Mathematics for the IB Diploma Applications and Interpretation SL 2 Worked Solutions [1 ed.] 9781925489842

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HAESE

MATHEMATICS

Mathematics

Application'sfand [nterpretation§SIE

Michael Haese

Mark{Humphries

Chris'Sangwin Ngoc Vo

with

[BiDiplomallRrogramme Bradley Steventon L CTIA'L

Joseph Small Michael Mampusti

SNOILLNTOS aaNd - [e]))

%

MATHEMATICS: APPLICATIONS AND INTERPRETATION Bradley Steventon Joseph Small Ngoc Vo Michael Mampusti

SL WORKED

SOLUTIONS

B.Ma.Sc. B.Ma.Sc. B.Ma.Sc. B.Ma.Adv.(Hons.), Ph.D.

Haese Mathematics 152 Richmond Road, Marleston,

SA 5033, AUSTRALIA

Telephone: +61 8 8210 4666, Fax: +61 8 8354 1238 Email: [email protected] Web:

www.haesemathematics.com

National Library of Australia Card Number & ISBN

978-1-925489-84-2

© Haese & Harris Publications 2020

Published by Haese Mathematics. 152 Richmond Road, Marleston, First Edition

SA 5033, AUSTRALIA

2020

Artwork by Brian Houston. Typeset in Australia by Charlotte Frost. Typeset in Times Roman 10. This book 1s available on Snowflake only. This book has been developed independently of the International Baccalaureate Organization (IBO). This book 1s in no way connected with, or endorsed by, the IBO. This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, research, criticism or review), no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher. Enquiries to be made to Haese Mathematics. Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers 1t has given a remuneration notice to Copyright Agency Limited (CAL). For information, contact the Copyright Agency Limited. Acknowledgements: While every attempt has been made to trace and acknowledge copyright, the authors and publishers apologise for any accidental infringement where copyright has proved untraceable. They would be pleased to come to a suitable agreement with the rightful owner. Disclaimer: All the internet addresses (URLs) given 1n this book were valid at the time of publication. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.

FOREWORD This book gives you fully worked solutions for every question in Exercises, Review Sets, Activities, and Investigations (which do not involve student experimentation) 1in each chapter of our textbook Mathematics: Applications and Interpretation SL. Correct answers can sometimes be obtained by different methods. In this book, where applicable, each worked solution 1s modelled on the worked example 1n the textbook. Be aware of the limitations of calculators and computer modelling packages. Understand that when your calculator gives an answer that is different from the answer you find in the book, you have not necessarily made a mistake, but the book may not be wrong either. Please contact us 1f you notice any errors 1n this book.

BS

e-mail:

[email protected]

web:

www.haesemathematics.com

JS

NV

MM

TABLE OF CONTENTS

Chapter 1 Chapter 2

Chapter 3 Chapter 4

Chapter 5 Chapter 6

Chapter 7 Chapter 8 Chapter 9

Chapter 10 Chapter 11

Chapter 12 Chapter 13 Chapter 14

Chapter 15 Chapter 16

Chapter 17

APPROXIMATIONS AND ERROR LOANS AND ANNUITIES FUNCTIONS MODELLING BIVARIATE STATISTICS QUADRATIC FUNCTIONS DIRECT AND INVERSE VARIATION EXPONENTIALS AND LOGARITHMS TRIGONOMETRIC FUNCTIONS DIFFERENTIATION PROPERTIES OF CURVES APPLICATIONS OF DIFFERENTIATION INTEGRATION DISCRETE RANDOM VARIABLES THE NORMAL DISTRIBUTION HYPOTHESIS TESTING VORONOI DIAGRAMS

25 51 105 133 180 252 278 334 374 420 468 505 551 582 623 660

Chapter 1

o O

81 ~ 80

aAa

852~ 90

O

128 ~ 130

{round up, as 8 is greater than 5}

162 =~ 160 104 ~ 100

{round down, as 2 is less than 5} {round down, as 4 is less than 5}

635 =~ 640

{5 is rounded up}

1822 ~ 1820

-

®

86 ~ 90

699

{round up, as 6 is greater than 5}

{round down, as 1 is less than 5} {5 is rounded up}

ERROR

W

AND

S

APPROXIMATIONS

=~ 700

{round down, as 2 is less than 5} {round up, as 9 is greater than 5}

{5 is rounded up}

O ®

26341 ~ 26300 8365 ~ 8000

{round down, as 3 is less than 5}

3500 ~ 4000

{5 is rounded up}

{5 is rounded up} {round down, as 4 is less than 5}

o O

31722 km? ~ 32000 km?*

A

4749 598 people ~ 4 750 000 people

O

85512 people =~ 85500 people

7692000 km? ~ 7700000 km? 5443 kg ~ 5400 kg

S5

W

8848 m =~ 8850 m

00

19210 ~ 19000 {round down, as 2 is less than 5} 19650 ~ 20000 {round up, as 6 is greater than 5} 114823 ~ 115000 {round up, as 8 is greater than 5}

-

®

O

A

3950 ~ 4000

o

215~ 200 {round down, as 1 is less than 5} 264 ~ 300 {round up, as 6 is greater than 5} 3750 ~ 3800 {5 is rounded up}

O

A

O

o

3045 ~ 3050

{round up, as 8 is greater than 5} {round up, as 7 is greater than 5} {5 is rounded up}

{round down, as 1 is less than 5}

{round up, as 9 is greater than 5}

{round down, as 4 is less than 5}

16950 km ~ 17000 km 384 400 km = 400000 km

{5 is rounded up} {round up, as 8 is greater than 5}

428 240515 people ~ 428 000 000 people

{round down, as 2 is less than 5}

Chapter 1 (Approximations and error)

6.181 ~ 6.2

Exercise 1A.2

{round up, as 8 > 5}

-

0.1940 = 0.2

W

6.181 ~ 6.18 {round down, as 1 < 5} 3.25 ~ 3.3 {5 is rounded up} 17.403 ~ 17.40 {round down, as 3 < 5} 2.13158 ~ 2.132 {5 is rounded up} 0.0972 ~ 0.10

5

®

O

A

O

1

o

6

{round up, as

102.382 ~ 102.38

9 > 5}

{round up, as

{round down, as 2 < 5}

2

9.58 seconds ~ 9.6 seconds

3

1435 m=~

4

0.012cm =~ 0.0l cm

144 m

7 > 5}

{round up, as 8 > 5}

{5 is rounded up} {round down, as 2 < 5}

5

a b ¢

3.14159~3.1 {round down, as 4 < 5} 3.14159 ~ 3.142 {5 is rounded up} 3.14159 ~ 3.1416 {round up, as 9 > 5}

6

a

0.263157895~0.3

b

0.263157895 ~ 0.26

¢

0.263157895 ~ 0.263158

7

{roundup,as

{round down, as

{round down, as

¢ V23~ 4.795831.... ~ 4.796

e

v/—15~

{round up, as

a

2 < 5} 8 > 5}

b V5~ 2.236067.... ~ 2.236

~ 1.587

f

{round down, as 4 < 5}

~ 7.663 b

{round down, as 0 < 5}

16.8+12.4 x17.1 =16.8+ 212.04

= 499.32

¢ 127+9-5="2 -5

= 228.84

d 127+(9—5)=%7

~ 9.11

e

37.4-16.1+ (4.2 —2.7) =374

—16.1

1.5

= 31.75

¢

2084

7.9+ 11.2

~ 26.67

214 186 741 3.2 16.1

0 < 5}

V450 ~ 7.663094 ....

{round down, as 2 < 5}

(16.84124)x17.1=29.2x17.1

{round down, as

d /4~ 1.587401....

—2.466212....

~ —2.466 8

3 < 5}

{round up, as 8 > 5}

a V2~1414213... ~ 1.414

6 > 5}

_ 1084 19.1 ~ 0.8%

o

h 22170 8.6

=00y 8.6 ~ 5.93

Chapter 1 (Approximations and error) 0.0768 + 7.1

Exercise 1A.3

7

7.1768

18.69 — 3.824

14.866 ~ (.48

40 goals

23 games

~ 1.74 goals per game

Kerry scored an average of 1.74 goals per game. 10

While 2.45 m =~ 2.5 m to the nearest integer, so

a

128 ~ 130

T

1is correct, Wang should have used the original value of 2.45 m to round 2.45 m =~ 2 m.

{2 significant figures}

This 1s the second significant figure, so we look at the next digit which is 8. The 8 tells us to round the 2 up to a 3, so we convert the 8 into 0.

8342 ~ 8300

{2 significant figures}

T

This 1s the second significant figure, so we look at the next digit which is 4. The 4 tells us to keep the original 3, so we convert the 42 into 00.

2.568 ~ 2.6

T

{2 significant figures}

This 1s the second significant figure, so we look at the next digit which is 6. The 6 tells us to round the 5 up to a 6 and delete the remaining digits.

0.0134 ~ 0.013

{2 significant figures}

T

These zeros at the front are place holders and so must stay. The first significant figure is 1, and the second significant figure is 3. The 4 tells us to leave the 3 as 1t 1s and delete the remaining digits.

163870 ~ 160000

T

{2 significant figures}

This 1s the second significant figure, so we look at the next digit which is 3. The 3 tells us to keep the original 6, so we convert the 3870 into 0000.

1.086 ~ 1.1

T

{2 significant figures}

This 1s the second significant figure, as 1t lies between two non-zero digits. The 8 tells us to round the O up to a 1 and delete the remaining digits.

3958 &~ 4000

T

{2 significant figures}

This 1s the second significant figure, so we look at the next digit which is 5. The 5 tells us to round the 39 to 40, so we convert the 58 into 00.

6.611 =~ 6.6

T

{2 significant figures}

This 1s the second significant figure, so we look at the next digit which is 1. The 1 tells us to leave the 6 as it 1s and delete the remaining digits.

8

Chapter 1 (Approximations and error)

a

83064 ~ 83100

Exercise 1A.3

{3 significant figures}

T

This 1s the third significant figure, so we look at the next digit which 1s 6. The 6 tells us to round the O to 1, so we convert the 64 into 00.

10044 ~ 10000

{3 significant figures}

!

This 1s the third significant figure, so we look at the next digit which 1s 4. The 4 tells us to keep the original 0, so we convert the 44 into 00.

0.10526 ~ 0.105

{3 significant figures}

T

This 1s the third significant figure, so we look at the next digit which is 2. The 2 tells us to leave the 5 as it is and delete the remaining digits.

31.695 ~ 31.7

T

{3 significant figures}

This 1s the third significant figure, so we look at the next digit which 1s 9. The 9 tells us to round the 6 up to a 7, and delete the remaining digits.

70.707 ~ 70.7

T

{3 significant figures}

This 1s the third significant figure, so we look at the next digit which i1s 0. The O tells us to leave the 7 as it 1s and delete the remaining digits.

4.0007 =~ 4.00

T

{3 significant figures}

This 1s the third significant figure, so we look at the next digit which 1s 0. The O tells us to leave the original O as it is and delete the remaining digits. 0.036 71 ~ 0.0367

{3 significant figures}

T

This 1s the third significant figure, so we look at the next digit which 1s 1. The 1 tells us to leave the 7 as 1t 1s and delete the remaining digits.

19.989 ~ 20.0

T

{3 significant figures}

This 1s the third significant figure, so we look at the next digit which is 8. The 8 tells us to round the 19.9 to 20.0 and delete the remaining digits.

16.382 ~ 16.38

1

{4 significant figures}

This 1s the fourth significant figure, so we look at the next digit which 1s 2. The 2 tells us to leave the 8 as it is and delete the remaining digits.

438.207 ~ 438.2

!

{4 significant figures}

This 1s the fourth significant figure, so we look at the next digit which 1s 0. The O tells us to leave the 2 as it 1s and delete the remaining digits.

6873681 ~ 6874000

T

{4 significant figures}

This 1s the fourth significant figure, so we look at the next digit which 1s 6. The 6 tells us to round the 3 up to a 4, so we convert the 681 into 000.

Chapter 1 (Approximations and error)

d

0.028885 =~ 0.02889

Exercise 1A.3

{4 significant figures}

T

This 1s the fourth significant figure, so we look at the next digit which 1s 5. The 5 tells us to round the 8 up to a 9 and delete the remaining digits.

L

a

96257 people ~ 100000 people

{1 significant figure}

T

first significant figure, next digit 1s 6, so we round up

b

96257 people ~ 96 000 people

T

{2 significant figures}

second significant figure, next digit 1s 2, so we round down

¢

96257 people ~ 96 300 people

T

{3 significant figures}

third significant figure, next digit is 5, so we round up

5

a

V7~2645751.... ~ 2.65 {3 significant figures}

¢

36+-17~2.117647.... ~ 2.12 {3 significant figures}

d

517 x 3802 = 1965634

~ 1970000 e

b

27 ~6.283185.... ~ 6.28 {3 significant figures}

{3 significant figures}

(0.986)° ~0.931932.... . ~ 0.932 {3 significant figures}

¢

163268 3.1

13.62 3.1

~ 4.393 548

~ 4.39

g

v5.4—218 =+/3.22 ~ 1.794435 ....

~ 1.79

{3 significant figures}

h 298 ~5795145.... V2.8

6

~ 5.73

{3 significant figures}

32 rows x 28 seats = 896 seats

~ 900 seats

{2 significant figures}

there are about 900 seats 1n the theatre.

7

30.1 m x 8.5 m = 255.85 m?

~ 256 m*

{3 significant figures}

the area of the ballroom is about 256 m?Z.

8

$752.25

= $188.0625

each person receives $188.0625

a

$188.0625 ~ $188

b

§$188.0625 ~ $188.06

{3 significant figures} {5 significant figures}

{3 significant figures}

9

10 9

Chapter 1 (Approximations and error)

Exercise 1B

1 hour = 60 min — 60 x 60 = 3600s

-

in one hour, sound will travel

343 ms~!

x 3600 s = 1234800 m

~ 1200000 m 10

{2 significant figures}

a

Eric has rounded each answer to 3 significant figures before using i1t in the next calculation, rather than using exact values.

b

Diameter of circle = /62 + 62 = V72 cm -

radius of circle = 3@ ~area

of circle = 1 X

i\

cm (@)

2

cm?

__ 72w 2 = B | cm

~ 56.548 667 .... cm?

~ 56.5 cm®

1

a

32 X 6 ~ 30X

b

D8 X 7

6

~ 180

e

487 x 50

207 x 3

~ 00 X 7

~ 80 x 30

~ 200 x 3

~ 420

~ 2400

~ 600

f

6117 x 4

g

48 x 23

h

61 x 42 ~ 60 x 40

~ 25000

~ 24 000

~ 1000

~ 2400

103 x 47

i

3125 x 18

k

422 x 307

|

3818 x 27

~ 100 x 50

~ 3000 x 20

~ 400 x 300

~ 4000 x 30

~ 5000

~ 60000

~ 120000

~ 120000

~

3 x

n

1

86

H.36 x 0.68 ~

~ 3

o

H x 0.7

+— 3

b

~ 60+ 5

~ 30

~ 12 f

~ 4000 = 20

~ 200

0.628 =3

~ 1000 j

C

~ H0) =5

~ 0.2

~ 10

d

610 =43 ~ 600 = 40

20 g

=+ 80

46.1 = 5.2

~ (0.6 =3

H12 =21 ~ H00 = 20

7180638 = 82 ~ 80000

30 X 6

~ 180

64 — 5

~ 90+ 3

28.37 X 6.13 ~

~ 3.5

4182 =19

i

d

~ H0 x 20

2.7 x 1.15

e

81 x 30

~ 6000 x 4

m

a

C

~ H00 x 50

i

2

{3 significant figures}

318 =+ 62

~ 15 h

47 320 = 193

~ 300 = 60

~ 50 000 = 200

~ 5 k

631.7 = 0.29 ~ 600 = 0.3

~ 2000

~2 250 |

18.7 - 3.86 ~ 20=4

~ 5

Chapter 1 (Approximations and error)

C

8 kg x €2.80/kg ~ 8 kg x €3/kg ~ €24

b

7 tickets x $87.30 per ticket ~ T tickets x $90 per ticket ~ $630

d

4.2 h x 63 kmh™* ~4hx60kmh!

b

3 tickets x $213 per ticket ~ 3 tickets x $200 per ticket ~ $600 55 L x £1.49/L ~ 60 L x £1/L ~ £60

~ 240 km

14 years x 365.25 days/year ~ 10 years x 400 days/year ~ 4000 days there are about 4000 days in 14 years.

Brodie will travel about 240 km. 423 tonnes 18 trucks

N

Exercise 1C

400 tonnes

d

20 trucks

~ 20 tonnes per truck each truck will carry about 20 tonnes.

18.2 hours per week x 12 weeks ~ 20 hours per week x 10 weeks ~ 200 hours 200 hours x €21.50 per hour ~ 200 hours x €20 per hour ~ €4000 the part-time worker will earn about €4000 over the 12 weeks.

d

The tape measure is accurate to :I:% cm.

b

The measuring cylinder is accurate to i% mL. The beaker 1s accurate to

2

i%

x 100 mL = +=50 mL.

d

The set of scales is accurate to

:I:% x 500 g = +250 g.

e

The thermometer is accurate to

i% x 0.1°C = +0.05°C.

The scales have accuracy i% kg

the range of values 1s 68 + % kg.

So Ront’s actual weight 1s in the range 67.5 kg to 68.5 kg. 67.5 kg < w < 68.5 kg 3

a

The range of values 1s

27 + % mm.

So, the actual value 1s 1n the range 26.5 mm to 27.5 mm.

b

38.3 cm = 383 mm The range of values 1s 383 & % mm. So, the actual value 1s in the range 382.5 mm or

to 383.5 mm

38.25 cm to 38.35 cm.

11

12

Chapter 1 (Approximations and error) C

Exercise 1C

4.8 m = 480 cm

d

The range of values 1s

480 + % X 10 cm

1.5kg=1500¢g The range of values 1s

= 480 4+ 5 cm.

= 1500 £ 50 g.

So, the actual value 1s in the range

So, the actual value 1s in the range

475 c¢cm to 485 cm

or

1500 + % x 100 g

1450 g to 1550 g

4.75 m to 4.85 m.

or

The range of values is 25+ 3 g.

1.45 kg to 1.55 kg.

f 3.75keg = 3750 g

So, the actual value 1s in the range 24.5 g to 25.5 g.

The range of values 1s

3750 + % x 10 g = 3700 £ 5 g.

So, the actual value is in the range 3745 g to 3755 g or 4

The thermometer reads 36.4°C, so it must be accurate to

3.745 kg to 3.755 kg. i%

x 0.1°C = +0.05°C.

the range of values 1s 36.4 £ 0.05°C.

Tom’s actual temperature lies between 36.35°C and 36.45°C. 36.35°C < T < 36.45°C 5

For distances less than 10 km, the exercise watch 1s accurate to

0.01

iT

km = +0.005 km.

For distances between 10 km and 100 km, the exercise watch 1s accurate to

The range of values 1s

1.06 £ 0.005 km,

or

:I:% km = +0.05 km.

1.055 km to 1.065 km.

if the watch displays 1.06 km, the least distance Joanne could have run 1s 1.055 km. The range of values 1s

9.72 4+ 0.005 km,

or

9.715 km to 9.725 km.

if the watch displays 9.72 km, the least distance Joanne could have run 1s 9.715 km. The range of values 1s

10.1 +£0.05 km,

or 10.05 km to 10.15 km.

if the watch displays 10.1 km, the least distance Joanne could have run is 10.05 km. The tape measure 1s accurate to £0.05 m for a measurement of 6.1 m, the range is

6.05 m to 6.15 m

for a measurement of 6.4 m, the range 1s

6.35 m to 6.45 m

for a measurement of 6.0 m, the range 1s

5.95 m to 6.05 m

for a measurement of 6.1 m, the range is

6.05 m to 6.15 m

6.4 m 1s likely to be the incorrect measurement, as its range of values is furthest from any of the other measurements.

6.05 m, as it 1s in the range of values for a measurement of 6.0 m or 6.1 m. 10 cm graduations, as all of the measurements were given to the nearest 0.1 m, or 10 cm. 2.4 m = 240 cm

The range of values is

240 £ 5 x 10 cm = 240 &= 5 cm.

So, the actual length of a rope is in the range

230m

< [ =z and y = 22 pass through (0, 0) and LS

y? = x is a rotation of y ==z 2 clockwise through 90° about the origin or y? = x is a reflection of y = z° in the line

b

|

The part of the graph of y? = x

ii

y = +/x

y = .

in the first quadrant corresponds to

is a function as any vertical line cuts the graph at most once.

y = /.

Chapter 3 (Functions)

a

Exercise 3B

Both curves are functions since any vertical line will cut each curve at most once.

b

Yo

=z 1

Y Y

1

iES

Jx

f(x)=3x+2

a

f(0)=3(0)+2

—0+2

{replacing z with (0)}

= 2

¢

b

f(2)=3(2)+2

— 6+2

{replacing x with (2)}

= &

f(=1)=3(—1)+2 =-342

{replacing = with (—1)}

f(—-5)=3(—5)+2

{replacing z with (—5)}

=1

d

— 1542 = —13

e

f(—3)=3(—3)+2

{replacing z with (—%)}

=—1+42 =1

2

f(x) =3z —x*+2

0° +2 a f(0)=3-(0)

b f(3)=3(3)—32+2

¢ f(=3)=3(=3)— (=3)* +2 = —9—9+42 =—16 e f(5)=3(3)—(8)°+2

d f(=7)=3(=7)—(=7)?+2 — 21 — 49 +2 = —68

=0-0+2

o 3

=9—-9+2

3§+

g(:r:)::fl—%

a g(1)=1-7=-3 € o(TIAS

4 s it

33

bimn:¢—§=3 4 —5 —4d g(-4)= =3

54

L

Chapter 3 (Functions)

a b

Exercise 3B

i f(2)=1

zy

i f(3) = -1

y=f(z) | 2

\

x=-4.

y=f(x)=4,

When

42

| [ | 2 1.#""1

-y 5

G(z)

=

\

g

;

\

2 + 3 T —4

a

1 G(2)= 2(2) +3 2 —4

i

G(0) = 200) +3 0—4

iii

G(—3) =

2(—3)+3 %)

=

Y

9 2

=)

_ 49 b

G(x) =

2r +3

.

e o

i undefined when

x —4 =0 .

So, when ¢

G(xr)=-3,

so

s

T —4

=4

x =4,

G(x) does not exist.

3

co2r4+3=-3(x—4) L 2c+3=-3x+12 bhr =9 -.

6 f(x)=1-3z, a

— 9

$_5

g(z)=+vz+5

f(=1)=1-3(-1)

b

—1+3

f(x) = g(4) .

=

1-3x=+v4+5

1—3z=+9

g(11) = V11 +5

1—-3xz=3

= /16

—3z =2 . $:_%

==y

L f(=1) =g(11) 7

flz)=7—-3z a

b

fla)=7-3a

f(—a)=T7-3(—a)

¢

=7+ 3a

fla+3)=7—3(a+3) =7—3a—9 = —3a — 2

d

f(2a) =7—3(2a) =7 —

ba

e

flx+2)=7T-3(zx+2) =7—3x—0 =1— 3z

f

fle+h)=7-3(x+h) =7—3xz — 3h

Chapter 3 (Functions)

8

55

F(r)=22>+3z—1 a

F(z+4)

=2(z+4)*+3(x+4) -1

b

= 22% + 16z + 32 + 3z + 11 = 22° 4+ 19z + 43 C

F(—x)

ol

— 2(—;1:;*)2 +3(—x) — 1

= 22° — 3z — 1 e

F(2-z)

=22—-x2)*+32—-1z) -1

=2(4 -4z +2°)+6 -3z —1 =8 — 8z +2x° +5— 3z = 22° — 11z + 13

= 2(2° +8x+16) + 3z + 12— 1

F(z?)

= 2(z%)? + 3(z?) - 1 = 2z* + 32% — 1

F(3x) = 2(3x)? + 3(3z) — 1 = 2(92°%) + 9z — 1

f

= 182" + 9z — 1 9

Exercise 3B

F(x + h)

=2(z+h)>+3(x+h) -1 = 2(z* +2zh + h?) + 3z +3h — 1 = 22% + 4xh + 2h* + 3z 4+ 3h — 1 =22° + (4h + 3)x +2h° +3h — 1

fl(z)==x"

d

2f(z—1)+5=2(x—1)*+5

=2(z* —2x+1)+5 — 2% —4x+2+5

= 2% — 4z + 7

o

1

f(%@

T

§£U

_ 1 (3)

_2 oz

C

2f(m)+3:2>5,

P(t) =5+ P(3) =5+ = 35

and hence

b=1—2(—2)

—2

= 5.

f(z)= —2x+ 5.

10¢ 10(3)

There are 35 L of petrol in the tank after 3 minutes.

b

When

P(t) =50,

then

5+ 10t = 50 10t = 45 .

t=4.5

After 45 minutes, there are 50 L of petrol in the tank. ¢

When

Samantha started to fill the tank, time

Now, when

t =0,

¢ = 0.

P(0)=5+ 10(0) =5

There were 5 L of petrol in the tank when Samantha started to fill it. 15

a

H(30) =800

After 30 minutes, the balloon is 800 m high.

i

b H(t) =600 when t =20 or 70.

After 20 minutes and after 70 minutes the balloon 1s 600 m high.

¢

A H(metres)

o 600

The height of the balloon was recorded for

=

0 < t < 80 minutes.

d

/

\\

[ e

The range of heights recorded was 0 m to

[j

[ [T

R

)J

1|/

900 m.

i

=

¥

%

40

60

%»

t (minutes)

Chapter 3 (Functions)

16

Exercise 3B

f(z) = az+2 where f(1)=1 and f(2)=5 J

So,

a(l)+i =l | (1)

and

a+b=1 b=1—-a

{1(2)+i =5 (2)

QQ—FE _r 2 . 4a+ b =10

... (%)

da+ (1 —a) =10 . 3a=19Y

{using (*)}

C.ooa=3

17

Substituting

a = 3 into (x) gives

So,

b= —2.

a=3,

b=1—3 = —2,

T(z) = az® + bz + ¢ where T(0)=—4, So,

T(1)= -2,

and T(2) =6

a(0)*+b(0) +c= —4 .oc=—4

Now

a(1)*+b(1)—4= -2

and

a(2)*+b(2)—4=6

a+b=2

.

b=2—a

.. (¥

4a+ 20 =10

C. 2a4+b=5 20+ (2 —a) =5 a=3

18

Substituting

a = 3 into (x) gives

So,

b=—-1,

a=3,

and

c=

b=2 -3 = —1.

—4.

V(¢) = 9000 — 900¢ a

V(4)=9000—900(4) = 9000 — 3600 = 5400 V(4)

is the value of the photocopier in pounds after 4 years.

the value of the photocopier 4 years after purchase 1s £5400. b

V{(t) =3600,

so

9000 — 900t = 3600 . 900t = 5400 t =06

After 6 years, the value of the photocopier is £3600.

¢

The original purchase price 1s when Now,

¢ = 0.

V(0) = 9000 — 900(0) = 9000

The original purchase price was £9000.

d

Wehave So,

t >0

and

V >0

9000 — 900t > 0 9000 = 900¢ t +3(0)—5

the y-intercept is —5. [EXE]:Show coordinates

Yi=x*(4)

L

&5

-]

4ߤl?3)+x2+3

%

-3 -2

Of

[EXE]:Show coordinates

5

Yi=x*(4)

2

A=-1.1 155'15&543

344

5

E

7

8

129

X=3.597632209

Y=0

[EXE]:Show coordinates

Yi=x*(4)

4fl§fl'3}+x2+3

A

1

2

[EXE]:Show coordinates

B

5

Yi=x"(4)

6

7

8

41‘3‘5?3)+x2+3

5

IIIIIIIIIH

X=-0 .401'?—'?3;235

3)J4

5

':4_53_'2\55'

.........}‘

1

41:!‘5?3)+x2+3

9

V’..

i 76 783181 X=0.68

MIN Y=-5.7b68421271

2

B

IIIIIIIIH

3)J4 5 6 7 &8 49

MAX Y=-3.541244021

[EXE]:Show coordinates

Yi=x*(4)

L

4

]

41‘555’3]+x2+3

-3 -2 i

]

X=2.713939849

Using technology, there are minimum turning points at (—0.402, —5.76) and a maximum turning point ¢

As

r— o0,

d

y—oo

andas

AY

(—0.402, —5.76)

x —

—o0,

Yy — 0.

+

3.60

1.12&” (0.688, —3.54) L

at (0.688, —3.54).

&£

y=zr—423 + 2?2+ 32 -5

(2.71, —15.2)

1

2 3J4

B

5

6

7

8

3

H

MIN

Y=-15.20033471

and

(2.71, —15.2),

Chapter 3 (Functions)

e d

Therangeis L]

[EXE]:Show

63

{y |y > —15.2}. coordinates

Y1=(6a4(x-3))-2

[l

q11I¥

[EXE]:Show

coordinates

Y1=(6a4(x-3))-2

11y

Y-ICEPT

ROOT

The y-intercept 1s —A4. b

Exercise 3D

The z-intercept 1s 6.

The denominator is zero when

= = 3

there 1s a vertical asymptote at x = 3. As

x — 400,

So,

y = —2

C

the graph gets closer to the line

y = —2.

1is a horizontal asymptote.

AY

g

d

The domainis

{z | x # 3}.

The range is {y |y # —2}. -

P

i

—4

Y d

I

x =3

[EXE]:Show coordinates

[EXE]:Show

Y1=(1122)x(x—4)(x .......

[EXE]:Show coordinates

Y1=(112)x(x—4)(x+3¥ /

3

coordinates

9

&

3

0

Y1=(122)x(x—4)(x+3¥ /

ROOT X=-3

o

. 4

2

0

ROOT

=0

The z-intercepts are

,

SR

S

5

ROOT

=0

=4

-14¥=0

—3, 0, and 4.

Y-ICEPT

The y-1intercept 1s 0. ii

[EXE]:Show

coordinates

[EXE]:Show

Y1=(112)x(x—4)(x+3Y¥

g

X

o

-9

MAX ¥=-1.6894254194-14¥=6.2986800504

¥=2.36092083

There is a maximum turning point at (2.36, —10.4).

il

coordinates

Y1=(1a2)x(x-4)(x+3¥

The graph has no asymptotes.

/

g

X

o MIN -14¥=-10.37287458

(—1.69, 6.30)

and a minimum turning point at

64

Chapter 3 (Functions)

Exercise 3D

iv.

The domainis

V.

(-1.69,6.30)

{z |z € R}.

Therangeis

{y |y € R}.

AY Y = %:1’:(:1: —4)(z + 3)

-}

4

—3

v b

'

g

[EXE]:Show

Yl=x*(3)-2x2+8

-

T

(2.36, —10.4) B

coordinates 131/

......... -9

Ot

X=-1.,34025083]

R

:

°N=0

9

ROOT

[EXE]:Show

Y1=x"(3)—-2x2+6

s

[E]

13¥

=0

131¥

:.

:

—¥=6

.

8

&

Y-ICEPT

The y-intercept is 6.

coordinates

JtoF

coordinates

[

X=0

The z-intercept 1s = —1.34. [E]

[EXE]:Show

Y1l=x"(3)-2x2+6

[EXE]:Show

Y1=x"(3)-2x2+6

coordinates

oo

3

5

MAX

SN=06

¥=1.333333348]

13}y

ot

3

MIN

S¥=4.814814815

There is a maximum turning point at (0, 6) and a minimum turning point at (1.33, 4.81). iili

The graph has no asymptotes.

iv

The domainis

v

{x |z € R}.

Therangeis

{y |y € R}.

AY

y=ax> — 22+ 6

(1.33, 4.81) 134

1

|

E

LS

>

Y C

i

E

[EXE]:Show coordinates

Y1=4(x2+4)

5 X=0

13}¥

_ofF

DE{

=2

NI CER

.

&

The y-intercept 1s 2. There are no z-intercepts.

il il

There is a minimum turning point at The graph has no asymptotes.

(0, 2).

Chapter 3 (Functions)

iv.

The domainis

v

{x | x € R}.

Therangeis

Exercise 3D

635

{y |y > 2}.

AY

y=+vz’+4

2 =3

&

£ £

Y i

[EXE]:Show coordinates

Y1=4(x2-4)

[EXE]:Show coordinates

13}¥

Y1=4(x2-4)

3

;

5

ROOT

=0

13}¥

=2

:

=

The z-intercepts are —2 and 2. There is no y-intercept. The graph has no turning points. The graph has no asymptotes.

iv v

The domainis

{z|x < —2

or = >2}.

Therangeis

{y |y = 0}.

AY

-

i

—2

S Y

[E] [EXE]:Show coordinates Y1=4(9-x2) of¥

——

[E] [EXE]:Show coordinates Y1=4(8-x2) ggi'

[EXE]:Show coordinates Y1=.(9-x2) of¥

|

a'fl-\# t

-aRkr=13

The z-intercepts are —3 and 3.

There is a maximum turning point at

ili

The graph has no asymptotes.

iv.

The domainis

{x| -3 0}.

AY

«/{

-

£X

Y k

[

[EXE]:Show coordinates

Y1=(4

B

)x“(4}+5x“(3)+5x%flw;fi

.

&

[EXE]l:Show coordinates

i

2

£

E

i

.

-3

s

ROQT X=-4,972844666

VY=0

_ac

The z-intercepts are

[EXE]:Show coordinates

?1={4.‘}x“{l)+5x“{3}+5xw

~ —4.97

5 i

X=-1.651243416

and

Y=0

=~ —1.55.

_3g

Yi=(4p)x"(4)+bx~(3)+bx

i

.

5

ROOT

3

L&

Y-ICEPT X=0

The y-intercept 1s 2.

Chapter 3 (Functions) [EXE]:Show

[E] [EXEl:Show coordinates Y 1=(d~i}xfl{4)+5x“{3}+fixh

¥=-3.882634012

[EXE]:Show

Y1i= (di}x“ (4)+5x~(3)+ B¢

T Of

-

coordinates

2

V=-33:47552503

=

MIN

T

3

7

¥=-0.8048659800

There are minimum turning points at turning point at (—0.805, 2.97).

-1

Exercise 3D

Y1=(4

Qgf

2

Y=2.967773567

5

MAX

(—3.88, —33.5)

coordinates

B)x"*(4)+bx*(3)+bx2%

%

4

3

2

¥=6.065607862e-8

and

69

(0, 2),

1

¥Y=2

of

1

2

-3¢

and a maximum

The graph has no asymptotes.

The domainis [l

{x | -6 0,

+

_|_

> T

(4)(1) > 0,

so we put a + sign here.

As the factors are single, the signs alternate.

|-

4y

+l /1

-


I

i

iy

so we put a + sign here.

¢

75

-

Y

—

-

Ay

Exercise 3E

=

_'2

When

z =1

{|]

we have

T

>

(1)(3) > 0,

so we put a + sign here.

As the factors are single, the signs alternate.

>

o

76

Chapter 3 (Functions)

e

Exercise 3E

(2z+1)(xz —4)

has zeros —3 > and 4.

f

—(x+1)(x—3)

has zeros —1 and 3.

_|_

|

==

1 When

|

A

2

z =5

4

we have

|

—

>

|

-4

(11)(1) > 0,

—1 When

so we put a + sign here.

—(3xz—2)(x+

1)

When

z =1

|

4

2 3 we have

h

T

(22 —1)(3—x) -4

|

—(1)(2) 0,

so we put a + sign here.

As the factor 1s squared, the signs do not change. -

-

-

T

As the factor is squared, the signs do not change. -4

-+

|

+

Chapter 3 (Functions)

¢

—(x—4)*

2(x +1)?

has zero 4.

-

| 4 When

=5

A

we have

| —1

—(1)? < 0,

When

so we put a — sign here.

|

=

4

—3(z +4)?

|

When

= =0

+

-

we have

r—

1

undefined when -

When

= = —2

x = 0 we have

:

13

X

*

we have

|

-+

T+

—

O

-+

.

1

.

1s zero when

undefined when

= = —1

Since

(x+ 1)

+

i —5

T

we have

and

When

are single

>

=+ ~1

£ =

4

0 1 we

have

-+

|

;

s

are single factors,

—

|

—3

r—2 20+

-+

-

0

.

1s zero when

1

> 1

%}{).

Since z and (x + 3) the signs alternate.

T

and

|

—3

= = 2

and

x = —%. _I_

.:

-

|

% > 0.

(x +5)

and

= = —3.

’

-

undefined when

=5]

factors, the signs alternate. -

>

+

-5 z =0

3

= =0

_I_

& = —5.

.

When

T+

U

2

.

§ > 0.

;

—2 1

Er

|

>

1s zero when

undefined when

1

—%(5)2 < 0,

|

5

£

and

Since (z+2) and (x — 1) factors, the signs alternate. -

> X1

As the factor is squared, the signs do not change. -

z = 1.

—2 z =2

L

so we put a — sign here.

>

|

When

has zero —2. I=

—3(4)% < 0,

|

1s zero when

>

2

As the factor 1s squared, the signs do not change.

T+ 2

+

—1

-

so we put a — sign here.

-

2(1)? > 0,

|

=(2z 4+ 5)* A

—4

we have

> T

As the factor is squared, the signs do not change. -4

has zero —4.

-

x =0

4

so we put a + sign here.

As the factor is squared, the signs do not change. i}

has zero —1.

-

-

Exercise 3E

11 When

are single

Since

£ = 3

(z —2)

4

l

9 we have

and

|

>

% > (.

(2x + 1)

are single

factors, the signs alternate. -¢

L

. —1 1

| 9

+

> U

77

Chapter 3 (Functions) 2

e

Exercise 3E

-

222

s zero when

4 —x

undefined when

= — —%

= = 4.

|

-

= =5

$

A we have

»

|

STt

g

:I‘:

N

5

undefined when

When

x = 3

=0

,

4

we have

(x+1)°

undefined when

i|

-

x =0

~1 When

x = 1

0 we have

r = —1.

ML

—1

el

0

Since

= = 2

(x — 1)?

(z+2)(xz —1)

and

3—=x

+

are single

i2 =1

4

|

and

x = 0.

L4

_I_

>

1 we have

%}U.

1is a squared factor, the

1s zero when

or 1 and undefined when »

T

E}D.

Ll

i—ll < 0.

sign stays the same about x = 1. Since x 1s a single factor, the sign changes about i ="l

Since 4x 1s a single factor, the sign changes about = = 0. Since (z+1)? is a squared factor, the sign stays the same about

we have

0 When

=z = —1. |

Qfli*w

;

T

e

2

E

z = 2.

1s zero when

£L

_|_

-~

-

(z —1)°

and

are single factors,

1s zero when

and

Since (4rx — 1) and (2 — x) factors, the signs alternate.

% > 0.

:

————

= =3

i

5:

e 0L

|

T

z = %

i

-l

undefined when

Since 3x and (x—2) the signs alternate.

.

are single

x = 2.

0|

-

T

When

-

4

1s zero when

is zero when

DO

2

3

*

.

2 —

i—gl < 0.

Since (2x +3) and (4 — x) factors, the signs alternate. -

—1

undefined when :

N3 When

and

4

s f—

78

B

|

—2 When

= =4

|

1

we have

= = —2

x = 3.

N SR

3

LE_E? < 0.

Since all of the factors are single, the signs alternate.

Chapter 3 (Functions) —

i

o (2z + 3)(z — 2)

1s zero when

and undefined when

¥

a

_% When

|

9

= =4

|

= = _% or 2.

e

a

-

= = 3

we have

3

4

B

v

e

— Lo 2

—1)

>

3

.

(z+5)(z—1) (z + 2)2

-

>

(11__)%(%‘:0'

L

Transformations of graphs i1s zero when

or 1 and undefined when

=

?

—9

Since

(x +5)

Since

(z + 2)*

1

a

GRAPHS OF THEFORM

1

> T

|

%1}0.

(z — 1)

factors, the signs alternate about and = = 1.

are single x = —5

is a squared factor, the

: sign stays the same about -2

=z = —2.

1

y — f(z) + d

f(z)=3x

i f(z)+2=3z+2 b

have

and

—H

PART 1:

|

—2

= = 2 we

x = —5

= = —2.

e

|

When

Since all of the factors are single, the signs

alternate. +

Investigation

Ay

i

f(z)—-3=3zx-3

i

f(zx)+6=3x+6

i

f(x)—3=x°-3

i

f(zr)+6=2°+6

A

f(z)

(z)+2

79

80 3

Chapter 3 (Functions) For e

PART 2: 1

y= f(x) =d, If d>0

Investigation

Transformations of graphs

the effect of d is to translate the graph vertically through d units.

1t moves upwards.

GRAPHSOFTHEFORM

o

y —=p f(z), p >0

AND

1t moves downwards.

y = f(qz), ¢ > 0

flz)=x+2

a

i 3f(z)=3(z+2)

b

i

3 f(z) =3(z+2)

iii 5 f(xz)=>5(z+2)

Y y=5f (m)f[ /A‘jm) 7 ~ B

A:%f('x)

—

.

Y+ 3f(ir)//

Y

¢

For the transformation from the z-axis.

d

(—2,0)

y = p f(x),

p > 0,

each point becomes p times its previous distance

is O units from the z-axis, so it becomes

all transformations of the form

2

If d 0 do not move the point

(—2, 0).

flex)=x+2 a

i

f(2x)=2x+2

i

f(3z) = 32 +2

i

:

ol

f(dx) =4z + 2

b

(2;1‘)/ 4x)

Y .

¢

For the transformation

d

from the y-axis. . . . . (0,2) is 0 units from the y-axis, so it becomes

:

y = f(qx),

all transformations of the form

g > 0,

f(qx),

each point becomes

:

:

:

— times its previous distance q

1 . . 0 x — = 0 units from the y-axis. q ¢ > 0 do not move the point (0, 2).

Chapter 3 (Functions)

PART 3: GRAPHS OF THEFORM 1

y — —f(z)

Investigation

AND

Transformations of graphs

y = f(—x)

f(z)=2x+3 a

i

—f(x)=—-2x+3) =

—2r—

i

f(—z)=2(—xz)+3

3

=

—2x + 3

A\ S AY

b

'D

3

y=f(—=z)

ilI “L7

A

y=—f(z)

a

A reflection of the graph in the x-axis moves

y = f(z)

to y= —f(x).

b

A reflection of the graph in the y-axis moves

y = f(x)

to y = f(—x).

81

82

Chapter 3 (Functions)

1

a

Exercise 3F

The graph of y = f(z)+5 translating

y = f(x)

b

5 units upwards.

|\

The graph of y = f(xz) — 2 is found by translating y = f(x)

AY

) f('j:)

—I_

5

s

A

_3

2 units downwards.

T ¥

AY

y:

‘—!‘

is found by

a,

;

““u

z

\

-

\

#"

_3:

3

y=f()

y=flz) -2 '

':*

_‘%!‘

y= f(z)

:

Y ‘o

Y

The graph of y = g(x)+ 1 1is found by translating y = g(x) 1 unit upwards.

%\

L

--"“':');""-"‘:-.

i

T

31): —

.

1;.'{.’1?] +11

\/:

3

WL

y = g()

=

The graph of y = g(x) — 3 1is found by translating y = g(x) 3 units downwards.

