775, 785 Multiplicative inverse, 591 Natural numbers, 3, 74 Natural-number exponents, 15 definition of, 76 Negative exponents, 18 Negative numbers, 6 Negative powers, 20 Negative reciprocal, 212 Nth root of nonnegative number, 31–33 Number line, 6–8, 12 Number of Roots Theorem, 512 Number problem, 96 Numbers. See also specific types. Numbers, relationships of sets of, 5 Numerator definition of, 63 rationalizing, 35–37, 47 Objective function definition of, 632 finding maximum value of, 633–634

finding minimum value of, 634–635 maximum or minimum of, 633 Odd function, 315–316, 399–400 Ohm’s Law, 139 One-to-one correspondence, 6 One-to-one function creating by restricting function domain, 388–389 definition of, 382, 407 determining, 382–383 finding inverse of, 386–387 graph of, and inverse function graph, 388 properties of, 385–386, 408 Open interval definition of, 9, 74 identifying, 316 Order combinations and, 766 permutations and, 766 in radical expressions, 31 Order of operations, 20–21 Ordered pair, 190 Origin (rectangular coordinate system) definition of, 192 test for symmetry of, 236, 270 Output value definition of, 190 equations and, 279 Parabola. See also Quadratic function. definition of, 235, 658, 711 finding equations of, 659–661 finding equations of two, 661 finding vertex of with general form equation, 299–302 with standard form equation, 296–297 as graph of quadratic function, 295–296, 395 graphing, 662–663 solving application problems involving, 663–666 standard equations of, 658–659, 660 vertex of, 300 writing equations of, 657–662 Partial fraction decomposition definition of, 613–614, 648 with denominator with distinct linear factors, 615 with denominator with distinct quadratic factors, 615–616 with denominator with repeated linear factors, 617 with denominator with repeated quadratic factors, 618–619 with numerator degree equal to or greater than denominator degree, 619 strategy for, 614 Pascal’s Triangle computing a combination using, 768 definition of, 778 expanding binomials with, 724–725 PEMDAS (order of operations), 21 Perfect cube, 34 Perfect power, 34 Perfect square, 34 Perimeter, 96 Periodic interest rate, 420 Permutation, 768–769, 784 Permutation formulas, 763–764, 784 Permutation problem, 762–765 pH of a solution, 465, 493 pH problems, 465–466, 469 Piecewise-defined function definition of, 400 graphing, 317–318 Plain text, 599 Plotting, 7 Point distance between two finding on a number line, 12, 75 finding the formula for, 199–200

graphing a function by plotting, 284–286 missing, on graphs, 358–359 on a number line, 6 Point-slope form, 218–219, 225, 267 Point-slope formula, 218 Polynomial adding, 42–43, 78 definition of, 41 Distributive Property and, 41 dividing, 48–50, 79 dividing with synthetic division, 504–506 like terms of, 41 multiplying, 43–46, 78 similar terms of, 41 subtracting, 42–43, 78 Polynomial equation approximating roots of, 545 confirming roots of, 523, 526 definition of, 500–501, 537 factoring, 141–142 finding all solutions of, 523–524 finding rational roots of, 520–523 finding roots of, 527 with real coefficients, 513–514 solving with synthetic division, 507–508, 538–539 Polynomial Factorization Theorem, 511–512, 540, 616 Polynomial function. See also specific types. applications for, 324–325 basic, 309 characteristics of, 310–312 degree of, 309, 398 evaluating using synthetic division, 506–507 graphing, 312–315, 399 long division and, 501–502 in one variable, 309, 398 Population doubling time, 454, 491 Population growth model, 431 Population growth problems, 454, 481 Positive numbers, 6 Power of exponential expressions, 15 finding with calculators, 16 Power over root, 30 Power Rules of exponents, 17 Prediction equation, 227 Present value, 420 Prime number, 3, 74 Prime polynomial, 53 Principle of Mathematical Induction definition of, 755–756, 783 proving formulas using, 757–759 Probability, 772–774, 784 Product of functions, 366, 406 of matrices, 583, 645 Product Rule for exponents, 17 Proportion definition of, 255, 272 property of, 255, 272 solving, 255–256 Pythagorean Theorem, 105, 120 Quadrant, 192 Quadratic equation choosing easiest solution strategy for, 113 with complex roots, 137 definition of, 106, 176 with a double root, 514 general form of second-degree, 710 solving, 106–113 solving application problems with, 178 types of, 113–114 Quadratic formula definition of, 112, 177 solving quadratic equations with, 111–113