A’ Y

\

1“‘

b

X

6 =

h

A Y t“‘

i?z

_

AN

\

—3\/”'@'“\

L 5

i

a

i

2

[/ 1] =/ N\ \_/ / Woooyy=r@) The graph of y = f(z) So, g(x)= f(x)+ 3.

has been translated 3 units upwards to result in y = g(z).

.

6

:

X

Chapter 3 (Functions)

L

a

The graph of y = g(x)

i1s found by translating

Log(x) = fx) —4

. g(x)=2x+3)—4 . glz)=2x—-1

b

{since

The graph of y = g(z)

4 units downwards.

is found by translating y = f(x)

glx)=(—2*+5x—T7)+3 . g(x) = —z*+5z—4

The graph of y = g(x)

83

f(z)=2x+ 3}

. 9(x) = flz) +3 ¢

y = f(x)

Exercise 3F

{since

3 units upwards.

f(z)=—2*+5x—T}

is found by translating y = f(z)

6 units downwards.

Log(x) = f(z) -6 _

g($):($3—|—2—i)—6

{Since

f(iE)Z:IIS—I—Q—i}

g(z) =a® —4— =

£

3

a

YA

-

fle)=73

P

;

Y

b

i

g(x)=f(z)+k

i

1

;.

g(x):;Jrk

i

As = — 400, as

= —0 I

l-l-k—?k‘,

x — $oo,

£

the horizontal asymptote of g(z) is y = k.

The graph of y = 2 f(x) is a vertical stretch of y = f(x) with scale factor 2.

b

The graph of y = f(3x) is a horizontal stretch of y = f(x) with scale factor 3.

Ty

AY

.

-l

/‘:

E}“L.\

\ *

y—fl[fi)*

’ iy

F

*

'lr‘

*,p L]

|

%

;JF'

/

* e

£ N

Y

\

’

..

/'-r -7

fi‘

b

b'ln.......d"‘ ‘

:

-

#

*

;.

.

-

fif T

r:-

x

-l

/Ty

%

“y — f(fE)

fi‘

',p L]

]

"

f(3j_fj:

:'*

*

‘

5¢

Y

=Y

/\

L'l

a

o

6

84 7

Chapter 3 (Functions) a

The graph of

Exercise 3F

y = %h(:z:) L

stretch of y = h(z)

is a vertical

b

"

with scale factor 3.1

The graph of y = h(%)

stretch of y = h(x)

with scale factor 2.

AY

.': =

AY

“x_

;

// “

13

t“

3

| -,

is a horizontal

o

,

.':

4—/

T

\*

j

\‘Q\

—3

y=zh(x)

)

t,,

3

“‘.\'\

L,

'-‘u

\

y

I-‘.

1

T

b

\

—3

P

L

P S~

M'\:H*

y=g(x)

>

3

L

T y=J(2) N\,

_3

\

Y

\

When

y = f(z)

i '11

-

1/ /

:

Lo

"'

:'I.‘

%

'I '

b

\\% L

‘*

So, g(z) =2 f(x)

Let the original linear function be

' "

\'\

“‘ -

The graph of y = f(x) has been vertically stretched with scale factor 2 to give y = g(x).

9

-

fy=1@) )

: \

o

] ]

A

\u 1;_‘-‘

Y

The graph of y = f(x) has been horizontally stretched with scale factor 3 to give y = g(x).

So. gla) = £(2).

y = f(x) = mx + a.

is vertically stretched with scale factor ¢, the resulting function is

Yy = cmx + ac

So, the resulting line has gradient cm.

The graph of y = g(x)

. g(z) =2 f(z) o .

g(x) = 2(x* +2) g(x) =222 +4

is a vertical stretch of y = f(x)

{since

f(z)=2*+2}

’

Y=9(@)

y = c(mx + a)

a

>

//

y = c(f(x))

10

:h(%)

v y=h(z)

A

a

I

1“

v

8

>

with scale factor 2.

Chapter 3 (Functions)

The graph of y = g(x)

is a horizontal stretch of y = f(x)

Exercise 3F

with scale factor 3.

g(x) = f(%) . g(x)=5-— 3(%)

{since

f(z)=>5— 3z}

. glx)=5—=x The graph of y = g(x)

is a vertical dilation of y = f(z)

with scale factor 7.

. g(z) = f(x) .

g(z) = 3(2° + 82% — 2)

{since

f(x) ="+

8z% — 2}

. g(x) = 32° + 227 — is a horizontal dilation of y = f(z)

. g(z) =2(22)* + (22) — 3 . g(x) =82% + 2z — 3

The graph of y = —f(x) is found by reflecting y = f(x) in the z-axis.

y=f(x) _

=

o.

|3

1.

-

el 1| -4

—1

P

O------0, |

v

y=< flx) | 2

\

/ \

T

y=f(z), eee=” ’

y=—f(z)

The graph of y = f(—=x) is found by reflecting y = f(x) in the y-axis. YA

)

AY

b

[\

o

-

>

\

y=f(-z)

b

-€T

/

-

/

— ¥

s

YA

f(x)=2z"+x — 3}

1-"-.

11

{since

.-

®

L g(x) = f(22)

with scale factor .

Fil LA

The graph of y = g(z)

,-':; -

835

86

Chapter 3 (Functions)

14

a

Exercise 3F

The graph of y = g(x)

L g(z) = —f(z) . . b

glx)=—(bx+7) gle)=-bx -7

The graph of y = g(x)

L g(z) = f(—=) N ¢

G

E =27 Led SIneen

The graph of y = g(x)

L g(x) = —f(z)

. g(x) = —(22* +1) og(x) = —22° — 1 d

The graph of y = g(x)

is a reflection of y = f(x) {since

f(x)=>5x+T7}

is a reflection of y = f(x)

in the y-axis.

itz h=2"} is a reflection of y = f(x)

{since

in the x-axis.

f(z)=22*+ 1}

is a reflection of y = f(x)

Log(z) = f(—) Log(x) = (—a)* = 2(—2)’ = 3(—x)* +5(—x) - 7 {since

in the z-axis.

in the y-axis.

f(z)=z*—22° —32*+5x— T}

o g(z) =a*+22° —32° —bx —7

Chapter 3 (Functions)

1

(=3,7),

(0,4),

(7, =3), 2

a

b

and

(2, —6)

(4,0),

and

lie on f.

(—6, 2)

lie on f1.

Aninverse function exists since f(z)

y = f(x)

passes through

y = f~'(x)

passes both the vertical and horizontal line tests.

(—1, —5)

passes through

(2, 4).

Exercise 3G

and

(-5, —1)

(4, 2)

Ay

and

il

T

J(4.2) r*

].)

. y=J(z)

3 (_1&

3

5

(=1,4) .

(1. 3‘,

;_....:...,

Y

y=J{T) &

0}

1s

{z | x>0}

Range 1s

{y|y eR}

Domain

-

{J-f

- 0

=

f(z) =+/x —2 has domain and range {y |y = 0}.

.

{z |z > 2} -

=T -4—

e ¢

Sy=

8

i

6

—2

6 «

—6

—6

_L

6l

v

f(x) has domain {x |z > 2} and range {y |y = 0}. *. f~Yz) has domain {z |z > 0)

d

f3)==x

o

{y |y > 2}.

and range

If f~'(z) =3, then y = f~!(x) passes through (z, 3). So, y = f(x) passes through (3, x). r=4+3—2 r=+v1=1

f '(z)

has domain

{z|z > —3}

and range

{y |y > —1}.

f(xz) has domain and range

b

If f(z) =2, then through (z, 2). So,

y= f~Y(z)

Now

(2.3) (—1,2)

{z |z > —1} {y |y > —3}.

y = f(x)

passes

passes through

(2, x).

f~'(2)=3

-

=Y

a

{1 LS

9

/(_31__1)

{from the graph}

="

10

a

AY

b

1/

5

/

>T

-

>T

Y

The

graph

horizontal line tests. the

passes and

function

Inverse.

the

vertical

The

an

fails

the

horizontal line test. .

has

graph the

function does not have an inverse.

The

graph

fails

the

horizontal line test. the

function

does

not have an inverse.

90

Chapter 3 (Functions)

11

a

Exercise 3G

f(-2)=(-2)*-2(-2)+5

f(=1)=(=1)* —=2(—=1) +5

=13

= 8

£(0) = (0)* —2(0) +5

f(1)=(1)*-2(1) +5

=5

=1—-2+45 =4

f(2)=(2)°-2(2) +5 =4—4+5 = 9 @

—2|1-11|

f(x) | 13|

0

1

8 | 51

2

4|

5

There are two points with the same y-coordinate, (0, 5) the horizontal line test and hence is not invertible.

12

f(x) =3z +6

and

b

f(0)=3(0)+6

(2, 5),

so f(x) does not pass

Y04 f(z)=32+6

Y=z

6

the y-intercept is 6. p

When

f(z) =0,

3z+6=0 -

C.

2

3x = —6 L=

':

;T

A

!

—2

;;"'

f_l(:‘[:)

6

—

/

the x-intercept is —2.

1

Y

f(x) is a line through (—2, 0) and (0, 6). f~1(x) is a line through (0, —2) and (6, 0). The line through

(0, —2)

and

(6, 0)

has gradient

) 6_5_02) — % A

and the y-intercept 1s —2.

3

f~Hz) =32 —2.

13

f(x) =mzx+c .

When

f(z) =0,

ML

f(0) =m(0) + ¢ the

e y-intercept

max+c=0

La=-Zs {m#0)

1s c.

the x-intercept 1s i

T

f(x) is a line through f~(z)

——C

(—i, D) Tri

is a line through

and

(0., —i)

(0, c). and

(c, 0).

Chapter 3 (Functions)

- (4)

has gradient

C —

|o

(c, 0)

o

T

3=

m

and

—

(0, —i)

91

o]

The line through

Review set 3A

and the y-intercept is -

™

f_l(x):i:c—i (i

T

::E_C,

% 20

m

1T

2

a

{(1,5), (4, 2), x-coordinate.

(2,5),

b

{(—4,0), (3,2), (0,—2), (3,5)} 1is nota function since two of the ordered pairs, and (3, 5), have the same x-coordinate 3.

is a function since no two ordered pairs have the same (3, 2)

f(z) =2x —x°

a f(2)=2(2)—2°

b f(—3)=2(-3) — (-3)

=4 —4

—

3

(6, —1)}

=—-6-9

O

—

a

¢ f(—-3)=2(-3)—(-3) =—1—4

—15

AY

o

b

&

——

=

-

L __4-

-

P

Yy

i Domainis

-

=

Y

{z |z € R}.

il

Rangeis

i

Each vertical line cuts the graph at most once, so the graph shows a function.

!

AY -

-

5

{y|y> —4}.

I Domainis

{z |z € R}.

ii

Rangeis

il

Each vertical line cuts the graph at most once, so the graph shows a function.

{y |y =2}

92

Chapter 3 (Functions)

C

\

AY

Review set 3A

/f

d

1 =

a7

i(/

>

vl

N

I

Domainis

{z |z € R}.

ii

Rangeis

il

All vertical lines cut the graph more than once, so the graph does not show a function.

{y|y 1}

I

Domainis

{x |z € R}.

il

Rangeis

il

Each vertical line cuts the graph at most once, so the graph shows a function.

{y|—5

0

r 0}. 8

a

AY :fflfifi\fi&\\\h‘/€5 ;;

-l

9

a

v —9

=

0

1+

D

(3z+2)(4—z) B

|

—3 When

X

has zeros —% and 4. [

2

Lo

4

= =5

b

we have

(m+—2)2 L

undefined when

.

(17)(—1) —=

|

and

> 0,

(©)?

sO we put a + sign here.

Since

=

(x — 3)

is a single factor, the sign ,

changes about z = 3. Since (z+2)*

:

is

a squared factor, the sign stays the same

about -

10

=z =3

z = —2.

_:2 When

As the factors are single, the signs

L

is zero when

r = —2.

Ll

E

—2

n|

|

3

T

- T

[EXE]:Show coordinates

Y1i=x"(3)-7x2¥4x+12

—rs

ol T

XAt

e Y=-ICEPT

The y-intercept 1s 12. [E]

[EXE]:Show coordinates

[EXE]:Show coordinates

Y1=x"(3)-7x2F4x+12

—rEEgot T AT X=-1

Y1=x~(3)-7 /

X

e p e |

|

Y=0

The z-intercepts are

Y1=x"(3)-7x2¥4x+12

Ax+12

/..'-‘5

ROOT

ROOT 23

[EXE]:Show coordinates

X=2

—1, 2, and 6.

ROOT

X=6

Chapter 3 (Functions) [E]

[EXE]:Show

coordinates

E]

Y1=x*(3)-7x2¥4x+12

4

-3

-2

ipf

Y1i=x"(3)-7

1

3

j X=0.306/7488287

[EXE]:Show

4

&

7

B

(4.36, —20.7).

coordinates Ax+12

MAX

MIN . 746349016

turning point at

The domain is {x | x € R}.

(0.306,

Therangeis

12.6)

and a minimum

{y |y € R}.

i

12.6)

y = x° —7z?+ 4z + 12

=Y

B

fant

EI

AY (0.306, 12.\

(4.36, —20.7) [EXE]:Show

Y1=4(x2-4)

coordinates

[E]

[EXE]:Show

Y1=4(x2-4)

13¥

\ X=-9

X=2

-Si=0

The z-intercepts are —2 and 2.

coordinates

13t¥

SK=0

There 1s no y-intercept.

The graph has no turning points.

EI

The domainis

{z |z < —2 AY

101

9

Y=12.598720101

There is a maximum

Review set 3B

or = >2}.

Therangeis

{y|y > 0}.

turning point at

102

Chapter 3 (Functions)

Review set 3B

N(t) = 100t x (1.3)~*

(20, 10.5) :

20

0 b

[2

e

t (days)

[EXE]:Show coordinates

Using technology, when

¢t =1,

N ~ 76.9.

about 77 people are infected after 1 day. [E]

[EXE]:Show coordinates

Using technology, the maximum value of NV is &~ 140, when t ~ 3.81. the outbreak was greatest after about 3.81 days, when about 140 people were infected. d

[E]

[EXE]:Show

coordinates

Using technology, N reduces to 15 when ¢t ~ 18.3. the outbreak was contained to 15 people after about 18.3 days.

12

a g(z)=—J(=z) = —(z? — 3x) = 31 — z*

b

g(z)=J(z)+2 =14 — 2+ 2 =16—=x

¢ g(x) = f(37) = 1(lz)+2 . = =xr + 2

Chapter 3 (Functions)

13

a

The graph of y = f(2x) 1is found by horizontally stretching y = f(x) with scale factor .

b

¢

3

\

H‘l-

103

The graph of y = 2f(x) is found by vertically stretching y = f(x) with scale factor 2.

‘y

T

Review set 3B

AY

T

X3

T

RSO \

;

v

;!_3

3 |« I

The graph of y = f(—x) is found by reflecting y = f(x) in the y-axis. Ay

14

a

The graph of y = g(z) = —f(x) AY

is found by reflecting y = f(x)

in the z-axis.

4

-

‘y=f(z)

15

- -

(1,4)

b

The maximum turning point of y = g(x) is (1, —4).

a

The graph of y = f(—xz)

is (3, 0). The minimum turning point of y = g(x)

is found by reflecting

y = f(x)

If y= f(x) has z-intercept 4 and y-intercept —1, then y = f(—x) has z-intercept —4 and y-intercept —1.

b

(4,0)

and

(0,4)

(0, —1) and

y = f~'(x)

lie on the graph of y = f(x).

(—1, 0) lie on the graph of y = f~1(x). has z-intercept —1 and y-intercept 4.

in the y-axis.

104

Chapter 3 (Functions)

16

a

. J(=3.5)

17

a,b

A

Review set 3B

Y

_ y=2

f(z) =8 — 2x f(0) =8 —2(0)

When

f(x)=0,

—_—

8 —2x =0 C. 2 =8 ".

the y-intercept 1s 8.

» Y "

..

3_‘,"

—_—

4

the z-intercept 1s 4.

Yy=1a

] I,

Y

¢

y=f'(z)

1|L“'L‘.f(:fl)zg_Qx

isa line through

(8, 0)

and

(0, 4), which has gradient

4—0

1

0—8

—_8

and y-intercept 4.

18

a

(—2,2)

@AY

b

2|

1

1)

l

}

1

1

i

hasrange

{y|00

0 0 160 32

it 1s reasonable to apply this model for

;;

0 < ¢ < 32.

T

|

& '

A =160—5(15) =160 — 75 = 85 and A= 160 —5 = ot < L


(30 — 2x) cm

ceefenn

and

132

Chapter 4 (Modelling) b

When

=2,

Review set 4B

V = 1352

x (cm)

1352 = a(2)” + b(2)* + ¢(2)

= =5,

D

10

V (em®) | 1352 | 2000 | 1000

8a + 4b + 2¢ = 1352

When

2

V = 2000

2000 = a(5)” + b(5)* + ¢(5) 125a + 25b + 5¢ = 2000 When

= =10,

V = 1000

. 1000 = a(10)® + b(10)* + ¢(10) 1000a + 100b + 10c = 1000 8a + 4b + 2¢ = 1352 125a + 250 + 5¢ = 2000 1000a + 1006 + 10c = 1000

So, we have the system of equations

an X+bn Y+Cn a

b

Z=dn

c

d

an X+bn Y+Cn ¥

1

8

4

2

1352

‘n{

2

125

25

5

Eflflfl|

Z

3

1000

100

10

1000

‘:|

Z=dn

=120

800

-

REPEAT

Solving these equations simultaneously using technology, we find that a =4, c = 900. ¢

The volume of the rectangular prism = length X width X height

= (30 —2z) cm X (30 — 2x) cm X x cm

= (900 — 120z + 4z*) x = cm”

= 42° — 1202° + 900z cm” which is in the form

V = ax® + bz? + cx +d

cm?,

where

d = 0.

So a model of this form 1s reasonable.

d

We must make a cut greater than 0 cm otherwise the sheet cannot be folded. We also require

30 — 2z > 0 as the side lengths must be positive 2x < 30 Lo

d

-

There 1s a weak, positive,

There 1s a strong, negative,

linear correlation with no outliers.

linear correlation with one outlier.

Ay

o

o

e

o ®

Ay o

¢

o ©

s

o

o

o

.I o

®e o

o

o

.l.

®

..

°

There 1s no correlation.

o

f

o

0

..I

o

°

o

O

» I

- L

There 1s a strong, negative, non-linear correlation with

There 1s a moderate, positive, linear correlation

There 1s a weak, positive,

one outlier.

with no outliers.

no outliers.

2

Day

non-linear correlation with

Mon | Tue | Wed | Thu | Fri | Sat | Sun

Hours worked

8

4

5

10

8

3

6

Number of customers

9

6

D

12

7

1

D

a

Hours worked 1s the explanatory variable. Number of customers 1s the response variable.

b

A number of customers 12

o o

8

Q

O

o

4

()

o

o

hours worked 0

2

4

6

8

10

fos-

Tiffany worked the same number of hours (8 hours) on Monday and Friday. Tiffany had the same number of customers (5 customers) on Wednesday and Sunday. d

The more hours that Tiffany works, the more customers she is likely to have, so we would expect a positive correlation between the variables.

134

Chapter 5 (Bivariate statistics)

3

Competitor | P

a

Q

R

Judge A

5 | 6.5

Judge B

6

3

7 | 85

4

|75

5

4.5

b

s 7

There appears to be strong, positive, linear correlation between Judge A’s scores and Judge B’s scores. This means that as Judge A’s scores

e

increase, Judge B’s scores increase.

o

¢

No, an increase in Judge A’s scores are not likely to cause an increase 1n Judge B’s scores. It 1s much more likely that both scores are related to

2

the quality of the ice skaters’ performances. JudgeA

0

0

4

2

4

6

Job

Al

Time (hours)

b

[

3

1

2

10

75

8

2

|25

9 |

3

|

Job G took the longest to complete (10 hours).

il

Job C involved the most workers (8 workers).

time 1 01 (hours) ]

]

¢

s

3

There is a strong, negative, non-linear correlation between number of workers and time.

the

‘—I—O 1

2

0

J

o

4

a

10

B|C|D|E|F|G|H

4 | 6

6O

5

b

8

Number of workers | 5 |

a

Y

5

|

o)

W

7

o

4

U

20|

) o

S

8

1045, dge B

6

Exercise SA

0

2

4

6

o

g

10

x = the number of apples bought by customers

y = the total cost of apples bought We expect strong, positive, linear correlation. This corresponds to D.

number of workers

b

1z = the number of pushups a student can perform in one minute y = the time taken for a student to run 100 metres

We expect moderate, negative, linear correlation. This corresponds to A. A l’y. 0, 0°

I....' ®

©

o e

o

o

©

o

o »L

Chapter 5 (Bivariate statistics)

¢

x = the height of a person y = the weight of the person

d

o %o o

©

o

:

S % o

°

e

y = the height of the student’s uncle We expect no correlation. This corresponds to €. ¢

o

‘y

o

a

b

¢

o

°

°

°

There 1s a moderate, positive, linear correlation between the number of hours of study and the marks obtained.

604 marks =

As the test is out of 50 marks and the outlier is greater than 50, we can assume it is an error and discard fit.

40

Yes, this 1s a causal relationship as spending more time studying for the test 1s likely to

20

O

o

©

* wi o

o

0 ®

0

°

o

®e

-

L

)

30

cause a higher mark.

o

o

e oo

6

e

. .

o

%o °

2

®

e

3

o

o

~

°

10

number of hours of study 8

10

12

o

6

Not causal, dependent on genetics and/or age.

O

4

Not causal, dependent on the size/location of the fire.

an

2

-

Causal, an increase in advertising is likely to cause an increase in sales.

O

0

Causal, the childrens’ adult height 1s determined by the genetics inherited from their parents to a great extent.

e

1

135

1z = the distance a student travels to school

We expect moderate, positive, linear correlation. This corresponds to B. Ay

Exercise 5B

Not causal, dependent on the population of the towns.

r=0.556 There 1s a weak, positive correlation between the number of employees of a company and 1ts export earnings.

04 export earnings ($ million) o

4

°

3 o

o

2

o

e

o

1

0

10

o o

o

20

30

number of employees

40

50

60

70 R0

136 2

Chapter 5 (Bivariate statistics)

Exercise 5B

a

b

P

Ay

;f

;

C

AY

.

.

l"!

i-'.



Y

B

A

ay

T

E

DedFornd (dFc)Real

D

Y

Ay

> T

C

-

r=1

e

Y

>

Y

»=0.6

d

3

-t

f;‘-

5

6

=

So, 7~ 0.786.

There is a moderate, positive correlation between = and y.

COPY ||[DRAW |

Chapter 5 (Bivariate statistics) x|

3

8

5

|14

y | 17 | 12 | 15| El

SUB

List

1 | List

2 | List

1

3

17

3

5

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2

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i

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3 | List

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4

=

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6

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i

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=-

r

=-1

b

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(ax+b)

=20

r2=1

MSe=0

8

y=ax+b

o

4

il

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y

)

12

0

e

12

14

(D

119 | 10 | 16

6

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Exercise 5B

o

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r = —1.

o | I

0

4

8

12

16

B

20

There is a perfect, negative correlation between z and y. r |

3

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8

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(sl e

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1 | List

Re

6

8 | 10

719

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1

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o

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y=ax+b

4

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o

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a b

°

6

0

i

X

6

8

10

r ~ 0.146.

=z

>

12

There 1s a very weak, positive correlation between x and .

137

138

Chapter 5 (Bivariate statistics)

Iy

Student

Exercise 5B

A|lB|C|DI|E|F

Phone calls received

gl

]

Text messagesreceived | 6 | 9 | [Deg)(Norm1]

SUB

List

1 | List 2 | List

1

4

B

3

1

2

2

7

4

3 | List 4

2

text

o

received

8

)

2]

2|

&5 |

8 | 4 |

b

®

Dedlforl) (fed ( E.’L‘C‘Fb)

a b

0

r2=0.66595518

MSe=2.66539923 =ax+b

(COPY](DRAW]

0

:

]Phfll‘lf: calls received o

=0.98479087 =2.54372623

r =0.81606077

¢

o

7

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4

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o

104

2 | 2 |

y

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a

0|

|G|

1

2

3

4

5

6

-

7

S0,

7~ 0.816.

8

¢

There is a moderate, positive correlation between phone calls received and text messages received.

d

Those students who receive several phone calls are also likely to receive several text messages and vice versa.

5

Athlete

A|B|C|DI|E|F|G|H]|I

Age (years)

12116

J | K|

L

(16 [ 18 | 13 | 19 | 11 | 10 [ 20 | 17 | 15 | 13

Distance thrown (m) | 20 | 35 | 23 | 38 | 27 | 47 | 18 [ 15 | 50 | 33 | 22 | 20

= SUB

List1 | List 2 | List 3 | List 4

1

12

20

16

23

2

16

4

18 [IIES

3

b

y

35

38

GRAPHJ CALC J TEST] INTR J DIST JII=N So,

k)

0

o oo

a

B i

o°f

=

o

LinearReg (ax+b) a

b r

=3.28947368

=-20.342105

=0.91730097

r2=0.84144108

MSe=23.2447368

y=ax+b

COPY

r =~ 0.917.

There i1s a strong, positive correlation between the age of the young athlete and the distance thrown. In general, the higher the young athlete’s age, the further they can throw a discus.

Chapter 5 (Bivariate statistics)

Temperature (x °C)

20 [ 3212713935

Exercise 5B

24 | 3036 ]| 29|

35

Drying time (y minutes) | 100 | 70 | 95 | 25 [ 38 [ 105 | 70 | 35 | 75 | 40 Degl(orm1]

SUB 1 2 3 4

List 1 | List 2 | List 3 | List 4 26 32 27 39

GRAPH)

d

100 70 95 25

CALC J TESTJ

120

25

INTR J DIST JI=EE

(S

A drying time (y minutes) o

o

p

80

o

I

40

e

»

o

temperature (x °C) a

24

26

28

30

32

34

36

¢

v=ax+b

So,

>

38

40

There 1s a very strong, negative correlation between femperature and drying time. In general, the higher the temperature, the lower the drying time.

COPY

r~ —0.987.

Number of supermarkets | 5 | 8

|12 |

7 | 6

|2

|15

(10|

7 |

3

Number of car accidents | 10 | 13 | 27 [ 19 | 10 | 6 | 40 | 30 | 22 | 37 E SUB

[Deg)[Norm1) List 1 | List 2 | List 3 | List 4

1 2 3 4

5 8 12 7

10 13 27 T

[Deg)[Norm1]

y

-

H 1

)

bg

=

B

g

O

w

GRAPH]

CALC ] TEST J INTR J DIST /I

CALC

oy number of car accidents o

40

30

°

20

o 0

a

H

10

°o o 0

2

4

6

8

number of supe nnprket_s_ 10 12 14 16 18

139

140

Chapter 5 (Bivariate statistics)

b

Exercise 5B

— Lingflffi%gg'fi*g)w b r

¢

The point (3, 37), which represents 37 car accidents in a town with 3 supermarkets, is

=8.61052631 =0.5715717

an outlier.

r2=0.32669421

MSe=106.752631

y=ax+b

So,

d

(COPY)(DRAW)

r~ 0.572.

|

Deaia] [fea

LinearReg(ax+b) a

=2.675

b

=-1.7333333

r =0.92757522 r £Z

Y

This scatter diagram has 2 data points above the least squares regression line and 3 points below the regression line. This pattern is shown in the residual plot in B. A residual

-+

-

X

Y

This scatter diagram has 3 data points above the least squares regression line and 2 points below the regression line. This pattern is shown in the residual plot in C. A residual

o

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o

o

X

o

>

€T

v C

°

v

AY o

°

-

-+

Qo

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o

= Y

€I

This scatter diagram has 2 data points above the least squares regression line, 2 points below the regression line, and 1 point on the regression line. This pattern 1s shown in the residual plot in A. A residual

°

5

.

o

E'

150

2

Chapter 5 (Bivariate statistics)

204y

15

:

10

:/,:J./ °

5

o

080

1

A

!

U'E_

~0.5 ~1 B

et

4

S

%

4

.

Y

Aresidualg e 2 o 0_ 2 _i

The values on the x-axis do not correspond to those

on the scatter diagram.

154 20

15

3o 4 . 5 * ’

o

1

2

athletes

3

4

o

>~

5

s

.

3

1% e

—r

The residuals in this plot indicate that all values are within about 4+3 of the regression line. The scatter diagram shows that the values are within about +3 of the regression line. So D is the correct residual plot.

e

e

corresponding These were

A, E, B, F, and I

¢

The residuals in this plot indicate that all values are above the regression line. The scatter diagram however shows that there are values below the regression line. So € is not the correct residual plot.

to

the

points

above the z-axis threw the discus further than expected.

b

] ° ie

.. °

Aresidual g’ L ' . 0 —1 '

The

°

e

_QI

a

T

The residuals in this plot indicate that there are values which are more than 4 units from the regression line. The scatter diagram however shows that the values are within about +3 of the regression line. So B is not the correct residual plot.

™

A residual

02O

3

So A is not the correct residual plot.

o

10 5

D

z

]

Y

C

-

D

e ’

Residual plots

o

o

2

Aresidual

Activity 2

athletes H, G,

The athlete closest to the z-axis is A. So athlete A performed closest to what the

linear model predicted. No, it is not possible to determine which athlete threw residual plot

between

the

the discus furthest. The only shows the difference actual

predicted distance.

10

A residual

distance

and

the

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oo —10

Y

-

1‘; T L

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e

’ c°

IS

D

20

h?}%;m)

Chapter 5 (Bivariate statistics)

L

a

When

z=3,

y=0.92(3)+ 1.8 =4.56

When

=z =7,

y=0.92(7)+ 1.8 =8.24

When z =4,

y=0.92(4)+ 1.8 = 5.48

)

Activity 2

AY p

i

So, the table is: ,

L | Yobs | Ypred

residual =

e

b

4

Yobs — Ypred

—0.72

3

6

4.56

1.44

4

D

H.48

—0.48

7

8

8.24

—0.24

Residual plots

//

AL A

“Y T Y

/

2 |

/

/

Y

/(yzg'QQ‘E +1.8

2|

6 |

8

L

2 residual o

1

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8

o

4

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i

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0

2

4

6

8

10

oM -

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LinearReg (ax+b)

a =1.64705882 b =0.36764705 r =0.98664054 rz2=0.97345956 MSe=1.25735294

y=ax-+

COPY

Using technology, the least squares regression line is

i

We finq Ypred fOr each data point by evaluating y ~ 1.65x + 0.368 for each of the z-values.

Z | Yobs |

y ~ 1.65x + 0.368..

Yored | residual = Yobs — Yored

2 3 3.66 D 9 8.60 7 | 13 | 11.90 10 | 16 | 16.84

—0.66 0.40 1.10 —0.84

151

Chapter 5 (Bivariate statistics) v

1.5

Activity 2

Residual plots

residual

ya

152

15 17

10 | 12 | 13

16 12

2 ]

4

10

12

14

16

DeglornI] (d/c)Real

Line arReg (ax+b)

a

b I;

r MS

=0.7425742 3 =5.465346 23 =(0.984162 14 9=fl.96857g 11 e=0.481384818

v=ax+b

Using technology, the least squares regression line 1s y ~ 0.743x + 5.47.

i

We find ypeq

y =~ 0.743z + 5.47

evaluating

o

for each data point by

for

each of the z-values.

B

B

0

8

7.69

0.31

7

10 | 10.66

—0.66

8

12 | 11.41

0.59

111

13 | 13.63

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151

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L

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4

6

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4

i

10

T

12

obs —

3

0.8y residual 0.4

e

14

16

0.40

s

Ypred

Chapter 5 (Bivariate statistics)

Activity 2

Residual plots

16

0

Il

2

4

6

8

10

12

14

16

Vs

o

12

LinearReg (ax+b)

a =-0.9468479 b =17.750309 r =-0.8602832 r2=0.74008728

MSe=10.6131025 y=ax+b

Using technology, the least squares regression line is

lii

v

y ~ —0.947x + 17.8.

>

s

We find yueq for each data point by evaluating y ~ —0.947x + 17.8 for each of the x-values.

g I | 18 | 16.80 14

6

12 | 12.07

—0.07

15

2

BB 1]

—1.59

4

10 | 13.96

—3.960

12

6

—0.39

9.23

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0

4

6

o

8

10

12

o

14

16

O

—2 _,;1

Q

Y

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9

64 residual

2

R

>

T

A

153

154 6

Chapter 5 (Bivariate statistics) a

Activity 2

Residual plots

1004 giastolic blood pressure (mm Hg)

95

o

87.5

°

°

5

85

-6 |

80 75 70 -1’ \ a0

b

1

0

20)

weight (kg»)l

90

100

From the graph, the residual for the point From the equation, when

= = 82,

110

(82, 87.5)

is about 2.5.

ypreq = 68.5 + 0.2 x 82

= 84.9 the residual = yobs — Ypred = 87.5 — 84.9 = 26

il

From the graph, the residual for the point From the equation, when

= = 103.3,

(103.3, 81.4)

is about —8.

Ypreqa = 68.5 4+ 0.2 x 103.3 = 89.16

.

the residual = yobs — Ypred = 81.4 — 89.16 = —7.76

8

34 residual

4

.

o

60

o 70

80

90

100

120

o o

—4

o ©

Y

o

o

_

weight (kg)

Chapter 5 (Bivariate statistics)

PART 2: 1T

Activity 2

Residual plots

155

ANALYSING RESIDUAL PLOTS

A

residual

5

B

A

o

residual

.

Qo

o

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U

10

0

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2+

20

30!

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51

o

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The residual plots for A, B, and D show points randomly scattered about the x-axis, with no obvious pattern.

C

Iflflidual 10'. 5l

0

e

®

2

3

o

—10"

°e

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==

—5

The residual plot for € however shows a clear, non-random pattern. So the residual plot for € shows a regression line which 1s not a good fit for the data.

0°

7 , °

°

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5

15‘y

©

10

-/

Y

a

A

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1

0

¢

1

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10

o

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Most of the residuals in this plot are above the x-axis. The scatter diagram however shows that only three data values are above

the regression line.

|

4 residual 1

1

Most of the residuals in this plot are above

°

0

o

T

'

So A i1s not the correct residual plot.

o

o

?

.

10

the

z-axis.

The

scatter

diagram

however

shows that only three data values are above

}j

the regression line.

I

So B 1s not the correct residual plot.

Chapter 5 (Bivariate statistics)

156

C

Activity 2

Residual plots Most of the residuals in this plot are below the x-axis, with only three residuals above it. The scatter diagram shows that only three data

A residual

1

0

D Y

10 .

0

. T

values are above the regression line, and the

values on the x-axis correspond to those on the scatter diagram. So € is the correct residual plot.

o

Y

This is very similar to residual plot €, except the values on the z-axis do not correspond to those on the scatter diagram. So D is not the correct residual plot.

A residual s

o

0

o

H

-

1

10

-{f O

Y

b

The residual plot does not appear to be random, so a linear model may not be appropriate for the data.

3

A residual 3 )

o

10

o

1

shots

0

takeE

5

—1

10°

°15

3

uniform numbes_ 5

5

15

®

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Y

L

o

D 0

o

—2

®

a

Yes, the points in both plots appear to be randomly scattered.

b

A linear model 1s most appropriate for the points scored vs shots taken data set. The residuals in this plot are generally smaller, which means that the points are generally closer to the least squares regression line.

a

T

1

y

i

2

3

4

H

332925242018 321

6

7

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AY O

24

°

o

16

8

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e T

O

9

4

6

8

-

8

9

9 | 8

Chapter 5 (Bivariate statistics)

Il

[

Activity 2

Residual plots

Goes @okea

LinearReg

a

(ax+b

=-3.1666

Using technology, the least squares regression line 1s y ~ —3.17x + 35.7, r~

il

157

and

—0.994.

Using technology, the residual plot is: 1.6 4 residus

1.2

0.8 0.4 Y=

0 —0.4 —0.8 —1.2

!

167"

There is very strong correlation and the residuals are randomly scattered, so the least squares regression line 1s appropriate to model the data. x | 22|

3.7

9.5

y | 3.6 | 7.1 | 22.5

6.2 13.3

14 1.9

3.9 7.6

7.5 16.8

8 18.2

2.9 11.5

|2y, 20 16 12

V=

iv

COPY

Using technology, the least squares regression line 1s y ~ 2.54x — 2.11, r ~ 0.999.

and

158

Chapter 5 (Bivariate statistics) il

Activity 2

Residual plots

Using technology, the residual plot 1s: 0.64 residual O

0.4

S L

U

Vs

0.2 et

—0.2

=

04" iv.

C

The residual plot shows a clear pattern and does not appear random. regression line is not appropriate to model the data. i o | y | 13|

I

9 1

1 6

1126 | (14| 2

91 9

9 |10

7 | 2 | 8 | 4 [

So the least squares

10 11

16ly o

12

o o

&

o o

4

o O

0

i

?

0

2

@

4

6

8

T

10

12

b

o

LinearReg (ax+b) a

=0.49094202

b

=4.5597826

r =0.38491211 r

e °

°

o 15

)

time

20

25

(

(min

2-

30

There 1s a moderate, positive, linear correlation between time in the store and money spent.

Chapter 5 (Bivariate statistics)

Spray concentration (x mL perL) | 3 | 5 Yield of tomatoes per bush (y)

6

8

9

Review set SA

173

11 | 15

67 | 90 | 103 | 120 | 124 | 150 | 82

ot y (tomatoes) o

120

©

o O

80

¢

o

40 0

b

x (mL per L) 0

2

[

o

4

0

8

10

12

14

-

16

o

LinearReg(ax+b) a =2.3938053 b r

=85.6504424 =0.33958865

r2=0.11532045

MSe=851.567256

y=ax+Db

So,

r ~ 0.340.

There is a very weak, positive, linear correlation between spray concentration and yield.

¢ d

Yes,

(15, 82)

is an outlier which is affecting the correlation.

B Degorni) (d7c)Real LinearReg(ax+b) a =9.92857142 b =39.5 r =0.99427855 rz=0,98858984 MSe=11.9464285 v=ax+b

COPY || DRAW

So,

r~ 0.994.

Yes 1t 1s now reasonable to draw a least squares regression line.

€

B

rohm Rk

LinearReg(ax+b) a

b

=9.92857142

39.5

0.99427855 0.98858984 MSe=11.9464285 r 1"2

v=ax+b

COPY || DRAW

Using technology, the least squares regression line 1s y ~ 9.93x + 39.5.

f

The gradient of the least squares regression line ~ 9.93. This indicates that for every additional mL per L the spray concentration increases, the yield of tomatoes per bush increases

on average by 9.93 tomatoes. The y-intercept of the least squares regression line ~ 39.5. This indicates that 1f the tomato bushes are not sprayed, the average yield per bush 1s approximately 39.5 tomatoes.

174

Chapter 5 (Bivariate statistics)

g

| When

=7,

Review set SA

y=9.93(7)+39.5

~ 109 If the spray concentration 1s 7 mL per L, the yield will be approximately 109 tomatoes per bush. il

When

y =200,

200 = 9.93x + 39.5 9.93x ~ 160.5 r~16.2

It the yield 1s 200 tomatoes per bush, the spray concentration would be approximately 16.2 mL per L. h

In g1, this 1s an interpolation, so this estimate is likely to be reliable.

In g ii, this 1s an extrapolation, so this estimate may not be reliable.

7 | Age(zyears) | 3| 9 | 7| 4 | 4 | 12| 8 | 6 | 5 | 10| 13 Height (y cm) | 94 | 132 | 123 | 102 | 109 | 150 | 127 | 110 | 115 | 145 | 157 d

170‘

" |(Cflfl)

150

°

130

o]

110

I

°

90

,

0 b

2

3|

6

8

10

12

14

LinearReg(ax+b) a =b5.97980613 b =79.966882 r =0.98285952 r2=0,96601284 MSe=15.7322742

y=ax+b

Using technology, the least squares regression line 1s y ~ 5.98x + 80.0. ¢

The gradient of the least squares regression line ~ 5.98. ogrows taller by an average of 5.98 cm.

d

When

z =05,

This indicates that each year, a child

y=~5.98(5)+80.0 ~ 110

So, a 5 year old child would be approximately 110 cm tall.

e

When

y =140,

140 = 5.98x + 80 H.98x ~ 60 xr ~ 10.0

A child would be expected to reach 140 cm in height at age 10 years.

Chapter 5 (Bivariate statistics)

Review set 5B

175

a

The variables are likely to be negatively correlated, as prices increase, the number of tickets sold 1s likely to decrease. This 1s a causal relationship as less people will be able to afford tickets as the prices increase.

b

The variables are likely to be positively correlated, as ice cream sales increase, the number of shark attacks is likely to increase. This 1s not a causal relationship as both of these variables are dependent on the time of year.

a

AY

00 :.: O o - 8o ©

Q0

°

In general, as x increases, y 1ncreases. So in general, as the rank of x increases, the rank of y increases. The ranks are positively correlated. The rank correlation coefficient is r, = 0.7 (€).

°

-

£

b

AY

i

o

As x increases, y always increases. So the ranks are perfectly positively correlated. The rank correlation coefficientis g =1 (A).

. 0°

o i'.'. ) -

£

C

AY

In general, as x increases, y decreases.

L

.

So 1n general, as the rank of x increases, the rank of y

=13 PO A

:

decreases. The ranks are negatively correlated. The rank correlation coefficient 1s r, = —0.4

.

(B).

o 2 -

xr

Student

A|lB|C|D|E|F|G|H|[TI]]I

Mathematics test | 64 | 67 | 69 | 70 | 73 | 74 | 77 | 82 [ 84 | 8 Art project a

g

8 | 82 | 80 | 82 | 72 | 71|70

(71

(62|

66

aAn project mark O

e

80

°

o

?0

o

o

o o o

60 i\/\

b

60

70

80

» Mathematics test mark

90

There is a strong, negative, linear correlation between the Mathematics and Art marks.

176 ~ Chapter 5 (Bivariate statistics)

¢

Review set 5B

Fdles] [0k

LinearReg(ax+b)

a =-0.9751295 b =146.74715 r =-0.9296181 rz=0.86418995 MSe=9.0126943

v=ax+b

So,

COPY J|[DRAW|

r~ —0.930.

Speed (v kmh™1)

10

20

30

40

o0

60

70

80

90

Stopping time (t s)

1.23

1.54

1.88

2.20

2.92

2.83

3.15

3.45

3.8

a

10 + 20 + 30 + .... + 80 + 90

v=

9

= 20

the mean point b

-

12341544188+ ... +3.45 +3.83 9

~ 2.51

(7, £) is (50, 2.51).

At (s) ;&‘Zl ............ ;.';.:;;;;;;1..; ................................... y'k' mean poit , (50, 2.

4

I 2 1 0

¢ d

0

20

;

40

5560

80

» v (kmh™!

100 110~

)

|

We estimate that the stopping time for a speed of 55 kmh™!

il

We estimate that the stopping time for a speed of 110 kmh™!

is about 2.7 seconds. is about 4.4 seconds.

The estimate in ¢ i 1s more likely to be reliable, since it 1s an interpolation.

5 | Area (x cm?) | 100 | 225 | 300 | 625 | 850 | 900 12 | 13 | 24 | 30 | 35

6

Price (£y)

g —0

30 y=0.033bx + 3.27

20

o

10 0

b

0

100

200

[g

Lineazgegégfi:EE%S b

=3.28797385

r =0.99422161

rz=0.98847662 MSe=1.87254901

y=ax+b

So,

r =~ 0.994.

300

_— _—

400

500

_— o

_—

600

700

¢

d

800

z (cm?) . 900

There 1s a very strong, positive correlation

between the area of a canvas and its price. :

B

The regression line 1s y =~ 0.0335x + 3.27.

Chapter 5 (Bivariate statistics) e

When

x=1200,

Review set 5B

177

y = 0.0335(1200)+ 3.27 ~ 43.42

We estimate that a canvas with area 1200 cm? will cost about £43.42. This is an extrapolation however, so it may be unreliable. Number of waterings (n) | 0 | Flowers produced (f)

a,d

/

100

3

4

D

6

7

18 [ 52 | 86 | 123 | 158 | 191 | 228 | 250

3004 200

1 [ 2

pal

//

07/ 0 2

4

n

6

&

There is a very strong, positive correlation between number of waterings and flowers produced. b

Deg)[Norm]]

LinearReg(ax+b)

a =33.9761904 b =19.3333333 r =0.9990374 r2=0,99807573 MSe=15.579365 y=ax+b

The regression line 1s [ ~ 34.0n + 19.3. ¢ e

Yes, plants need water to grow. So it is expected that an increase 1n watering will result 1in an increase in flower production. I

5 times a fortnight = 2.5 times a week

When

n = 2.5,

f=34.0(2.5)+19.3

When

n =10,

f ~ 34.0(10)+ 19.3

~ 104 Violet can expect from this bed.

il

about

~ 359 104

flowers

Violet can expect from this bed.

about

359

flowers

The estimate for

n = 2.5

1s reliable as it is an interpolation.

The estimate for be a problem.

n = 10

1s unreliable as it 1s an extrapolation and over-watering could

178

Chapter 5 (Bivariate statistics)

Review set 5B

7 | Price ($p) | 2.50 | 1.90 | 1.60 | 2.10 | 2.20 | 1.40 | 1.70 | 1.85 Sales (s) | 389 | 450 | 448 | 386 | 381 | 458 | 597 | 431 a

As

600 -

b

°

Yes, the point

(1.70, 597)

is an outlier.

[t should not be deleted as there is no evidence that 1t 1s a recording error.

500

450

T e

o

e

400

T 1.40

c

1.80

2.20

p(d) 2.60

-

Dedfiornd (d/c)Rea)

LinearReg

(ax+b)

a =-116.46005 b =664.50197 r =-0.5884164 rz=0.34623396

MSe=3722.76173 v=ax+b

COPY

The regression line 1s

s =~ —116p + 665.

d

The gradient of the least squares regression line =~ —116. This indicates that for every additional dollar the price increases by, the number of sales decreases by 116.

e

No, the prediction of sales of Supa-fizz if it was priced at 50 cents would not be accurate, as it 1s an extrapolation well beyond the range of data values given.

8

Time (t hours)

3

6

|9

12 | 15 |

18

21

24

o

=

W

=

Ol

Sy

=]


—5

has

a=2,

Now b _ 0 2 2(2)

c= —3.

= Jon

¢

b| L

L

=z = —2.

= = % b=0,

and

=0

the axis of symmetry 1s x = 0.

¢c = —5.

c= —T.

207

208

Chapter 6 (Quadratic functions)

f y=-52°+7r

has

Exercise 6F

a=-5,

b=7,

and

c=0.

the axis of symmetry 1s x = 1—7{'}.

b

Now

——

2a

= —

c=4.

and

b= -1,

a=-3,

has

y=-3z°—xz+4

g

—1

2(—3)

_ 1

6 the axis of symmetry is = = —z. h

y=10x —3z%

has

b

a= -3,

b=10,

and

¢ =0.

10

Now - =2 = T3

.

.

the axis of symmetry is

i

f(z)=4%2°+2—1 Now

b

has

.

5

x = 3.

a=3,

b=1,

and

c=—1.

11

2a

2(g) T

R

1

= —4 the axis of symmetry 1s = = —4.

a

y=0

when

(z+2)(x—5)=0 o x=-—2

or

r=295

the x-intercepts are —2 and 5.

lies halfway between —2 and 5, so the axis of symmetry 1s x = 3.

ol

_2; 2= 2

b y=(z+2)(z—5) = 2° — 5z + 2z — 10 :$2—3$—10

has

(1,:1jl

b:_3gl

the axis of symmetry i1s = = 3. Y

5

and

¢ =

—10

Chapter 6 (Quadratic functions)

6

f(z)=azx’>+6x—4

has a=a,

The axis of symmetry 1s

=z = —2,

b=6, oy

S5

209

and c= —4. b

so

Exercise 6G

a

= —2

2a

4a =6

a=3

a

i y=222+52x4+1 2a

has

a=2,

b=5,

y=22°+5x+7 b

Now

——

2a

c=1.

and

c=7T.

2(2)

the axis of symmetry 1s = = — i

and

has

a=2,

e| O

7

b=5,

=2

2(2)

_5

4 the axis of symmetry 1s

ili

y=22°+5x—-4 b

Now

——

2a

has

= = —%.

a=2,

b=5,

and c= —4.

5

= ——

2(2)

_2

4 the axis of symmetry is x = —%. The axes of symmetry are the same. b

The value of ¢ only affects the vertical translation of the graph. symmetry (since it i1s a vertical line).

1

a

y=z°—4zx+2 has

Now

a=1,

b= -4,

and

¢ = 2.

the vertex

a > 0,

y=2%—4(2)+2

= =2,

— —2

1s

(2: _2)

so the shape is \/

the vertex (2, —2) turning point.

1is a minimum

b

= =

b

—2 —_"2_9o 2(1) 2a the axis of symmetry is = = 2.

When

Hence

5o

It does not affect the axis of

depends only on a and b.

a

y=(z+3)(z—-1) has

x-intercepts —3 and 1.

the axis of symmetry is _ . s 3+l 3 | y=(-1+3)(-1-1) When z=-1, .

= (2)(—2)

= th CVCrrtexcx

a > 0,

1 1s

( —1,1

—a). 4)

so the shape is \/‘

the vertex

(—1, —4)

turning point.

1s a minimum

Chapter 6 (Quadratic functions)

210 ¢

Exercise 6G

y=2x°+14 has

a=2,

Now

y = —3x° + 1 b=0,

and

c=4.

—i=—i=0

2a

has

Now

2(2)

the axis of symmetry i1s = = 0. When

z=0,

y=14

2a

z=0,

y=1

the vertex

(0, 1)

has

has

z-intercepts 4 and —2. the axis of symmetry i1s

a =—-1,

y=(1—4)(1+2)

b= -4,

b

Now

= (=3)(3)

=20 = 21

When

== -2,

y=—(—2)*—4(-2)—-9 =—4+8—-9 = —95

g

the vertex is (—2, —5).

so the shape is \/

a < 0, so the shape is /\

is a minimum

turning point.

y=2z%+6x—1 has

a=2, —i

NOW

2a

b=6, —

and

—i

2(2)

¢ = —§

—_

the axis of symmetry

When o =—3,

—1.

the vertex (—2, —5) is a maximum turning point.

y = —2(c +3)(z — 4) has

x-intercepts —3 and 4.

the axis of symmetry 1s

2

_

—3+4

= = —%

i1s

y=2(-3)"+6(—3

When

z =3,

59 —9—1 2

11 the vertex is (—3, —%5).

11 (—3,3 —%)

minimum turning point.

the vertex is (3, 3). a < 0, so the shape is /\

so the shape i1s \/

the vertex

y=—2(

_249

— il

a > 0,

c= —9.

the axis of symmetry i1s « = —2.

the vertex is (1, —9).

(1, —9)

and

—4

— g

the vertex

is a maximum

turning point.

y=—x°—4z —9

a > 0,

= ()

2(—3)

y=(x—4)(x+2)

When z=1,

c=1.

a < 0, so the shape is /\

(0, 4) is a minimum

2

and

the vertex is (0, 1).

turning point.

e

s

When

so the shape 1s \/

the vertex

b=0,

the axis of symmetry 1s = = 0.

the vertex is (0, 4). a > 0,

a=—-3,

isa

the vertex

(%, %)

isa

maximum turning point.

Chapter 6 (Quadratic functions)

Exercise 6G

211

I y=—22’+2-5 has Now

a:—%.}

b=1,

and

¢=

—5.

has

b

1 =1 —— = e 2(—3)

—7,

and

¢ =

Now —— =——" =14 aa

the axis of symmetry is

When z=1,

a:%,,

x = 1.

2(7)

the axis of symmetry 1s x = 14.

y=-1(1)2+1-5

When z =14,

y= 1(14)* —7(14) + 6 = —43

the vertex is

the vertex is (14, —43).

(1, —3).

a > 0,

a < 0, so the shape is /\

the vertex

(1, —3)

the vertex

isa

a

(14, —43)

minimum turning point.

maximum turning point.

2

so the shape is \/

isa

i When =0, y=(—-1)(-7)=7 the y-intercept 1s 7. When y =0, (x—1)(x—7)=0 o x=1orT7 the z-intercepts are 1 and 7.

il

The axis of symmetry is halfway between the x-intercepts 1 and 7. So, the axis of symmetry 1s

iii

When

x =4,

Y

= = 4.

v

y=(4—-1)4-7)

= (3)(—3)

V4, —9)

The domainis {x | x € R}. The range is {y |y = —9}.

= —9 So, the vertex is b

y=-—2°—-—6x—8

i

When

z =0,

(4, —9). has

a=-1,

b=

-6,

y=—8

the y-intercept 1s —8. When y =0, —x° —6x—8=0 Using technology, x = —2

and

c=

8.

iv

or —4

the x-intercepts are —2 and —A4. as

!

b

—6

_EE_"_m—U'__3 So, the axis of symmetry i1s x = —3.

lii

When

z = -3,

y=—(-3)"—6(-3) -8 =—-9+4+18—38 e=pl

So, the vertex is (—3, 1).

v

The domainis {z |z € R}. The range is {y | y < 1}.

212

Chapter 6 (Quadratic functions) c

y=6bxr—=zx °

When

has

a=—1,

=0,

Exercise 6G b=6,

and

c =0.

y=20

iv

the y-intercept is 0. When

y =0,

6z —x° =0 Using technology,

= =0

or 6 -

the x-intercepts are 0 and 6. b

—_——

.

2a

——

6

2(—1)

:3

So, the axis of symmetry is When

= = 3.

v

= = 3,

y = 6(3) - 3°

ot

The domainis {z |z € R}. The range is {y | y < 9}.

=18—-9 =9

So, the vertex is (3, 9). When

=0,

y=—(—1)(—2) = -2

v

AY

the y-intercept 1s —2.

When

y =0,

—(x—1)(x—2)=0 ox=1

or 2

the z-intercepts are 1 and 2. The axis of symmetry is halfway between the z-intercepts 1 and 2. So, the axisi of symmetry 1si = —= 33.

So, the vertex is (%, +).

v

The domainis {z |z € R}. The range is {y |y < 5}

Chapter 6 (Quadratic functions) e

y=22°+4x—-24 i

When

=0,

has

a=2,

b=4,

and

y=-24

c= —24. iV

y=22%+42 - 24

the y-intercept is —24. When

r=—14Y

y =0,

= ~ —4.61

.

e

2(2)

i

When

z = —1,

~

or

—4.61

~

2.01

: A—24

i

So, the axis of symmetry 1s

%

;

or =~ 2.61

the z-intercepts are ~ —4.61 ~ 2.61.

.._E__j_

I

B

22> + 4z — 24 = 0 Using technology,

Exercise 6G

V(—1,—-26): (=1, -26):v

v

=z = —1.

The domainis {x |r € R}, The range is {y | y > —26}.

y=2(—1)°+4(-1) — 24 =2—-—4—-24

= —26

So, the vertex is (—1, —26). When

=0,

a= -3,

b=4,

y=—1

and

c= —1. v

the y-intercept 1s —1 When

wll—*

I

has

wlm

y=-32°+4x—1

y =0,

—3z° +4rx—1=0 Using technology,

= = % or 1

the z-intercepts are + and 1.

izl oa

o8 2(—3)

23

i

When

z = %

y=—3(3)" +4(3) — 1 =—3+3-1 _3 1

So, the vertex is (

8% v

So, the axis of symmetry is x = %

b

f

,5)-

The

. domainis

The range is

4r

]

{z |z € R}. {y |y < %}

213

214

Chapter 6 (Quadratic functions) g

y=22°—5x+2

I

has

Exercise 6G

a=2

b= -5,

and

When =0, y=2 the y-intercept is 2. When

c= 2.

iv

y =0,

22° — b +2 =0

Using technology,

= = % or 2

the z-intercepts are = and 2.

i

-

2 —_ "9 _> 2a

2(2) 4

So, the axis of symmetry is x = 2. iii

When

v

z = 2,

The domainis

{z | z € R}.

The range is {y |y > —%

:

y=2(3)" —5(3) +2 _ 25

25

=§ ~ 7T T2 _9

8

So, the vertex is

h

1

(%, —%

.

When z=0, y=(-5)(1)=-5 the y-intercept i1s —5. When

iv ~

y =0,

2z —5)2z+1)=0

$=%Or

B

:

_%

y= (2

— 5)(2z + 1)

the x-intercepts are % and —%.

Il

i

The axis of symmetry is halfway between the x-intercepts % and —%.

So, the axis of symmetry is = = 1. When z =1,

y=(2(1) —=5)(2(1) + 1)

= (—3)(3) =9

So, the vertex is (1, —9).

v

The domain is {z |z € R}, The range is' {y | y = —95. 9}

Chapter 6 (Quadratic functions) y:—im2+2m—3 I

has

a:—i,

b=2,

and

When =0, y = -3 the y-intercept 1s —3. When

Exercise 6G

c= —3. iv

y =0,

—}13:2 +2r—3=0 Using technology,

© =2

or 6

the x-intercepts are 2 and 6. -

b

2

P

4

(¢

So, the axis of symmetry 1s il

When

= = 4.

v

The domainis

{z |x € R}.

The range is {y | y < 1}.

z =4,

y=—1(4)°%+2(4) -3 =-—-44+8-—-3 =1

So, the vertex is (4, 1). 3

y=z*—8r+5 Now

_i

20

—

has ——_8

2(1)

a=1, —

2 =4,

and

c= 5.

4

the axis of symmetry is

When

b=-8,

x = 4.

y=4*—-8(4)+5 =16 —-32+5 = —11

So, the vertex is (4, —11). a > 0,

so the vertex 1s a minimum turning point.

the minimum value of y = 2° —8x +5

has

y=—x2—10z+4 Now

S

2a

N

2(—1)

z = -5,

b=—10,

y = —11

and

when

z = 4.

¢ =4.

= =9

the axis of symmetry 1s

When

a=—1,

is

x = —5.

y=—(-5)*—10(-5)+4 = —25+ 950+ 4 = 29

So, the vertex is a < 0,

(—5, 29).

so the vertex 1s a maximum turning point.

the maximum value of y = —2° — 10z +4

is y =29

when

z = —5.

215

Chapter 6 (Quadratic functions)

216 ¢

y=32°+32z—2

roye

Now

2a

-

has

a=3,

L

S8

2(3)

Exercise 6H b=3,

and

c= —2.

2

the axis of symmetry 1s = = —%

11) So, the vertex i1s (—35,1 —=-).

a > 0,

so the vertex 1s a minimum turning point.

the minimum value of y =32% +32x —2 d

y=

—%$2+7:1:—4

Now

_2 2a

has

71

=17,

b=7,

and

when

= = —%.

c = —4.

=

2(_2)

the axis of symmetry is

When

a = —%,,

is y = —%

x = 7.

y=—-4(7)°+7(7)—4

=—-224+49—-4 _ 41 2

So, the vertex is (7, 4. a < 0,

so the vertex 1S a maximum turning point.

the maximum value of

1

a

y = —%CUQ +7x—4

15

y=

%

when

= = 7.

Since the z-intercepts are 1 and 2, y = a(z — 1)(x — 2). When

=0,

AY

y=4

4= a(~1)(-2)

4

a=2

The quadraticis

y =2(z — 1)(x — 2).

-

P

>

Y

b

The graph touches the z-axis at = =2, When

z =0,

so

y = a(x — 2)*.

AY

y=12

12 = a(—2)?

12

a=3

The quadratic is y = 3(z — 2)2.

-

>

>

>

Chapter 6 (Quadratic functions)

¢

Since the x-intercepts are 1 and 3, y = a(x — 1)(x — 3). When

z=0,

Exercise 6H

217

AY

y=3

3 =a(-1)(-3)

3

oa=1

The quadraticis

y = (xz — 1)(x — 3).

B

P

1IN

_3

T

Y

d

Since the z-intercepts are —1 and 3, y = a(x + 1)(xz — 3). When

=0,

AY

y=3

3

3=a(l)(—3) oa=—1

The quadraticis

y = —(x + 1)(x — 3).

—1/

-

'/

1y

3

"

T

Y

e

The graph touches the z-axis at z =1, When

=0,

so y = a(z — 1)2.

y= -3

AY T

R

]'

.

|

3/\

—3 =a(-1)* oa=—3

The quadratic is y = —3(xz — 1)2. Y AY

f Since the z-intercepts are —2 and 3, y = a(z + 2)(z — 3). When

=0,

12/\

y=12

12 = a(2)(—3)

-

oa=—2

2/’-:}

The quadratic is y = —2(z + 2)(x — 3).

a

The axis of symmetry

= = 3

lies midway between the z-intercepts.

the other xz-intercept 1s 4.

the quadratic has the form y=a(x —2)(x —4) where But when

=0,

y

Y

a >0

'

=12

12 = a(—2)(—4) a,

——

3

-

3

=3

The quadratic is y = 3(z — 2)(z — 4).

1 unit:1 unit -«

P

¥

\ 2

/ '

4

T

\

S,

L

218

Chapter 6 (Quadratic functions) b

Exercise 6H

The axis of symmetry x = —1 lies midway between the x-intercepts. the other z-intercept 1s 2. the quadratic has the form . AY y=a(xr+4)(x —2) where a + 4t +3

metres,

t >0

a H(0)=—4(0)*+4(0)+3 =3 the pier 1s 3 m above the water. b

For the quadratic function H(t), Since

a < 0, :

the shape is

a =—4,

b=4,

and

¢ = 3.

/\

:

b

The maximum height occurs when

¢ = ——

2a

= —

4

2(—4)

1

= —

2

So, it takes Andrew 0.5 seconds to reach the maximum height of his dive.

¢ H(3)=-43)*+4(3)+3 =—1+4+2+3 =4 So, Andrew 1s 4 m above the water at his highest point. d

Andrew hits the water when

Bl (et DegomT

H(t) =0

B

aX? +bX+c=0

—

a

4t 4 4t +3=0

b

TS

= —% or 2

aX? +bX+c=0

c

31[-@}

B REPEAT

Since ¢ must be positive, Andrew hits the water after 1.5 seconds. G

—x2 4+ 20x dollars,

0+0+ 1.7 — 1.7 the javelin was released 1.7 m above the ground.

¢ H(20) = —0.015(20)* + 20+ 1.7 = —6+ 20+ 1.7 = 15.7

the javelin 1s 15.7 m high after travelling 20 m horizontally. d

For the quadratic function H(z),

Since

a = —0.015,

and

¢=1.7.

a < 0, the shape is /\ :

:

The maximum height occurs when

Now

b=1,

b

z = ——

2a

= —

1 2(—0.015)

= 100 3

H(19) = —0.015(429)% + 182 4 1.7 100 = - 50 + == + 1.7

~ 184

the maximum height reached by the javelin 1s about 18.4 m.

e

The javelin hits the ground when

H(x) =10

—0.0152° +

[l

[athlDegForm)

aX9+bX+c=E

(d7c]Real

a

[-0.015

[l

—1.66

[MathDegForml)

aX? +bX+c=0

c

(d7c)Real

¥1

1

HE[-I.EEE}

[SOLVE]PIIM3(CLEAR](EDIT ] r~

4+ 1.7 =0

1.7

68.32539593

[REPEAT)

or 68.3

Since x must be positive, the javelin hits the ground after travelling about 68.3 m horizontally. it 1s reasonable to use this model for

f

0 < o < 68.3.

From e, the javelin travels about 68.3 m before hitting the ground. .

the distance recorded for the throw ~ 68.3 — 3.4 ~ 64.9 m

Chapter 6 (Quadratic functions) a

|

When

=0,

Exercise 6]

231

y=28

8 = a(0)* + b(0) + ¢ c=38

il

The axis of symmetry is halfway between the r-1ntercepts —3 and 3. So, the axis of symmetry 1s x = 0.

b 2a

b=20 Wi

When

z=-3,

y=0

. 0=a(-3)* +0(-3)+8 .

0=9a+8

9a = —8 . {1:_%

So, the quadratic model is y = —%mz + 8. b

The semi-trailer is 4 m wide, so we use the equation in a to find the height of the tunnel when it 1s 4 m wide.

When

z =42,

y=-3(2)?+38

= —32 +38 :%

~ 4.44 m

For heights greater than 4.44 m, the tunnel 1s less than 4 m wide. But the semi-trailer 1s 5 m high. the semi-trailer will not fit through the tunnel.

-

Ay (m)

(20,15.25) (10, 9)

Y

a

-

(30, 19)

fence

75m

>

Let the quadratic model be

When z =10, y =9 When =20, y=1525 When z =30, y =19

T_fl m

y = ax? + bz + c.

o 9=a(10)2+b(10)+c .. 15.25=a(20)2+5b(20)+¢c s, 19 =a(30)* +b(30) ¢

We solve the system of equations

technology.

(m)

100a 4+ 10b +¢c =9 400a + 20b 4 ¢ = 15.25 900a + 306 + ¢ =19

or or or

100a+ 10b+c =9 400a+ 20b+c = 15.25 900a + 30b+ c =19 simultaneously using

232

Chapter 6 (Quadratic functions) G

an

X+bn

a

[d/c]Real

Y+CE

[E

Z=dn d

100

10

1

2

400

20

1

900

30 R

a = —%.},

Y+Cn

[dic]Real

Z=dn

¥

9

?[

15.2b

1|

Z

19

[SOLVE)(ENERCLEARIEDIT] We find that

[MathlDegformi)

an X+bn

c

1

3

Exercise 6J

0.2%

:

REPEAT

b=1,

and

e c = fi.

So, the quadratic model is y = —g52% + = + +. b

For the quadratic model y, a = —55, Since

a < 0,

the shape i1s

and

¢=

1.

/\

The maximum height occurs when

Now, when

b=1,

= = T

b a

I

=

3

=40

2(— =5

y = —%(40)2 +40 + 1

= =40,

= —20+40+ 3 = 20.25

the maximum height reached by the ball is 20.25 m.

z=75,

¢ When

y=—5(75)°"+75+ 1

= — 122 4+ 75+ & = 4.9375 So, the ball 1s 4.9375 m above the ground after travelling 75 m horizontally. the ball will clear the boundary fence.

10

a

The top of the cliff is 60 m above sea level, so

H(0) = 60

60 = a(0)* + b(0) + ¢ c = 60

b

The vertex has t-coordinate 2

..

—2£ =) L

..

—b=4a

da+b=0

H(2)=80

.. 80=a(2)*+b(2)+60

or

4a-+2b=20 2a +b =10

¢

_ We solve the system of equations El

[Math)DegNorn]]

an X+bn Y=CE a

2

E

MathDegNornl)

_ _ simultaneously using technology.

([d7c]Real

H|:

i mmm

¢ = —5

2a + b =10

an X+bn Y=Cn

c

a 10

We find that

4da+b=0

and

REPEAT b = 20.

-9

Chapter 6 (Quadratic functions)

d

Exercise 6]

233

H(t) = —5t* 4 20t + 60 H(3) = —5(3)* +20(3) + 60 = —45 + 60 + 60 =75 After 3 seconds, the stone 1s 75 m above sea level.

e¢

aX2 +bX+c=0

s -5 c

= (t) =0 0 c. —=5t2 +20t+6 o

—

TR

The stone hits the water when H(t) =0

aXZ +bX+c=0

0,

&

n € Z.

5

5

>

Let the quadratic model be y = ax? + bz + c. When

=0,

y =50

50 = a(0)* + b(0) + ¢ c = 9

The vertex L has x-coordinate 50

..

—Qi

s

= 50

—b = 100a 100a +b6 =20

When

z =50,

y =30

.. 30=a(50)*+b(50)+50

or

2500a+ 50b = —20 250a + 5b = —2

_ We solve the system of equations E

HathDegMormi]

an X+bn

1[ 2l

E

Y=CE

100 250

SOLVE|(MARFIICLEARI[

We find that

100a +b=0 { 9500 + 5b — —2 HathDegMorni]

an X+bn

c

Y=Cn

[d7c)Real

K[E} vL -0.8

1 0 5 m] EDIT

¢« = —125

and

b= —2.

s

e

So, the quadratic model is y = 3z2° — £z + 50.

, , simultaneously using technology.

Chapter 6 (Quadratic functions)

b

Point P has z-coordinate

When

z =80,

Activity 3

Projectile motion

235

100 — 20 = 80.

y= —3=(80)° — £(80) + 50

= 222 — 64 + 50 = 37.2 the tightrope 1s 37.2 m above ground level at point P. ¢

The platforms are 100 apart. the quadratic model is valid for

0 < o < 100.

e 1

2=20t,

a

y=—-49t*4+14.7t+1,

1 When t=1,

t=>0

2=20(1)=20

and

y=—-49(1)*>+14.7(1) +1 = —4.9+14.7+1 —10.8

After 1 second, the ball has horizontal component 20 m and vertical component 10.8 m.

ii When t=2,

2=20(2)=40

and

y= —4.9(2)%+14.7(2) +1 = —19.6 +294+ 1 = 10.8

After 2 seconds, the ball has horizontal component 40 m and vertical component 10.8 m. b

|

For the quadratic function y,

Since

a < 0, the shape is

a = —4.9,

b =14.7,

and

c = 1.

/\

The maximum height occurs when

¢ = _ 2 p e e g 1.5 2a 2(—4.9)

So, the ball 1s at i1ts highest point after 1.5 seconds.

ii When

t=15

y=—-49(1.5)°+14.7(1.5)+1 = —11.020 +22.05 + 1 — 12.025

the maximum height reached by the ball 1s 12.025 m.

iii

When

t=1.5,

z=20(1.5)=30

the ball travelled 30 m horizontally before reaching its maximum height. ¢

1

The ball hits the ground when

3y =0

—4.9t2 +14.7t+1 =0 [Hath) Beg) Hornd

[Hath)BeglFornd) (d7c)Real

aX2+bX+c=E a

[

-4.9

aX2 +bX+c=0 c

Hl

14.7

HE[—U.HEE]

[SOLVE]PIIN3(CLEARI[ EDIT ]

t ~ —0.0666

1

REPEAT

3.066550871

or 3.07

Since ¢t must be positive, the ball hits the ground after about 3.07 seconds.

Chapter 6 (Quadratic functions)

236

ii

When

t~3.07,

Review set 6A

z~20(3.07)~614

the ball travelled about 61.4 m horizontally before hitting the ground.

2

x=30t, a

y=—-49t*4+20t+5,

When

After

t=1,

t=>0

z=30(1)=30

and

y = —4.9(1)%> +20(1) +5 = —4.94+20+5 = 20.1

1 second,

the stone has horizontal

component 30 m and vertical component 20.1 m.

10m

-

For the quadratic function y, Since

a < 0,

the shape is

a = —4.9,

t~2.04,

and

c¢ = 5.

/\

The maximum height occurs when

When

b =20,

120m

¢ = SR 2a

e

ety 2(—4.9)

2.04

y~ —4.9(2.04)% +20(2.04) + 5 ~ 25.4

the maximum height reached by the stone is about 25.4 m. The stone has travelled 120 m when

x = 120

. 30t = 120 . t=4

When t =4,

y=—4.9(4)?+20(4)+5 = —7844+80+5 = 6.6

Since the castle 1s 10 m high, the stone will hit the castle. 3

A trajectory of 45° appears to give the greatest range for the cannon.

1

f(z)=2°—-3z-15

a f(0)=0%-3(0)—15 — —15

b f(1)=1*>-3(1)-15 =1—-—3—15

— 17

»

distance

Chapter 6 (Quadratic functions)

¢

f(z)=3

when

Review set 6A

237

z°—-3z—15=3 > — 3z —18=0

2

MathDegMornd

aXE+bX+c=E C

[(dic)Resl

E

aX2 +bX+c=0

-3

i;[-i

-18

[SOLVE] NENEI3(CLEAR](EDIT ] r=—3

2

[d/

P

f(0)=(5)(—6)

the y-intercept 1s 9.

¢

2

—4x+

—30 the y-intercept 1s —30.

y=—(—4)3 = —16

the y-intercept 1s —16.

H

a

When

y=0,

Bxz—5)(z+2)=0 T = % or —2

the zeros are 2 and —2. b

When

y=0,

—3(z—7)"=0 L=

the zero 1s 7.

¢ £

9

When _I_ 433

f(z)=0, =

12

—

£ Embafmn) R aX2 +bX+c=0

O

r——6

B EabDedlfor) (d7c)eal

a

or

2

C

b

1

aX2 +bX+c=0

o

¥1

4

the zeros are —6 and 2. [SOLVE) DIENA{3([CLEARI(EDIT]

xz[

=N

-5

2

Review set 6A 4

f

AY

\

*‘

a

o

]

--""""-‘—

5

Chapter 6 (Quadratic functions)

0

238

A

‘t‘

\

9

/

"”

-

4

."" t’*

-

19

';r

/

T

\

.

y

T

e’

:’EJ

2

>

4

Y

b

.

AY S *

|, "‘

—4

—2

4

K

Sy =af r

il

'r

e 9--.."‘

2

>

4

x b2

-

4

6

a

The z-intercepts are 2 and 6. 4 1s halfway between 2 and 6, so the axis of symmetry 1s x = 4.

b

The z-intercepts are —3 and 4.

= is halfway between —3 and 4, so the axis of symmetry IS

The only z-intercept is —2, so the axis of symmetry is r = —2.

_9i

.

AY =Y

¢

© = %

7

y=—-2(x—1)(x+3) a

Since

a = —2

which is < 0, the parabola has shape

the parabola opens downwards.

/\‘

Chapter 6 (Quadratic functions)

b

When

z=0,

y=-2(—1)(3) = 6

the y-intercept 1s 6.

¢

When

y=0,

—2(x—1)(z+3)=0 o x=1or

=3

the x-intercepts are 1 and —3. d

AY

6

—3

-

1o\1 N

I

y=—2(z—1)(z + 3) Y 8

a

y=0

when

(z+1)(x—4)=0 '

r=-1or4

the x-intercepts are —1 and 4. % is halfway between

y=2*—-7x—3

has

a=1,

the axis of symmetry is

¢

f(z)=-22°+3x—-5 Now

2a

has

b= -7,

a

a=-2,

_3 4

has

a=-1,

¢ 2(—1) =]

the axis of symmetry i1s x = 4.

=4,

and

¢ = 5.

S

2 2a

When

b=3,

2(—2)

y=-2°+8+5

Now

c¢c= —3.

y=—4°>+8(4)+5 =—16+32+5 = 21

the vertex is (4, 21).

1s

x = %

the axis of symmetry i1s = = 9

and

s |G

b

—1 and 4, so the axis of symmetry

b=8,

and

c=5.

= = %

Review set 6A

239

240

Chapter 6 (Quadratic functions)

b f(z)=(z—-5)(z+4)

Review set 6A

has z-intercepts —4 and 5.

the axis of symmetry 1s

= =

445

1 5

f(3)=(-5(G+4)

= (=3)(3)

-3

the vertex is (3, —5F).

¢

f(x)=3z*+12x —4 Now

o

2a

has

a=3,

b=12,

and

¢ = —4.

2(3)

= —2 the axis of symmetry 1s

=z = —2.

f(—=2) =3(-2)*+12(-2) -4 — 19 — 24 —4

— 16 the vertex is (—2, —16). 10

y=2°

-2z —15

When

z =0,

y=-—15

the y-intercept 1s —15. When

y =0,

El

aX2+bX+c=E

x> — 22 —15=0 r=—3

or5

the z-intercepts are —3 and 5.

C

1

SOLVE |[MaRxI3|CLEARJ{

aX2 +bX+c=0

o

C

a

-2 IE

EDIT

[MHathlDegMormi)

lE

REPEAT

1 1s halfway between —3 and 5, so the axis of symmetry 1s = = 1.

iv When

z=1,

y=1°—-2(1)—-15 —1-2-15

= 16

the vertex is (1, —16).

[d7c)Real

Chapter 6 (Quadratic functions)

11

a

The vertex has x-coordinate 2.

.

\

the axis of symmetry is z = 2. =

Tl'{e axis of symmetry lies midway between the r-1ntercepts.

. ',

the other xz-intercept 1s —1. the quadratic has the form y=a(r+1)(z—5) where

But when

z =2,

AY

Review set 6A =2

241

/ h T

3ltf1it5 ?3Uflit5 5 —1 \/

02, —20) vy

a >0

y=—20

—20=a(2+1)(2—-5) —20 = a(3)(—3) flzg

20

The quadratic is y = 2*(z + 1)(z — 5). The axis of symmetry

= =4

lies midway between the x-intercepts.

the other x-intercept is 1.

the quadratic has the form

y=a(r—1)(r—7) But when

AY

where

a 0. But when

z =0,

y=2

b

:.

2 =ya(3)° S

il—g

The quadratic is y = 2(z + 3)%. Since the z-intercepts are —1 and 2, the quadratic has the form When

=0,

y=—6

L

—6 =a(1)(=2)

The quadraticis

AY

a=3

y =3(z + 1)(x — 2) =3(z* —z —2) — 322 — 32— 6

9 E

‘/{::}

B

. X

Y

y =a(z + 1)(x —2),

a # 0.

242

Chapter 6 (Quadratic functions)

b

Review set 6A

Since the graph touches the z-axis at 4, the quadratic has the form

When

z =2,

y = a(z — 4)*,

a # 0.

y=12 12 = a(2 — 4)7 12 = a(—2)? Loa=3

The quadratic is

y = 3(z — 4)?

= 3(z* — 8z + 16) = 3z° — 24z + 48

13

(7'

8

or

(_1&3)

a—b=-5

) +b(3) +8

or

9a + 3b = —9 o

¢

3a+b=

We solve the system of equations 2] an X+bn

Y=Cn

2

1 2[

1 3

b

il

SOLVE |[MaRF|3

We

!

o

—_ b = —9H

simultaneously using technology.

-3

Y=Cn

]3

¥

1

-3

3a+ b=

g an X+bn

EDIT

find that

{

(3.,—1)\

-

(

1$nhh‘

WL

Fan

—1)?4+b(—1)+8

f(z)=az?+bx +c

REPEAT

a = —2

and

b = 3.

So, the function is f(z) = —222 + 3x + 8. 14

Let the equation of the quadratic be When

=1,

y= -2

When

z=2,

y=6

When

z =3y , =20

y = az® + bx + c.

—Z—fl()

()

b(2)+

20—{1()

()+c

a+b+c= —2 da+2b+c=6

We solve the system of equations

9a +3b+c E

an X+bn a

Y+Cn

[d/c)(Real b

Z=dn

E] c

an X+bn

d

1

1

1

-2

2

A

2

1

B

3

9

3

1

SOLVE]PRNRI3([CLEAR][EDIT |

We find that

a =3,

Y+Cn

¥

1

20

b= —1,

or or or

b(1)+

1‘:"|:

-1 |

Z

-

(d7c]Real

Z=dn

[REPEAT]

and

¢ = —4.

So, the quadratic has equation y = 3z% — x — 4.

=20

a+b+c=-2 4a+2b+c=6 9a+3b+ c= 20 simultaneously using technology.

= T

Chapter 6 (Quadratic functions) 15

We

graph

Yl

=

XZ

—

3X

and

E

2

YQ

=3X"—-5X

set of axes.

on

—24

The graphs intersect at and (4, 4).

16

the

Same

[EXE]:Show coordinates

¥Y1=x2-32%

Y2=3x2-bg-24

Y2=3x2-5)%-24

ettt

55 Y

X=-3

X=4

h = —4.9t*> +19.6t + 1.4 metres

a When t=1,

h=-49(1)>+19.6(1)+ 1.4 = —49+19.6+1.4 = 16.1

After 1 second, the ball is 16.1 m above the ground.

b

For the quadratic function h,

Since

a < 0,

a = —4.9,

the shape is

When ¢t =2,

b =19.6,

and

¢c=1.4.

/\

The maximum height occurs when

¢ = B . 2 2a 2(—4.9)

h=—-4.9(2)"+19.6(2) + 1.4 = —19.6 +39.2 + 1.4 =21

the maximum height reached by the ball 1s 21 m. 17

a

Let the other side of a field be

shown.

¥y m long as

ym

|

So, the total length of fencing required is 3x + 4y. [f there 1s 2000 m of fencing available, then

TmT

3r + 4y = 2000

.

4y = 2000 — 3z .y

=500 — %m

B

()

The total area of the fields

A(z) = z(2y) = 2y m

Substituting equation (%),

2

A(z) = 22(500 — 2x)

= (1000 — 22%) m? 2 b

The area A(z) is a quadratic with a = —2, Since

a < 0, the shape i1s :

The maximum

When

z = 103&,

= 500 — 250

b

r = ——

y =500 — 2 (&3?0) = 250

b= 1000, and ¢ = 0.

/\

area occurs when

[EXE]l:Show coordinates

Yi=x2232

55X

(—3, 18)

E

Review set 6A

2a

= —

1000

QQ_%

{using (x)}

=g

1000

243

244

Chapter 6 (Quadratic functions)

Also

Review set 6A

A (£2) =1000 (£2) — 3 (L:z?fi)z _ _

1000000 3

500000 3

500000 __ — 500000 _ 166 66622

the maximum

possible total area of the two fields 1s 166 666%

m?

when the fields are

250 m by 3335 m. 18

a

§583)=14+2+3=6

b

S(n)=an’*+bn+c S(1)=1

.

S2)=3 S3)=6

1=a(1)*+b1)+c

or

.. 3=a(2)*+b2)+c .. 6=a(3)?+b(3)+c

a+b+c=1

or or

4a+2b+c=3 9%a+3b+c=6 a+b+c=1

We solve the system of equations

da + 2b+ c =3

simultaneously using technology.

9a +3b+c =06

[fatfi Deg flornd

an X+bn Y+CE a

Z=dn

HatiiDeglffornD) [d7c)Real

p

C

an X+bn Y+Cn

1

1

1

1

1

X m| ?|: 0.5

-

4

2

1

3|

L

3

9

3

1

6

We find that a =3,

¢

Z=dn

0

l

2

b= 3,

REPEAT

and

c=

S6)=1+2+3+4+5+6=21 The model in b is

S(n) = in® + in

S(6) = 3(6)* + 3(6) =18+ 3 = [

d

S(60) = 1(60)* + %(60) = 1800 + 30 = 1830

e S(23) =3(23)" +3(23)

B+

= 4.375 This result 1s not meaningful as we cannot find the sum of the first “two and a half” positive integers.

f It is appropriate to use this model when

n € Z*.

Chapter 6 (Quadratic functions)

Review set 6B

245

REVIEW SET 6B —2(1)% +13(1) — 4 =-2+13—4 =7

2

9

does not satisfy the function

[f g(z)=25

f(z)= —2z% + 132 — 4.

then

5

aX2 +bX+ec=0

r°—5r—9=2>5

a

% —5x — 14 =0

-

rx=-—2

o0

L

a

z=0,

]

‘43— 5

™ "

y=(5)(—1) = —9

the y-intercept 1s —9.

When

y =0,

(z+5)(z—1)=0 o x=—0

or 1

the x-intercepts are —5 and 1.

b

When

=0,

y=-2(-3)4) = 24

the y-intercept 1s 24.

When

y =0,

—2(2x—3)(z+4) o

the z-intercepts are 3 and —4.

0 %or

—4

Al

x1[‘}

-5 I

fl L

A

When

P

1

aX?2 +bX+c=0

c

g -14

=

S~ —4"\

b

x=17

Qo

L

o

=

(1, 1, 5) 5)

P

1 f(1)

REPEAT

& il

246

Chapter 6 (Quadratic functions)

Review set 6B

¢ f(0)=-3

the y-intercept 1s —3.

When f(z) =0,

821’:2 —

3

920 —3=0

-3

1

3

or

C

1

the z-intercepts are —% and 3.

°

e

aX2+bX+ec=0

c

X1 IOE

-2 HEE

X2l

;

(SOLVE) PN CLEAREDIT]

AY

y =

—%:{:2

than

y = z°.

y = 3x* y = x°.

=Y

5

5 Walalen) @6

aX2 +bX+c=0

.

€r =

PR

opens

-0.5

%

(REPEAT downwards

and 1s “wider”

opens upwards and is “thinner” than

y = —2x 2 opens downwards and 1s “‘thinner’ than y = 2°. Y

6

a

The graph of y = z*

is B.

¢

The graph of y = 3z*

is A.

b

The graph of y = —22?

is €.

d

The graph of y = —2z°

is D.

y=23(x—2) a

Since

a =3

whichis > 0, the parabola has shape

\/

the parabola opens upwards.

b When

z=0,

y=3(-2) =12

the y-intercept 1s 12.

¢

When

y=0,

3(z—2)"=0 Lo

=2

the x-intercept 1s 2. d

7

a

AY

y=0

when

(z—2)(xz—9)=0 r=2

or9

the z-intercepts are 2 and 9. 11 2

is halfway between 2 and 9, so the axis of symmetry i1s x = %

2

Chapter 6 (Quadratic functions) b

y=—22+8—1 Now

has

o

2a

a=-1,

b=8,

and

Review set 6B

247

c= —1.

2(—1)

= the axis of symmetry 1s x = 4.

¢

y=2z*—x+3

has

b

Now

a=%,

b=-1,

and

c=.

—1

——=-—— 2a 2(3)

Lt

(W

— 1 the axis of symmetry 1s = = 3. 8

Let the other z-intercept be a. x-intercepts a and —3, we have

Since the axis of symmetry

x = 6

lies halfway between the

a+(738) _ 6 a—3

=12 a=

15

So, the other z-intercept is 15. 9

y=-32°+8x+7 b

Now

has

a=-3,

8

b=28,

and

c=7.

4

—— = — = = 2a 2(-3) 3

the axis of symmetry 1s x =

When

z =13,

4

3

y=-3(5)"+8(3)+7

=447 3

7

the vertex is (5, 30). a < 0,

so the shape 1s

the vertex 10

a

/\

(%, %)

1S a maximum turning point.

y=—-2°4+10z—9

Now

A

2a

has

L

2(—1)

a=-1,

b=10,

and

c = —09.