I3

squared variables in formulas and, 114 Quadratic function. See also Parabola. applications for, 306–308 characteristics of, 296 definition of, 294, 309, 395 finding vertex of with general form equation, 299–302 with standard form equation, 296–297 graphing by plotting points, 295 process for, 396–397 standard form of, 297–299 using a calculator, 296 solving maximum problems with, 303 solving minimum problems with, 304 standard form of an equation of, 297 standard form of equation of, 396 Quadratic inequality, 155–159, 182 Quotient definition of, 48 of functions, 366, 406 Quotient Rule for exponents, 19 Radical conjugate, 46 Radical equation definition of, 141 graphing with a calculator, 241 solving, 143–144, 181 solving applications of, 146–147 Radical expression adding, 35 definition of, 31, 32, 77 division properties of, 33 imaginary numbers and, 129 multiplication properties of, 33 multiplying, 35 properties of, 77 rationalizing denominators of, 36–37, 47, 79 rationalizing numerators of, 36, 47 simplifying and combining, 33–35 rationalizing denominators and, 37–38 strategy for, 32–33 Radical sign, 31 Radicand, 31 Radioactive decay formula for, 430, 486 problems in, 429–430 Radius definition of, 243, 270 finding, 244 Range of an exponential function, 416–417 of a function definition of, 279, 393 identifying, 285, 394 stating, 388–390 of a logarithmic function, 439 of a rational function, 345 Ratio, 255 Rational equation definition of, 90, 174 solving, 90–91 writing in quadratic form and solving, 115–116 Rational exponents definition of with exponents of 1, 27–28 with exponents that are not 1, 29–31 summary of, 77 negative, 30 rule for, 29, 77 Rational expression adding, 68–79, 81 definition of, 63 dividing, 67

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

I4

Index

Rational expression (continued) multiplying, 66–67 simplifying, 65–67, 69, 81 subtracting, 68, 81 Rational function characteristics of, 345–347 definition of, 344, 403 finding horizontal asymptotes of, 349–351 finding the domain of, 344 finding vertical asymptotes of, 347–348 graph of, 346–347 graphing, 352–358, 405 graphing with a missing point, 358–359 graphs of, 360 solving problems modeled by, 359–360 Rational inequality, 159–161, 182 Rational numbers, 3, 74 Rational Root Theorem, 520–521, 542 Real numbers definition of, 4, 74 graphing on the number line, 6 identifying sets of, 3 inequalities and, 181 inserting arithmetic means between, 741–742 Power Property of, 142, 180 properties of, 5 Real part of complex number, 130 Reciprocal function, 346 Rectangular coordinate system, 192, 263 Reduced row echelon form definition of, 571 examples of, 572 Reflection, graphing functions with, 331–333, 402 Regression equation, 227 Relations in correspondences, 288 Remainder Theorem, 501–503, 537 Restriction (variable), 86–87 Richter scale, 452, 490 Right triangle problem, 120–121 Rise (slope), 208 Root approximating real, 534–535 complex, 137 definition of, 86 discriminant and, 114–115 integer bounds on, 516–518, 541 irrational, 531 nonreal, 515–518 of polynomial equations confirming roots of, 523, 526 finding, 523–525, 527 finding rational, 520–523 strategy for finding rational, 521–522 rational, 520–526, 542–543 Root of multiplicity k, 512 Roster method, 3 Row echelon form definition of, 566, 644 examples of, 567 reduced, 571–574 writing matrices with, 567–569 Row equivalent, 566 Row operations determinants and, 604–606 writing matrices with, 566–569 Run (slope), 208 Sample space definition of, 772 showing, 773 Scalar definition of, 580 matrices and real number properties and, 587 multiplying a matrix by, 581 Scalar multiple, 581 Scattergram, 227

Scientific notation definition of, 76 simplifying computations with, 23 writing numbers in, 22 Secant line, 702 Second-degree equation. See Quadratic equation. Second-degree polynomial function. See Quadratic function. Sequence. See also specific types. defining recursively, 732–733, 780 definition of, 731, 779 function notation and, 732 solving problems involving, 743–744 Series (sequential) definition of, 733, 779 solving problems involving, 743–744 Set of composite numbers, 3 definition of, 2 of integers, 3 of prime numbers, 3 Set-builder notation, 3 Setup costs, 98 Shared-work problem, 98–99, 124–125 Sign graph quadratic inequalities and, 155–158 rational inequalities and, 159–161 Signum function, 325–326 Similar radical, 34 Simultaneous solution, 551 Slant asymptote, 351–352, 404 Slope definition of, 206 finding given its equation in standard form, 208 given two points, 207 of horizontal lines, 210–211 of parallel lines, 211–212, 266 of perpendicular lines, 212–214, 266 using formulas, 224–225 of vertical lines, 210–211 of horizontal lines, 210, 266 of nonvertical line, 206, 266 solving application problems with, 208–210 of vertical lines, 210, 266 Slope-intercept form, 219, 225, 267 Solution definition of, 86 of a system of equations, 551 Solution set, 86 Solve (equation), 86 Solving the system, 551 Special product formulas, 44, 78 Square root definition of, 31 solving radical equations with, 143–144 Square root function, 285 Square Root Property definition of, 107, 177 solving quadratic equations with, 107–108, 111 Squaring function, 285 Standard form of equation of a circle, 244, 710 of equation of a quadratic function, 297 of linear equation, 193, 225, 267 Standard notation, 22–23 Step function, 320 Subscript notation, 199–200 Subset definition of, 2 graphing on the number line, 6 Substitution equation, 704 Substitution method of solving systems of equations, 553–554, 642, 718 Substitution Property, 87