=9

the axis of symmetry is x = 5.

When

z =5

y=—-5>+10(5)—9 = —290+900—-9 = 16

So, the vertex is

a < 0,

(5, 16).

so the vertex 1s a maximum turning point.

the maximum value of y = —2?

4+ 10z —9

is

y =16

when

= = 5.

Chapter 6 (Quadratic functions)

243 b

y=22*—-2x+5

has

Review set 6B

a=2,

the axis of symmetry is

= =

b=—-2,

and

c= 5.

E.

=1 -1+5 So, the vertex is ( il2° 92/ a > 0, so the vertex 1s a minimum turning point. when

o =

y=—z*+7z—10 a

|

When

=0,

y=-10

the y-intercept 1s —10.

i

When

y =0,

E aX2+bX+c=E

—x? 4+ Tr—10=0 r=2

E: S S

or H

the z-intercepts are 2 and 5.

iii

SOLVE Jpang)s

E [MathDeg/Norm]] aX2+bX+c=0

o

c

7

EDIT

|

REPEAT

£2 is halfway between 2 and 5, so the axis of symmetry is z = 5

z =2,

y =15

s

—-9=a(-1)?+b(-1)+c s

5=a(l)* +b(1)+c

15=a(2)?+b(2) +c

or

a—b+c=-9

or

a+b+c=5

or

4a+2b+c¢=15

a—b+c=-9 We solve the system of equations

a+b+c=5 4da +2b+c

E MathDegMorml [dic)Real an X+bn Y+Cn Z=dn a

b

1

1

z

1

3

4

SOLVE|(MaNA3(CLEAR|{

We find that

c

-1

= an X+bn

d

¥

1

-9

1

1

5

2

1

EDIT

a =1,

't{ i

15

b=7,

Y+Cn

=15

Z=dn

7 | -3

REPEAT

and

simultaneously using technology.

¢ = —3.

So, the quadratic is y = z? + Tx — 3.

1

250

Chapter 6 (Quadratic functions) d

Review set 6B

The vertex has xz-coordinate 3

..

—Qi = (1

..

—b=6a

6a + b= 0

When

z=3,

y=15

When

z=1,

y=7

..

15=a(3)?+b(3)+c

or

T=a(l)*+b(1)+c

or

s

9a+3b+c=15 a+b+c="7

6a +b=0

We solve the system of equations

E

[MathlDeglNorml) [d7c)Real

an X+bn Y+CE

1 2 3

=l

Z=dn

6 9

1 3

1

We find that

X 1{ Z

0 15

1

EDIT

simultaneously using technology.

(d/c]Real

Z=dn

12| -3

REPEAT

a = -2,

So, the quadratic is

14

an X+bn Y+Cn

d

0 1

1

SOLVE |(NaRFIF|CLEARI{

9a + 3b+ ¢ =15 a+b+c=7

b=12,

and

¢ = —3.

y = —222 + 122 — 3.

Let the quadratic function be y = az? + bz + c.

When

=2,

y=>5

When

=6,

y=—-1

When z=2,

..

y=-3

5=a(2)’+b2)+c

or

4a+2b+c=5

—1=a(6)°+b(6)+c

or

36a+6b+c=—1

.. -3=a(2)°+b2)+c

5

[HathlDeglMorml [d/c]Real

an X+En

Y+Cn

a

b

Z=dn

c

2

1

?

36

B

1

=

4

Z

1

CLEARI

an X+En

d

4

MRS

4da+20+c=95 36a + 60+ c= —1 da +2b+c= -3

(Hath)Deg/Norml]

1

SOLVE

—

We solve the system of equations

5]

No

or

Y+Cn

4a+2b+c=-3 simultaneously using technology.

[d/c)[Real

Z=dn

Solution

-1

-3

EDIT

REPEAT

We find that there are no solutions for a, b, and c.

So, there 1s no such quadratic. The x-coordinate 2 corresponds to two different points, which is not possible for a quadratic function. 15

We

graph

Yl

and Yo = 2X set of axes.

—

—

X2

3

The graphs intersect at and (—2, —7).

-

13X

-+

15

on the same

(—9, —21)

[l

[EXE]:Show coordinates

Y1=x2+13x+15

Y2=2x-3

[l 33E3

[EXE]:Show coordinates

Y1=x24+13x+15

¥2=2x-3

33

Chapter 6 (Quadratic functions)

zm

16

Review set 6B

b If 8+ 9y =600, then y= 1 ym

The area of each pen1s

Substituting

251

.

A = zv.

y = 99 =82

into A4 we get

A — oy (600 — Sm) A

There are 8 sect}ons of fence w.ith length r m, and 9 sections of fence with length y m, with total length 600 m.

A=

600x

_ dz*2

9

9

_% z2 1 % r

m2

8x + 9y = 600 The area A 1s a quadratic with

Since

a < 0,

a = —%.f,

/\

1ts shape 1s

:

The maximum area occurs when

When

.

z =2,

y=

b=

b

=z = ——

2a

600 — 8(22) :

@.}

and

3

= ———

= —

2

A=—§(7_25)2+%(%)

600 — 300

_ 300

75

2(_5)

and

¢ = 0.

= —1250 + 2500

)

= 1250

9 __= 5~ 100 __= 3339 l the maximum possible area for each pen is 1250 m? when the dimensions are 1 17

If $x is the price increase of the sunglasses, then the retailer will lose % So, the revenue

R = cost of sunglasses X number of customers

= (45 + x) (50 — %)

dollars per day

R = (45 + z) (50—%) = 2250 — 30z + 50z — 2 = —22° + 20z + 2250 For the quadratic function R,

Since

a < 0,

The maximum

When z =15,

the shape is

a = —%,

b= 20,

and

c = 2250.

/\

revenue occurs when

z = —Qi

a

= —

R = —2(15)" + 20(15) + 2250

(202 2W_ % 3

=]

5

— —150 + 300 + 2250 — 2400

the maximum daily revenue 1s $2400 when the sunglasses are sold for

$45 + $15 = $60 each.

37% m by

customers.

Chapter 7 DIRECT AND

AY

D indicates that y is directly proportional to z, as it is a straight line which passes through the origin.

=Y

1

INVERSE VARIATION

il

The points lie on a straight line which passes through the origin, so y o< .

The gradient of the line — f;“ =5 the proportionality constant

il

k£ = 5.

The points lie on a straight line which does not pass through the origin, so y 1s not directly proportional to .

Chapter 7 (Direct and inverse variation)

Exercise 7A

253

7.9

4.5

=Y

o

.

=

Q0

il The points do not lie on a straight line, so y 1s not directly proportional to z.

30

30 iil

The points do not lie on a straight line, so y 1S not directly proportional to .

=Y

P

et

P

o

D

et

Ay -

20

Weight (w kg)

Cost ($C)

30

4 C($)

45

60 b

The graph is a straight line which passes through the origin, so C' is directly proportional to w.

¢

The gradient of the line =

N o

o o

a

15

|__I.

15

o

e D

C = 15w

15—0 1—0

Time (t hours) | Cost (€C)

When

t =1,

the cost 1s €6. the cost 1s €12.

(ig=st =21

6

When

t = 2,

]

2

12

When

t = 2.5,

Zaali= "3

18

When

t =4,

3

When

d=6cm,

V=40 mL,

where k is a constant.

so o

40 =k x 216 40

216 So a

5

40 =k x 6° D

27

V:%d? When

d=4

cm,

V:%xf

b

When

V = 30 mL,

30:%.%3

: c.od=27T%6

:32—270"“11.9mL

d = 3v/6 ~ 5.45 cm

The volume V is directly proportional to the cube of its side lengths I, so

a

Ifl is increased by 5%, then [ 1s multiplied by 1.05

. 1? is multiplied by (1.05)° = 1.157625 -,V is multiplied by 1.157625 {as V o [°} So, the volume is increased by about 15.8%. b

V' is multiplied by 2 . 1% is multiplied by 2

. 1 is multiplied by

{as

V o [°}

v/2 ~ 1.260

So, the side length is increased by about 26.0%. FE ocm

with proportionality constant

k£ =

b

FE «v?

with proportionality constant

k =

sv~. 2

b

a

o=

4

Exercise 7B

sm.

V o< [°.

Chapter 7 (Direct and inverse variation) ¢

Exercise 7C

259

If v is decreased by 10%, then

v 1s multiplied by

1 —-0.1 =0.9

. v? is multiplied by (0.9)? . E is multiplied by (0.9)* =0.81

{as E xv*

from b}

So, the kinetic energy decreases by 19%. d

We

assume

that

the

amount

of heat

generated

is directly

proportional

to

the

stopping

distance d.

So,

dxE

and

FE ocv®

{from b}

d o< v°

alz

|2 |12

y

|3|4]6 8 | 6 | 4

r | 1| 2] y 2010

xy | 24 | 24 | 24 | 24 ry = 24

xy | 20 | 20 | 24 | 20

for each point.

xy 18 not the same value for each point. x and y are not inversely proportional.

r and y are inversely proportional, and

Yy =

%

AY

*

I

\ o

19

\.\*

!

VT2 ¢

3

£

10 |

D

=

12

6| 5

12012

y

6

1

ry = 60 for each point. r and y are inversely proportional, flnd

y

—

@

4] 5 6 | 4

YA

20

A

\

12 \

8 4

I

Y Y

8

12 &

260

Chapter 7 (Direct and inverse variation)

2

YA

-+

3

a

Exercise 7C

B indicates that ¢ 1s inversely proportional to x, as the graph is a hyperbola.

o

v

No,

I

0 =90 — ¢.

There is no constant £ such that

6 x ¢ = k.

6 and ¢ are not inversely proportional. b

tanf ==

and

Y

tanf =

tan@

4

yocl

a

t&ll';f):g

T

!

tan¢

and tan¢

are inversely proportional.

XL

b

If x 1s doubled, then

x 1s multiplied by 2

If x i1s divided by 7, then x 1s multiplied by %

-, y is multiplied by 7.

. y is multiplied by % ",y 1s halved. ¢

d

If z 1s multiplied by %, then

If x is increased by 30%, then r 1s multiplied by 1.3

y 1s multiplied by %.

-,y is multiplied by % ~ 0.769 . y is decreased by about 23.1%. 5

C’ocl

L

X 3 P

a t

6 | 18

Ol

15

b

0

C

|15 | 20 __

x2

C is multiplied by % . %

t 1s multiplied by 3

- Cis multiplied by 1 {as C

%}

:C:%:E}

6

t

-. t is multiplied by & .

-

S0, B

The time taken ¢ 1s inversely proportional to the number of gardeners n, so

t =6 hours

when

n =25,

so

6:E k=30

Sowhen

n=3,

t=

30

e 10 hours

it would take 3 gardeners 10 hours to do the task.

{as C

t = —. T

%}

Chapter 7 (Direct and inverse variation) 7

The object’s acceleration a is inversely proportional to its mass m, so

a=15ms™?

when

m =5 kg,

so

Exercise 7D

261

a = —. e

1.52% k=17.5

So, a

a:E

Tri

When

7.5

m =2 kg,

a:?:3.75ms

=

.

b

When

a=10ms

o

75

~°,

—

Tt

kg m=0.75

. 8

a

=10

The amount in Wendy’s savings account 1s constant, and the cost of the total number of shares is the share price multiplied by the number of shares. This is a relationship of the form xzy = k, so number of shares is inversely proportional to share price.

b

1

If the share price drops by £0.15 from £6.25, then it has been multiplied by

6.25 —0.15 6.25

122 125

:

,

o

125

This means the number of shares Wendy can buy is multiplied by —55° SO she can now buy

il

125

—n

shares

122

(rounded down).

For Wendy to be able to buy 1.5 times as many shares, each share must be 1 1.5

ol

2

.

:

times the price.

This means the price must decrease by 33%%.

1

yx

a

1

—

If z 1s doubled, then

b

r is multiplied by 2

- 2 is multiplied by 22 =8

. = is multiplied by 1 i

.y is multiplied by .y 1s divided by 8. ¢

If y 1s multiplied by 64, then

ks

e

— s multiplied by 64

. z” is multiplied by & . x is multiplied by

. x 1s divided by 4.

¢/ fiid =1

If x is multiplied by %, then

7 is multiplied by (2)” = 22

.

1

—3

-

12

1s multiplied by 2—;’

l_

. y is multiplied by 422

262 2

Chapter 7 (Direct and inverse variation)

Exercise 7D

1

yx —

a

X3

x 1s multiplied by 3

N

r | 8 | 24

.z

is multiplied by 3% =9

Y

S5

18

2!

N multiplied

by 3 1

le

.

.

.

:

1

P}

{as y

. y is multiplied by 5 " Yy=2Tx5=3

[z

b

|8

y | 27 | 7O A

75 i

y is multiplied by Z_i B 2_;)

1

i

1s multiplied by %

=

{as

P}

y

. x* is multiplied by = {as

% B %

x 1s multiplied by

s=8x}=%

= > 0}

1

a

:

j_l%.\

-

T

c 18 multiplied by

8

W

(2)° = £

-, ¢ is multiplied by

12138

1 by -27 s multiplied

Lo1

M | 64

2

3

‘a3

.

.

. M is multiplied by 2°

{as M

1

.

. M =6.4x 2 =216 b

c M

M is multiplied by =;

|12

|1

|64

~

1

Lz

L

a

The volume of a cylinder

o5

1.

n

.

s multiplied by ¢z

{as

M

|

3

-, ¢ is multiplied by 64 -. ¢ is multiplied by c.e=12x4 =48 V = wr?h,

where h is the height and r is the radius

V

1

T

T2

h=—X— 1

h o< — T

v/64 = 4

:

{since V and 7 are constants}

So, the height of the can is inversely proportional to the square of its radius.

Chapter 7 (Direct and inverse variation)

Exercise 7E

263

b Ifr is multiplied by =2 then 3.04 2

h is multiplied by (%)

{as hoc —)

.

oy

.

3.04

.. the height 1s

1

1s multiplied by

d 5

the radius 1s

~ 10.4 cm.

then

15.3

12.9

{as

%

fas

. 7 is multiplied by .

°

12.9 x (fi)

¢ If h is multiplied by g 3

T

3.04 x

%

1

h

> 0)

~ 2.79 cm.

The cans would otherwise be too narrow or too wide for practical use.

The tidal acceleration a 1s inversely proportional to the cube of the distance d, so

a

a

1

-

If d is increased by 10%, then d 1s multiplied by 1.1

*. a 1s multiplied by

1

~ (0.751

(1.1)3

{as

a x d—lg}

So, the tidal acceleration is decreased by about 24.9%.

b

[If a is tripled, then a 1s multiplied by 3

Loyil

ol

s multipli.ed by 3

{as

1

a x =

-, d” is multiplied by % .

.

.

.

. d 1s multiplied by

1

~

75 N 0.693

So, the distance is decreased by about 30.7%.

CEEO y xx, so y = kxr where k 1s a constant. From the graph we see that y = 7 when = = 6. 7T =k

x6

and so

k:%.

So, the equation of the variation model 1S Yy = %:1:.

«

4

/

/

//

_,/ —-

Vs

b

AY OO

The graph of y against x 1s a straight line which passes through the origin.

=1

a

&

1

Chapter 7 (Direct and inverse variation)

H = kd*

where k is a constant.

From the graph, we see that

H =1

when

d = 4.

1=k x4° 4

k= 16

S

a

so

H

H xd?

So, the equation of the variation model 1s d2

el

2

Exercise 7E

b2

264

"{f‘ When

d =16,

-

2 | 4|

1 (k 8 |

6|

H = - x 16° = 16.

So, Teresa 1s 16 m above sea level.

m o [?, so m = kl® m = 0.4

when

where k is a constant.

[ =20,

so

0.4 =k

x 20°

0.4 = 8000k L

1

k = 55600 s . So, the variation model 1s

When [ =50,

m =

[3 : 20 000

3

m =

— 20000

_

125000

;

A (kg) m

/

:

sz

1

(20.0.4)

//

!(cm)

1 L ¥

r?y =5 .

1]

.

for every data point 5;5

k=25

10

| 20 |

1 8t

I

30 | 40 G

i Rov?,

5 T (s

so R=kv?®

Using the first point,

where k is a constant 0.5 =k x 10°

g

s

200

[ s v 200

. i

When

v=20,

202 R=—=2+#4

When

v =30,

302 R=— =4.5+#£13.5

200 200

R= —402 = 8 # 32

When v =40,

200

this model is incorrect.

v | 10 | 20 | 30 | R|lo5 ]| 4 |135] eH3

=

. | 2000

— = 3p0p

2000

|

1or every

1 | 2000

datadata

£L

s2? =10

5

Y

— =05

40 32 2000

point, s0 R == 5= 5355 V" 13-

point, so

=10

{as x>0}

Chapter 7 (Direct and inverse variation)

266

¢

When

v =50,

Exercise 7F

R=--2000 x50’ =

125000

695

2000

So, when the car’s velocity is 50 kmh™!, its air resistance is 62.5 units.

i

1

2

y | 0.6

97|

3

1

488 | 153.5

The correlation coefficient r 1s very close to 1, so the fit is 11

1

|

=

PDWErREgGD215882 a =0,

b =3.99982755

The power 1s very close to 4, so it 1s reasonable to conclude O : 4 that y 1s directly proportional to z*.

r =0.99999778 rz=0.99999556 MSe=3.8454E-05

:

=a-x"b

The model is y ~ 0.602z*. i

2

3

6

Des)orm) (d/c)Real

9

y | 100 [ 29.6 | 3.7 | 1.1

The correlation coefficient r 1s very close to —1, so the fit 1s excellent. _ = The power 1s very close to —3, so it 1s reasonable to conclude

a =798.571493

b

=-2.9986991

r =60égggggg$

that y 1s inversely proportional to x3.

MEEEE "B837-086 v=a*'*'Xx

: 799 —. ~ y 1s The model

COPY]

£

z|

35|

P|50

7

|8]

10

1892|745

=

We expect inverse variation between the variables, as the points appear to lie on a curve which 1s asymptotic to both axes.

+P

?

= 30 50 i\t

o

10

°o o

=

-

4

Y The correlation coefficient r 1s very close to —1, so the fit 1s excellent. ' -

The power is very close to —2, so it is reasonable to conclude that P is inversely proportional to x2. :

451

The model 18 P ~ —-. I

When

=4,

P~

451 — 42

~ 28.2

B

PowerReg

6

8

[@c)Rea)

a =450.652452

Itf ::% : ggéggg rz=0.99999418

MSe=E.3335E-DB y=a-x”

1

=

T

Exercise 7F

267

e

Zach only charges his phone when the battery has completely run out, so when C' = 0. Hence the graph of C against ¢ should pass through the origin.

¢ = 0,

e

We expect the charge to increase as time increases.

Chapter 7 (Direct and inverse variation)

Time (t minutes)

8

29

32

45

Percentage charge (C%)

10

31

40

00

The model 1s

¢

When

C ~ 1.25¢.

t =56,

Deglfornd] [d/c]Real

PowerReg a =1.2544452

b r r2 MSe S

v=a-'X 4

My

=t

The power 1s very close to 1, so it 1s reasonable to conclude that C' 1s directly proportional to .

B

B

The correlation coefficient r 1s very close to 1, so the fit is excellent.

Qo OO O WO X O B WO 300 .00 NN

b

We would expect direct variation between C' and ¢ because:

T=O00

a

COPY J|IDRAW |

C ~1.25x 56 ~ 70

So, after 56 minutes, the phone will receive about 70% charge.

The percentage charge C'% is always between 0% and 100%, and so

05x%

C

V:lfi—g@?’,

a

yoaz?,

When

so yxz? so

and

V xa®

so y=kx*

z =8

k=5 and

b

P==—=2n%

so Pxn*

and

k=32 -

k:li—g

where k is a constant.

and y =30,

30 =k x &°

3064 _1532

So the model is y = 3222

b

1 When z=4,

y=22x4° = 15

ii When y =150,

22z =150 = pt—320)

—75

.

xz=8/5

{as

z >0}

~ 17.9 a

7

2

4

D

3

y |20]10] 8 | 5

xy | 40 | 40 | 40 | 40

xy = 40

for each point.

r and y are inversely proportional,

and

4

y = —0 £

i

3

gy |20

D

8 | 10

12| 8 [ 6

xzy | 60 | 60 | 64 | 60

xy 1s not the same value for each point. x and y are not inversely proportional.

272 7

Chapter 7 (Direct and inverse variation)

Review set 7A

The frequency f 1s inversely proportional to wavelength A, so

k

[ = T

where £ 1s a constant,

When f =500 THz, A = 600 nm, so 500 — % So,

300000

=

When

A

k = 300000

.

A =480 nm,

300 000

f =

480

= 625 THz

the frequency of the blue light wave 1s 625 THz. 8

The resistance r is inversely proportional to the square of the diameter d, so

k

r = =

where £ 1s a

constant.

F=- 094 ohms when d—0Adem,

so

(.24 Besy (0.44)2 k= 0.24(0.44)" = 0.046 464

r=-20>

a When d=03cm,

(0.3)2

~ (0.516 ohms b

When

r =0.45

0.046 464

Oth,

T

=

2

(.45

0.046 464 045

d~0.321

9

a

yxaz?®, y=24

so y=kx* when

b When z=11,

10

a

where k is a constant.

x =6,

The model is y =

2

so

24 =k x 67

24

RT3

£z*

y=2x11° 242

e

AD | ) 10 \ = LI

20

10

Ay

\

/

40

2

"

/

20

10

)

/

/

(6,24)

i

0f—= 0 b

2

4

O

)x

8

The points appear to lie on a curve which 1s asymptotic to both axes, which 1s what .

1

we would expect to see if D o —;.

\

p

R

\ \LT

“TY

cm

123

4

p

Chapter 7 (Direct and inverse variation) C

2

1.5

D

40 | 22.5 | 144 | 10 | 5.625

p?D | 90 |

p*D =90

25 |

d

P

90

3

90 | 90

When

p=5,

Review set 7A

273

90

D=

52

::1%:::}6

5

90

at every point

k=90 11

alxz| 2| 4 D it y | 12 ] 96 | 188 | 514 The correlation coefficient r is very close to 1, so the fit is excellent.

The power 1s very close to 3, so it 1s reasonable to assume

that vy is directly proportional to x°. The model is y ~ 1.50z°. b

El DeglNorni) [d7c)Real PowerReg a =1.5006033%7 b =3.00001458 r =0.99999951 rz2=0,99999903 MSe=3.6613e-06 v=a‘x"b

|xz| 09|14 22 | 2.5 y | 762 | 130 | 21.3 | 12.8 The correlation coefficient r 1s very close to —1, so the fit is excellent.

|

WEE

N

=499.727434 T el s,

g

The power 1s very close to —4, so it 1s reasonable to assume = : 4 that y 1s inversely proportional to x*.

500

The model is y ~ —-.

r =-0.9999998 r

Y

£I

The area of the garden bed 1s A = 7r?, A

o

where r is the radius

r?

The amount of compost is directly proportional to A. the amount of compost 1s directly proportional to the square of the radius. b

2C =3.6 ;. C=18A

Let C be the amount of compost used.

If r is increased by 15%, then r 1s multiplied by 1.15

. C is multiplied by (1.15)* = 1.3225 . C 1is increased by 32.25%. So, Tamzin needs

{as

250 x 0.3225 = 80.625 kg

C x r*

from a}

extra compost.

Chapter 7 (Direct and inverse variation)

¢

Review set 7B

275

If C 1s increased by 40, then (' is multiplied by

A

250

. r% is multiplied by 22

25

{as

. r is multiplied by

C o r?

/52 = @

from a}

~ 1.0770

{as

r > 0}

r 1s increased by about 7.70%.

So, Tamzin can extend the radius by about 5

1

a

yx—, I

—16

so

y=—

k

which 1s about 23.1 cm.

:

where £k is a constant.

£

when zr ==6, 0, soSO 16 _6_3 = —

Yy =

wnen

k=16 So the model 1s

b

3 x 0.0770 ~ 0.231 m

x 216 = 3456

y = 3426. L

i When =4,

y= >0 43

i When y=—2,

= 54

520 _ _o 3 cooxd = —1728 r=

6

—12

The number of workers n is inversely proportional to the number of days ¢ to paint the silo, so k

n = B

When

.

where k 1s a constant.

n =3, t =18,

so

3=i

18

k=54 So

n:E. L

When

n = &,

8=5t_4

54 __ t=5 _ =6.75

it will take 6.75 days for 8 people to paint the silo. The radius r 1s inversely proportional to the square ot the orbital speed s, so

If r is increased by 20%, then r 1s multiplied by 1.2

. Siz is multiplied by 1.2

. 52 is multiplied by % _ ', s 1s multiplied by

{as r o si?} Sh |

7

\/% ~ 0.9129

{as

s >0}

So, the orbital speed decreases by about 8.71%.

1

r x —. S

276 8

Chapter 7 (Direct and inverse variation) a

1

yx-—,

k

so

y=—

The points

(4, 7)

Hi

Review set 7B

:

where k 1s a constant.

I

and

(14, 2)

Z

are marked on the

\

graph'

Using

1.0)

(4,7),

y=7

when

z =4,

so

\

7= %

\

)

\\

k = 28 y

Check: b

9 | h|

When

When

=z =14,

4

y= e 28

z=0.1,

2

28

y= e

P

N

28

—

& 2.

T

10

v

280.

6

V132|256 | 864 a

If the height of a regular pyramid increases, then each side length also increases. This means the pyramid gets larger in all 3 dimensions as the height increases, so we should expect that V is directly proportional to h?.

b

Voh?

so

V=32

when

V==kh?®

where k is a constant.

h=2,

so

32=Fkx?2°

k=228 =04

V=2h’ ¢ When

h=4,

V=2x4° _= £2

w

When

h=6,

64

=

=128 _9256 So the model

d

V = %h?’

i When h=8,

|z |

2 | 3 |

6

yla5[20]

_ 2 512

a

5 | 28|

i When V =50,

excellent.

The power 1s very close to 2, so it 1s reasonable to assume s o . 2 that y is inversely proportional to x~.

180

y ~ —

V

2hr° =50

b =125

cooh=

18

Hent

The model 1s

216

10

The correlation coefficient r 1s very close to —1, so the fit is

.

X

satisfies every data point.

V=2x8

8 |

2

5

=22 =864

= 1022 —204.8 10

V =% x6

r2

PowerReg

a

=180.178485

E ::333555%3

rz=0.99999818

MSe=4_ 4379g-08

et

Chapter 7 (Direct and inverse variation)

©

When

Review set 7B

277

180

z =5,

Yy~ =

~ 7.2

1M1

alz|25]

5 [ 75

]10]

125

12017

We expect direct variation between the variables, as the graph of y against x apppears to be a curve which passes through the origin.

.

80 60 T

o

g

£

-+

D

y

b

The correlation coefficient r 1s very close to 1, so the fit is

excellent.

The power is very close to 2, sagit is reasonable to assume

O L85 1529456 Bt

eas]

y=a-x~

Distance (d m)

1

5

10

15

COEY)

20

Sound intensity (I Wm~2) | 63.7 | 2.55 | 0.637 | 0.283 | 0.159 I xd

63.7 | 12.75 | 6.37 | 4.245 |

3.18

The values of I X d are not constant, so / and d are not inversely proportional, and Abbas i1s Incorrect.

The correlation coefficient 1s very close to —1, so the fit 1s excellent. The power 1s very close to —2, so it 1s reasonable to assume

that [ is inversely proportional to d?. The model 1s

63.7

[ =~ —

If d is increased by 40%, then d 1s multiplied by 1.4 . 1 1s multiplied by

E

DeglNorm1] [d/c]Real

PowerReg a =63.7330543 b =-2.0003923 -0.9999999 . 9999999 L 9379e-07 Ay = N ™ 2@ U1 oo

a

B

P 2=0.99992508 be 0

5

12

15

g

di ' that y is1s directly proportional to x~.

The model 1s- y 2~ 0.753x~. 2

10

>

1

(1.4)2

~ 0.510

. I is decreased by about 49%.

1

{as I x =

from b}

COPY

Cha

pter 8

EXPONENTIALS AND

LOGARITHMS

y = 3%

' 1s an exponential function as the variable x appears in the exponent.

O an O

f(r) =4?* —1 y = x> — 2

®

1

o

EXERCISE 8A

g(x) = v/xr+5

is an exponential function as the variable x appears in the exponent.

is not an exponential function as the variable = does not appear in the exponent.

f(z) =7—2"7

is an exponential function as the variable = appears in the exponent. is not an exponential function as the variable & does not appear in the exponent.

f(z) = —3 x 52

& 2

is an exponential function as the variable x appears in the exponent.

a

f(2)=2°-3 =43 =1

b

f(1)=2'-3 =2-3 = —1

d

f(-1)=2"1-3

e

f(-2)=2"%-3

=1 2 _

¢ f(0)=2°-3 =1-3 = —2

== 1 —3

= —2 or —23%

= —+L or —22

3 f(zr)=5x3" a f(1)=5x3!

b

=9 XxX3

=9

d f(—4)=5x3""*

X 27

¢ f(0)=5x3’ = O

e f(—1)=5x3""1

=5>°

~ 2.48

7

~ 4.31

b

y=23+1 =9

f(x) = 5**

f(1) =52 the point

the point (3, 9) satisfies y = 2% +1.

¢

¢ h(5)=3x(1.1)°

flz)=3""—2 . f(0)=3" -2

(1, 5)

f(z) =5

d y=6x2° When z=—1,

y=6x 21

— _ th

e

_3

int (0, —1 - po_mg_m( ) )

f(z) =

o

tisf] satisties

_ the point =16

flz)=—-4x3"+1

f

f(2)=—-4x3*+1

B

the point

(2, —13)

f(x) =—4x3*

+ 1.

027,

s

(—1,

3)

. satisfies

y=4"5%4+2 When

z=—1,

y=4"3"Y

does not satisfy

66

the point

(—1, 66)

satisfies

42,

flx)=3"—1

f(0)y=3"—1 =1-—-1 =0

The function passes through the origin

(0, 0),

42

— 64+ 2

y=4"5%

8

does not satisfy

so the axes intercepts are both zero.

279

280 9

Chapter 8 (Exponentials and logarithms)

Exercise 8B

f(r)=2"7-8

a f(0)=2°-8

b f(—-3)=2"("3 _38

S = —7

=25 — 8 =8 —38

the y-intercept 1s —7.

= () the x-intercept 1s —3.

EXERCISE 8B y=47

alx|-3|-2|-1(0|1]

yl b

&

||

5

2 ]|

[1]4]16

3

|64

|

If z 1s increased by 1, the value of y 1s quadrupled.

Il

If x is decreased by 1, the value of vy is divided by 4.

¢

e

i As

z— o0,

Fromdii,as

=Y

(J

>

-

T

n o

AY

i

d

y—o0.

£ — —o0,

i y— 0.

Do

Q2

ol

3

o= | =

—

o -

=

B

ol e

2

)

| =

.-—"""""r

=

b

o

the horizontal asymptote 1s y = 0.

o

1

¢

The graph of y = (%)m

1s decreasing.

As

¢ —o0, — y— 0.

Chapter 8 (Exponentials and logarithms) d

e

i

As

> 00,

Fromdi,as

y— 0.

£ —

—oo,

i

281

£ —o0, — y — oc.

y— 0T .

the horizontal asymptote is

a

As

Exercise 8B

y = 0.

y=2+3

b

y=3*—-4

z| -3 [ -2|-1[0f1][2]3 y|—-32|-32|-32|-3|-1|5]|23

h

43

42

41

7

1

2

)

T

T3

9

1

Y

""Il-..--

&

+

| b

&)

L~y

M~

D

)

|__L

z|-3|-2|-1({0|1]|2] 3 y |32 |33 |32 |4|5]|7]11

P

5

3

o

—H Y

¢c

y=47° z|

d

3|

-2|-110|1]

2 1

1

y=5x2*

e 1

z|

3| 5

Ay

1

2

3

AY

oy

M

-

?;24—13

-2|-1]0] 5 b

-

3

-

£

Y

a

i y=27°

|

-3

i

y=(3)

-2]-1l0f1]2]3

z|-3]-—2]-1[0of1]2]3

|1|3]|%]%

y| 8 | 4| 2 |1|L]L]L

y| 8 | 4|2

The corresponding y-values are the same for each function.

b

277 =(271)

- ()"

¢

\

Ay 1) L&

282

a

Investigation 1

Exponential graphs

AY g y=10"

y=3" y=2" y=1.3"%

i:}

L £Z

Yy

b

For

y=a*

where

a > 1:

I

The function 1s increasing.

ii

As

il

x — —oo,

y — 07,

so the horizontal asymptote is y = 0.

a affects how steeply the graph increases or decreases. Increasing a makes the graph increase more steeply, and decreasing a makes the graph increase less steeply.

il Wb

Yy

For I

y=a"*

ili

where

0 0,

a =+ 1:

or y=a""=1.

the y-intercept of each graph is 1.

€

ii

For For

a>1, as 00, y=a *— 0", andas * — —o0, y=a* — 0. 00 and a > 1.

have

both graphs lie above the horizontal asymptote y = 0 and are increasing. y = 10" 1s steeper than as 10 > 2.

y = 27

y = 2% corresponds to €, and y = 10® corresponds to B.

¢

y=-—5"

has

k 1.

the graph lies below the horizontal asymptote y = —5*

d

y=(%)

y=—(3)

a

and is decreasing.

y = 0

and is increasing.

corresponds to A. has k 4}.

8

a y=27+1 i When

z=0,

y=2"+1 =141

il

The horizontal asymptote is y = 1.

==

the y-intercept 1s 2. iii

When

When

z=2,

z=-2,

y=2°+1 =441

iv

y=2+4+1

=5

y=22+1

(2,3)

~_5 144

4

1

v

The domainis

{x |z € R}.

AY

Therangeis

{y|y > 1}.

2

(2,5)

Chapter 8 (Exponentials and logarithms)

b

Exercise 8C

y=3%+4

i When

=0,

y=3"+14

i

=1+4 =5

The horizontal asymptote is y = 4.

the y-intercept 1s 5.

When 2=2,

y=32+14

v

= o1 +4 O|=

iii

+9

(—2,13)

36

37 .y.f.il ........... é‘:?gz-f—)—»

L

37 9

AY

When z=-2,

y=3""+4 =

y=3"("2 44

o

o

T

Y

=3%+4

=944 — 13 v

The domainis

¢ y=(3)

i When 2=0,

{z |z € R}.

Therangeis

y=(2)" =1

{y |y > 4}

iil

The horizontal asymptote is y = 0.

the y-intercept 1s 1. v

e

b

I

b

y=

I

Sl R S

o

I O

o

| B

=z = -2,

fi

When

v

The domainis

{x | z € R}.

The range is ly |y > 0}.

AY

287

288

Chapter 8 (Exponentials and logarithms)

d y=(3)" -3 I

When

=0,

y=

Exercise 8C

(%)U — 3

il

The horizontal asymptote i1s y = —3.

=1-3 = —2 the y-intercept 1s —2.

i When z=2,

y=(1)"-3

iv

12

=53 — 1 _ 12 4

—

1l

o

When

= = —2,

-+

1

y:(%

2 _3

12

=— -3 2

=5-3

=43 =1

v e

The domainis

{x |x € R}.

Therangeis

{y |y > —3}.

y=2-2° i

When

z=0,

y=2-—2° =D=

i

The horizontal asymptote 1s

=1

the y-intercept is 1. iii

When

=2,

When

z=-2,

y=2-—2°

iv

=24 = —2 y=2-—2"2 =2_2_21

_8_1 4 4 _4 T v

The domainis

{x |z € R}.

Therangeis

{y |y < 2}

y = 2.

Chapter 8 (Exponentials and logarithms)

f

Exercise 8C

y=4""+3 i

When

=0,

y=4"+3 =1+3 =4

il

The horizontal asymptote is y = 3.

the y-intercept 1s 4. iii

When

=2,

y=4"°+3

_=

AY

( —2,19 )

+3

_ 1, When z = -2,

iv

48

= 18 y=4"02 43

2

-

~3

4 (213)

= ) RS '---,...._.i...

&

~

=16 + 3 =19

v

The domainis

g y=3-27" i When

=0,

{z |z € R}.

Therangeis

y=3—2° =3-1

{y |y > 3}.

il

The horizontal asymptote is y = 3.

= 2

the y-intercept 1s 2.

iii

When

=2,

y=3—2"" | — 3 — = — 12

—

7T

_ 1

iv

_

AY

1

1

4

When z=-2,

y=3—2"(2

=322

)

=3 -4

v

The domainis

{x |z € R}.

Therangeis

{y|y < 3}.

289

290

Chapter 8 (Exponentials and logarithms)

h

Exercise 8C

y=—2x37+1 I When

=0,

y= —% x 3Y +1 1

il

The horizontal asymptote is y = 1.

— 1

2

the y-intercept is %

iii

When

=2,

y=—2x3"°+1

iv

_ i L0 =—3 X g3+

et

1.1

=-—3xXgtl _

z=-2,

1 ., 18

18 T I8

_ 111 When

e

=

y=1

AY

R

i 2 /4

1

)

-

=—2x37%41

X

(~2.—3) y

y= —% x 37(=2) 11

1

(21_7)

o2

X941

9 | 2

—2 13

— _7

v 9

a

The domainis

2

{x |z € R}.

Therangeis

{y|y < 1}.

y=kx2+4c

Substituting

AY

(0, —5)

into the equation gives

(2,10)

0 —b5=kx2"+cC

k+4+c= -5 Substituting

(2, 10)

o

-

into the equation gives

10=kXx2°+c

B

Ak + ¢ = 10

Solving the system of equations

E

[HathlDegMNorml]

an X+bn Y=CE

b

y=5x2%—-10

When

z=6,

y=25x2%—10 =9 X 64 — 10 = 310

v

Y=Cn

FatibegNorn] ([@/c)Rea) C

1 ‘]

fln:x445n1v=4:n

¥

4

?[-fl

c= 3.

7

[SOLVE)PRNAI3(CLEARJEDIT )

1

[REPEAT)

the exponential model 1s y = 2% 4 3.

b

y=kx3"+c Substituting

(—1, 3)

into the equation gives

3=kx31+c

%k+c=3

Substituting

(1, —13)

into the equation gives

—13=kx3' +¢ 3k+c=

—13

Solving the system of equations 1 k

_|_

3

[g

oReal

8n X+Dn a

=

=

3

1 [ 0.3333

3k+c=-—13

simultaneously gives €= 9

Y=CE

2

k = —6,

the exponential model 1s

3

TdRed

1

1 fl]

[SOLVE]PRIAIACLEAR)(EDIT )

y = —6 x 3% + 5.

an X+bn ¥

C

3

‘E"[

_13

Y=Cn

IIIIHE]

REPEAT]

B

8

292

Chapter 8 (Exponentials and logarithms) c

Exercise 8C

y=kx27%+4c

Substituting

AY

(—2, 9) into the equation gives

9=1Fkx2"("2

(—2.9)

4

=k x 22 +c

-

dk +c=9

Substituting

(1, —5)

Y

into the equation gives

—5=kx21+¢ _ +c = —

1

Ek

Solving the system of equations 4k

>

L

T4 ¢

Y= kx2

2

_



e

Y

b

y=kx4"+c Substituting

(1, —4)

into the equation gives —4 =k x4' +c¢ 4k +c= —4

Substituting (2, 2) into the equation gives

2 =k x 4° + ¢

16k +c =2 Solving the system of equations

{

4k +c=

[§ wHbglen) @k an X+bn

—4

16k + ¢ = 2

simultaneously gives c = —6.

k =

Y=CE

A

bo |=

12

B c

2

the y-intercept is —=-. d

The horizontal asymptote of y = % X 4* —6

15 y= —6.

an X+bn

1 X

Y=Cn

ERel

Chapter 8 (Exponentials and logarithms)

13

Exercise 8D

293

f(z)=kxa*"+c The graph has horizontal asymptote

The y-intercept is 10, so

y =2,

so

¢ = 2.

f(0) = 10

kxa’+2=10

k=28

The graph passes through

(5, 258),

so

f(5) = 258

. 8xa’+2 =258 -

8a° = 256 a’ = 32 ar="

So, the exponential function is f(xz) =8 x 2% + 2. 14

f(x)=kxa *+c The graph has horizontal asymptote

The y-intercept is 1, so

y =4,

so

c¢ = 4.

f(0) =1

kxa’+4=1 k= -3

The graph passes through

(1, 2), so —3xa

f(1) =2 '14+4=2

S =3 (]l

—

pICa

a

So, the exponential function is f(z) = —3 X (%)

44,

33X 1

a

'

From the graph:

I i

2 =3 2*=0.6

when when

z~1.6. =~

—0.7.

y=2"

294

Chapter 8 (Exponentials and logarithms)

b

i Wegraph

Y; =2*

and

Exercise 8D

ii We graph

Yy = 3

on the same set of axes, and find their point of intersection. El

Y2 = 0.6

point of intersection. [E]

y

¥2=3

and

on the same set of axes, and find their

[EXE]:Show coordinates

Y1=2"%(x)

Y; = 2%

[EXE]:Show coordinates

Y1=2"(x) ¥2=0, g

¥

/

fi N

=

2

1

0

X=1.584962501

[¥=3

The solutioni1s

2

a

2~

1.58.

-2

The solution 1s

We graph

V

The solutioni1s

= ~ 4.32.

[EXE]:Show

Y1=3"(x) ¥Y2=30

v

0

Y, =30

on

coordinates

d

J x INTSECT

1

¥=0.6

We graph Y; =4* and Y2 = 100 on the same set of axes, and find their point of itersection.

The solution 1s

and

i

= ~ —0.737.

[EXE]:Show

the same set of axes, and find their point of intersection. [E]

b

coordinates

Y; =3*

-1

¥=-0,7369655942

Wegraph Y; =2* and Y2 =20 on the same set of axes, and find their point of intersection. [EXE]:Show

¢

—INTSECT

coordinates

= ~ 3.32.

We graph Y; = (1.2)*

and Yo =3

the same set of axes, and find their point of intersection. [E] [EXE]:Show Yi=1.2"(x)¥ ¥Y2=3

coordinates '

'

B

-

2ur

The solution1s

e

g

V=30

5

INTSECT

= ~ 3.10.

set of axes,

point of intersection. [E]

and

find their

[EXE]:Show coordinates

Yi=1.04~(x) Y2=4. 2334

10

__IJ!,

g

10

X=36.819B61249

The solution 1s

50 Y=4,238

= ~ 36.8.

.

2

0 X=6,02568b6103

The solution 1s

Wegraph Y; = (1.04)* and Yy = 4.238 on the same

J

E

= X=3.005903274

/

30 INTSECT

on

2

-

% ke INBSEL]

5§

1 ¥=3

= ~ 6.03.

f Wegraph Y; = (0.9)* and Y2 = 0.5 on the same set of axes, and find their point of intersection. [E]

[EXE]:Show coordinates

Yi=,0"(x)

¥2=.5 \

¥

15

=

K=6.578818478

The solution 1s

2

1

¥=0.05

= ~ 6.58.

&

a

INTSECT

X

Chapter 8 (Exponentials and logarithms)

a

Wegraph Y; =3x2%

and Yo =93

on

the same set of axes, and find their point of intersection. [E]

[EXE]:Show

Y1=3x2"(x)

[V

¥2=03 | .

coordinates

J

75

b

on

the same

set of axes,

[EXE]:Show

j/

’f1=4flx.5“£xn ¥ Z2=10 E

¥

-

\

X=4.8564

-1

Qé%l

The solution 1s

¢

I

Y=83

The solution 1s

[¥

Sy

_F'

X=2,464973821

= i

¥=120

4

INTSECT

on the same

on the same

=

34

The solution 1s

and

f We graph

set of axes, and

Yo

54

V2=470

A=6.9b3636888

We graph

A =

and

1

-

-l

| 1

0

1

X=3.4640735621

The solution 1s

/

L

2

Y=560

= ~ 3.46.

Yy = 50

and find their

7

o

20

€

The solution1s

|¥

50

E ot

3

:

4

2

e

Qg

X¥=b,3b0615443

z ~ 6.95.

[EXE]:Show coordinates

Y2=50

£

I

on the same set of axes, point of intersection. Y1=3*(x)+5

'1HTSE$T :

¥=3b0

Y, =3%+5

/

——

2

b

and

on the same set of axes, and

Y1=260x1. 126~(x) [¥ E

S

= 250 x (1.125)*

find their point of intersection.

-

a00

The solution 1s

coordinates

[EXE]:Show coordinates

“{x)y

-

set of axes, and

o ~ 9.88.

Y;

= 470

[EXE]:Show coordinates

v2=360

[E]

Yo

and

x

z ~ 2.46.

Y1=60(

g

Y; =21 x (1.05)*

find their point of intersection.

find their point of intersection. E]l

We graph

[EXE]:Show

We graph Y; =500 x (0.95)* = 350

= ~ 6.21.

/,

40

The solution 1s

Y=10

: IHT‘&Q

1

%

iy 80

Y,

d

[EXE]:Show coordinates

¥2=120

2

0 X=6.212567438

x ~ 4.95.

Y1=8x3"(x)

e

==

JLIHH_EC_-E

Wegraph Y =8 x 3% and Yy = 120 on the same set of axes, and find their point of intersection. El

N

.

4 =

//

find their

coordinates

=i

=

and

point of intersection.

F=rp

25

295

Wegraph Y; = 40%(0.8)* and Y5 = 10

‘?‘r

=

3

Exercise 8D

h

i

2

1

x

= ~ 5.36.

We graph

Y; = 60 + 10 x (1.5)*

Y,

on the same

=

80

10

i

=1.7006 The solutioni1s

O

and

set of axes, and

find their point of intersection. 2=80

AINTSECT

INTSECT

Y=470

[EXE]:Show coordinates

/

|

=

= ~ 1.71.

296

Chapter 8 (Exponentials and logarithms) i

Exercise 8E.1

We graph

Y; = 20+ 80 x (0.75)*

and

Y,

on the same

and

=

30

set of axes,

find their point of intersection. [E]

[EXE]:Show

coordinates

Y1=20+80x.75"(x) W

¥2=30

EDK &0

\"h

1

|

N

T[]

20k

B 6 4 -2 X=7.228262518

The solution 1s

L

The graph of

2

Y=30

= ~ 7.23.

y = 10 — 8 x (0.5)*

horizontal asymptote

y < 10

p

lies below its

AY

y = 10.

for all z.

So, the equation

10 — 8 x (0.5)" = k has:

a

1 solution for

b

no solutions for

k < 10 k£ > 10.

EXERCISE 8E.1 1

A(t) =3 x (1.08)"

square metres

The initial area covered by the weed was 3 m?.

b

The multiplier is 1.08, so the area increases by 8% each day.

¢

i A(2)=3x(1.08)* ~ 3.50 The area covered after 2 days i1s

about 3.50 m?.

il

A(30) = 3 x (1.08)°" ~ 30.2

The

area covered after 30 days 1s

about 30.2 m?.

i A(10) = 3 x (1.08)" The

~ 6.48 area covered

about 6.48 m?.

after

10 days

1s

Chapter 8 (Exponentials and logarithms) d

A A (m?) 1

30

é

30, 30.2

(30,

Exercise 8E.1

y

(1.08) )" 0| AA(t) =3=3 x (1.08

*

)

X

—

15

/

/

10

8}

(2, 3.50)

5;"/{456'

0

> { (d

4

0 5 10 15 20 25 30 [EXE]:Show coordinates

[EXE]:Show coordinates

vi=3%

08" (x)

V1=3=F,

7. 08°(x)

-—fl; I.-JEII.JI

a2

|

2

a2

J

L

32

|

L

&

x=0 O

|

L

B

3

|

L

2

|

L

a2

J

L

&R

I

-

]

y=3

X=

2 O}

[EXE]:Show coordinates ¥1=3=F. 08*(x)

#:

x=1P}

iy

[EXE]:Show 7. 08*(x)

P RN R

V=CALE

Y=6.476774992

W(t) =100 x (1.07)"

-

25

-5

¥=3.4992 coordinates

....... fi.-r:ggjfigs v=30.18797067

grams

a W(0) =100 x (1.07)° = 100 x 1

= 100

the initial weight was 100 grams.

b

The value 1.07 means that the weight is increasing by 7% every hour.

¢

i W(4)=100x (1.07)*

W (10) = 100 x (1.07)*°

~ 131 The weight 131 grams.

i

~ 197

after 4 hours

1s about

W (24) =100 x (1.07)** ~ o07 The weight after 24 hours is about 507 grames.

d

A W (grams)

400

200

(24,

/(t) =100 x (1.07)Y_~

507)

o

el -

e e |_—— (10,197) (% 1o/l) 0 0 10

t (hours) = 20

The weight after 10 hours 1s about 197 grams.

297

298

Chapter 8 (Exponentials and logarithms) [EXE]:Show coordinates

Exercise 8E.1 [EXE]:Show coordinates

P1=100x(1.07)"(x)

Pl=100x(1.07)"(x)

-

1

2

L

]

10

L

1%

E=4

[EXE]:Show

coordinates

2

1II:I

15

20

E=10

3

25

a

a0

L

I“

1

Ix

flE—CflL

coordinates

PWl=100x(1.07)"*(x)

d0

.QE;CHL

I

¥Y=196.71b51357

P(n)=Fy x (1.23)™

i

29

¥=131.079601

[EXE]:Show

P=100x(1.07)*(x)

L

20

2

10

15

20

=24

:

i

i

25

30

fi.E_cflL

Y=507.2366953

possums

The initial population is 50 possums, so

P(0) = 50

Py x (1.23)° = 50 Py =50 b

P(n)=>50x

(1.23)"

i P(2) =50 x (1.23)°

i P(5) =50 x (1.23)°

~ 75.6

~ 141

After 2 years, the expected population is about 76 possumes.

i

After 5 years, the expected population is about 141 possums.

P(10) = 50 x (1.23)* ~ 396

After 10 years, the expected population 1S about 396 possums. 500

A P (possums) 400

3001 200

piny =50 x (1.

d

From

the

graph

» 1 (years)

m

2(2’ 76i

00

n

C,

1t

appecars

that

1t

would

take

about

11 years for the population to reach 500. When P =500, 500 =50 x (1.23)". Using technology, n ~ 11.1.

It will take about 11.1 years for the population to reach 500.

[EXE]:Show coordinates —

Chapter 8 (Exponentials and logarithms)

L

V(t) =5 x

a

(1.03)"

299

units

i V(0)=5x(1.03)°

i

— 5

V(20) =5 x (1.03)%° ~ 9.03

The speed of the reaction at 0°C 5 units. b

Exercise 8E.1

is

The speed of the reaction at 20°C is about 9.03 units.

Percentage increase in speed from 0°C to 20°C = (V(Q?’(U)V(U)>

x 100%

~ (9‘03_5) x 100% ~ 80.6% So, the percentage increase in reaction speed from 0°C to 20°C is about 80.6%. €

10

AV (units)

ot

V(t) =5 x (1.03)"

8

oo

//

6 - — 4

’

t (°C)

00 d

D

When

10

V(t)

—

15

15,

15

.

Using technology,

-

20 =5

X

(103)t

t ~ 37.2.

El

[EXE]:Show coordinates

Y1=px1.03"(x)

¥2=15 10

'J&_//

T;

_—

..-E---""-"-_# ol

5

10

-~

| 15

X=37.16700967

@

25

¥=1b

2D

EEEETEETJTE

The speed of the reaction will reach 15 units when the temperature is about 37.2°C. 5

N=4x1.332",

t>0

a When t=0,

N =4x1.332°

b When t =16,

=4 x 1 =4 the

number

..

of people

the number

initially

N =4 x 1.332'° ~ 393 of people

infected after

16 days was about 393.

infected was 4. ¢

When

-

N

=

1200,

1200

Using technology,

=

4 x

¢~ 19.9.

1332t

El

[EXE]:Show coordinates

YI=4%1.332°(x)

¥2=1200

0

5

X=19.80688111

4

10

i5

¥=1200

INTSECT

It will take about 19.9 days for everybody in the school to catch the flu. d

Since there are 1200 people in the school and it takes about 19.9 days for all of them to catch the flu (from ¢), 1t 1s only reasonable to use this model for

0 0}

The bear population is increasing by 10% every year. ¢

2018 1s 20 years after 1998, so

t = 20.

B(20) = 200 x (1.1)*" ~ 1350

The expected bear population in 2018 1s about 1350 bears. d

2008 is 10 years after 1998, so

t = 10.

B(10) = 200 x (1.1) Percentage increase from 2008 to 2018 = (3(20;(15(10))

x 100%

~ (200 x (1.1)%0 — 200 x (1.1)10 - ( 200 x (1.1)10

R

~ 159%

B(t) =2000,

2000 =200 x (1.1)".

Using technology,

El [EXE]:Show coordinates %J.a@flxl . 17(x)

¢~ 24.2.

[T

e When

2=R000

AR

1500

jfi

1000t

-

/

10

Y=2000

= e

P

E

¥x=94.15885793

15

20

It will take about 24.2 years for the population to reach 2000.

7

a

b

A(t) =5000 x (1.08)"

euros,

where t is the number of years since Kayla deposited €5000.

i A(2)=5000 x (1.08)°

i

A(5) = 5000 x (1.08)°

— H&32

~ 7346.64

So there was €5832 in the account after

So

there was

2 years.

after 5 years.

€7346.64

in the account

Chapter 8 (Exponentials and logarithms)

‘

10 000

i (€)A(t)

—

3000

a

. .4 5000 xX (1.08)¢

/'(gm
0

700 000‘ V($)

The value increases by 7.5% each year,

b

S~

500 000

V =k x (1.075)¢ t=1,

/

600 000

so the multiplier a = 1.075. When

301

/

6000 4000

Exercise 8E.1

400 000 7"

V = 387000

k x (1.075)! = 387000

PUSURY
@

— 00

b

I

a

(1+3) a

10 100 1000

2.593 724 46 2.704 813 829 2.716 923 932

10000

2.718 145927

100 000 2.718268 237 1000000 | 2.718280469 10000000 | 2.718 281693

el ~ 2.718281 828 This appears to be the value of

(1 + l) (1l

a

as

a — oQ.

3

1

izt Peg)ora]

]

eIy

Uy = upe"t,

ug = 1000,

r =0.06,

9 718281828

ang):

L.

t =1

u, = 1000 x €991 ~ 1061.84 The final amount is $1061.84.

This 1s the same (to the nearest cent) as the final amount when interest was compounded by the second (in 1 e) or by the millisecond (in 1 f).

312

Chapter 8 (Exponentials and logarithms)

Exercise 8F

YO y = —e”

1s the reflection of y = e®

in the z-axis.

Yy =€

is the reflection of y = e®

1n the y-axis.

=Y

AY

AI—‘ w1/

1

2

3

—

b

AY

y=e*

>

p

> £

The y-intercept is 3.

The y-intercept 1s —2.

The horizontal asymptote 1s y = 2.

The horizontal asymptote 1s AY

d

C

Y =

e’

y =

e

>

B The

y-1intercept 1s 3.

The horizontal asymptote 1s y = 0.

y = —3.

=

v The

%E'T

= £

y-intercept is %

The horizontal asymptote 1s

y = 0.

Chapter 8 (Exponentials and logarithms)

Exercise 8F

313

w

Oo

f

The y-intercept 1s 1.

The y-intercept 1s 1.

The horizontal asymptote 1s y = 0.

The horizontal asymptote 1s y = 0.

e* ~ 7.3891

b

e’ ~ 20.086

e’? ~ 1.6487

e

1~ 0.36788

e

231 ~10.074

e 231

e829 ~ 125.09

h

e *%%Y 2 0.007994 5

50e 01764 ~ 41.914

80e~0-6342 ~ 42 429

k

1000e!-2042 a~ 3540.3

0.25e3-6742

e

2~ 0.006 3424

=10

We graph Y, = e* and Y5 = 10 and find their point of intersection.

The solution 1s

e

2~ 0.099 261

on the same set of axes

[EXE]:Show coordinates

Yl=a®

14

z ~ 2.30.

— 15

We graph

Y; = e?%

and

Yo = 15

and find their point of intersection.

on the same set of axes

[EXE]:Show coordinates

Vi=e~(2x) yo=15

-

|V

=

5 |

The solution1s

'15

i

o

}-

_..--""f

X=1.354026101

= ~ 1.35.

"™

V=16

A J

'

5

:

INTsECT

oeӴ =0.3

We graph

Y; = 5e™*

and

Y, = 0.3

axes and find their point of intersection.

on the same set of

[EXE]:Show coordinates

Y1=5(e"(-x))

vo=.3

..

¥

\

\

2

The solution 1s

= ~ 2.81.

:3

i

314

Chapter 8 (Exponentials and logarithms) d

Exercise 8F

% —4=-1

We graph

Y; =e’* —4 )

and

.

.

Y2 = —1

on the same set of

.

axes and find their point of intersection.

[EXE]:Show coordinates

Y1i=(e"(3x))-4

y2=-1

0.75 |

a

———

AY

moiozo

o s oo b

e

*"""5

=]

-0.25

e

L

D.'Eyfi

0.75

B

[¥=-1

|

INTSECT

e

y =10

—

=

&

f(x) =10 — €° Y

b

The domain of f(x) is {z | z € R}. The range of f(z) is {y |y < 10}.

¢

7

As

W)=

a

r—o00,

%2

y—

—00,

andas

r —

—oo,

y—

107.

grams

i W(0)=2¢ =2 _

2

il ¢=30 min = 3 hour x1

.

W(i)

i

W(l3) =2e*

~ 4.23

The weight of the culture is about

4.23 grams after 15 hours.

264

2

2 2.07 _ _ The weight of the culture is about 2.57 grams after 30 minutes.

. . The weight of the culture 1s 2 grams initially. 3

—

iv

W(6)=2e” ~ 40.2 The

i

____/(

X=0.3662040962

The solution 1s = ~ 0.366. 6

05

y

weight of the culture 1s about

40.2 grams after 6 hours.

|

Chapter 8 (Exponentials and logarithms)

8

I(t)="75e a

1"

315

amps

1 I(1)="T5e

"1

~ 64.6

b

(amps)

- q& 1, 64.6)

About 64.6 amps of current is still flowing after 1 second.

45 30

i 1(10) = 75¢~ 13

\_\ I(#) = 75e=0:15¢t N (10, 16.7)

L5

~ 16.7

T

02

4 6 81012141618

t (seconds)

About 16.7 amps of current 1s still flowing after 10 seconds.

¢

Exercise 8F

When I(t)=1,

1=75e"",

Using technology,

[EXE]:Show

coordinates

¢~ 28.8.

[t will take about 28.8 seconds for the current to fall to 1 amp.

A=kxe "' +c kL a

The horizontal asymptote is A =3, A

When Lk %

t =0,

o

kXB

—0.5¢

so

¢ = 3.

12 K A (kL) 10

+3

A=10

8—0.5(0]

+3

=10

6

k+3=10 k

—

b

\

:

7

So, the exponential model is

A:kXE_U'Et_l_C

8

N

o T

A = 7e~ 9>t 4 3.

0

;fi-l-;-;i """""""""""""""""" i

t (min) 0

2

4

6

8

10

The hole is in the side of the tank as the tank never completely empties.

¢ When

t =2,

A=7e %532

3

=Te ' 4+3 ~ H.H8

There 1s about 5.58 kL of water 1n the tank after 2 minutes.

There is initially 10 kL of water in the tank, so the amount of water in the tank after losing 6 kL 1s 10 —6 =4 kL.

¢~ 3.89.

2N

[EXE]:Show

coordinates

3 1=7(e*(-.bx))+3 ‘H.\-\

==

i 3

oy

Using technology,

4="7Te "5 43,

...

A =4,

d kI

When

5

:

g —

1

3,801820298

|17

III:I

d

o

9

V=4

5

g IH—'I'—:';:EGIE

It will take about 3.89 minutes for the tank to lose 6 kL. of water.

|

Chapter 8 (Exponentials and logarithms)

316

10

Exercise 8G

V(t) =650(4+ 2 x e 0-1%) As

t — 00, e 01t ot the speed of the meteor 1s decreasing.

i V(0) =650(4 +2 x e 1%

b

i

V(120) = 650(4 + 2 x e~ 1120y = 650(4 + 2 x e~ %)

= 650(4+2 x 1) = 650(6)

~ 2600 ms™*

= 3900

The speed of the meteor after 2 minutes

The speed of the meteor when it was

was about 2600 ms—!.

first sighted was 3900 ms™—!.

¢

When

V(t) = 3000,

3000 = 650(4+ 2 x e 1),

Using technology,

[EXE]:Show coordinates

B50(4+2x(e*(-0.1x))) 2=3000

¢t~ 11.8.

-11.78664906.

Y=3000

It will take about 11.8 seconds for the meteor’s speed to reach 3000 ms™

= 3

=4

log 10 = log(10")

log1 = log(10") =0

=1

log(10* x 100) = log(10* x 10?)

log(10™) =n

= log(10%"%)

1—m 10g(10—m) _= log(10"™™)

=1—-m

100 < 237 < 1000 log 100 < log 237 < log 1000

.

log 237 ~ 2.37

log(10%) < log 237 < log(10?) 2 < log237 < 3

We know that log1 =log(10°) =0 Also,

0.1 < 0.6

-

1

4

v

f

The y-intercept of y = 10" is 1. the z-intercept of y =logx 1s 1.

g

The domainis

{x |z > 0}.

Therangeis

{y |y € R}.

2 L 12 M _= 2log (5 a

When

x 10'°, F =6.2

b

When

M =5.1,

Using technology,

(6.2>1

In 40

8963x107° Al

101!:);3,‘(&953)

_ 10los(8.963)—5

Also, logk el

-

k>e

10'°8% < 10t .

k l[](l

=

m

B

[athlDeglNornd) [d/c]Real

Eq: 80=

d

In

0. 0

g(3) =377 -2 1

:3_3_2 | — 2 = ==

:_13_?,

f(x)=3x%x2%—-12

f(0)=3x2"-12 =3 —12 = —9 the y-intercept 1s —9.

b

X

l““fim . 284003157

k ~ 2.28.

The substance decays by about 2.28% each year.

a

about

98.7 years.

So, substance B’s half-life is

2

327

f(2) =3x2%-12 =3x4—-12 =12 —12

=0 the z-intercept is 2.

328 3

Chapter 8 (Exponentials and logarithms)

Review set 8B

y= % X 4% alx|

3| 19]

1

1

1

1

Yy

-2|-1]101]12] 395

]

5

%

8

3

b

AY 96)

32

Jo

, X

4T

=Y

=

E

I

—

4

a

The horizontal asymptote of y = —% X5

+3

18

y=3.

b

The horizontal asymptote of y =27"% —4

1s y= —4.

1 ¢ The horizontal asymptote of y =6 —8 x (3)" is y=6.

5

f(x)=-2x3"—-14

a

f(-2)=-2x32%-4

-22 -4

f(2)=—-2x3°—4

34

= —18—4

= 22

—42 2 b

i f(0)=-2x3"—4

ii

— 924 S

The horizontal asymptote is y = —4.

the y-intercept 1s —6 C

AY R

d

The domainis

=

{z |z € R}.

Therangeis

{y |y < —4}.

a y=Fkx2T4c Substituting

(0, 10)

Ay into the equation gives

10=kx2+c

10

k+c=10

Substituting (2, &) into the equation gives 11 _= k2 = + C 11

N

6

£

4

k—|—C=T

i

& v

(2,%) y=kx2""+c

Chapter 8 (Exponentials and logarithms)

Solving the system of equations

[ morgis o an XtbnY =Cn

k+c=10

a

{%k+c%

d o5

simultaneously gives c=4.

k£ =

6,

b

Review set 8B

329

B

an X+bn Y=Cn

c

x[m]

1 mmEE

S 5 5

SOLVE)RENEIB(CLEAR) EDIT

g

REPEAT

So, the exponential model 1s y =6 x 27% 4 4.

b

y=kxa*+c The horizontal asymptote 1s y =5,

so

¢ = 5.

y=kxa®*+5

Substituting

(0, 2)

into the equation gives

h

2=kxa’+5 k+5=2 -

=

(=1, 1)

)

_3

Y

y=-—-3xa*+95 Substituting

(—1, —1)

into the equation gives

—1=-3xa'+5

o

¥

(l

—ba = —3 =

It

So, the exponential model 1s

7

y = —3 X (%)J =+ 3.

y=kx5*+4c a

The z-intercept is 1.

The y-intercept is 8.

When

y=0

When

kx5 '+c=0

s

z=1,

z=0,

y =28

kx5 +c=38

E+c:O

k+c=28

5

Solving the system of equations L

{E

k

tec=0 —|— -

—

[ moogis

an X+bn Y=CE

1[

8

2

| i simultaneously gives c= —2.

0.2

1

k& =— 10,

When

z=-1,

y=10x5"("1 — 10 x 5t — 2 — 48

an X+bn Y=Cn

1

1

[I]

fl

8

So, the exponential model 1s y = 10 x 57% — 2. b

g

2

-2]

T

10

330

8

Chapter 8 (Exponentials and logarithms)

a Wegraph

Y; =4%

and

Yo =30

Review set 8B

on

the same set of axes, and find their point of intersection. [EXE]:Show

=2+ 2330

-

B

.

: .

: .

2

1

1

:

2

2p

3

05

The solution 1s

¢

We graph

coordinates

1

=

1p

: .

|

1

'

b

Wegraph Y; =2x (1.6)* and Yo =7 on the same set of axes, point of intersection. [EXE]:Show

VI=2x1.6"(x)y yo=7 ) &

'_-—I.fi

1

0.5

x=2.a&5ifi

= ~ 2.45.

'

/

'

/’;3-(

A3

The solution 1s

Y; =50 x (0.65)*

coordinates

and find their

05

1

15

¥=7

2

2.‘5‘51{‘13?‘5@_1?

|

= ~ 2.67.

and

Yo = 15 on the same set of axes, and find their point of intersection. [EXE]:Show coordinates

Y1=50%x.65%(x)

Y2=15

,D\\ Ao}

\

20F

s

X=2.794848076 | Y=15 The solution 1s

9

. mrsEeE

= ~ 2.79.

The graph of y = 4 x (0.3) — 2 horizontal asymptote y = —2. y > —2

4 x (0.3)* —2 =k

a

1 solution for

b

no solutions for

LY

t=0,

has:

k£ > —2 k£ < —2.

W = 1500 x (0.993)"

a When

lies above its

for all x.

So, the equation

10

S

grams

W =1500 x (0.993)° = 1000 x 1 = 1500

The initial weight of the radioactive substance was 1500 grams. b

|1

When

¢t =400,

W = 1500 x (0.993)*"" ~ 90.3

The weight remaining after 400 years was about 90.3 grams.

il

When

t = 800,

W = 1500 x (0.993)%% ~ 5.44

The weight remaining after 800 years was about 5.44 grams.

Chapter 8 (Exponentials and logarithms) C

331

A W (grams)

1500

\

!

\ W=1500

x (0.993)"

\

\

\

(40“.

.

()03}

KBG{}

400

When

5 44\

/ \ 2 / \’\=__._._>

Y0 d

Review set 8B

W =100,

300

;

(years)

100 = 1500 x (0.993)".

Using technology,

[EXE]:Show

t =~ 386.

coordinates

[t takes about 386 years for the weight to reduce to 100 grams. 11

P=kxa®

a

birds,

t>0

The population increases by 15% each year, so the multiplier a = 1.15.

b

40004 p

P =k x (1.15)" When

¢t =2,

P =k x {1;,

3000

/

/

P = 1058

/

k x (1.15)* = 1058 . _ 1058

2000

%

/

k = 800 The initial population was 800 birds.

¢ When

t=5

0

P =800x (1.15)°

0

2

4

6

! (years’)J

8

10

12

~ 1610

The population after 5 years was about 1610 birds.

a=e

Sooa>1 =0,

k0

2"

o

mai=ll

When

=0,

. .

AY

funf:tlon 1s above horizontal asymptote . . and 1s Increasing

y=3x2"°=3

the y-intercept 1s 3. the graph is E.

\Y 3 N7

-

When

b

Y=

y=—e” ke —1

)

a

y = ka® + c.

Y

Y

V=

Use the exponential function

A

12

No, this model is unlikely to be accurate for very large values of ¢ as the population of birds would be unrealistically large. The birds would eventually reach a limiting population.

Wl

d

332

Chapter 8 (Exponentials and logarithms)

¢ y=e 1 k=1

k>0

a=e

a}l}

Review set 8B

d y=3+=L—(3)

function 1s above . horizontal asymptote e yoP and 1s increasing

k=1

..

a = é

When

=0, y=¢€"+1=2 the y-intercept 'is 2. |

Zszrr}];tost{; y =1 &3 Lgnzonts) y the graph is A. AY

k>0

-

When

function 1s above

0

Y

Y e

’y

—

—f

k=-1 1

.

==

.. |

o When

=0,

1

(1)$

e’

€

k a reflection 1n the xz-axis if a < 0,

1

-y = sing

/fy::ein:c

N\

y = asin(bx) + d

the period is

N\

\180°

’

L

b > 0,

:

)A }/

~qy = sin 2

Y

b

pm—

——

O

o

:I-IE-

i

//

LANC

f

__,...-'""


,

.

1805

sinx

s

360°

)

downwards by 2 units.

—

5407

720° @

—2

34

Yy

=sinr — 2

Y

d

y= —2sinx 1is a vertical stretch of y = sinx In the x-axis. The amplitude 1s 2. y=sinzx

Yy=2sinx -

’ ¥

¥

-l"

h‘

am*

.

1802, .

e

y =sin3x

-

1-

%

1

-

-

*

%

*

P "'l'lll"

»

’

*

L]

%

#

.

360°

=

with scale factor 2, followed by a reflection

At

5409

:

*

The period 1s %

= 120°.

..

*

*

1

y=—2sinx

is a horizontal stretch of y = sinz

720° 7 P

S

f

¥

o

with scale factor .3

the maximum values are 120° units apart. y =sinx #is

Yy = sin 3x

720° »

f

y=sin$§

is a horizontal stretch of y = sinxz

.

.

The period 1s

360° —

.

= 720°.

@

o

with scale factor

— = 2. 2

.

. the maximum values are 720" o units apart.

2 _.“‘ Yy = SIn T .

5

—]_"

8

a

y= %casm

1."'

o

11'*

1‘-**'

.

>

:%60@7;200

m

1s a vertical stretch of y = cosx

u_,"

with scale factor %

So, a vertical stretch with scale factor % will map

b

y =cosx

onto

y = %cos .

y = —cosx i1s a reflection of y = cosx in the z-axis. So, a reflection in the z-axis will map y =cosx onto y = — cosx.

Chapter 9 (‘Trigonometric functions)

Exercise 9D

¢

y=-cosx+ 3 1s a vertical translation of y = cosx upwards by 3 units. So, a vertical translation of 3 units upwards will map y = cosx onto y = cosx + 3.

a

y=cosx—

d

y=

—% cosx

1

1s a vertical translation of y = cosz

1s a vertical stretch of y = cosx

in the z-axis. The amplitude is =.

343

downwards by 1 unit.

with scale factor %,, followed by a reflection

344

Chapter 9 (Trigonometric functions) e

y = cos2x

with scale factor 35L

1s a horizontal stretch of y = cosx O

The period is

f

Exercise 9D

y = cos %fi

= 180°.

..

the maximum values are 180° units apart.

is a horizontal stretch of y = cosx

:

.

The period 1s

360°

——

=240°.

.

with scale factor %

:

.

the maximum values are 240° units apart.

2

10

a

y=4sinx 1s a vertical stretch of y = sinx The amplitude 1s 4 and the period is 360°,

-

,

}

}

-

180°

—11

90°

LR

Sy

N

with scale factor 4.

360°

>

T

270°

(315°, —2V/2)

Y b

I

When

z =

150D,

Y =

4 sin ]_50':'

ii

When

r =

315435

y =

3

=4 x

()

&

= 2

4sin 315°

11

y = 3cosx

Now, a

has minimum value —3 and maximum value 3.

y =3cosx + d

1is a vertical translation of y = 3cosx

y=3cosx+d will lie entirely above the x-axis when than 3 units upwards.

d>3

by d units. y = 3cosx

has been translated more

Chapter 9 (Trigonometric functions)

b

y=3cosx+d will lie entirely below the z-axis when than 3 units downwards. d< —3

y = 3cosx

Exercise 9D

has been translated more

y = 3cosx +d will lie partially above and partially below the z-axis when been translated between 0 and 3 units upwards or downwards. —3 — SIN

> —2sInx

A reflection in the z-axis, then a vertical stretch with scale factor 2 maps y = —2sinx. 14

a

y = sinx

y = sinz

onto

a = 2, so the amplitude is |2| = 2. d =1,

so the principal axis 1s

We stretch y = 2cosx

y = 1.

y = cosax vertically with scale factor 2 to give upwards by 1 unit to give y = 2cosx + 1.

y = 2cosx,

..--® Y=2COST -

._1-' 1-

-----------------------------

—

Ee" e ‘.- --------------

--------

-

-

;+

*

-

-

==

-

----------------------------------------

then translate

346

Chapter 9 (Trigonometric functions)

b

360°

Exercise 9D

360°

b =2,

so the period is

d = 3,

so the principal axis 1s y = 3.

b

2

= 180°.

horizontally with scale factor 521 to give

We stretch

y = sinx

y = sin 2z

upwards by 3 units to give

y = sin2z,

then translate

y = sin 2z + 3.

y=sin2x + 3

¢

a= —2,

so the amplitude is | —2| = 2.

We stretch y = 2cosx

d

a =3,

b =3, We

y = cosx vertically with scale factor 2 to give 1n the x-axis to give y = —2cosx.

y =

2cosx,

then reflect

y =

%cos x,

then stretch

* sothe amplitude is* |35 1 | = 3.1

.

.

so the period is

stretch

Y = %cms x

y =

cosx

360°

=

360°

A

ey 120°.

vertically with scale factor % to give

horizontally with scale factor % to give

y = %CDS 3.

Chapter 9 (Trigonometric functions)

e

a =3,

sothe amplitudeis

b =4,

360° so the period is

360° = 90°. 4

d =7,

so the principal axis1s

y = 7.

We stretch y = sinx

vertically with scale factor 3 to give y = 3sinx,

y = 3sin4dx,

y = 3sin4x

upwards

AY y=3sin4dxr+7

4-

....................................................... .

¥

’"N

emmmmme- P

':""Jf

*y-?-%i-]:} g;{---:t

1:.___-1:

v

‘,‘*:

erm e

e

B

o~ y=3sin4dzx,--,

;.r

s

b= %,

then stretch y = 3sinx

then translate

71

a = —1,

347

|3|=3.

horizontally with scale factor % to give by 7 units to give y = 3sindx + 7. 10

Exercise 9D

"".,‘*‘

:‘

1|."1|I----.._”_'-r,,r

908

=

‘uh*;’

“‘

:‘

r*---“:l

180

~.

e

‘1-‘

_.'

t‘ -----

x>

’:[;'

¥ 7==270°""""" Y.-*" 360°

hfit""-'h

‘

: _________

fit“*‘

y=3sinx

so the amplitude is |—1| = 1. el

so the period is

000

=

360°

—

= 720°.

2

d =1,

so the principal axis 1s y = 1.

We reflect y = cosx

1n the xz-axis to give y = — cosx,

with scale factor 2 to give : give y _= —cos 51 + 1.

15

a

a =06, d = 10,

y = — cos %m,

sothe amplitudeis

then stretch y = — cosx

then translate

y = — cos %:1:

horizontally

upwards by 1 unit to

|6| = 6.

so the principal axis 1s y = 10.

We stretch y = 6sinx

y = sinx vertically with scale factor 6 to give upwards by 10 units to give y = 6sinx + 10.

y = 6sinx,

y=06sinx + 10 10 4 +---=-nmmmeeee- S meE {1,

e -« i T

.

e 180°

.

~+------- T y =

a2t f;36[]°

51Nn.r

e -

———== e H40°

=

-—p T ',-720':’

then translate

348

Chapter 9 (Trigonometric functions)

b

When

z =30°,

Exercise 9D

y=6sin30° + 10

=6 x 3+ 10 = It}

¢

From the graph, the maximum value of y 1s 16, which occurs when r = 90°

d

or

16

450°

From the graph, the minimum value of ¥ is 4, which occurs when r = 270°

d

iy

sinx = 1.

or

sinz = —1.

630°

(45D$5)

WXTT

f(z)=asin2x+d

J

.

\‘/f’f (135°, —3)

‘

£(135°) = —3

f(457) =5

asin(2 x 135°) +d = —3 . asin270° +d = -3 —a+d= -3

asin(2x45°)4+d=5 asin90” +d =5 a+d=5 [MathDegNorm1] [d/c]Real an X+bn Y=CE a

1[ 2

1 il

(1)

e

5|

EDIT

REPEAT

Solving (1) and (2) simultaneously using technology gives b

Pom ™ 7S Ay

and

d = 1.

xZr

(90°, —4)

£(30°) =

c.oacos(2 x30°%)+d =

1

E('L—l—d—

£ Hefibefon] [@JRa

2

1 1

SOLVE)DIENRB(CLEAR[ EDIT |

.

1

2

15

e

acos(2x90°) +d=—4 . acos180° +d= —4 —a+d= -4

=] an X+bn

Y=CE

0.5 -1

f(90%) = —4

1

acos60° +d = 12

1[ 2

a =4

f(z)=acos2x+d

v

an X+bn

. (2)

£l [MathiDeglorml [dic)Real an X+bn Y=Cn

c

1 1

SOLVE |(M3R3I3|CLEARI|

..

fl.5|

e

. (2)

Y=Cn

REPEAT

Solving (1) and (2) simultaneously using technology gives

a =3

and

d = —1.

Chapter 9 (‘Trigonometric functions)

Exercise 9D

349

f(z)=acosxz+d

£(60°) = 2

acos60° +d = 2

1[

.

Soo—a+d=5

([d7c] Redl

Y=CE 0.6

acos180° +d =5

iee CAB)

%{1+d=2 an X+bn

£(180°) = 5

an X+bn

1 ;}

Y=Cn

[d7c) Real

fl:-i}

0

[SOLVE)DEIRIA(CLEAR](EDIT]

=7

[REPEAT)

Solving (1) and (2) simultaneously using technology gives d

AY

(30°,3)

a = —2

and

d = 3.

(120°,5+3)

4—//_‘\ f(r)=asinz+d

-

.. (2)

-

= £

Y 8 3 £(120°) = 8£¥3

f(30°) =3 3 asinSUD—l—d=%

-

ke s Bt S

1 El

© fard=8

=3

(Hath)[Deg)[Form])

an X+bn 1[ 2L

]

Y=CE

a

1

1.2247

1

X

1.5|

Y=Cn

m|

?[1.3333

1.622008468 EDIT

1 3

REPEAT

Solving (1) and (2) simultaneously using technology gives

17

a

Ay -

P

a = 53

The amplitude 1s 1, so 1890

'

)

[(Wath)Deg)[Norml)

an X+bn

C

0.5

SOLVE |MaNIF|CLEARJ[

asin120D+d=8$6@

3690

'

>

B

..

The period i1s 360°, so

and

d = 3.

a = 1. 360°

= 360

o

==l The principal axis1s

y = —2,

so

d = —2.

The equation of the function i1s y = sinx — 2.

350

Chapter 9 (Trigonometric functions)

Exercise 9D The amplitude is 1, so

a = 1.

The period is 120°, so

360°

= 120° b=3

The principal axis1s

y =0,

so

d = 0.

The equation of the function is The amplitude 1s 2, and

a 0 a = 8. The hamster rotates through one revolution each second, so the period 1s 1 second.

360 _ | b

b = 360 d = kN (32+ 16)

The lowest position 1s 3 cm above ground level, so

d=11

So, b

When

H = 8sin(360t)° + 11 cm. t=4.7,

H =8sin(360 x 4.7)° + 11 ~ 3.39

After 4.7 seconds, P is about 3.39 cm above ground level.

ks

REVIEW SET 9B 1T

a

cos500° = cos(360° + 140°) — cos 140°

b

2

a

The function repeats itself over and over

sin(—100°) = sin(620° — 720°) = sin(620°)

in a horizontal direction, in intervals of

STRA™

length & units.

Y

I

YW | b 3

1

period =38

.

il

.

.

a

Theperiod of y=2sin3x

©

The period of y = —4cos5 —1

.

1s

N

.

18

)

= 120°.

360° —

=T720°.

i

2

4

a

The principal axis of y = —% sinx +5

b

The principal axis of y =2cos% —4

‘

6

‘

14

oW

22

TN

g

Y

maximum value = 5

360°

™ RATTTRATTTTTRAT

1S y = 5.

is y= —4.

il

minimum value = —1

Chapter 9 (Trigonometric functions) 5

y=sinbx,

—

h—

O

b

360

period = aiac .

15° —

b

.

360° 1080°

b= 3

O

¢

360 b

p o 360

has minimum value and maximum value

y==zcosz+1

period = 500 .90

—

o

360 b

.y 360

5

Qo

15°

;o b=24

y = 5sinx —3

d = —3,

367

b>0

period = Sfibfl 1080°

Review set 9B

©

s b=40

5(—1) —3 = -8 5(1) — 3 =2

{when {when

sinz = —1} sinx =1}

has minimum value

%(—-1)+1=2

{when

cosz = —1}

and maximum value

=(1)+1=%

{when

cosz =1}

so the principal axis i1s y = —3.

We translate

y = cosx

downwards by 3 units to give

y = cosx — 3. 0 >

b

a = —2, so the amplitude is | —2| = 2. We stretch y = sinz vertically with scale factor 2 to give y = 2sinz, thenreflect y = 2sinx in the x-axis to give y = —2sinx. y=2s8Inx

a = 4, so the amplitude is |4 | = 4. d = —1, so the principal axis1s y = —1. We stretch y = sinax vertically with scale factor 4 to give y = 4sinxz downwards by 1 unit to give y =4sinz — 1.

y = 4sinx,

then translate

Chapter 9 (Trigonometric functions)

3638

d

a=3,2

sothe amplitudeis

b = 2,

so the period 1s

We

.

stretch

y = 2 cosz

4.

|3 |=3.2

360°

y = cosx

Review set 9B

360°

==

180°.

vertically with scale factor % to give

horizontally with scale factor % to give

y =

%CDS x,

then

stretch

y = % cos 2. ,-8 Y=COST

d

vertical stretch

vertical translation

scale factor %

1 unit upwards

> = COS T

COsST

»zcosz+1

A vertical stretch with scale factor %, then a translation 1 unit upwards maps

y = cosx

onto

y =smx

onto

y=%c05$+1. horizontal stretch 2

reflection 1n

b

r-axis

sinz

scale factor %

» —sinx

> —sin S

A reflection in the z-axis, then a horizontal stretch with scale factor 32 maps Yy = —sin %m

a

a =3,3

* so the amplitude is.

b =3,

so the period is

We

stretch

Yy = %sinzfl

:

y =

.

sinxz .

| 5|3 | = 33

360°

360°

el

120°.

vertically with scale factor 5 to give .

.

horizontally with scale factor % to give

3

.

y = %sin 3.

y =

3 sin x, 3

then

stretch

Chapter 9 (Trigonometric functions)

b

a =2, b =3,

sothe amplitudeis so the period is

d = —1,

Review set 9B

369

|2| = 2.

36;

O

— 299"

o

_ 1900,

so the principal axis 1s y = —1.

We stretch y = cosx vertically with scale factor 2 to give y = 2cosz, y = 2cosx horizontally with scale factor % to give y = 2cos 3z, then translate downwards by 1 unit to give y = 2cos3x — 1. Y

Yy =2C0ST

then stretch y = 2cos3x

=

-

¢

b=4,

360°

so the period is

We stretch

y = sinx

360°

i

= 90°,

horizontally with scale factor % to give

y = sin4x.

Y

d

a=

—%,

b =2, We

so the amplitude is

so the period is

stretch

Yy = %casm

y =

cosxz

| —% | = %

O

O

P

T 180°.

vertically with scale factor % to give

horizontally with scale factor % to give

in the x-axis to give

y =

y = % cos 2z,

%CDS r, then stretch

then reflect

y = —% COS 2.

.

.

y=s5c082r

Y= —%ms?m

y = %CDS 2x

Chapter 9 (Trigonometric functions)

370

a

a =3,

sothe amplitudeis

Review set 9B

|3| = 3.

360°

360°

b = 2,

so the period is

d = 5,

so the principal axis 1s y = 5.

2

= 180°.

We stretch y = sinz vertically with scale factor 3 to give y = 3sinx, horizontally with scale factor % to give

by 5 units to give

y = 3sin2x,

then stretch y = 3sinx

then translate

y = 3sin2x

upwards

y = 3sin2x + 5. y=3sin2x + 5 Yy =JsIn 2x

b

When

z=30°,

y=3sin(2x30°)+5 = 3sin60° + 5

=3x¥3 45

= 3¥3 1 5~ 7.60 The amplitude 1s 4, so

a = 4. O

300" _ 360°

The period is 360°, so

b=1

The principal axis1s

y =6,

so

The equation of the function is

3

d, \

R

d = 6.

360°

=

The amplitude is S_T(_Li) = £, S0

y=acosbx + d

180°

360°

Y

y = 4sinx + 6.

ANVA

I80°

a= . 2

The period is 180°, so

007

1800 b=2

_4

..........

Y

The principal axis is y = _4; i SO

iT d B =

So,

. a=3,

e b=2,

and

1 d=e —3.

—2,

Chapter 9 (Trigonometric functions)

Review set 9B

371

y=asinx +d

M) (6 ) S

'

:

180°

v

=

360°

When z = 60°, y = Y3+

When z = 330°, y=2

asin60° + d = Y36

o asin330° +d = 3

g 4d=346 B

.

(1)

L

MathPedorn]) (d7c)Real

an X+bn a

1[

Y=CE

1

-0.5

HathDPedMornl)

an X+bn

C

0.866

2

E

X

1.933

“éJ

‘f|:

Y=Cn

For e

()

(d7c)Real

1 2

REPEAT a = %

and

d = %

H(t) = 10sin(30¢)° + 20: a =10,

so the amplitudeis :

e

b= 30,

e

the principal axis 1s 30

oTetd=g

1.