Subtraction Equality Property of, 87 property of fractions, 64, 87 property of inequality, 151 Sum of functions definition of, 366 evaluating, 370 process for finding, 406 of matrices, 579, 645 partial, of sequences definition of, 733 evaluating, 734, 736–738 of two cubes, 58–59, 80 of two squares, 137–138 Summation notation, 733–734, 780 Summation of a Constant Property, 734–735 Summation of a Product Property, 735 Summation of a Sum Property, 736 Supply and demand, 550 Symmetry graphing an equation using, 240–242 tests for, 236, 269–270 using to graph equations, 235–238 Synthetic division definition of, 538–539 evaluating polynomial functions using, 506–507 polynomials and, 504–506 solving polynomial equations with, 507–508 System of equations consistency of, 552 definition of, 550–551 dependence of equations in, 552 inconsistency of, 552 independence of equations in, 552 solution of, 551 solving with infinitely many solutions, 556 involving three equations in three variables, 557–558 using determinants, 606–609 using Gauss–Jordan elimination, 571–574 using graphing, 551–553, 641 using matrix inversion, 595–596 using nonlinear, 706–707 using the addition method, 554–555, 642 using the substitution method, 553–554, 642 solving application problems involving, 559 solving dependent using matrix methods, 570–571 solving inconsistent, 556–557 solving inconsistent using matrix methods, 570 solving nonlinear by addition, 705–706, 718 by graphing, 702–703, 717 by substitution, 703–705, 706–707, 718 Systems matrix definition of, 564–565, 643 writing in row echelon form, 567–569, 644 Table of solutions, 191 Tangent line, 702 Term (sequential), 732 Test number quadratic inequalities and, 155–158 rational inequalities and, 159–161 Tower of Hanoi, 760–761 Transformation (graph). See Translation (graph). Transitive Property, 151, 181 Translation (graph) applying, 337 definition of, 327

exponential functions and, 422–424 graphing exponential functions using, 422–423 horizontal, 328–329 summary of, 336 vertical, 327–328 Triangle, 609–610 Trichotomy Property, 150, 181 Trinomial definition of, 41, 78 difference of two squares and, 59 factoring, 56–58, 59–60, 80 factoring with integer coefficients, 56–57 Turning point (graphing), 313 Unbounded interval, 8 Uniform motion problem, 100–101, 121–122 Union, 2 Unit cost, 98 Variable definition of, 86 solving formulas for, 91–92 solving formulas for squared, 114 Vertex definition of, 235, 296, 397 formula for, 300 of hyperbola, 688 parabolic definition of, 658 finding with a graphing calculator, 302 finding with general form equation, 299–302 finding with standard form equation, 296–297 Vertical asymptote definition of, 346 finding, 347–348, 404 Vertical line test, 286–288, 394 Vertical stretching, 333, 402 Vertical translation, 327–328, 401 Voltage gain, 466–467 Weber–Fechner Law, 467, 493 Weighting, in calculations, 581–583 Whole numbers, 3, 74 Word problem, 95 x-axis definition of, 192 reflections about, 331–333 test for symmetry of, 236, 269 x-coordinate, 190 x-intercept definition of, 196, 263 finding, 234–235, 269 graphing an equation using, 236–238, 240–242 y-axis definition of, 192 reflections about, 331–333 test for symmetry of, 236, 269 y-coordinate, 190 y-intercept definition of, 196, 263 finding, 224–225, 269 of a graph, 234–235 graphing an equation using, 236–238, 240–242 Zero exponent, 18 Zero matrix. See Additive identity matrix. Zero of a polynomial function, 500–501, 537 Zero of multiplicity 2, 508 Zero polynomial, 41 Zero Row or Column Theorem, 603 Zero-Factor Theorem, 106–107, 177

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Properties of Logarithms

Parabolas

If b is a positive number and b 2 1,

A parabola is the set of all points in a plane equidistant from a line l (called the directrix) and fixed point F (called the focus) that is not on line l.

1. logb 1 5 0 2. logb b 5 1 3. logb bx 5 x 4. b

logb x

These distances are equal.