Solving (1) and (2) simultaneously using technology gives

a

_ 5

1 -E]

1.25

13

i

:

so the period is

|[10| =10 360

360

T

e 12 seconds

H = 20.

A H (m)

H(t) = 10 sin (30¢)° + 20 ' b

i

6

t

12

I

18

t

24

i

30

t

-

36 1 (s)

H(9) =10sin(30 x 9)° + 20 = 10sin 270° + 20 =10 x (—1) + 20 =10 So, the height of the blade’s tip after 9 seconds 1s 10 m above the ground.

¢

H(t) =10sin(30t)° 4+ 20 has minimum value 10(—1) 420 =10 So, the minimum height of the blade’s tip is 10 m.

{when

sin(30t)° = —1}

d

From part a, the period is 12 seconds, so it takes 12 seconds for the blade to complete a full revolution.

372

14

Chapter 9 (Trigonometric functions)

Review set 9B

y=acos(bt)° +d a

The graph is at a minimum when ¢ = 0 months, : . 1B and again when ¢ = 12 months, so the period is

14

12 — 0 = 12 months. 360 —

SO.,

12

b

b= 30

When

t =2,

y=10.75

s

X+En

Y=C

a

;[

y—qng{hf}flql—d

t =6,

3

12

1

y=14.5

.o

—a+d=145

... (2)

E MathiDegdForml) [dc)Real an X+ En Y=Cn

C

i

1n.?5]

fi[ 14.5

[SOLVE]RERAI3(CLEAR][EDIT ]

12} S

(REPEAT]

Solving (1) and (2) simultaneously using technology gives

¢

S~

acosl180° +d=14.5

.. (1)

[d7c]Res] E

u:f

N

acos(30 x 6)° +d=14.5 C.

za+d=10.75 dn

TN

1

When

acos60” +d = 10.75

MathDegMornd]

(6,14.5)

4

acos(30 x 2)° +d = 10.75

el

e

.z(‘(z“ 10.75)

9 b

e

-

R

=12

LY

a = —%

and

d = 12.

y=—2cos(30t)° + 12 September 1s 8 months after January. y= —2cos(30 x 8)° + 12 bojon

t =8,

b2

When

cos 240° + 12

= -3 x(-3) +12 = 13.25

The average number of daylight hours in Los Angeles in September is 13.25 hours.

15

d

The parameters b and d will stay the same since the number of daylight hours should be periodic over 12 months, with a mean of 12 daylight hours per day. The parameter a will change depending on the latitude of the city.

a

The paint spot 1s 1nitially at its highest point which 1s 2 m above the ground. The amplitude

a =1 m.

The period 1s 2 seconds, so

360

s 2 b = 180

The principal axis 1s y = = d=1

—2|—

= QLQO =1

Chapter 9 (Trigonometric functions)

The graph 1s:

A H (m) 2

................................................................................

1..

i

b ¢

v

2

H(t) = acos(bt)® +d H(t) = cos(180t)° +1m

4

6

|

8

e

t (s)

{using a}

H(3.6) =cos(180 x 3.6)° +1 ~ 1.31

The height of the spot after 3.6 seconds is about 1.31 m.

Review set 9B

373

Chapter 10

DIFFERENTIATION EXERCISE 10A.1 1

2

a

62 beats per minute means that in one minute, Kirsten’s heart 1s expected to beat 62 times.

b

Number of beats per hour = 62 beats per minute X 60 minutes = 3720 beats

a

Number of words = 14 x 380

— 5320 words error rate =

& errors 5320 words

~ 0.001 50 errors per word b

5320 words = 53.2 x 100 words *

error rate =

& errors

53.2 (100 words)

~ (.150 errors per 100 words 3

Niko’s hourly rate = a2

Marita’s hourly rate =

12 hours

= £12.35 per hour

D

13 hours

= £12.15 per hour

Niko worked for a better rate.

4

a

Tyre wear = 8 — 2.3 = 5.7 mm .

.

wearing

b

5.7 mm

rate = ——— ° 32178 km

: .

~ 0.000 177 mm per km

5

a

Time taken from 11:43 am to 3:39 pm

— 17 min + 3 h 39 min = 3 h 56 min

",

or

3%

32178 km = 3.2178 x 10000 km

b

_

wearing rate

5.7 mm

—

3.2178(10 000 km)

~ 1.77 mm per 10000 km

350 km = 350000 m

322 hours = 322 x 60 x 60 hours

average speed = L ' S 3% lours

~ 89.0 kmh™

= 14 160 seconds

', -

average speed = i D) &% 5P 14 160 seconds


h—0

. = lim h—0

i

.

{as

h # 0}

h

K10z + 5h)

=T

— éin}j (10z +5h) — 10x

{as h # 0}

398

Chapter 10 (Differentiation)

i

Investigation 3

Rules for differentiation

y=f@)=-2°

1o\ _ i @+ flz) = r},fln

D)= f(x) h

—2(+xh)* 2 — (—2x*) 2

pag ol h—0

h

. = lim

—2(x? + 2zh + h?) + 222

h—0

h

.

=247 — 4zh — 2h? + 242

h—0

h

= lim

1l

. KH(—4x — 2h) hems0 K

= lim (—4x —2h) h—0

{as

h # 0}

= —4x

b

¢

If f(x)=ca™

where cis a constant, then

| For y=4z*

1 j—y o

31

using the software we find:

21

1lol1]

432 | —128 | —16 | 0|

Now, using b, Check:

f’(z) = cna™'.

j_y — 4 x 4241 £

2 | 3 16 | 128 | 432

= 1623,

When z — —3,

j—y — 16 x (—3)% = —432

v

When z — —2,

j—y — 16 x (—2)3 = 128

v

When z — —1,

j—y — 16 x (=1)3 = —16

When 2=0.

¥dx _16x03=0

When z=1,

¥dx —-16x13=16

When z=2,

¥dx _16x2%=128

v

When 2=3

% _16x33=432

v

L

£

£

v v

v

Chapter 10 (Differentiation) For

y:—g

T

—3

dy .

02=5 >

Now, ow,

L

—1

= —2x

Investigation 3

Rules for differentiation

, using the software we find:

-2 | —1

|05

0

1

2

3

undefined | 2 | 0.5 | 0.25 _

2

R2

using b, b, —2 == _2x (—1)z~T "1 =272 ==2 = 2z

using

Check:

When

= N= —3,

dy 2 9 E_(—S)Q_fi

When

z _ = o—2,

dy x _ (22

2 _ 1 172

v

When

z 4 =

4y 2 9 dm—(_1)2—1—2

v

—1,

v

When z=0,

2 — 202 Whichis undefined dx

When z=1,

%dx _2_9 12

_

dy

2

o

When

‘JU—2,,

5—2—2—3—5

When

z B= 3,

dy _ 2 _ 2g

v

v

v

, using the software we find: T dy

o

— o

0.2=35

=l 2

0.75 |

dy

.

ii

1

2 —

6 | undefined | —6 | —0.75 | —0.2

=3 x (—2)x

—

Now, using b,

0

_o_1

6

_a

= —627°%

= —— £Z

Check:

When

z =_ —3,

When

_ = 9—2,

When

= =

When z = 0, o

—1,

dy %

6 (3p

6 - @8

9

v

Yyo _ __6 _3 (23 _ _681

v

=

V

13

= =06

j—y B —[% Hichis undefined £Z

When

z =1,

d'y

-

6

.

When

z =_ 2,

¢ —dy = T36 = 8=

When

z =_ 3,

—dy =

13 L

6

z=-06

i3

_ § =% 9 m6 =%

V

Vi

v

9

i

399

Chapter 10 (Differentiation)

400

a

Investigation 3

In Exercise 10F, question 1 ¢, we have

Rules for differentiation

f(z) =2z —1

and

The derivative of 2z 1s 2, and the derivative of —11s0.

In Exercise 10F, question 1 d, we have

2+0=2

f(z)=3—2

and

The derivative of 3 is 0 and the derivative of —x is —1. In Exercise 10F, question 2 a, we have The derivative of 22 is 2

y = z* + 2

y =3 — x 2

f'(z) = —1.

0+ (—-1)=-1

and

and the derivative of 21s 0.

In Exercise 10F, question 2 b, we have

f'(x) = 2.

%

V

=2,

XL

2x+0=2z d

and

&

= 922

dx

The derivative of 3 is 0 and the derivative of —z2 is —2z.

0+ (—2z) = -2z

In Exercise 10F, question 2 ¢, we have

dy _ 4x + 1.

The derivative of 2z

y = 22 +x

is 4z and the derivative of zis

In Exercise 10F, question 2 d, we have

and 1.

dx

4z +1=4x+1

y = —z° + 5z — 3

and

V

;fi_y = —2x + 9. £

The derivative of —x? is —2xz, the derivative of 5z is 5, and the derivative of —3 is 0. —2r+5+0=—-2x+5 V

In Exercise 10F, question 3 a, we have

f(z) =3z° —1

The derivative of 3z2 is 6z and the derivative of —11s

and 0.

f'(z) = 6x.

6z +0=6x

In each case, the derivative of the function is equal to the sum of the derivatives of each term.

If f(zx)=u(x)+v(x), wu(x) = x”,3

For

then

f'(z)=1u'(z)+v'(z).

using the software we find:

zx | -3|-2|-1]ofl1]2]3 W(z) | 27| 12| 3 |o|3]|12]27 For

v(z) = —2z2, 12

3|

v'(z) | 12|

using the software we find:

-2|-1101]

8 |

4

1

|0

2

3

-4 | -8 | —12

For f(z)=wu(x)+v(z) = x> — 22°, using the software we find:

z | -3|-2|-1]0f 1 ]2]3 Fl)| 39|20 7 |o]|-1]4]15 Now,

«'(=3)+v(-3)=274+12=39= f'(-3)

uw'(=2)+1/(-2)=124+8=20=f'(-2) V u(-1) 4+ (-1)=3+4=7=f(-1) v

V

u'(0)+v'(0)=0+0=0= f'(0) V (D) +(1)=3+(-40)=-1=f(1) v ' (2)+v'(2) =12+ (-8)=4=f'(2) vV w(3)+0(3) =27+ (-12)=15=f'(3) V

Chapter 10 (Differentiation)

j

m

fl(x)=2*+3x—5

k

flx) =2¢° 4+

P

- f(@) = 22+ 3(1)

—1

- f() = 2(22) +1 =4r +1

P

flz)=3z>—-7x+8

L) =30 —-7(1) =6z —7 q

o

LI RIS

flx) =7—z—42°

s

- @) =146 = —1—122°

t

I

- f(2) = 5(1)

flE) i= %3:4 — 62° f'(z) = 1(42®) — 6(22)

.

r

f(x) =bx — 2

o

Exercise 10G

401

f(x) =243

- f'(z) =22

f(x) =4 — 227

o fe) = —2(20) = —dx

f(z) = z° — 4z° + 6z f(x) = 32 — 4(2x) + 6(1)

= 32" — 8z + 6

f(m)z%a:S—%;nz—Z

1) = 3(32%) — §(22) _=fr° 3,2 _— 7 1o

fl@) = (20— 1)?

= 4z° — 4z + 1

oo

i)

=4(2x) — 4(1) = 8r —4

2

a Let f(z)= %

b Let f(z)= 3}5

¢ Let f(z)= mig

_ 2

— 5

_ 8

s fi(r) = 2277

s fi(x) = —5z7°

2

9

28 d

Let

f(:}:):%

L f() =3(-1272) = 34 3

8

b e

=3z7"

s fi(r) = 8277

Let

f(:}:)—;ig

20 f Let

= 47"

/(@) = 4(=327) = —1227* 12

f( ):_miz;

= Tz~ *

f'(x) = ~7(~42) = 28z7° 28

402

Chapter 10 (Differentiation)

g

Let

Exercise 10G

3

f(:r):2:12‘+m—2

Let

S JO fla)=a2 -3 =z° — 6z~

= 22 + 312

f'(x)

() =2(1) + 3(—2z~7) =2 — 6z °

g I Let

2% + 624

0

6

S

f(x)=9-— $—23 =9

[et

2773

f(x) = —2(—3z7%) — 6z~4

_5mal

i Let f(z) =5+

Let

—2r % 4+ 9z~4

fl(x) = 2(—2277) + 9(—427°) = —4z73 — 362° . Ml 0 o

3‘13

iII5

m Let f(z)=5-—+= 11:2

L

Let

—5—8x % + 43

f(x) = —8(—2x73) +4(—3z~%) — 1623 — 1224 16 12 z3 4 o

Let

f(x)=4x— -

Let

dx

_

— 4

o

=2z — 6(—1z2)

e

1,..—1

f'(x) = 4@1) — 3(-1277)

Chapter 10 (Differentiation)

r

Let

Exercise 10G

2r — 5

f(z)=

—

2z

D

T 22

22

S -

AL

1132

=2z ! — 5~

f'(z) =2(-127%) — 5(-227") = 22 % 4+10z° 2

10

-T2 T3 b

(2 =12(2)° -1

¢ f(0)=12(0)* -1

= 47

b =1—x _1_m_21

5

a

S

=" =i

¢3)=1——

¢

e

-=1—3 1

_ 3 — 7

— 8 9

b

y = 100z

y = T’

dy rate 7(2x)

= 100

y:6$+i

= 27X

d

£

y = 2.5x° — 1.42% — 1.3

Ydx _ 9.5(32%) — 1.4(2x)

= 6x + 5z~ 1

dy

g

L=

= 7.52% — 2.8x

=6 — bz~ ° zfi_m_g5

e

y =10(x + 1) = 10x + 10 dy L~ 10(1)

= 10

T T

=1—+

2dy _—100(1) C

¢(-2)=1—

f

y = 4dnz® dy =

Am(3x7)2 = 12722

403

404

Chapter 10 (Differentiation)

g

Exercise 10G

y=(z+1)(z—2) =z

h

— 1 —2

Wdm_2$ _o

y=(5-2)°

= 25 — 10z + z*

oy = b= s

1

10(1) + 2x

=2x — 10

|

y=(3+2x)(2—x) =6 —x —x°

Ydx _ 19 6

a

y—%t4—%t

1

b

y:7_2_t

d

d_? = 3(4t%) — 3(1)

= 2%~} o3

1

C

V:%?TT‘B dV

=

=7— 3¢t~

=317 d

2m(3r7)

= 4’

_

= 5t142

=

2t2

7

a

y = x°

dy _s

b

y =1z

dy o 2 =y 3

21

_

When z — 2, j—y:2(2):4

Lo

v _ So, the tangent has gradient 4.

— 5x + 2

At the point dy

5(1)

Zn

(3, 14),

— =3(3)? =5 =22 dx

So, the tangent has gradient 22.

_

y=—3

c

8

d

dy—Q:E 5.2 — 3+ 7 Y

dy

—

dx

=8(—2

—3

(=2z7)

=. —16x -3

At the point

£

(9, 8% ),

GyRCI LN KR gl de

93

729

16 So, the tangent has gradient —=5%. 729

=l

When

—.3

» = —1,

ay =4(-1) -3 So, 0, the the t tangent th has gradien dient —7.

Chapter 10 (Differentiation)

e

— —O

y =2

f

£

Yy =

dy

e o T

el 10

r

2(1) = 5(=127)

dy

dy B

(2, 2),

| Px

e

— 4(—1z7?) i

D

When

—94 5

o )

. .13 S0, the tangent has gradient 7

fe)=a2*+ b+

POz

8

_ — 8(—2z~°)

L= N 08 iQ 4 1_2

EZQJrQ_z

8

I

>

=z —4z~ ! — 8z~

__2+m_2O

At the point

405

x5 — 4z — 8

x4z

= 2x — 5x !

Exercise 10G

1Dz +2¢,

z = —1,

dy _

L

4

(2

16

=1 So, the tangent has gradient —11.

f(2)=4,

and

f/(-1)=2

f'(x) =2z + (b+1)

But f'(—-1)=2,

so

2(-1)+b+1=2 —14+6=2 b=23

So,

f(z)=z"+ (3+ 1)z + 2c

= x° + 4z + 2c But f(2) =4, so 2°+4(2)+2c=4 2c = —8 c=—4

9

y =4r — S I

=4x — 3z} 4

dx

?

£L

_ 4+ 32

=4+ —3

1s the gradient function of

y = 4x — 2 £I

from which the gradient of the tangent at any point

can be found. It is also the instantaneous rate of change of y with respect to .

10

a

S =2t +4t m R dt

dsS

%

At +4

ms1

is the instantaneous rate of change in position at time ¢. It gives the speed of the car at

time ¢, 1n metres per second.

b

When

t =3,

dsS

5:4(3)+4 — 16

ms "

The instantaneous rate of change in position at time of the car after 3 seconds 1s 16 ms™*. 1

¢ = 3 seconds

is 16 ms™,

or the speed

406

Chapter 10 (Differentiation)

1

Exercise 10G

C = 1785 + 3z 4 0.002z*

pounds

= =3+ 0.002(2x) £

= 3+ 0.004x

When

pounds per toaster

= 1000, f£—0 — 3+ 0.004(1000) £

=17

When 1000 toasters are being produced each week, the cost of production 1s increasing by £7 per toaster.

12

y=u1x—4dx+ 7T dy

.

B

the tangent has gradient 2 when

2z —4 =2 '

20 =20 =3

When

=3,

y=3°—-4(3)4+7 =9-12+4+7 =4

So, the tangent has gradient 2 at the point 13

a

(3, 4).

y =z + 5z + 1

b

y =322+ 11z +5

@=2$—|—5

dy = 6x + 11

the tangent has gradient 3 when

the tangent has gradient —7 when

2r+5

ox + 11 = —7

dx

=3

g o

When

6xr = —18

— 2

Lo

=—1

z=—-1,

y=(-1)>+5(—1)+1

—1-5+1 — —3

So, the tangent has gradient 3 at the point

(—1, —3).

When

rx=-3

z = -3, y=3(-3)°+11(-3) +5 =27—-33+5 = —1

So, the tangent has gradient point (—3, —1).

—7

at the

Chapter 10 (Differentiation)

d

Exercise 10G

f(x) = 3z® — br + 2 f'(x) =92° =5 the tangent has gradient 4 when 9z° — 5 =4

9z° =9 =

oz

==l

f(1) =3(1)” —5(1) +2 =3—95+4+2

So, the tangent has gradient 4 at the points (1, 0) and (—1, 4).

So, the tangent has gradient % at the point

(2,5 2

D/

f(z) = az”® + bx +c f'(z) = 2ax + b The tangent gradient 1s 0.

1s

horizontal

when

the

the tangent 1s horizontal when 2ax +b=10 2ax = —b

So, the tangent i1s horizontal at the point B

b

b2

2a°

da

—_

b2

- —+c]. 2{1_'_)

407

408

Chapter 10 (Differentiation)

14

Exercise 10G

y=—x°—6x—4

_op

YW _

a

—

£Z

YA

6

The tangent at P passes through The gradient of the tangent

(0, 12). =

y-step

_

:

(—6, 0) and

v

_

The tangent has gradient 2 when

v

—2z — 6 =2 —2z = 8 .

T=—4

y=—(—4)? —6(—4) —4 = —16+24 -4 = 4

So, P has coordinates

15

(—4, 4).

flx)=2°+ax+5 fl(x) =32° +a Now

f/(1)=10,

so

3(1)*+a =10 34+a=10 a="17

16

f(x) = =3z* +ax + b f(z) = -6z +a

Now

f'(=3)=9,

so

—6(=3)+a=29 184+a=9 a=—9

Also,

f(=3)=38,

so

B

E

y=—x*—6x—4

=2

== —4,

/

:

12 -0

0—(—6)

When

—06

@

—3(=3)*+(-9)(-3)+b=38 —27+27+b=28 b=8

Chapter 10 (Differentiation)

Exercise 10G

409

fl@) =20 +a+ >

17

— 2% +a+bx !

f'(x) =4z — bz~ b

Now

f'(1)= -2,

b

so

4(1)—1—2:—2 . 4—b=-2 =

Also, f(1)=11,

10P=26

so 2(1)+a+7 =11 24+a+6=11 8+a=11 a=3

18

a

The tangent line L passes through

AY

(=2, —5) and (3, 0).

y=2’+2°+ax+b

The gradient of the tangent

EgUrtieD

~(=2,-5)

xr-step

?

_ 0—(=5) - 3-(-2) At

=]

g

Y

So, the gradient line L 1s 1.

of the

tangent

y=z"+z°+ar+b &

dx

3z° +2z+a

Now, when

= = —2,

j_y =1,

so

3(-2)*+2(-2)+a=1 12—4+a=1 8+a=1 Loa=-—T

Also, when

=z = -2,

y= -5,

so

(=2)°+(=2)*4+(-7)(-2)+b= -5

£

—8+4+14+b=

-5

1I0+b=—-5 .

b=

-15

410

Chapter 10 (Differentiation)

Review set 10A

REVIEW SET 10A 1

:

AY

4

/

j‘/

A/

-

2

a

b—a

- 2—(-1) — 4

=3

)

The graph of height against time 1s a straight line. . the height increases by the same amount each time interval. the

ski-lift 1s increasing

constant rate.

b

_ f(b) = f(a)

1-(-3)

>

‘

v/ Y

ly=f(z)

average rate of change in f(z) from A to B

rate of change = (40

.

—

in height

A height (metres) 200 400

at a

jd

300

200

0) m

(250 — 0) s

100

— 1.6 ms —1

b

)

0

3

a

Wecan make

¥

/

/

pd

/

S/

P

//

i

time (seconds)

50

100

150

200

250

300

6x — 7 as close as we like to —1 by making z sufficiently close to 1.

lim (6 —7) = —1

r—1

2 _

b h—0 lim 2"

_

_ h—0 iy 22D = %imfl (2h — 1)

{as

h # 0}

=1 4

The tangent to y = f(x)

gradient

S A 0—8

F® =4

2

at the point where

x = 3 has

Chapter 10 (Differentiation)

5

a

Review set 10A

f(x)=2%4+22 h—0

=

lim

h—0

h

(= + h)* +2(z + h)] — [z° + 24 h

_ Iim 2% + 2zh + h2 + 26 + 2h — 22 — 24 h—0

—

h

.

2xzh+ h? +2h

h—0

h

|lim

[

K2z + h + 2)

h—0

}f

:}]ifl‘h(2$+h+2)

{as

h # 0}

— 20 + 2

b

y=f(z)=4-3z"

dy _ . f@+h) -~ f(@) dx

h—0

~ i

h

4 —B 3(x+ h)*]2N —A[4 — 3z~]2.2

h—0

.

lim

h

4—3(x% + 2zh + h?) — 4 + 32

h—0

—

.

|lim

h

A—34%7 —6xh —3h? — A+ 342

h—0

h

. H(—6x — 3h) = hl—% M

= lim (—6xz — 3h)

{as

h—0

h # 0}

= —bx

6

a

y=f(z)=2z>-1 dax

h—0

h

e+ h)2 o[22

h—()

—

1] 2 [222—1]1

—1| — h

.

2(x®+2zh+h%)—1—-22%+1

h—0

h

lim

T

nt 262 + dzh +2h%2 — Y — 262 + X h—0

— }{12}] (4x +2h)

h

{as h # 0}

= 4x

b

The gradient of the tangent to

¢

The gradient of the tangent i1s equal to —12 when

y = 2z® — 1 at the point where 4z = —12 r=—3

=z =4

is 4 x 4 = 16.

411

412

7

Chapter 10 (Differentiation)

a

Review set 10A

f(z)==z°—3z° h—0

h

[(z + k)% — 3(z + h)?] — [z — 3z?]

= lim

h—0

h

iy 20+ 32%h + 3ah? 4 b3 — 347 — 6ah — 3h% — 25 + 347

'y

h—0

h

.

32°h + 3zh? 4+ h? — 6zh — 3h*

h—0

h

= lim

. K(3xz? + 3xzh + h? — 6x — 3h) = lim h—0

A

= lim (3z% + 3zh + h* — 6z — 3h)

{as h # 0}

= 322 — 6z

b

Y

The illustrated tangents pass through the point where = = —1 and the point where = = 3. The gradient of the tangent at © = —1

=

3

f'(=1) =3(=1)* = 6(-1)

%

1is

=9

f(z) =z — 322

and the gradient of the tangent at * =3

1'(3) = 3(3)* - 6(3)

1s

=9

Since

f'(—1) = f’(3),

the gradients of the

tangents are equal. This means the tangents are parallel.

8

a C

f(x) = bz’ f'(x) = 1527 flz) = Tz? — 2 .

€T

— 14z + 3z~2

9

+

flx) =3x — =

8

3

=3+ —

::3_2

f(z) = —x* + 4z — 2

fl(x) = —2x+4

At the point (-3, —23),

d

4

f'(z) =3 —4(-227") =3 +8z7°

f'(z) = 7(2z) = 3(—277) 14x

f(x) = z° — bz f'(x)=6z> -5

=3z — 4z~7

= 72?2 — 3z~!

—

b

f'(=3)=—2(-3)+4 = 10

So, the gradient of the tangent is 10.

Chapter 10 (Differentiation)

10

a

s

f(x) =7+

— 3z°

b

fFB)=T7+3-3(3) =10 — 27 = —17

11

s fl(x)=1—6z s f1(3) =1-6(3) =1—18 = —17 AY

_ —2x + 8

™

The tangent has gradient —4 when

—2x + 8 = —4 c. 2z =-—12

.

When

z=6,

-

T =

y=—(6)°+8(6)—7

Y

— 364487

y=—2"+8s—17

=) So, P has coordinates 12

(6, 5).

y=azr® — 3z +3

dy -

Jax” 9 — 3

Now, when z = 2, ? — 21, so 3a(2)?—3=21 L

12a — 3 = 21 12a = 24 (=52

13

a

(z) = 2° — 32

b ;.

fi(x) =22 -3

So, the gradient of f(z) 1s —9.

fl@)=x+2 N

at

x = —1

So, the gradient of f(z) is —12.

d

o

/

=3z

2

—2x

s

=1-

12 !

So, the gradient of f(x) is —1.

F@)=1-=

So, the gradient of f(x) 7 1S g

at

= = 3

= = 2

-1

o2

2

at

fla)=a2®—a2®—z—2

1

fz)=1-2x /

f(x) = =32 + 4 fli(z) = —6x

- F(2) = —6(2) — 12

f'(=1)=2(-1) -3 = -5

¢

413

f(x) =7+ z— 32"

y=—x>+8x—7 dx

Review set 10A

f(0) =3(0)7—2(0) —1

S|

at

=z = 0

414

Chapter 10 (Differentiation)

Review set 10A

14

S(t) = 0.3t° — 18t* + 550t grams S’(t) = 0.9t* — 36t + 550 grams per second S’(t) represents the instantaneous rate of change in weight of sand in the bucket, in grams per second, for a given value of ¢.

15

a

The tangent shown on the graph passes through

and

(5, 0).

the gradient of the tangent is

: —

s —1,

Also, since the tangent passes through

(0, 5),

‘

y — 5

equation

i

ek

L

so

it has

-t

1

0

.

=

v

Y—90=-—=I

S Sowhen

Yy=—+93

=3,

y=—-3+5=2

the point of contact is (3, 2), and hence b

(0, 5)

f(z) has the form

f(3) = 2.

f(z) = az® + bx + c

The y-interceptis 14

.

f(0) =14

a(0)* +b(0) + ¢ = 14 c=

Now

14

f(3)=2

{from a}

9a + 3b = —12

... (1)

a3 +b(3)+14=2 Also f'(3) = —1 and f'(z) =2azx+b 2a(3) +b=—1 6a+b=—-1 =]

dn

X+En

... (2)

(d/c]Red)

El

Y=CE

a

dn X'l"En

c

Y=Cn

(d/c)Real

x|:

e

s — 1’2

[REPEAT]

Solving (1) and (2) simultaneously using technology gives

So,

f(x)=z*— Tz + 14.

1

a =1

and

b = —7.

Chapter 10 (Differentiation)

Review set 10B

415

REVIEW SET 10B 1

AY

The tangent at x = 2

13

//G/

// /

a

has gradient

o

1

o

==

f{:fl:}

3

)

1

1S

E.

v

/

y oo 2

1—20

the instantaneous rate of change in f(x) at = = 2

o y

)

)

average rate of change in temperature from 7 am

to noon

A temperature (°C)

=

_ (20—-10) °C

20

(6 —1) h

g

N\

7

— % °C per hour = 2°C per hour b

"

average rate of change in temperature from 3 pm

DL

[DpapTY

6am

~ (20 —25) °C

ol 9am

12pm

3 pm

6pnl‘r

(11 —=9)h

— _?5 °C per hour — —2.5°C per hour

:

. h3—3h ] hoo R

8

. H(h?* —3) = ljm —~—~/ hoo A :Ajmb(h?—?))

) {as

h # 0}

. 3x% —3x . lim —— = lim pol 71 o — ]im1

— 3

4

=3

\ | AY \v=f(z) \\

B

a b

9

\ 12 }_}/ =_ =2

Y

T

53\\

\

\

v

f(—1) is below the z-axis, so f(—1) is negative. f/(0) is the gradient of the tangent to f(x) at the point where

= = 0.

f'(0)

1is positive.

¢

f(2)

is above the xz-axis, so

d

f/(3)

is the gradient of the tangent to f(x) at the point

f'(3)

1is negative.

—

\

3z(z—T) ———= (z—T) 3x {as z # 1}

where

= = 3.

Since the curve is increasing at

f(2)

= = 0,

is positive.

Since the curve 1s decreasing at

= = 3,

416

5

Chapter 10 (Differentiation)

a

x f@)y=22-2|

\

Review set 10B

3| -2-1] 0 1]2 7|2 |-1|-2]|-1]| 2

/(@)

\

&

N

=Y

P

e

\

M

-

—

f@) =% =2\

o

IH\YEP ) /|

Y

b

The tangent to f(x) = 2* — 2 at the point where

x = 2 has gradient

the instantaneous rate of change in f(z) = x* —2 c

f!(z):hm

when

=z =2

6; is 4.

f(2+h)_f(2)

h—0

h

_ o @4 R)22 9]-2 02[2 _ 9] h—0

.

=

lim

h

4+4h+h*—-2-2

h—)

.

—

lim

h

4h + h?

h—0

—

.

K4+ h)

hlE}]

A

]

=;11H}3 (4+ h)

{as h # 0}

=4

6

a

f(zx)=z*-22 ff(ill‘)

—

lim

f(m_l_h)_f(fl:)

h—(

h

— i @R h—0

4

2@+ h)] — (2 — 21 -

ir=d ol

h

+ 622h? + 4xh3 + ht — 26 — 2h — 2% + 24 = lim 2% + 423h h—0

h

.

4x3h + 6z2h? + 4zh3 + h* — 2k

h—0

)

— lim

.

(423 + 62%h + 4zh? + B3 — 2)

h—0

K

— lim

= lim (42° + 62°h + 4zh? +h* —2)

{as h+#0}

= 43 — 2

b f(-2)=4(-2)%-2 —

—34

The gradient of the tangent to y = f(x)

at the point where

= = —2

is —34.

e g = 4.

Chapter 10 (Differentiation)

7

a

Review set 10B

417

y=f(x)=2x"4+5xr—2 W_ dill'

i, flEth) — f2) h—10

h

2 — lim (z+ h)*+5(x+ h) — 2] — [z°2 4 5z — 2] h—0

h

— Tim 22 +2zh+h2+56+b5h—2— 22 — 54+ 2 h—0

— =

h

.

2xh+ h? +5h

h—0

h

lim

s

K(2x + h +5)

lim

h—10)

}'L’

zflin}](QerthE))

{as

h # 0}

=2x+ 5

b

The tangent to

f(x) = x* + 5z — 2 has gradient —3 when

f’(z) = j_y = -3 £Z

20 4+ 9=

-3

20 = —8 r=—4

Now,

f(—4)=(—4)*+5(—4) -2 = —0

So, the tangent has gradient —3 at the point 8

f(t) =452 — 4.8t*

a

metres,

(—4, —6).

0 —-Tzx+4

b

—%dy = 3(2z) r- 7(1) ‘

YT

TS

419

y = 2x° — 62 + Tz — 4

.dy

2l

o= =2(327) — 6(2x) +7(1)

=6z — 7 3

Review set 10B

=6z° — 12z + 7

5

dy =r~“)—5(—3x"7) e Y dm—S( = -3z % + 15z ¢ 3

15

— oL

2

flz)

=222

B

:

T

=+ 2z * fl(x)=1—-2x""

//

-

/

3

i

i f()=1-

*

\

o2

b

:

f{=z)

-

2

P

713

— —1

gradient= —1

3

&

gmd'ent:l

gradient of tangent = —1

-3

2

i f(=2)=1-—5 (—2) =1-2 2

B

/A

2

V|; f(a)Y 7 =222 x Y

lag 2

gradient of tangent = % y = 2z° +ax +b

dy _ .b6x° 2 + a i

Since the gradient of the tangent at (—2, 33)

is 10, then

6(—2)% +a = 10 24 +a = 10 Loa=—14 oy =2z — 14z + b

Since the curve passes through

(=2, 33),

then

33 =2(—2)° —14(-2) +b = —16+28 40 b=21

e

Cha

pter 11

PROPERTIES OF CURVES

a b

f(z) =2 — 4z fl(z)=2x —4 The point of contact is (1, —3).

£'(1) = 2(1) — 4

=Y

1

= —2

So, the tangent has equation

a

y = —2(z — 1) + (—3) Loy=—2x—1

y=z’ When

b = =4,

y=4°?

When

=16,

so the point of contact is (4, 16).

dy =2z,

Now

—

=z = -2, d

Now

=4 — 3z%, dx

Ydy — =_9(4) 2(4) = 8

The tangent has equation

y = 12(z — (~2)) + (-8) = 12(r +2) — 8

= 8xr — 32+ 16

= 12x +24 -8 = 12z 4+ 16

= 8¢ — 16

3r—1 =

When

£

4

=z = —1,

Y 3

so at T

—

—1]

dy _ P dy

—_— =

=

o.—2_ O —

3

12

The tangent has equation

y=—3(z—(-1))+(-3) =—-3(x+1)-3 =—-3r—3—3 = —3r —6

4

=—=-4_3

3

When

y:_—1:—3,

so the point of contact is (—1, —3). Now

x = —2,

dx

y=8(x—4)+ 16

y=

soat

W _ 3(-2)? =12

The tangent has equation

c

y=(-2)°= -8,

so the point of contact is (—2, —8).

soat x =4,

dx

3

y==z

—

z = 2,

yz%zgz

so the point of contact is (2,

3— —

L

3

Now SO at

xr = 2,

2%dx —_19;4=_12x4 dy 12

12

3

dz

16

4

24

The tangent has equation

y=—5(x—-2)+3 = —%:1: + % T % =—%:1:+2

Chapter 11 (Properties of curves) e

z =1,

y=12+5(1)—-4=2,

so the point of contact is Now

d—y=2m—|—5, dx

When z = -2,

y=2(—2)2+5(—2)+3 — 1,

(1, 2).

so at

x = 1,

so the point of contact is Now

dy =2(1 — —— =2(1)+ =T

ay

=4x+5,

dx

(—2, 1).

soat

r =

—2,

Wdx 4(—2)+5=—3

The tangent has equation

y="Tx—1)+2

The tangent has equation

y=—3(z— (~2)) +1

=T — 7+ 2 =7

421

f y=22°+5z+3

y=z°’+5x—4

When

Exercise 11A

= —3(z+2)+1

—9

=—3r—6+1 = -3 y =x° + 2z

When z =0,

y=03+2(0)=0,

so the point of contact is Now

-2 =3z*+2,

(0, 0). soat

x=0,

y=m

When

2

+4=;1:+4:13_1

L

y= ( 111+4 = —9,

so the point of contactis

(—1, —5).

Now &dx —1_dp2=1-2T soat

y=2(x—0)+0

_1)2

=z = -1,

W 30)> +2=2 The tangent has equation

—95

dy

r=—1,

4

&=1—(_1)2

= -3

The tangent has equation

= 2x

y=—3(—(-1)) + (-5) = —-3(x+1)—5 =—-3r—3—9 = -3z — 8 y = 2% — bx

y=x—22°+3

When

2 =2,

y=2-2(2)2+3= -3,

When

z =1,

y=1%—-5(1) = —4,

so, the point of contact is (2, —3).

so, the point of contact is (1, —4).

Now

Now

@zl—él:c, dx

so at

dy =1—-4(2) — = (2) = — The tangent has equation

y=—T7(z—2)+ (-3 =—Tx+14—-3 = —Yz+ 11

=7

x = 2,

@:3:1:2—5, dx

dy —

—

=3(1)"

=3(1)y

2

—H

soat

=5

=

The tangent has equation

y=—2(z—1)+ (-4 = 2xr+2—4 =27

=

=9

x =1,

422

Chapter 11 (Properties of curves)

k

y

3 22

Exercise 11A

1

€T

I y:33:2—l:33:2—3:_1 I

iIIZ

=3z ' — a7? Now

@

dax

— _—3x

=—= dy

3

When

y=3(-1)"—(_1\2

242577 2

+ —5» 3

SO at

Now

(-1)3

=—-3—2

ay

£

dy

-

1

6(—1)+ = —1

b

L fi(z) =22 +6

x = 3

—

D

f/(2)=0

when

—-2z+6=0 —2z = —6 T —"3

1s horizontal.

y=2z°+kz*—3

a Ydx _6s2 4 2%z When

z = 2,

b @:4 dx

4+ 2k(2) =4 L 6(2)° 4k = —20 k=

Since

k=

—5,

When

z= = 2,

y = 2x° — 5% — 3 y = 2(2)° —5(2)* — 3 = —7

24 + 4k =4 -5

So, the point of contact is (2, —7). The tangent has equation

y=4x—2)+(=7) =4x — 15

5

x = —1,

y=—5x—(-1))+4 = —5(x+1)+4 = —9r —9+4 = Eort.1

flx) = —2° +6x+4

The tangent at the point where

soat

The tangent has equation

= —5(zx+1)—4 =—o9r—90—14 = —oxr — 9

h

— 6z + 22

—_—

y=—5(z—(-1)) + (—4)

¢

=4

£

The tangent has equation

a

-1

:6::c+i2,

= —5H

3

_

so, the point of contact is (—1, 4).

(—1, —4),

2

(D2

= = —1,

y=1— 3z + 122° — 8z j—i=—3—|—24:1:—24:132

When z = 1, j—y — 34+ 24(1) — 24(1)? £

So, the tangent at

= 3

(1, 2)

has gradient —3.

The tangents to the curve have gradient —3 when

—3 4 24x — 24z* = -3 24x° — 24x = 0 24x(x — 1) =0 r=0or]l

Chapter 11 (Properties of curves)

Exercise 1A

423

So the other z-value for which the tangent to the curve has gradient —3 is x = 0,

and when

=0,

y=1-—3(0)+12(0)* —8(0)° = 1.

the tangent to the curve at This tangent has equation

The tangent to the curve

(0, 1)

is parallel to the tangent at

(1, 2).

y = —3(x —0) + 1 = —3x +1 y = x*+ax+b

at the point where

x =1

is 2z+y =6

or y = —2x+6.

the tangent has gradient —2, and the point of contact is (1, —2(1) + 6) whichis Now,

y=x°+ax+b dy

E—2$+a

When

x =1,

j—y:—Q

and

£Z

y =4

21) +a = —2

© Pta(l)+b=4 s

Ot aq= —2

S

... (%)

Ca=—4

a

f(x) =2° + =

b

L

=

1+ (—4)+b=4 b=7

Horizontal tangents have gradient 0, so 8

% 4+ 42

2$—m—3=[]

f(z) =2z — 8z ° el = a8 Fel 3

224 -8 3 2zt — 8 = 21t = rt =

¢

When

$:\/§,

r=+v2

V2) f(VE) == (V2)? (VO ++ —

4

_2+§ 4 the horizontal tangent at

When

== —v2,

(\/5, 4)

is

y =4.

9

4

f(—v2)=(-v2)*+

o)

=24 1 =4 the horizontal tangent at

(—\/§, 4)

The tangents are the same line,

y = 4.

{using (x)}

1s

y=4.

(1, 4).

424

8

Chapter 11 (Properties of curves)

a

Let

b

When

y=f(z)=a"+1

fi(z) =22

Now

z=1,

y=(1)*+1=2,

The tangent has equation

When

a

=0,

y =2(x — 1) + 2 =2 — 2 + 2 = 21

y = 2x.

y=2(0)=0,

so the tangent passes through the origin.

f(z)=22>-5z+1

i f(—1)=2(-1)% -5(-1)+1

i We use technology to find where the

— 21541

tangent meets the curve again:

=4

EI

So, the point of contactis Now

g

..

(—1, 4).

[EXE]:Show coordinates

Yomxit |

f/(z) =6x% —5

_

f(_]-)—fi(_]-) —

o

6

2

E:_,

AN —

9

5

=1

.

The

the tangent has equation

y=1(x — (—1)) +4

(2,7)

A5 1

a3

=2

.K_,_,'

.

j

]

/

]

—1.;:,‘/:. -2;{_”

tangent meets

3

4

x

5

INTSECT

the curve

5

2, 7).

.

again at

5

=x+1+4

=x+5 y=:172+2—|—2=:1:2+3$_1—|—2 I

i

When

z=3,

y=3"+ g + 2 94149 =

19

So, the point of contact is (3, 12). Now

@:2:1:—3m_2 dx

3 I

i

When z =3,

3

1 _ 17

=b0-3=3 the tangent has equation

y = (z—3)+ 12

Ll

— 17+ 12

r— 9o

We use technology to find where the tangent meets the curve again: [EEF]:SE;; coordinates 15-l

¥2=(4\743)x=5

\[

T

T

T

o\ TN

K=—fi}.3333333€33

—-==203)- 5 _

Il

‘[?1 .

.

wl|g

b

w|g

9

so the point of contact is (1, 2).

f/(1) =2(1) =2

the tangent has equation

¢

Exercise 11A

T

1 H

INTSECT Y=-6.888888889

The tangent meets the curve again at

about (—0.333, —6.89).

Chapter 11 (Properties of curves)

¢

425

flx)=z3+5

i f(1.5) = (1.5)° +5

ii We use technology to find where the

— 3.3754+5 =

tangent meets the curve again:

8.37H *

5]

*

So, the point of contactis

Now

,ff(@ = 32° f

(15)

(1.5, 8.375).

3(15)

_

3

2

+5

12=6.75x-1.75

2

—

[EXE]:Show coordinates

:'Ilfx“(fl

a

¥

LT /:,-- o f

]

-

2

3

j?-fz

225

| the tangent has equation

q

|

)Z:;/

— 0.75

uu?=-22

=

_

INTSECT

The tangent meets the curve again at (—3, —22)

y = 6.75(x — 1.5) + 8.375 — 6.75z — 10.125 + 8.375 — 6.75z — 1.75 d

Exercise 11A

y:$3+1:a’:3+m_1 I

.

i

When

z=-1,

1

B

y=(-1)7°+ — 1.1 —

Il

_2

We use technology to find where the tangent meets the curve again: Elj?:j]:;fmw_ coordinates E‘l:x

So, the point of contact is Now

d

—Z

dm

(—1, —2).

1dx)

#?‘

=

— =3z%— 22

s\

2.5 2 -15 -1 0.

ol :

X=

— 3r° — ig

7/

|

!

/

_E-

,B

‘=2

05E !

1

1.5]

2

25 =

INTSECT

|

xX

When

7 — —1

dy dx

3(—1)2 — =3 -1 = 2

the tangent has equation

y=2(z—(-1)) -2 = 2(:}: -+ 1) — 2 = 2r 42— 2 — 2

1 (—1)2

The tangent meets the curve again at (1, 2).

426

Chapter 11 (Properties of curves)

e

Exercise 11A

f(z)=3z>+22"—x+2 i

£(0.5) = 3(0.5)° +2(0.5)* — 0.5 + 2

il

We use technology to find where the

=3%x0.125+2%x025+1.5 =

2.375 L

tangent meets the curve again: el

.

So, the point of contactis (0.5, 2.375). Now

ff(fli')

—

9$2

+4r

[EXE]:Show coordinates

Y1=3x"(3)+2x2—x+2

y2=3. Q%Ln.j_g\fi‘/

—1

25 2 ~1f‘/—1

£/(0.5) = 9(0.5)* + 4(0.5) — 1

=9x0.25+2— 1

e

= 3.29

o

05

1

15

2

25 4

LA | ;,

7 =_a. 000008007

1

:é]ju:an(ge?.tfil;:eejg;. curve again at

the tangent has equation y = 3.25(x — 0.5) + 2.375 = 3.20x — 1.625 + 2.375 = 3.25x + 0.75 10

a

Consider the tangent to

When

z =2,

Now

—-2= =3z?

y = 2°

y =23 =8

£

at

x = 2.

so the point of contactis

andsoat

(2, 8).

z =2,

Wdx _ 3(2)? =12 So, the tangent at

(2, 8)

has gradient 12 and its equationis

y = 12(x — 2) + 8 = 122 — 24 4 8

= 122 — 16

We use technology to find where the tangent meets the curve again: [E [EXE]:Show coordinates Y1i=x"(3) 275¥ Y2=12x-16

INTSECT A=-4

The tangent meets the curve again at Consider the tangent to

When

z = —1,

Now

j_y — —32z° +4x

y = —2° +2z* +1

at

y=—(—1)> +2(—=1)*+1 =4

£

dy i

(—4, —64).

andsoat

= = —1.

and so the point of contact is (—1, 4).

x = —1,

o 3(—1)*+4(-1) 132 L ==y -7

So, the tangent at (—1, 4)

has gradient —7 and its equationis

y = —7(z — (—1)) +4 =—7(z+1)+4 = —T7r—T7+4 = —7z—3

Chapter 11 (Properties of curves)

Exercise 11A

We use technology to find where the tangent meets the curve again: [EXE]:Show coordinates

Y1=-x"(3)+2x2%1

2c:¥

¥Y2=-7x-3

IIIIIIIIIIIIIIIIII

INTSECT

X=4

The tangent meets the curve again at (4, —31). :

1

Consider the tangentto

z =1,

Now

y=2a"' —

y= L

d

ml

== dx

=——

1

L

dy

dx

y = - — — r x?

1

When

1

1

at = = :

:

0 and so the point of contact is (1, 0).

2 .

724278 2

+—

andsoat

AL

z =1,

1

4+ =2 =1 12 * 13

=

So, the tangent at (1, 0)

has gradient 1 and its equationis

y = 1(z — 1)+ 0 =z —1

We use technology to find where the tangent meets the curve again: [EXE]:Show

coordinates

Yi=(1ax)—(11(x2)) Y2=x-1

o[¥ [

2

g INTSECT

The tangent meets the curve again at (—1, —2). 11

y=2x°+3z°

When z=-1,

—x+4

y=2(-1)>+3(-1)?—(-1)+4

=-2+3+1+4 = 6

So, the point of contactis Now

When

z _= —1,

(—1, 6).

@=6$2+6$—1 dx dy T

2 s 6(—1)°+6(—1)—1

__

=6—-6-—1

|

the tangent has equation

y = —1(x — (—1)) +6

= —(r+1)+6 = —x—1+4+6 = —xr+95

427

428

Chapter 11 (Properties of curves)

Exercise 11A

The tangent meets the x-axis where

y =0

—x+5=0 r =29

So, the point where the tangent meets the x-axis is (5, 0). b

y=2x"+5

dy

o

dr

9 (-2,

At

_3) —3),

%

2

Wd:::_g( _ 3 922)° _ =12

the tangent has equation

y = 12(x — (—2)) — 3 =12(x +2) — 3 =12+ 24 — 3 = 12z + 21

The tangent meets the line

y =2

where

12z + 21 =2 122 = —19

So, the point where the tangent meets the line

y =2

is

19 1

b

==

(—13, 2).

y==-+1=2z""1+1

c @

dx

(—2 (=2,0),

At

—

_23}_2

—

_i

2

dy

2 =—29 =—3 1 — = — 2 4 (—2)2 dx

the tangent has equation

y = —%(

—(=2))+0

= —2(z +2) — —%m We y

_

il

use technology to find where the tangent meets the line 2$

L

R

3:

So, the point where the tangent meets the line

T

E’lfi-fl;fix-l"

y =2z — 3

is

(0.8, —1.4).

—"T——.

Chapter 11 (Properties of curves)

d

y=322

When

Exercise 11A

-2z +1

z=1,

y=3(1)-2(1)+1 =3—-2+1 = 2

So, the point of contact is

When

(1, 2).

Now

¥dm _gz2_9

=1,

%dx —9(1)2 -2 —9_9

. the tangent has equation

y = 7(x — 1) + 2 =(rx—7++2 =T7r—29

The tangent meets the y-axis where

x = 0

y =1(0)= —9

So, the point where the tangent meets the y-axis is

AY

flx)=a*—22° — 2% + 4 () = 42° — 62° — 22

At PoL,6), f(-1) = (-1 — 6(-1)? 21 — 4642

— _8

P B

A&

12

(0, —5).

the tangent at P has equation

y=—8(z—(-1))+6 =—8(x+1)+6

/

— 87— 846 = —8x — 2

At Q(2,0),

|

f'(2)=4(2)°-6(2)" —2(2) =32—-24 -4 =4

the tangent at Q has equation

y =4(x —2)+ 0 =4 — 8

We use technology to find where the tangents intersect:

B [EXE]:Show coordinates YI=4x-8 \ y yo=-szo2N\\

/

INISEC]

X=0.5

So, the tangents intersect at R(0.5, —6).

429

430

13

Chapter 11 (Properties of curves)

Exercise 11A

f(z)=z*—-222+22+3

a f(1)=1*—2(1)>+2(1)+3 =1—-2+4+2+43 = 4

So, the point of contact is (1, 4).

Now

f/(z) = 4a® — 4 + 2 oo (1) =4(1)° —4(1) + 2 =4 —-442 =2 the tangent has equation

b

y =2(z—1)+4 =2z —2+44 = 22 + 2

We use technology to find where the tangent meets the curve again: [E] [EXE]:Show coordinates Yi=x"(4)22x2+2x+3[y Y2=2x+2 BF

I z INTREU]

The tangent meets the curve again at

¢ At (-1,0),

(—1, 0).

f'(-1)=4(-1)°—4(-1)+2 = —4+4+2 = 2

the tangent has equation

y =2(xz —(—1))+0

=2(x + 1) = 22 + 2

So,

y = 2x + 2 is the tangent to the curve at the point

(—1, 0)

also.

Chapter 11 (Properties of curves)

1

Yy==x

y =z

— 5

@zlflc

When

=z = -2,

a

dx

When

z = 3,

e

So, the normal at (3, 9)

has gradient —%.

y=(-2)°—-5(-2)+2 = —8+10+2 =4

So, the point of contactis Now

the normal has equation

y=—2(—3)+9

(—2, 4).

@:33_@2—5 dx

z —_9 = —2,

When

=—sr+35+9

431

+ 2

(3)h=26

dx

Exercise 11B

W dm—S( _g_92_ 2)°

=5

=12 —95 =7

= —2x+3 So, the normal at 15

(—2, 4)

has gradient

the normal has equation

y=—%(x—(-2))+4 = —1(z+2)+4 = —%3; — % + 4

_= 1.,—7TT 263 C

y=21° —3x+1

y:l+2 £

=z

142

dx

h When

=« = -1,

——

dy

y=2(1)*—-3(1)+1

—2-3+1

=0 So, the point of contact is (1, 0).

@__33_2

o =

When z =1,

Now

1

—==s = = —1

When

x =1,

fl:Ei:;r:2—3 dx

@=6(1)2—3 dx

So, the normal at (—1, 1) has gradient 1. the normal has equation

y=1(x—(-1))+1 —r+1+1

=z4+ 2

=6—3 =3

So, the normal at (1, 0) has gradient —z. the normal has equation

y=—3(x—1)+0 —

=

b

1

—3% * 3

432

Chapter 11 (Properties of curves) e

Exercise 11B

y=zx°—-3z+2

When

=3,

y:3:1:—|—l—4=3$—|—3:_1—4 I

y=23°—3(3)+2

When

=9—-9+2

z =1,

y:3(1)+%—4 =3+1-14 =0

=3

So, the point of contact is (3, 2). Now

o z _= 3,

When

So, the point of contact is (1, 0).

@=23:—3 dx

Now

dy - _=2(3) B —3 =6—-3 =3

So, the normal at (3, 2)

has gradient a5 3

When

z =1,

-

dy =3 —x

2

dx

dy

1

dx

12

the normal has equation

y=—%($—3)—|—2

So, the normal at (1, 0) has gradient —3.

= —ir+1+2

the normal has equation

= —z2 +3

y=—5(—1)+0 =

_ 2 _9p!

2

&£

dy

—

= -2

dx

a

&

&9

e ] . 1

At

P(2,

2.1)1,

dy —_—

3=

2 = ——

2

-

2

-

1

O

the tangent at P has equation

y=—5(r—-2)+1 = —2z+1+1 =—%:1:—|—2

ii

The normal at P(2, 1)

has gradient 2.

the normal has equation

y =2(zx —2) +1 =2z —4+1

= 2z — 3

1 ! —3%T+3

Chapter 11 (Properties of curves)

Exercise 11B

b

y=2r—3

3

f(a:):mZ—%:mQ—Sm_l

f'(x) =2z + 8z~2 :233+;;2

2 f-D=_ (=92(27~ e 5B

b f(3) _32_8 =3 -

=444

.

So, the point of contact is Now

f'(—2)

.

=2(-2)+

F=2) ~ _(4 +)2

(—2, 8). :

(—2)°

8

19

3

E So, the point of contact is

Now

f'(3) =2(3) + 3%

= —2 y=—2(x—(-2))+38 = —2(x+2)+8 = -2z —4+4+8

= 6—92

So, the normal at (3, &%) has gradient 9 2 the equation of the normal 1s

= —2x+4

y=—z(r—3)+=

__9...,27

=

& flz)=(z—-1)(z—-4) f(xk)=0

when

(zx—1)(x—4)=0 . x=1or4

the z-intercepts are 1 and 4.

Now

f(0)=(0—-1)(0—4)=4 the y-intercept 1s 4.

b

flz) = (z —1)(x —4) —=z° —bx+4

s

filz) =2z -5

{from a}

=6+3

the equation of the tangent 1s

a

1 (3, 3-).

—m Tr

19T3

433

434

Chapter 11 (Properties of curves)

¢

Exercise 11B

f(-1)=(-1-1)(-1-4) =10 So the point of contact is (—1, 10). f'(=1)=2(-1)—-5= -7 So, the normal at

(—1, 10)

has gradient %

the normal has equation

y = %(m —(—1))+10

=1(z+1)+10

=2z + 2 + 10

jo+ 3

e

y=zx+ -

£

Nz — 4} Y

d

1 =4 y= 2

y=a°

£

)

dx ::fg(?)

Eg%

::f2,

4

S{}‘NTKflJ

y

::14

So, the normal at (2, 4) has gradient — 7. the normal has equation

y=1(x the


—-—12x+2

When

== -2,

y = z°

y=(-2)°—12(-2)+2

= —84+24+2

Y _ a2

az

]

So when

Now

So, the normal at (—1, —1) 1 -3 the normal has equation

j_y — 372 — 19 & o dy So when z = —2, dr

o/

3(—2)

5\2

12

R — ()

i

—

—3(x+1) -1 1.1 ? i

the normal must be a vertical line.

* gl

Since the normal passes through (—2, 18), it has equation =z = —2. The normal meets the z-axis at

3

The normal meets the line _% e % — 3

(—2, 0).

. -

Y="=l-10 £

y=2z3

When

=z ' -3

z=1,

=0

d

1

dx

(—1)2

=L =—

= —

So, the normal at (—1, —4) has gradient 1.

So, the point of contact is (1, 0). Now

% _ 6z — 3 dx

So when

the normal has equation

z =1,

y=1z—(-1))—4

&y _ 6(1)% — 3 dx

—6—3 =3

=r+1-—-4 =xr—3

So, the normal at (1, 0) has gradient — .

We use technology to find where normal meets the line y = —2x + 1:

the

[EXE]:Show coordinates

Y2=x-3

at

=2-3+1

2

Y1i==2+1

y =3

y=2(1)°—-3(1)+1

Y_ _g2-_1 z=-1,

where

-3z +1

.

Sowhen

y =3

_l,_ 13 T3 3 r=—13

The normal meets the line (—13, 3).

dx

has gradient

y=—3(z—(-1) -1 )

So, the gradient of the normal at (—2, 18) is undefined.

d

3(-1)°

d_y =

z = —1,

=18 So, the point of contact is (—2, 18).

r

/

|

the normal has equation

y=—=z(x—1)+0

— 4+

The normal meets the y-axis where

* =0

y=-3(0)+3 X=2

1

¥=-1

The normal meets the line at about (1.33, —1.67).

y = —2z + 1

— 3 So, the normal meets the y-axis at (0, %)

Chapter 11 (Properties of curves)

Exercise 11B

437

£

=5 —qaz ! d—y — ax 2

dx

s

e So when

1

dy

z = -2,

—

de

a

=

a

(—-2)2

So, the normal at the point where pB

= —

4

x = —2

has gradient —i. 1

1

= —4 a

i

— 1,3 _

T~

Yy=3&

3 2 5dy _— 3T

2 +axr + 4

_ 1.3 _ .2 y=s5r"—2°+axr+4

47

2z 4+ a

At point P, with x-coordinate 2, the gradient is 1. 2(2)° 2 —2(2)+a=1

6 —4+a=1

P

a=-—1

i

y=s23—2?—x+4

So, the coordinates of P are

b

(2, 2).

The normal at P(2, 2) has gradient —1. the normal has equation

¢

j

y = —1(x — 2) + 2 = —x + 2+ 2

We use technology to find where the normal meets the curve again: 2]

[EXE]:Show coordinates

Y1=0.5x"(3)-x2cx+P V2=—x+4 -

f

+

€ 5 4P

/

E|.

—? 0

X=0

The normal meets the curve again at the point Q(0, 4).

£

Y

438

Chapter 11 (Properties of curves)

d When o=0,

Exercise 11C

-2(0)-1 T

S|

the tangent has equation This 1s the same

y = —1(x —0) +4 = —r+4

line as the normal to the curve at P,

!

3, —1 [NCERY

I

The graph is increasing for

= > 0.

il

The graph 1s never decreasing.

I

The graph is never increasing.

il

The graph is decreasing for —2 < x 2.

e

I

The graph is increasing for all

il

real x. The graph is never decreasing.

f

(1?

_1)

yU e

MY

WL

r=4

(25 4)

€L :

Y

I

The graph is increasing for

i

2 4. f(x) is decreasing for 0 < z < 4.

_20__

v

flz)=23—622+10

) £

440

Exercise 11C AY

flx) =2z +1

a

fi(z) =2

/f(m)2m+l

which has sign diagram:

-

-

/()

-

r

So, f(x) is increasing for all = € R.

-

\f

f(z) = -3

which has sign diagram:

-

-

> T

AY

flr) = —-3x+ 2

b

/\

P

(r)=—3x+2

f'(x)

>

XL

_

)

So, f(x) is decreasing for all = € R.

-

-

£

Y

f(z) =a°

AY

f'(x) = 2

f(z) =a?

which has sign diagram:

-+

-

[l

-

0

T

So, f(x) is increasing for x > 0, and decreasing for

x < 0.

Y

fla) = —2° f'(x) = ~3a°

AY

which has sign diagram:

—

0

£

- W T

i

So, f(x) is decreasing for all = € R. f(x) =22° + 3z — 4

f(x) =42 +3

f(x)=2x%+3z—4

which has sign diagram: -4

_3

-

X

So, f(z) is increasing for = > —2, and decreasing for

= < —%.

=Y

d

-

AN

L

Chapter 11 (Properties of curves)

Chapter 11 (Properties of curves)

Exercise 11C

fl@) == =27

L maioe » Mgl fiz)=—27° = — 2

which has sign diagram:

-

-4

-

.

0

-

T

-

&

£

f=)

So, f(x) is decreasing for all z # 0.

v

T WL

YA ]

8

m| o

|

Lo

H|

o)

1

e

Il ]

fl@)= —=2 = -z which has sign diagram: il

-

i+ :

fla)= -

[z

-

-

0

x

So, f(x) is increasing for = > 0, and decreasing for x < 0.

fi#

f(x) = z° — 627 f'(x) =3z — 12z = 3z(x — 4) which has sign diagram:

t]




i

\/

f(x) =23 —627

BJ@®

z

So, f(x) is increasing for x > 4,

and decreasing for

x < 0

and

Now

f’(z) =0

when

—6z°+4 = &

P2 =3

2

3

ks

r=44/2

|

Y

AY

..

=

'

0 < x < 4.

f(z) = —22° + 4z fl(z) = =62 +4

o

>

/(@)

e

441

442

Chapter 11 (Properties of curves)

5

i

Exercise 11C

f(x) = —4a® + 152% + 18z + 3

f'(z) = —122* + 30z + 18

i

We find the zeros of f’(x) using technology: E [HathiDegfornd] (d7c)Real

El MathDegMornl [dic]Real

aXE+bX+c=E [

-12

aX?2 +bX+c=0

30

E;[ 18

[SOLVE]REXR{3(CLEAR]_EDIT f'(x)=0

So, il

-u.a}

— z=—5 1 or z=3.

when

f'(x) has sign diagram: i

-

fi(zx)

_ 1

3

>

So, f(x)

i

|

£

is increasing for

—%

< o < 3,

and decreasing for

and

f(z) =2* —62° + 32 — 1

fl(x) =32° — 122+ 3

i

We find the zeros of f’(x) using technology: E [MathlDeglorni) [(d7c)Res] aX2+EX+c=E a

C

3

E [MathDegMorml) (d7c)Rea] aX2 +bX+c=0

c

xl

-12

xz[u.zsm}

3

[SOLVE]MH3M3(CLEAR] EDIT ]

So, il

f'(x) =0

when

2+43

[REPEAT]

x =~ 0.268 or 3.73.

f’(x) has sign diagram: i}

+ 1

~0.268

-

| +

~3.73

J'(=

>


3.73,

f(z) = x° — 32° + bz + 2 f'(z) =3z* — 6z +5 b

= < —%

We find the zeros of f’(x) E

using technology:

HathRadNorml [d7c)Real

aX2 +bX+c=0 a

L

b

3

E

a

HatRadBorml [dic)Real L

c

-6 IE

No

Real

Press: SOLVE | (NaRIF|CLEARJ{

EDIT

So, f'(z) #0 forall z € R.

SOLVE |(anFIF|CLEARI[

Roots

[EXIT] EDIT

and decreasing for

= >

Chapter 11 (Properties of curves)

Exercise 11D

4

f'(xz) has sign diagram:

+

f'(z) £

So,

f'(x) >0

C

80

forall

z € R

Ay

which means that f(x) is increasing for all z € R. As we can see from the graph of y = f(x), f(x) is increasing for all = € R.

/ /y:$3——3:1:2+5$+2

_1L'U

_—-——/F

= —2//€:’_1

2

/20

3

4

5

;

/L4

Fy 7

a

f(z) =2z + g

b

I

— 97 4+ 8¢~1


—12z—13

f'(-2) =0

¢

L 3(=2)°+a=0

Now

f(z)=0

when

3z%—12=0

- 372 =12

12+a=0

- ox2 =1

Loa=—12

Also,

s

f(—2)=3

=49

the sign diagram for f’(x) is:

(=2)° —12(=2) +b =3

84+ 244+b=3

N

b= —13

_9

\If 9

_I_

(2)

2V~

9

Now f(2)=—-29, f(—2)=3, so there is a local minimum at (2, —29) and a local maximum at (—2, 3).

10

P(z) = ax® + bx* +cx +d

.. (1) o, Pl(z)=3az® +2bzx +c Now (0, 2) lieson P(x), so P(0)=2 a(0) +b(0) + ¢(0) +d = 2 Lod=2

The tangent at (0, 2) is y =9z +2,

so

P'(0)=9

3a(0) +2b(0) +c=9

e=9

There is a stationary point at (—1, —7), o

3a—2b+c=0

So, using (2), 3a—2b=—-9 Finally,

(=1, —7)

lies on

... (3)

P(x)

oa(=1)° +b(=1)° +¢(~1)+d= -7 —a+b—-9+2=-7 b—a=70 a=~o

So, using (3),

3a — 2a= —9 C.ooa=-—9 L

so P'(—1) = 0.

{using (1)}

. 3a(=1)*+2b(—=1)+c=0

a=b=-9

P(z) = —9z° — 92° + 9z + 2

.. (2)

Chapter 11 (Properties of curves)

11

a Let Now

f(r)=2°>—-122-2, f'(z) = 3z% — 12 f'(z) =0

when

for

Exercise 11D

—-3

and a local minimum at

Critical value (x) | f(x) —2 (end point)

—16

0 (local maximum)

4

2 (local minimum)

0

3 (end point)

4

The greatest of these values is 4 when

* =0

or

The least of these values i1s —16 when

= = —2.

= = 3.

= = 2.

451

452

Chapter 11 (Properties of curves) ¢

Let

Exercise 11D

f(;r):mzfi—E:mE—l—lG:E_lj

for

£

1—5r+2

dy 280 D d$—3(2) P

=12-95

(2,0)

=7 The tangent has equation

y="7x—2)+0

Y

=Tx — 14 C

Y =

MY

1 — 2z -

-

e

2T

_3?2

£

=g % -2z} d

L~ dx

9y

_

_

S 4 2x?

—EJri

T3 dy

iz

2

2

1B

— —2+2

1-2z

Yy=—2

T L

1

Review set 11A

soat

(1, —1),

2

12

The tangent has equation

y=0x—1)—1

= =]

>

7

454

2

Chapter 11 (Properties of curves)

Review set 11A b

The tangent to the curve

f(z) =a— — 2 £

at (-1, —1)

is y=—6x — 7.

the tangent has gradient —6, and the point of contactis

(—1, —1).

b

Now,

f(z)=a— 2 =a—bz™"

L f@) ==

f'(=1)=—6

and

2 _ g (—1)3

9=

o, T

—6

b=3 .. So,

a=2

3

and

f(=1) =-1

b

T

s.oa—3=—1

(%)

Sooa=

{using (%)}

b= 3.

y = 2x° +4x — 1

dy.

6x“ 2 +

—

4;

4, soat

(1, 1 5),

dy rrde 6(1)° 2 + 4 = 10 the tangent has equation

y = 10(x — 1)+ 5 =10z — 1045 = 10z — 5

We use technology to find where the tangent meets the curve again: [l

[EXE]:Show coordinates

YI=2x7(3)+hx-1

7

Y2=10x-b

E— 4

8

2

1

3

‘551“55

X=-2

//

-

#

INTSECT

[¥=-256

_

The tangent meets the curve again at

(—2, —25).

Chapter 11 (Properties of curves)

L

a

£

y=2°+3(2)—2

When

=8+6—2

z =1,

'y:2+%+3(1) =2+1+3

=T

So, the point of contact is Now

When

=6

(2, 12).

So, the point of contact is (1, 6).

@:3$2+3 dx

Now

(2)° 2 4+ 3

dy T,

x _= 2,

h When

(2, 12)

_ L 15°

Y — 7% £

=z =1,

dy

1

==-—— e 12+3

has gradient

the normal has equation

= —1+4+3

So, the normal at (1, 6) has gradient —3.

yz—l—é(m—2)+12

the normal has equation

y=—3(x—1)+6

= —I—Effl =t % + 12

= — ém + % +6

5

=

y=z?—-Tzx—44

When

z=-3,

1

13 3T+ 3

y=(-3)*—-7(-3)—44 =94+21—-44 = —14

So, the point of contact is (—3, —14). Now

ay _ 20 — 7 dr

z _= o —3,

When

Y& 5=2(-3)—7 oy

S k18 So, the normal at

(—3, —14)

has gradient %

the normal has equation

y = 5(z — (—3)) — 14 ==(z+3)—14 =%:}:+%—14

We use technology to find where the normal meets the curve again: ]

[EXE]l:Show coordinates

?1=—x2='?fi1|& Y2=(1413)x-

'

10

5

v

"?Q_ll 3)

0

—otF

X=10.07692308

'

_l|¥

/

+3

== 1 +3

=12+3 =45 So, the normal at

455

b y=2+1+3z=2+2"1+30z

y=2x°+3cx—2

When =2,

Review set 11A

:J

INTSECT

}’=-12.99dflé284

The normal meets the curve again at about

(10.1, —13.0).

456

6

Chapter 11 (Properties of curves)

a

Review set 11A

y=2u+—==2r+4a> €I

dy dm_Qo

8 o.—3_o_ ° =2 8 .

The tangent has a gradient of 1. dy dx —_—

8 3

—_:1

1=F

8

> =8 =

When

z = 2,

4

y=2(2)-|—2—2 =4+1 =9

So, the coordinates of P are

(2, 5).

b

The tangent has equation

y=1(x —2)+5 =x—2+409 =z+3

¢

The tangent cuts the x-axis where

y =0

r+3=0

= -3 So, the tangent cuts the z-axis at

d

(—3, 0).

The normal at P(2, 5) has gradient —1. the normal has equation

y=—1(z —2)+5 =—2+24+595

= —x+7 7

a

AY

(—1,3)

—/N

b

R

I

V@ The function is increasing for = < —1 and = > 4, and decreasing for —1

T

f(x) is increasing for = > —1. f(x) is decreasing for x < —1. () has has sign f'(x) sign

diagram: diag

. \|f L

-

8

@) -

1

x

flx) =z* —42° —82* +5

f'(x) = 42> — 122° — 162

We find the zeros of f’(z) using technology: Bl MathRedMorn] [dic][Redl

aX3 +bX2 +cX+d=0 a

b

C

4

B c

-12

aX3 +bX2 +cX+d=0

d

%1

-16

:{2{ X3

SOLVEIPENRI3([CLEARI EDIT ]

So,

f'(z) =0

MathRadMorml] [dic]Red

REPEAT]

when

z = —1, 0, or 4.

/() f(@)

has sien gn diagram diag :

a

f(x) is increasing for

—1 < o
=3(=1)* —=9(—1)+50 =55 and f(3) =3° —3(3)* — 9(3) + 50 = 23 So, there is a local maximum at (—1, 55)

10

and a local minimum at (3, 23).

f(x)=2°+ Az + B

fl(z) =32+ A a

f(x) has a stationary point at (1, 5) o

ARG

3(1)° +A=0 3+A=0 ‘A =38

Also, f(1)=5

.. (1)°=3(1)+B=5 1-34+B=5 B =7

b

Since

A=-3

and

B=7,

f(z)=2°>—-32+7

and Now

f'(z) =0

when

f'(z) = 32% — 3

3z°—-3=0

. 3z =3 r? =1 r==xl

—1

(@

=V

NN

+7

__

So f'(x) has sign diagram:

Now f(=1)=(-1)>=3(-1)+7=9 and f(3)=1°—-3(1)+T7=5 So, there is a local maximum at (—1, 9) and a local minimum at (1, 5). 11

f(z)=2°—42* +4x

a

f(0)=0°-4(0)+4(0)=0 the y-intercept 1s 0.

b

f(z)=32*—-8x+4 We find the zeros of f’(x) using technology: [Math|Rad[Norm1]

[d/c][Real

aX2 +bX+c=0 a

C

3

b

B

aX2 +bX+c=0

£ -s I

HW?IIIfl X2l 0.6666

4

SOLVE)RENERCLEARI(EDIT So, f'(x)

1) —=0 f'(x)

when

z _= %2 or2

has sign diagram:

465

466

Chapter 11 (Properties of curves)

¢

Review set 11B

f(z) is increasing for f(z) is decreasing for

d Now f(3)=(3)’-4(3)*+4(3) .

16

=27 "9 T3

3

_ 3227 and

f(2) = 2° — 4(2)* 4 4(2)

=8—16+38 =0 (%, 32)

and a local minimum at

=Y

So, f(z) has a local maximum at

12

Let

f(z)=2°—-3z°+5, for

f(x) = 3x* — 6x = 3z(x — 2)

which 1s O when

—10

forall

0 +0.0061z° + 18x + 14230

dollars,

a

C(0) = $14230,

b

C'(z) = —0.0000216x* + 0.01227 + 18

X

:

Isn pounds per 1tem produced.

0 < z < 1200

which is the fixed operation cost without producing any items.

This is the rate at which the production cost (in dollars) 1s changing per item when x items are made. It gives an estimate of the cost of making the (x + 1)th item each day.

¢ C'(300) = —0.000021 6(300)* + 0.0122(300) + 18 = 19.716 The cost of producing the 301st item each day is about $19.72.

d

C(301) — C(300) = —0.000 007 2(301)* 4+ 0.0061(301)* + 18(301) + 14230 — [—0.000 007 2(300)° + 0.0061(300)* + 18(300) + 14 230] ~ $20004.32 — $19 984.60 ~ $19.72 The actual cost of producing the 301st item is about $19.72.

L

C(x)=T800+ 7x — 0.0001z*

dollars

a

The marginal cost function is C’(x) = 7 — 0.0002x

b

(C'(220) = 7 — 0.0002(220) = 6.956

dollars per pair.

~ $6.96

This estimates the cost of making the 221st pair of jeans if 220 pairs are currently being made. ¢

(C(221)— C(220) = 7800 + 7(221) — 0.0001(221)% — 7800 + 7(220) — 0.0001(220)2] = 6.99509

~ $6.96

This 1s the actual cost of making the 221st pair of jeans. The answer 1n b is a very good estimate.

474

a

Cv)=3202+

.i

200

Exercise 12A.2

euros

U

1em2 | 200000 = £(50)" + 3

C(50)

= 500 4 4000 = 4500

§i

if the average speed is 50 kmh™!, the total cost of the journey is 4500 euros. C(100)

1qpam2 | 200000 = £(100)~ + T

— 2000 + 2000 = 4000

if the average speed is 100 kmh™?, the total cost of the journey is 4000 euros. 200000

g

v? +

euros,

v

v > 0

b

v — 20000002


0,

c

so 404 < x < 49596. 2

P(z) = 5z — 2000 — —

10000

12 _= or — 2000 — 155557 ()

=5

P@) =5~

— —=_

5500

This 1s the rate at which the profit earned is increasing or decreasing when producing x items per year.

5—

when

d P(z)=0

—

5000

= e

£

5000

x = 25000

Sign

L diagram

for

.. P’(x)

is:

P

-

N 5% 000

|+ |

/

> T

0

So, the profit 1s increasing when

8

0 < z < 25 000.

C(x)= —0.000004z> + 10z + 3000 dollars, a

0 < x < 600

The revenue is the amount of income generated, so

b 125001

R(z) = 24x.

Ay (S) R(z) = 24x

10000 t 7500 1

C(z) = —0.000 004z* + 102 + 3000

5000 1 25001

. x (items)

0 C

C($) '

=

100 R(ifl)

200

300

when

Using technology,

400 —0.000

= ~ 212.

500 OU4$3

600 +

-

102

+

3000

=

24zx.

E :f1='

[EXE]l:Show coordinates

gl

yo00004x~(3)+10x+3000

-

This 1s the breakeven point, when the revenue is equal to the | _ cost of producing the items.

—1

/fi/mu

flff/ "

X=211.57955G68

200

|

300 Y=

|

400

[NTPSECT

D??.QUESES

2

Chapter 12 (Applications of differentiation)

d

P(r)=R(x)—C(x)

e

P'(z)=0.000012z* + 14

Exercise 12B.1

477

— 242 — (—0.000 0042 + 10z + 3000) = 0.000 0042 + 142 — 3000 dollars

f P’'(120) = 0.000012(120)* + 14 — 14.1728 ~ 14.17

This means that the total profit is increasing at a rate of about $14.17 per item when 120 items are produced.

EXERCISE 12B.1 1

P(z) = —0.0222° + 11z — 720 P'(z) = —0.044z + 11

Now,

P'(z) =0

when

—0.044x + 11 =0 11 =0.044x 11

L= —

0.044

= 250 ’;-50\

2

l

So, P’'(x) has sign diagram:

_

Pl

> T

0

So, the profit is maximised when 250 items are made per day.

2

h(t) = 1.6 + 32t — 4.9t h'(t) =32 — 9.8t Now

A/(t)=0

when

32—9.8t=0 .

32 =9.8t

g 12

9.8

ot~

h'(t) has sign diagram:

"

3.27

s

the maximum height occurs when

/-.\

~ 3.27

—

T

P(a)

¢ ~ 3.27

h(3.27) ~ 1.6 + 32(3.27) — 4.9(3.27)7 ~ 03.8

So, the maximum height reached by the stone 1s about 53.8 m.

Chapter 12 (Applications of differentiation)

478

3

a

Exercise 12B.1

Let the remaining fence have length y m. The total length of the fence 1s 60 m 2z +y = 60 existing

y = 60 — 2z

The area of the enclosure

fence

A = width x length :my

= 2(60 — 22) m? the area of the enclosure is given by

A(z) = z(60 — 2x) m?.

A(x) = z(60 — 2x)

= 60z — 2z°

. Al(z) =60 — 4z So,

A'(z) =0

when

60— 4z =0 .

4x =60 r =15

|+ |

A’(x) has sign diagram:

0 The area 1s maximised when

z = 15

and

y = 60 — 2(15) = 30

existing fence

The area of the enclosure is maximised by constructing a fence with two sides of length 15 m, and one side of length 30 m. 4L

a

Now the area

A = 100 m? ry = 100

e So,

ym

100

I 1m +

A

L=2x+vy

existing fence

L=2¢+%

X

L =2z

a9 dx

+ 100z !

q100p2 =9

which is 0 when

— =2

W12

£Z

o

x? =50

. z=+/50 So,

dL

&

z

has sign diagram:

A=100m?

{z >0} ;

0

So, the length is a minimum when

_v

|

V50 x = /50.

T

Chapter 12 (Applications of differentiation)

When z = V50,

Exercise 12B.1

479

L = 2v50 + 20

/50

~ 28.3

So, the minimum value of L 1s about 28.3 m which occurs when

= = /50 =~ 7.07.

z =150~ 7.07 and y = o2 ~ 14.1 V50

So, the optimal situation is:

14.1m 7.07m +

existing fence

Inner surface area = area of square base + area of 4 rectangular sides S

108 =xXxrx+4d

H

XXy

i :

r° + 4y = 108

z? + dzy = 108 dry = 108 — x

< @)

+2(2:1: X @)

12

= 2(2:52 + 19

21 cm

12

@)

L

£

I

Az) = 422 + 22 em? £

¢

A(z) = 42* 4+ 600z~ Al (z) = 8x — 6002 600

A'(r) =0

when

8z — g

=0

£L

ST

=

600

;13_2

8z° = 600 x° =75

=75~

A’(x)

has sign diagram:

_ 0

M.

~ 4.22

4.22

e

-

T

So, the inner surface area of the box is a minimum when

A(V75) (V75) =

4(V75)° =4(V75)°

z = /75 ~ 4.22.

600

+ i

~ 213 So, the minimum

inner surface area is about 213 cm?, when

x = /75 ~ 4.22.

Chapter 12 (Applications of differentiation)

Exercise 12B.1

T = V75~ 4.22 2

= 2V75

p

~ 8.43

100 _

100

72

(m)i

s

!

0.62

So, the optimal box shape is:

(36 —2z) cm

V (x) = area of base x height V(z) = (36 —2x)* x x V(z) = 2(36 — 22)* cm?

The volume of the container is

o V(z) = 2(36 — 2z)7

= 2(1296 — 144z + 422) — 12962 — 14422 + 423

V'(x) = 1296 — 288z + 1227 V'(z) =0 B

when

1296 — 288z + 12z° = 0

HathiRadfornd) (d7)Real

aX2 +bX+c=0 a

C

12

B HatRedForn]) (d7c)Real

b

el

-288

SOLVE]JENR3(CLEAR[ EDIT

Using technology,

V'(x)

aX?2 +bX+c=0

c

1296

REPEAT

= = 6 or 18.

has sign diagram:

_|_

|

0 The volume 1s a maximum

So,

6 cm X 6 cm

when

el

18

g

6

x = 6.

squares should be cut out to produce the container of greatest capacity.

481

482 8

Chapter 12 (Applications of differentiation) Production cost

C(x) = fi:}cz + 8z 4+ 20

Selling price p(z) =23 — 2z Revenue Profit

Exercise 12B.1

pounds

pounds per blanket

R(z) = zp(z) = 23z — 52°

pounds

P(x) = revenue — cost

= (23z — 52°) — ($2° + 8z + 20)

= —32% 4 152 — 20 P'(z) = —5z + 15

Now

P'(z) =0

when

—35z+15=0

r == =10 2

, : / P’(z) has sign diagram:

A|

+

|

{I}

10

_

-Plx)

x

So, the profit is maximised when 10 blankets are produced per day.

9

a

volume of cylindrical can = 7r?h Now, capacity is 1 litre which is equivalent to 1000 cm?. So, the volume = 1000 ¢m®

7r*h = 1000 B —

10020

hcm o

mwr

b

---------------

Total surface area of cylindrical can A = 27r? + 27rh

A =27r* + 2???“(1000) ’?TT2

A =2mr? +

¢

A = 27r® +2000r1 G

dr

— Arr — 2000r 2 = 47r — 2020 -

gt dr

0 when

A4nr — T r2

0

drr = 20g0 B

47r® = 2000

3 _ 500 m

"

94 dr

pas sign diagram:

i

o0~ 7T

-

|+

~ H.42

5.42

dA

4 r

e

T

cm?

- ‘‘‘‘‘‘‘‘

-

Chapter 12 (Applications of differentiation) So, the total surface area 1s a minimum when When

r =

¢ @, dl

I 0

r =

2O

H.42.

T

o 5> ~ 10.8 ( 3 /500) T

So, the can should have dimensions:

0.42 cm

10.8 cm

----------------- : 10

a

Perimeter = 2[ + 2nx

m

""""" zm

- 400 = 21 + 27z

T

s 21 =400 — 27 .

[ m

l

[ =200 — mx

x>0

.....................

and

[ > 0

200 —mx

\—/

;

for the track to exist

> 0

.o

< 200 200

L

— m

So, 0

¢ (hours)

The mayor’s model seems reasonable; the attendance was always increasing until shortly after 1 pm, then 1t began to decrease.

494

Chapter 12 (Applications of differentiation) d

5

a|

From the graph in ¢, the maximum attendance was about 7470 people which occurred when t ~ 5.31, which was at about 1:19 pm.

Distance (r cm)

20

40

80

llluminance (I 1x)

450

112.3

28.1

Ir

9000

4492

2248

Ir?

130 000

179680

179840

Model B b

Exercise 12C

B SUB

s

3600000 | 7187200 | 14387200

(I = %)

best fits the data as Ir? is approximately constant.

T

Delfom) [@EIRed

B

List 1 | List 2 | List 3 | List 4

1 2 3

20| 40| 80|

4

Celfom) [@EFed

PowerReg

a

450 112.3 28.1

=180266.021

b =-2.0006414 r =-0.9999998 r2=0,99999969

]

_MSEfé . 1876 x150 6

NPIEDINDNES > ]

e

The correlation coefficient r 1s very close to 1, and the power is very close to —2, so it 1s reasonable to conclude that I is inversely proportional to 2. The model is I ~ — c

T

T

3 66,,

which agrees with our answer to a.

180 802 2 66 T

~ 1802667* I + 24x + 5400 dollars a

(C'(0) = 5400 dollars,

which is the fixed operating cost without producing any items.

b

C'(z) = —0.0006z° + 24 This 1s the rate at which the production cost (in dollars) is changing per lot of one thousand pairs of chopsticks when x thousand pairs of chopsticks are produced. It estimates the cost of producing the (z + 1)th lot of one thousand pairs of chopsticks.

¢ C’(100) = —0.0006(100)* + 24 = 18 The cost of producing the 101st lot of one thousand pairs of chopsticks is approximately $18.

d

C(101) — C(100) = —0.0002(101)? + 24(101) + 5400 — [—0.0002(100) + 24(100) + 5400] = 17.9398 ~ 17.94 The actual cost of producing the 101st lot of one thousand pairs of chopsticks 1s approximately

$17.94.

2

C(v)=10v+ 2 v

a

curos

i C(15) =10(15) + 22

i C(24) = 10(24) + o

= 156

= 243.75

So, the cost of running the barge for

So, the cost of running the barge for

2 hours at 15 kmh~! is

5 hours at 24 kmh™—1! is

2 x 156 = €312.

b

C)=10w+2

5 x 243.75 = €1218.75.

v

= 10v + 90v ™!

C'(v) =10 — 90v 2 90

= 10 — = :

/

90

i

| 0(10):1U—fi

i

So, the rate of change in the cost of

running

the

barge

at

10 kmh~!

€9.10 per hour per kmh—!,

is

/

90

C(G)zlo—fi—2

So, the rate of change in the cost

of running the barge at 6 kmh~!

€7.50 per hour per kmh~1.

is

496

Chapter 12 (Applications of differentiation)

Review set 12A

v

90

=

1=

o 3

“U2

9

—

V=

I

C’(v) has sign diagram:

10

{as v > 0}

N ;Ia. +

- (v)

0

So, the minimum cost per hour occurs when 3

a

Let

OD = x,

so C has coordinates

(z,

v =3 kmh™1.

9 — :1:2).

Area of rectangle ABCD = length x width

.

A(z) =22 x (9 — z°)

b

A(z) = 18z — 227

=

oo Al(z) =18 — 627 A'(x) =0

when

18 —6z° =0 .