5x

5. Product Rule: logb MN 5 logb M 1 logb N

(x, y) Vertex

6. Quotient Rule: M logb 5 logb M 2 logb N N 7. Power Rule: logb Mp 5 p logb M

Axis of the parabola

Directrix

Focus

Parabola

8. One-to-One Property: If logb x 5 logb y, then x 5 y.

Graphs of f 1x2 5 ex and f 1x2 5 ln x y

y

Parabola opening

Vertex at origin

Right Left Upward Downward

y2 5 4px y2 5 4px x2 5 4py x2 5 4py

x f(x) = e x

f (x) = ln x x

Natural Logarithm Properties 1. ln 1 5 0 2. ln e 5 1 3. ln ex 5 x 4. eln x 5 x

Theorems Remainder Theorem: If P 1x2 is a polynomial, r is any number, and P 1x2 is divided by x 2 r, the remainder isP 1r2 . Factor Theorem: If P 1x2 is a polynomial and r is any number, then If P 1r2 5 0, then x 2 r is a factor of P 1x2 . If x 2 r is a factor of P 1x2 , then P 1r2 5 0.

Binomial Theorem: If n is any positive integer, then

n! 1a 1 b2 n 5 an 1 an21b 1! 1n 2 12 ! 1

1

n! n! an22b2 1 an23b3 2! 1n 2 22 ! 3! 1n 2 32 !

1 p . 02 1 p , 02 1 p . 02 1 p , 02

• For a parabola that opens right or left with vertex at the origin, the directrix is x 5 2p and the focus is 1 p, 02 . • For a parabola that opens upward or downward with vertex at the origin, the directrix is y 5 2p and the focus is 10, p2 . Parabola opening

Vertex at V(h, k) 1y 2 k2 2 5 1y 2 k2 2 5 1x 2 h2 2 5 1x 2 h2 2 5

Right Left Upward Downward

4p 1x 2 h2 4p 1x 2 h2 4p 1y 2 k2 4p 1y 2 k2

1p 1p 1p 1p

. 02 , 02 . 02 , 02

• For a parabola that opens right or left with vertex at 1h, k2 , the directrix is x 5 2p 1 h and the focus is 1h 1 p, k2 . • For a parabola that opens upward or downward with vertex at 1h, k2 , the directrix is y 5 2p 1 k and the focus is 1h, k 1 p2 .

Ellipses An ellipse is the set of all points P in a plane such that the sum of the distances from P to two other fixed points F and F r is a positive constant. The standard equations of an ellipse with center 1h, k2 and major axis horizontal is 1y 2 k2 2 1x 2 h2 2 1 5 1, a2 b2

where a . b . 0.

n! an2rbr 1 c 1 bn r! 1n 2 r2 !

y F'(h – c, k) B(h, b + k) V'(–a + h, k)

b (h, k)

a

F(h + c, k) V(a + h, k)

B'(h, –b + k)

x

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

The standard equations of an ellipse with center 1h, k2 and major axis vertical is 1y 2 k2 1x 2 h2 1 5 1, b2 a2 2

2

where a . b . 0.

Permutations and Combinations

n! ; P 1n, n2 5 n!; P 1n, 02 5 1 1n 2 r2 ! n! C 1n, r2 5 ; C 1n, n2 5 1; C 1n, 02 5 1 r! 1n 2 r2 ! P 1n, r2 5

Geometry Formulas

y

Rectangle: A 5 l ? w; P 5 2l 1 2w

V(h, a + k) F(h, k + c)

a b B'(–b + h, k)

Width = w

B(b + h, k)

(h, k) F'(h, k – c) V'(h, –a + k)

Length = l x

1 Triangle: A 5 b ? h 2

Hyberbolas A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from point P to two other points in the plane is a positive constant.

a

The standard equation of a hyperbola with center at 1h, k2 and foci on a line parallel to the x-axis is 1y 2 k2 2 1x 2 h2 2 2 5 1, 2 a b2

c

Height = h Base = b

Circle: C 5 2pr; A 5 pr2

where a2 1 b2 5 c2 .

r y

V'(–a + h, k) C(h, k)

F'(–c + h, k)

V(a + h, k)

Rectangular solid: V 5 l ? w ? h

F(c + h, k) x

Height = h

The standard equation of a hyperbola with center at 1h, k2 and foci on a line parallel to the y-axis is 1y 2 h2 2 1x 2 k2 2 2 5 1, a2 b2

Width = w Length = l

Sphere: V 5

4 3 pr 3

where a2 1 b2 5 c2 . r y F(h, c + k)

V(h, a + k)

Right circular cone: V 5

C(h, k)

1 2 pr h 3

Height = h

V'(h, – a + k) x r F'(h, –c + k)

Pythagorean Theorem a 2 1 b 2 5 c2

Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.