6x° =18

{z>0}

r=v3 which has sign diagram:

i

:

0 So, the area is a maximum when

When

z =3,

a

fi

z = /3.

y=9—(v3)* =6

So, C has coordinates 5

T Sl

(1/3, 6).

perimeter = 2x + 2y + 7x . 200=2x 4+ 2y + mx 2y = 200 — 2z — x L y=100 —x — S

2rm

= () €L

=Y

= 18z — 2z°

Chapter 12 (Applications of differentiation)

b

area of lawn A = area of rectangle =2

Review set 12A

4+ area of semi-circle

Xy + %’fl'iflz

=22(100 — z — Zz) + Z2°

{using a}

= 200z — 2z° — wx? + %:1:2

= 200z — 22° — Zz° A=200z —2*(2+ %) m’ ¢

dA —=200—-4x — nx dx

Now

dA i

"i

=0

when

200—4z

—7x

=0

4o + ma = 200

x(4+ 7) = 200

200

4 ~+ T

T~

o8

dx

has sign diagram:

R

28.0

N

|

o

200

T

4 4+ 7

:

.

The area of the lawn 1s maximised when

x = ror

200 T

~ 28.0

ad y=100- 20 _T( 0 4 4 1

2\44+T7

~ 28.0

The dimensions of the lawn of maximum

area are:

/\

28.0m H6.0 m

a

capacity = 1 kL = 1 m’ volume of box = area of base x height 1= :523; 1

y=—, X

b

>0

area of steel needed = area of base + area of 4 sides

ke

= z° + 4zy

= 1° + 4w (i) 72

{from a}

4 £I

Steel costs $24 per m?, so total cost of steel = (3’:2 + E) X 24 I

C(z) = 2422 + 2 dollars X

497

498

Chapter 12 (Applications of differentiation)

¢

Review set 12A

Oz) =242% + 22 = 2422 1 962 £I

C'(x) = 48z — 9622 = 48z — — £

d

C'(z) =0

when

48z — % =0 £

96

.

48z° = 96 $3

—

9

r=v2

\/

has sign diagram:

ST

So, the cost is a minimum when

z = /2 ~ 1.26

C’(x)

When

z = /2,

-

1

y=

1+

,.

@

~ (.630

(V2)2

C(V2) + o== 24(V2)* (V2) == 24(¥2)? ~ 114.29 So, the dimensions which minimise the cost of the box are about The cost of the steel in this case would be about $114.29.

6

D(t) =+/24.01t* — 29413 + 93612 a

b

metres

The stone flies for 5.9 seconds before landing, so the domain of D(¢) is {t |0 — 8822 + 1872 [d%]Real aX3+bX9+c§+d=0 a

[ 96.04

-882

B [MathRadMornd (d7c)Real aX3 +bX2 +cX+d=0

] c

xl

1872 K

x2{3.3296| X3

0

[SOLVE]PHRIB(CLEAR] EDIT |

Using technology,

D

o So, D? is maximised when D? is maximised when

5.8b54011926

REPEAT

%(Dfi) — 0 when

2 (D?) has sign diagram:

¢

1.26 m x 1.26 m x 0.630 m.

t =0,

1+ é

3.33,

or 5.85. d

{ \

Y-+ ~3.33 ~5.85

5|9 [

(D7)

t ~ 3.33 seconds.

t =~ 3.33 seconds,

so D is also maximised when

D(3.33) ~ 1/24.01(3.33)% — 294(3.33)3 + 936(3.33)2 ~ 49.8 So, the maximum

:

distance between the stone and Max 1s about 49.8 m.

t =~ 3.33 seconds.

Chapter 12 (Applications of differentiation) d

The stone lands when

Review set 12B

499

t = 5.9 seconds.

D(5.9) = 1/24.01(5.9)% — 294(5.9)3 + 936(5.9)2 ~ 36.0 So, the stone lands about 36.0 m from Makx.

7

h(t) = at* + bt + ¢ metres a

The ball is thrown from a height of 1.6 m above the ground, so

b

The ball initially gains height at 16.4 ms™*, so Now

A(0) = 1.6 c=1.6

h'(0) = 16.4 h'(t) =2at +b b=16.4

¢

After 2 seconds the ball is falling at 3.2 ms™!, so

h/(2) = —3.2

2a(2) +16.4 = —3.2 .

d

h'(t)=-98t+164=0

when

4a = —19.6 oa=—4.9

—9.8t=—16.4 t

_

—16.4 —9.8

t ~ 1.67

h'(t) has sign diagram:

-

f(fl

ball

was

|+

h(t) = —4.9t° +16.4t + 1.6 h(1.67) ~ —4.9(1.67)° + 16.4(1.67) + 1.6 ~ 15.3 So, the maximum height reached by 1.67 seconds after the ball was thrown.

1

the

about

H(t) =15+ 19t — 0.8t* metres

a H'(t)=19—-1.6t ms™}! b H'(0)=19ms™* H'(10) =19 — 1.6(10) =3 ms ™' H'(20) =19 — 1.6(20) = —13 ms™! These are the instantaneous speeds at

¢ = 0, 10, and 20 s.

A positive sign means the ball 1s travelling upwards. A negative sign means the ball 1s travelling downwards.

15.3

m,

which

occurred

about

500

Chapter 12 (Applications of differentiation)

¢

H(t)=0

when

1.5+ 19¢— 0.8t =0 El

aX2+bX+c=E a

C

-0.8

c

¥1

19

xz[—n.n?s}

Using technology,

2

a

(W)

t > 0,

[MHathDegMorml

aX2 +pX+c=0

1.5

Since

Review set 12B

REPEAT

¢~ —0.0787

or

23.82868667

23.8 s.

the ball returns to the ground after about 23.8 seconds.

50 000 ==— + 20 v 2

dollars

642 50000 C(64) = T + — 204.8 + 781.25

= 086.05

the cost of running the train for 1 hour at 64 kmh~—! is $986.05

the cost of running the train for 5 hours at 64 kmh~! is $4930.25. b

2 C(”U):;—U—I—5UUUU v

euros,

v >0

= 550 + 500000~ C'(v) = 75v — 50 000y ™2

3

= 10V~

50 000 02

ei C(75)e = 15(75) — 50—000

ol ] 50 000 i C7°(90) = 5(90) — 502

__ 75 _ 80

=10

— 25 18 ~ —1.39

B

500

=9 — 51

9

_ 229 81 ~ 2.83

if the average speed is 75 kmh~1,

if the average speed is 90 kmh—1,

about $1.39 per kmh~1.

about $2.83 per kmh~1.

the rate of change in the cost of running the train 1s decreasing at

¢ (C(v) is a minimum when 1

the rate of change in the cost of running the train 1s increasing at

C’(v) =0 50 000

+v® — 50000 = 0 +v® = 50000 v® = 500000 v~

T79.4 kmh™!

S

C’(v) has sign diagram:

0

~79.4

v

the cost of running the train 1s a minimum when the average speed of the train 1s about 79.4 kmh—1.

Chapter 12 (Applications of differentiation)

3

Suppose the sheet i1s bent z cm from each end. To maximise the water carried we need to maximise the area of the cross-section.

A=2(24—22),

00 20 < 10 O dx

ii

F(3) — F(2)

16%

/

3 0

> dx

= F(3) — F(0)

i unitsz

—

E I

L

3 2

20%

(d/c]Real x3dx

uflit52

516

Chapter 13 (Integration)

n

g

3

Exercise 13C

ok x> dx

31 4

[]

SR LIAEA

d

We see that the property 2

/

b

a

f(x) dz + /

3

f$3d$+/ 0 2

L

N EE NPT D

a

f(x) dx

holds true, as

$3d$=4+16%=20%=/$3d$. 0

and G(x) be the antiderivative of g(x).

/af(i:) dr = F(a) — F(a) = /

b

f(x) dzr = /

c

3

Let F(x) be the antiderivative of f(x)

a

c

b

f(x) dx = area of the region under

the curve y = f(x) T = a.

between

x = a

The antiderivative of f(z) = k F(z) = kx.

. /bk‘di?: F(b) — F(a) " —kb—ka

and

= k(b —a)

This region has 0 width, so its area = 0.

/f ) dz = F(a)— F(b)

= —[F(b) — F(a)]

is

d

= F(z) = f(2)

L (kF() =k f(@ d

/f:r)d:r

k F'(x)

So,

1is the antiderivative of

/h k f(z) dz = k F(b) — k F(a) ! = k[F(b)— F(a)]

k/ f(z) da i F(x) = f(xz)

and

% G(z) = g(z)

% F(z) + G(z)] = f(z) + g(z) F(z)+ G(x) is the antiderivative of f(x)+ g(x).

o. [ 17(0) + g(o)] da = [F(b) + G(b)] — [F(a) + G(a)] = [F(b) — F(a)] + [G(b) — G(a)] :/:f(m)der/:g(m)dm

Chapter 13 (Integration)

t [ fa)das

a

2

(@) dz = Bt5] ~ Fla) + F(e) - Bt

The antiderivative ofx is %3}2.

é

Exercise 13D

b

/;1: dr = %3:2 + c

b

/:1:2 de = 32° + ¢

b

/:,c? dr = -z ' +c

(z°) = 32°

= (32°) =23 a 3

The antiderivative of z? is z2°. dim (x™1) = —x~4

& () =a a

The antiderivative of 72

is —x L.

= -

4

+c

XL

dii (z7%) = =223

& () =a a

The antiderivative of 73

is —%33_2_

b

/3:3

G Ta= —%3:_2 + = —#

5

a

In questions 1 to &4, we saw that the antiderivative of =" is This result does not apply when

n = —1,

So, we predict that the antiderivative of " b

Using a, the antiderivative of z° is %mfi.

d;i (g2°) = 5(62°) =2°

v

as

_11+ -

1s

%_H "t

+ c

- Jlr : g1

1s undefined.

forall

ne€Z

except

n=—1.

2177

518

6

Chapter 13 (Integration)

i

dax

T +

Exercise 13E

%) = 3x° + 27

/(3$2—|—2:1:)d$=a:3—|—:1?2+c

T (3z* — 22°) = 122° — 4o xI

/(12:1?3 — 4z)dr = 3z* — 22° + ¢ 4/(3$3$)d:}::3$42m2+c /(3:173 —x)dr = %:1:4 — %:122 +c

Chapter 13 (Integration)

o

[@+3)do

=5

+

2

/(4—:}3)(:53: 72

:%+3$+c

:4:1:—?+c

=%$2+3$—|—C

=4:1:—%:1:2—|—c

e

=

C

Exercise 13E

/(;1:2—2) dx 7

=

—3

=2 r+cC

—

13 3

_9r

1 ¢

+cC

%$2+$3—|—C

15L ,2

3

a

@=

@=4$2

C

dx

dx

y/fid:x:

y=[4$2

= 6bx + ¢

dx

1m—i—c

@—i=x_2 dx

T2

y/$_2d$ —1

43

S e

:m_—l—i—c

—%:{:3+c

1

= —=7TC £

d

Wdax _934 y/(?xgfil)dm 4

:2%—4$+c 1:1:4—4:1:+c

@=4$3+33:2 dx

y=[(4$3+3m2) dx 44

3z

dx

2

y/(Zmz)dm =3

— 3 Ty e

:2:1:—m_—1+c

=zt + 2% + ¢

:2:r:+l+c £

519

520

4L

Chapter 13 (Integration)

a

Exercise 13E

/(:1:2+3x2)d$ —_—

—

3 2

x> >

—+

b

/(255’23:1:+1)d3: 203 2

Tr+c

——2

:%m3+%$2—2$+c

C

A3 —

4

— —

3

1.4 411,'

e

4.3 +3$

d

1.2 37

3$+C

175

___

i

=3

_1_.5 _Eflj

fE3

1 —§$

.3

f

(L—i)dflfi 3z3

1

e

—f(fiifl

g tEte

a7 —§$

+3+4J”j

=%$2+%$3—|—%$4—|—C

SCZ

i

=

+2$‘|_C

D 2 —m——|—$++c 5 2

4 _1_.4

=U=ifn— 3

S51_.5 7

T

12 ST T

T

_.—3

1

i _/(433 5 4 == _

c

—2

—2z77)

dz

2$_1

pastriy:

R

R

12

%m—z

pOEFT

1 4 =T

x4

3

2

(z* —2° —x+2) dx —

c

/(%m + 2% 4+ 2°) do

+c

—3x

+ x4+

—

:%$3—%$2+m+c

/(—:1:3 +4x* — 3) dx e =+

322 >

S

—1

2

el

—2

2

+

1,3

4$)d$

4182

14

B

4

4

4o

=——4+iz—drl40

5

a

/(2:1:+1)2d:1: /(4m2+4m+1)d$

b

/(a:Jri)gd:x: /($2+2+%)d$ J

A

1

R

42

R

=22+ 2% +z+c

=/(:1:2+2—|—:1:_2)d$ 3

.

of: Liger X5 K A 3+$+_1+c 1

=%$3+2:13———|—c J

Chapter 13 (Integration)

C

/(3:8)2 dx

0

d

/(2+1)

T

2

Exercise 13E

dx

— /(41:‘4 +4x7% +1) dx -

4p—3

4

T

4p—1

-

= —— 33 — =+ $+

e

/(:1:+1)3 dz =/(:132—|—23:+1)(:13—|—1) dx

/(:1:3+:£2+23:2+2:1:+:B+1) da /(m3+3$2+3m+1)

dx

=—+Si+3i+m+c ="+t

+3° +r+c

/(:13— 1)?* dx

/($22$+1)($22$+1) i /($42$3—|—$22$3—|—4$22$+$22$—|—1)

=/(m4—4m3—|—6$2—4m—|—1) €T

5

=

4 —I—

G5

3

Aq?

=%$5—$4—|—2$3—23:

6

dx

dx

+ x +

+x+cC

a LiF)+G@)] = F(z)+ G'(z) dx = f(x) + g(x)

b

Using a,

/[f(:l:)—i—g(:fl)}dfl:—F(m +G(z) +c

ff(;r d:l:—l—/g(:r) dz

SLE TG +C

g2

Chapter 13 (Integration)

Exercise 13F

_

fl(z) =2z —1

b

f'(z) = 32% + 22

fx) = / (3z% + 2z) da

f(a:)/(?x—l)d;c 2

:2%—$+c =z

But

f(0)=3,

—z+c

=z° +z°+c

so 0—0+c=3

But

f(2)=05,

so 84+4+c=5 c=—7

c—.J

flz)=2" —2z+3

fx)=z3+z2 =7

f(z) =3 — 227

I

d

fl)=z-=

f(z) = / (3 — 222) d

f(x):f($2$2)

2

:3m—%+c

2

f(1)=1,

dx

201

=

=3:1:—%:1?3—|—c

But

2

2

=dir*+=+c

so 3(1) — (1)’ +c=1

£z

But f(1)=2 so 4(1)*+2+c=

%+c=1

%—I—G—Q

L1 C=73

_ 1.2, 2 1 f(z)=32"+—-—3 f

f'(x) =32° —4x +1 f(a:)f(3$24:§+1) 33

— 3

:%m"l—Qerc

But f(4) =0,

—x

But

so +(4)* —2(4)+c=0 64—8+c=0 c=

f(z) = 32* — 22 — 56

f(0)=12,

A2

g

e

-2+

so

dx

+e

0—0+0+c=12 c=12

—9ob0

f(z)=2° —22° +x + 12

Chapter 13 (Integration) dy dx

3

—1—3zx7?

dr

y=/(:1?—2:1?2) dz it

/(1— 3z?%) da

S

But the curve passes through so when

oy

y=4.

x =2,

(2, 4),

i

£L

But the curve has z-intercept 3,

sowhen

4¢ . 4=2-4

=3,

‘.3+§+c=0

f(z) =ax +1

f(z) = /(a$+l)dm flaflfig

=T+$+C

f(0)=3,

and

so

¢=3

f(x) = 2az® + x + 3 f(3)= -3

ta(3)* +3+3 = -3 2a+6=—3 %a:—Q a=

—2

@) =3(-2)a* +2 +3 =—z’+x+3

f'(x) = ax® + bx

f(z) = / (az? + bz) da Now

f(0)=1,

so c=1

f(z)= saz® + 5bz? + 1,

f(=1) = =2

ta(=1)°+ 1b(-1)* +1 = -2 —§a+%b+1:—2

—3a+ b= -3

-I—c

=$_|_E_|_C

A 2 _ 2093 - 4=302)"—350@) +c

Now

Exercise 13F

and

f(1) =4

szt 5b+1= coosa+3b=3

y=0.

523

Chapter 13 (Integration)

524

Exercise 13G

Solving the system of equations

{

(athRad|[Norn1]

an X+bn

—%aqt%b:—:}

gives T

a =9

Y=CE

1[—0.333 2L0.3333

za+3b=3 and

El an X+bn 0.5 0.6

—3|

e

Y=Cn

b =0. SOLVE (3|3

= %(9):{:3 + %(0):‘52 +1

EDIT

=3z° +1

d

/

1 0

3% dx

= [+*],

0

Properties:

b

/ a

/

b

C

flx)dx+ |

c

f(z)dx= /

b

c f(x) der = c/

b

f(x) dz

a

f(z) dz,

where c is a constant

Chapter 13 (Integration)

~0)

Exercise 13G

— (8 —2(9)) - (% - 2(4) - 6%

3

= (£ —2(9)) — (0-0)

-3

— 2%

d:t:+/b

f(x) dx =

b

a

(z) dx

1(—3:1:) dx

C

/1(:122 — 3z) dx

525

526

Chapter 13 (Integration)

Exercise 13G

2 /(3m2$+6)dm 0 BIC =10 g

C

_ 322 _ 2 6w 3 2 =

2

S*Jg—m—-l-fiil‘?]

— (8—2+12)—0

4 /(3+2m$2)dm 1 ' > 3714

d

= ) |30+ 222 _ Z 3

0

2

=

0

-

a

514 33:+3:2—$—}

3

14

), -240 ) —(34+1-1 (3(4)+42 3(4) + 3 (3+1-3)

—

_ 1%

1

=12+16—-% -3—-1+ 3 =3 1

2

3

=/(:1:2—|—6:1:+9)d$

o +

E-gl

]

|—'|.

1 E..g

|

el

I

1

2

1

|

—fl

2 /(:1?+3)2d:1:

o

e ||

—1

f

=

1

m| 8 4

e

% dx

|

e

3 3 /—2d$:/

Chapter 13 (Integration)

6

a

/

Exercise 13G

527

(3z2 + 2) dox = 72

0

[2° + 22] " =72 m°e +2m —0 =72

m> +2m —72=0

Using technology,

B

Eites) ok aX3 +bX2 +oX+d=0

aX3 +bX2 +cX+d=0

=4

m

TR

C

e 1 0

x1 C I

2 IEEE

-72 2m.

f

b

4

REPEAT

(2 —1)dx =4

Tre. [:1-:2

o

‘I:]

2m Tri

4

_

Tri

(4m? — 2m) — (m? —m) =4

3m?—m—4=20 ke B aX2 +bX+c=0 = )

Using technology. or = m —= _1ord

C

3

FtRdNen] @Rl aX2 - +bX+c=0

]

xz[

D

-1

7

a

];

1 %3

5

b

dx

MORE

AR P R BN RS T

6 /2

—

1

Var —3

;

G

D

0

i

/—3

d

1

Le"dx

P

Using technology, 1

/ 0

e® dr ~ 1.718

D

AS

V1—x dr o

1

=~ 4.667

Eiion) Wk

fflxel"‘dx

1.718281828

B EVEES

R RS

AR

Using technology,

dr =29

[

e 4.666666667

[]

Using technology,

3

[REPEAT]

0 f_aJl—x dx

2

[]

B

-4

[SOLVE)PENRI3(CLEAR] EDIT ]

—1}

[

0.7182818289

D Using technology, 1

/ 0

rel% dr ~ 0.7183

528

Chapter 13 (Integration)

e

B

3

Illn

Exercise 13G

f

xdx

B

I_le"de

1.295836866

]

[]

Solve|d/dx|d%dx? J dx [SolveN[EI=N

1.493648266

R e e e

Using technology, 3 / Inz dr ~ 1.296 1

a

1

>

Using technology, 1 . f e" dr~1.494 —1

/;f(flz)dm—i—/:f(:v)d:r:/:f(m)d

b [f(x)dx/;f(x)dx/jf(m)dm+/:f(m)dm =f:f(m)dm C

/139(:1:)d:fi+/jg(m)dm+/:g(m)dm/lgg(a:)da:

a

/lgf(x)d$+/:f(m)d:r:fifif(m)dm

| /:f(x)dn;=/fif(m)dm—/13f(g:)d —3

/

2

d$+/ f(z) d$+/ f(a:)d:c—/ f(a h /Qf(:r)d:r:/(] f(g:)d:r—/fl_‘fif(:r)dm—/jf(m)dm =7—(=2)=5

10

a

/1_

f(x) dx = —

f(x) dx

b

(2—I—f($))d$=/_12d$—|—

f(x) dx

Chapter 13 (Integration)

dx

d

f

1 L

kf(x)de=17

k|

flx)de=T b — —

When When

z=1, z=2,

==

k(—4) =7

y=4(1)=14 y=4(2)=38

Area = area of trapezium 2

— 6 units?

b

Area= f

_

IS

2 1

[QI

4x dx

[9,.2]2 ]1

= 2(2)* — 2(1)° — 6 units®

2

a

AY -t

-

y="5

; :

6 ;

I Area=6 x5 o

— 30 units?

il Area= / NF fl

E

—

0

—6

[5$]

5 dz fl

6

= 5(0) — 5(—6) — 30 units?

Exercise 13H

530

Chapter 13 (Integration)

Exercise 13H

b

I

Area = area of trapezium

=sY

2

3

a

Area of blue shaded region 2

/

222 dx

\

0

)

\ \\

b

Area of green shaded region 3

/

2702 dx

2

.

r2.313

—

[§$

__

54 3

]2 3

= 122 units”

/

= Y

/

/

——

Chapter 13 (Integration) b

Exercise 13H

531

AY

| ?

7

Y

Y

1 Area = f

4 2

0 _[1.311 _[3$]D 1

dr

Area = /

3

>

1 1 414 :[Zm]l

=30 1 =

E

dr

=641

ite? units

- 63%

| units?

-

L

A and B are the z-intercepts of y = —x? + = + 6.

When

y=0,

AY

—z2°4+2+6=0

-

(dZc)Real aX9+bX+c=E a

C

-1

E (drc] Real aX2 +bX+c=0

c

¥1

1 I

xn[

6

Using technology,

Ais

(—2,0)

REPEAT)

=z = —2 or 3

and Bis

A

(3, 0).

-2}

3

532

Chapter 13 (Integration)

b

Area:/

3

Exercise 13H

(—z* +z +6) dr

=2

= [—%;1:3 + %:1:2 £ 6:‘{:]3_2

= (-94+2+18) — (2 +2—12)

= 134 - (74 — 20%

When

y=0,

—22°+2x4+4=0

MathRadHormi]

[d/c]Real

aX2 +bX+cec=0

b

a

[

-2

aX2 +bX+c=0

c

2

xl[l} X2 -1

4

Using technology,

area = / —

2 —1

(d/c]Real

2

REPEAT

© = —1 or 2

AY

y=—22°+ 2z +4

(—2z° + 2z + 4) dx 3

[—%fl;‘

+ x

2

-

-|-4:1?]_1

b

P

5

=Y

6

units?

= (—%(8)+4+8) — (—-3(-1)+1—4) — 9 units?

7

a

When

v

y=0,

—(x—2)(x—5)=0 5

area=/ 2

/ =/ —

2

2

Ay

r=2or5

—(xz — 2)(x — 5) dzx

5

-4

A

\n

—(2* — Tz + 10) dx 5

y=—(z—2)(z—5)

(—x* + Tz— 10) dx

[—;mg + %:1:2 — 10:1:];

= (—22% + 15 —50) — (—& 4 14 — 20

Y

2

5

E’

Chapter 13 (Integration)

b

When

y=0,

3—2°=0 . 22 =3 "

V3

area — / —

—V3

m::I:\/g

(3 — 2%) dx

[3$ — %mg

}\/5 —/3

_ (3\/5- . (\/3)3) - (—3¢§- ! (-fi)g) — 2v/3 — (—2V/3) — 4+/3 units? 2

8

1

Areazf

—

dx

€I

1 2

=f

1

T % dr 112

e

-}~ (-1 =3

1

. 9 units

2

9

a

Area/

a

—2d$

1

£

2

8:/

1

ax”2 dr

g — [—ax_1]2

Exercise 13H

533

534

Chapter 13 (Integration)

b

Exercise 13H

Area = /

(z° +2) d

6a = [%&?3 —|—23L’]a_fl ba =

(%afi +2a,)

— (-%(13 — 2@)

ba = %ag—l-fila

6a units?

%a3—2a=0

-}

,I\

|

T

soa’—3a=0

a(a2—3)

=0 a="01 +/3

but ¢

a=13

a > 0,

When

y =0,

a’ —

% =0 iITZ:{IZ

r=a

das x>0}

1

area = / 18 =

0

(a® — 2%) dz

[{123: — %3:3]

18 units?

a

0

18 = (a’ — 3a’) —0

2a” 3 =18 a® =27 a=3

d

Area= / 4 =

0

(z° — 62* + dax) dx

AY y=1>—

Ha:‘rl — 213 + 2{13_':2]

4=%a4—2a3—|—2a3—0

0

6x°+ dax

Fy i

-

a* = 16

a=2

10

a

i

When

{as a >0}

y=0,

v

3—x2=0 —

area =z

X 3 X 3

=4%units2

:I;.

Chapter 13 (Integration)

Area = /

il

3 0

Investigation 2

AY

(34 2z — 2°) dx

— 30+ 22— 1]} =(9+9-2) -0 — 9 units?

b

When

=0,

3—z=3

and

3+ 2z —z°=3,

and when

=3,

3—x=0 3

and

3+ 2x—z°=0. 3

shaded area = /

0

(3 + 22 — 2?) dz — /

—9—44

0

(3 —x) dx

— 4% units?

=0 b

Since the curve lies on or above the x-axis for 0 < x < 1, the first integral in a is the area bounded by y = 23, the z-axis, and the vertical lines =0 and = = 1. The second integral in a does not give an area as the curve lies on or below the z-axis for —1
2) ~0.998

20% of 25 =5

P(test cancelled) = P(X > 6) ~ 0.807

6

Let X be the number of successful shots from the free throw line.

n=20,

so

X =0,1,2,3,..,0r20,

and

p=85%

= 0.85

X ~ B(20, 0.85)

a P(X =20)=(5,)(0.85)*°(0.15)" ~ (.0388

b

[ o

BinomialCD(18,20,20,>

0.404896278

|']

AR

RG]

P(X > 18) ~ 0.405 7

¢

[0 GotEfen &HED BinomialCD(14,17 20, , 0.5731686211

[]

ARG

P(14 < X < 17) ~ 0.573

For Jelena to win a set of 6 games to 4, she must win 5 of the first 9 games, and then win the 10th game. Let X be the number of games Jelena wins in the first 9 games. n=9 so X=0,1,2,3,..,0or9 Now, Martina beats Jelena in 2 games out of 3, so the probability of Jelena winning a game is o

2

1

p=1-5=3. X ~ B(9, %) So,

P(J wins 6 games to 4)

= P(J wins 5 of first 9 games) x P(J wins 10th game)

=P(X =5) x 3

~0.1024 x 1 ~ (0.0341

B HatDegMonl) [d/c]Real

BinomialPD(5,9,1+3)

Ansx (1+3) -

LA RN

0.1024234111

0.03414113702

568 8

Chapter 14 (Discrete random variables) Let X

Exercise 14E

be the number of heads.

n=200,

so

X=0,1,2,3,..,0r200,

and

p=5

X ~ B(200, 3) a

BinomialCD(

[

[ eonies

oina

BinomialCD(96,104, 20>

)

0.4753776443

[]

' Bpd | Bed [InvB.

' Bpd [ Bed [InvB|

P(90 < X < 110) = 0.863

P(95 < X < 105) =P(96

< X < 104)

~ 0.475

9

a

P(rolling double sixes) = & x & = 37

b

Let X be the number of double sixes rolled.

n=>500,

so

X=0,1,2,3,..,0r500,

and

p= Bl

£ Watileglforn]) (bRea) 0.8461199997

X ~ B(500, %)

P(10 < X < 20) ~ 0.846 10

ARRN

Let X be the number of traffic lights Shelley has stopped at. n=15 so X=0,1,23,..,0orl5 and p=0.6 X ~ B(15, 0.6)

a

P(Shelley will be late) = P(X > 11)

—1-P(X < 11) ~ (0.0905

b

El Math[RadMormi) [zbc)Real 1nnmlaICDi 11,15, 9D94980978 1-Ans

-

0.0905019024

P(Shelley will be on time) = P(X < 11) ~ (0.909

El [Math[RadMormi) (b7 Real 1nnmlaICDi 11,15,

P(Shelley will be on time all 5 days) = [P(X < 11)]’

Ans®

~ 0.622

L]

9094980976

0.6223131455

AR RN

¢

P(Shelley will be on time) = P(X < 12) ~ 0.973

P(Shelley will be on time all 5 days) = [P(X < 12)]° ~ (0.872

yes, the probability that Shelley 1s on time for work each day of a 5 day week is now about 87.2%.

El

[MathRadMNorni] [zb/c[Real

BinomialCD(0,12,15, .> 0.9728859992

Ans®

L] IR

0.8715850405

Chapter 14 (Discrete random variables)

11

Investigation 2

569

Let X be the number of solar components which fail. n=20

so

X=0,1,2,3,..,0r20,

and

p=0.85

X ~ B(20, 0.85)

a

P(hot water unit fails within one year) = P(all 20 components fail) = P(X = 20)

= (0.85)%" ~ 0.0388

b

P(hot water unit with n components fails within one year) = (0.85)" .

P(hot water unit with n components is operating after one year) = 1 — (0.85)" .

we need to find the smallest integer n such that

Using technology,

1 — (0.85)™ = 0.98

n ~ 24.1 components.

1 — (0.85)" > 0.98

when

El

Y1=1=0.856%(x)

¥2=.98

at least 25 solar components are needed. Check:

When

n = 25,

the probability that at least one

component will still work is 1 — (0.85)25 ~ 0.983 > 0.98

v

=24507118369

Consult the graphics calculator instructions by clicking on the icon 1n the Investigation box if you need help obtaining the result shown.

s

10

" — 30 o

_ 5

e

p = 0.25

=il

=295

=i

="

i — 05

1

el

L

=1

=05

©w=12.5

i

0 ~2.7386 |

=20

=2

o~

E;_

FLOAT

¥=0.98

AUTO

REAL

X=749 2x=7.5 2x2=61.875 Sx= 0 —Ppox=2.371708245 n=1 minx=0 $Q1=6

c~09487 | 0 ~1.3693 | 0 ~1.5811 | o ~ 1.4491 og~16432 | c~23717 |

Pe

VL

MORHMAL

p =5l

N

: //

1 X ~ B(30, 0.25)

2

[EXE):Show coordinates

2.5100

=

c~21213 | 0 ~3.0619 | 0 ~3.5355 | o~ 3.2404

RADIAN

et

HMP

I]

Chapter 14 (Discrete random variables)

570

np =

Exercise 14F

R

=t

np=7

vnp(l—p) | /np(1—p) | /np(1—p) | /np(1—p)

~ (0.9487

~ 1.3693

np =3

~ 1.5811

np ="7.5

~ 1.4491

R

=l

Vrp(l=p) | y/np(L—p) | /np(1—p) | /np(1—p)

~ 1.6432

~ 2.3717

np =

~ 2.7386

RRI=NS

~ 2.5100

np = 25

np = 39

vnp(l—p) | /np(1—p) | /np(1—p) | /np(1-p)

=22

~ 3.0619

= 0

Our results in 2 and 3 agree with the formulae

1

a

~ 3.2404

y =np

and

o = \/ np(1l — p).

X ~ B(6,0.5)

i p=np =06x05

o = /np(1 —p) =6 x0.5x 0.5

—

~ 1.22

1

3

D

6

0.0938 | 0.2344 | 0.3125 | 0.2344 | 0.0938 | 0.0156

0.2 0.1

_I

0 i

1

2

—

—

=

N

3

1

5

The distribution 1s symmetric.

b X ~ B(6,0.2)

i p=np =6x0.2 = 1.2

o = v/np(1 —p) =6 x02x%0.8 ~ (0.980

6

=

Chapter 14 (Discrete random variables) ii

0

0

1

2

3

4

Exercise 14F

5

571

6

P(z;) | 0.2621 | 0.3932 | 0.2458 | 0.0819 | 0.0154 | 0.0015 | 0.0001 A probability 0.4 0.3 0.2}

—

—

0.1}

=

-

0

il

¢

0

1

2

3

|

'

4

D

-

0

T

The distribution is positively skewed.

X ~ B(6,0.8)

| p=np

o = /np(1 —p)

=06Xx0.8

=6 x 0.8 x 0.2

= 4.8 il

~ 0.980

T;

0

1

2

3

4

5

6

P(z;) | 0.0001 [ 0.0015 | 0.0154 | 0.0819 | 0.2458 | 0.3932 | 0.2621 A probability 0.4 0.3 0.2

-

-

R

0.1

-

—

R

= iii 2

1

2

3

1

5

6

=

The distribution is negatively skewed, and is the exact reflection of the distribution in b.

X ~ B(10, 0.5)

mean

(= np

=10 % 3

and

variance

o° = np(l — p)

=9

3

a

X ~ B(30,0.04) px = NP = 30 x 0.04 — 1.2

ox = \/flp(l —p)

=10x 4 x 1 =25

b

Y ~ B(30, 0.96) Hy = Tp = 30 x 0.96 = 28.8

oy = \/np(l —p)

= /30 x 0.04 x 0.96

= /30 x 0.96 x 0.04

~ 1.07

~ 1.07

Chapter 14 (Discrete random variables)

572

Review set 14A

b X ~ B(30, 0.13) o= /np(l —p)

= np — 30 x 0.13 — 3.9

— /30

x 0.13 x 0.87

~ 1.84

5 X ~ B(38,0.75) a

b

o =+/np(1 —p)

u=mnp =38 x 0.75 = 28.5

— /38 x 0.75 x 0.25 ~ 2.67

u— o0 =28.5—2.67

1+o =~ 28.5 4+ 2.67

~ 25.8

~ 31.2

El Hath)Degl(Norm]) _(zb/c] Real 75 , & Binumialcg(26,31, . 7401514723

[]

. Plu—o 3)=P(X =3)+P(X

=4)

= 0.10 + 0.05

= (.15 ¢

Since

P(X = 2)

is the greatest probability, 2 is the mode of this distribution.

d E(X) =) zp 1=1

= 0(0.10) + 1(0.30) + 2(0.45) + 3(0.10) + 4(0.05) —0+0.3+0.9+0.3+0.2 — 1.7 5

a

X i1s a discrete random variable because it has a set of distinct possible values.

b

X=0,1, or 2

C

1st draw

2nd draw 2

3

areaG G< 4

5

Outcome

X

GG

2

Probability

2x2=2 SR 20

1

.

1P

5

x

0

|1

P(z) | & | 2

Chapter 14 (Discrete random variables) —

E(X)

d

Z

i=1

LiPi

= (0 &) + (1% 2) + (2 —

Review set 14A

o) 10

6

5

= 1.2 green balls

X has probability table: 4

Z LiPDq i=1

~1(3) +3(3) +4(3) +6(3)

| S

P(x)

3 2 6

E(X) ==

— 23 T

e

D= | =

xZ

Sy

6

~ 3.83

7

a

Let X denote the amount of money Lakshmi wins from one roll. X has probability table: i

214168101

Pzl E(X)

1

1

1

s1sl5ls| =

Z

1=1

1

12 1

5|

1

3

LiPq

= (@x )+ (Ex3)+(6x )+ (8 x3) +(10x3) + (12 3) =7

Lakshmi can expect to win $7 from one roll of the die.

b

Expected gain = $7 — $8 = —$1. So, Lakshmi should not play many games as she would lose $1 per game in the long run.

8

The number of trials 1s

n = 5.

The probability of success (kicking a goal) is p = 80% = 0.8. Let X be the number of goals scored, and G represent scoring a goal. X ~ B(5, 0.8)

P(3 goals then misses twice)

— P(GGGG'G)

— (0.8)%(1 — 0.8)° = 0.02048

3 goals and misses twice) 3 goals)

575

Chapter 14 (Discrete random variables)

P(z) =a(z* — 8x) where a X has probability table:

=0,

1, 2, 3, ..... 8

T

0

1

2

3

4

D

6

7

8

B

0

—Ta

—12a

—15a

—16a

—15a

—12a

—Ta

0

If this is a probability distribution then

> P(z;) =1 i=1

. 0+ (=7a) + (—12a) + (—15a) + (—16a) + (—15a) + (—12a) + (—7a) + 0 =1 —84a =1 .

=

&

) +3(2) +4(29) +5(¢ ) +6(82) +7() +8(0) +2(2 ) +1(& ) b E(X)=0(0 — 4 marsupials M

9

Review set 14A

square spinner

pentagonal

spinner

¢

i X ~B(10, ) 1

P(X =1)= (1) ()" ()" ~ 0.00416

PX = 9) = (1) ()" (%) ~0.0207

it 1s more likely that exactly one red will occur 9 times.

—%g7

Chapter 14 (Discrete random variables) 11

Review set 14B

Let X denote the number of players who turn up to a game. n=9

so

X=0,1,23,...,0or9,

and

p=75%=0.75

X ~ B(9, 0.75)

a

1 P(X=9)=(3)(0.75)%(0.25)°

i

~ 0.0751

[g

=

BinomialCD(0,5,9,.75>

-

0.165725708

LR

P(forfeit) = P(X < 6) =P(X 0.3336442014

i

o Fad orad

BinomialCD(50,60,175>

[]

0.09307652644

O

P(X < 40) = P(X < 39)

P(50 < X < 60) ~ 0.0931

~ (0.334

11

a

coin

bowling

toss

1 4

-

2 strikes

4

2 heads

.

9—

1 strike ,

0 strikes

1 2

1 head \

- 1 strike '

2

e3

%

0 heads

b

N 1 1 P(X=0)=5+3+73

1

.

¢

L

1

X

% X

5

orel Rl=

36

% ==

%

1

2 X5=3

|

LN5 X i3=e ]3

: 0 strikes

51

0 strikes

% X 1=

X

£2w=3 ]

oS P(X=1)=2+4

— 25 — 36

T

probability

% 1

P(X =2) =

|

85 — 18

0

1

2

2D

o

1

The expected return per game i1s

E(X)=(0x2)+(10x %) + (20 x 5=) — 13—0 dollars

~ $3.33

d

Suvi’s expected gain per game ~ $3.33 — $5 ~ —§$1.67 Suvi should not play the game many times as she is expected to lose $1.67 per game on average.

Chapter 15 THE NORMAL

DISTRIBUTION

e

e

Distributions B, D, and F are symmetrical and bell-shaped. B, D, and F appear to be normally distributed. 2

Most measurements 1n each situation will be centred about the mean, with random variation about the mean explained by some of the factors listed below.

a

b

¢

3

a

The diameters may be affected by: e

the type of lathe used

e

e

the operating speed of the lathe.

the steadiness of the woodworker’s hand

The scores may be affected by: e

the time spent studying

e

general knowledge.

e

natural ability (for example, memory, learning ability)

e

walking speed

The times may be affected by: e

weather conditions

e

ftraffic.

The variable 1s not likely to be normally distributed as i1t is more likely that there would be more people younger than the mean age than there are older. The distribution may be positively skewed.

b

e

physical fitness

The variable is likely to be normally distributed as the long jumper 1s likely to jump the same distance consistently, but it will vary due to factors such as the speed at which the long jumper runs before the jump, and the positioning of their body before hitting the sand.

Chapter 15 (The normal distribution)

Investigation 1

The variable 1s not likely to be normally distributed as each number has the same chance of being drawn. The distribution should be uniform. A

d

Properties of the normal curve

583

The variable is likely to be normally distributed as the lengths of the carrots will be generally centred around the mean, but will vary due to factors such as soil quality, different weather conditions, harvest times, and so on.

The variable i1s not likely to be normally distributed. People are most likely to be served quite quickly. The distribution is likely to be negatively skewed.

The variable is not likely to be normal as it 1s a discrete variable. Each egg has the same probability of being brown, so the distribution 1s binomial. A

0

g

The variable i1s not likely to be normally distributed as it 1s a discrete variable. Most families will have 0 - 2 children, and there will be much fewer families with more than 2 children. The distribution will be positively skewed.

1

2

3

4

B 5

6

7

8

9

10

11

127

The variable is not likely to be normally distributed as there will tend to be many more shorter buildings than tall buildings in a city. The distribution will be positively skewed.

A

1

u controls where the which 1s reasonable o controls the shape which 1s reasonable

centre since p of the since o

of the distribution 1s. As p changes, the curve 1s translated horizontally, is the mean and a measure of centre. curve. As o increases, the curve becomes flatter and more spread out, 1s the standard deviation and a measure of spread.

2

The curve has a vertical line of symmetry

3

The function 1s never negative. be negative.

z = pu.

This is important because a probability density function can never

584

Chapter 15 (The normal distribution) x — 400,

Exercise 15A.2

4

As

the curve approaches zero from above. The x-axis 1s a horizontal asymptote.

5

The area under the curve should remain constant as we change 1 and o, as the area under a probability density function must be 1.

A and B have the same mean, and C and D have the same mean.

The mean of A and B 1s lower than the mean of C and D. B has a greater spread, and hence a larger standard deviation than A. Similarly, C has a larger standard deviation than D. a

pu=2>5,

=2

corresponds to B

b

u=15,

0 =0.5

¢

u=29>,

=1

corresponds to A

d

u=15

o =3

2 | Distribution | Mean (mL) | Standard deviation (mL) A

29

O

B

30

2

E

21

10

corresponds to D

corresponds to C

Chapter 15 (The normal distribution)

Exercise 15B.1

EXERCISE 15B.1 1

X ~ N(30, 52) a

I

The value which 1s 2 standard deviations above the mean =30+ — 4()

il

The value which 1s 1 standard deviation below the mean = 30 — 5 — 25

b

I

35=30+5

39 1s 1 standard deviation above the mean. N

20=30—-2x%x5

20 1s 2 standard deviations below the mean. i

45=30+3x5 45 1s 3 standard deviations above the mean.

2 x 5

585

586

Chapter 15 (The normal distribution)

Exercise 15B.1

C

a*

.

:

15

i

20

d

:

25

:

30

E

.

30

40

45

|

a

ERTN

-

X

34.13%

e

s

15

s

20

20

30

39

40

45

X

About 34.13% of the values of X are between 25 and 30.

15

20

25

30

35

40

5

X

About 13.59% of the values of X are between 35 and 40. the probability that a randomly selected member of the population will measure between 35 and 40 1s approximately 0.1359. 2

a

b

u=20

o0=4

|

About 34.13% of X values are between 20 and 24.

3

12

16

20

24

28

32 About 13.59% of X values are between 12 and 16.

Chapter 15 (The normal distribution)

Exercise 15B.1

About 2.15% + 0.13% = 2.28% X values are greater than 28.

8

0.13%

-4

N

12

16

20

24

28

34.13%

2.15%

13.59% h9 — 8

67 — 8

79 —8

— 51

— 59

— 67

of

39

34.13% |

2.15%

587

(=175

0.13%

|

75+ 8

83+ 8

91 +8

= &3

=91

=99

L

-

examination score

About

13.59% + 2.15% + 0.13% = 15.87% of students would be expected to have scored more than 83.

51

59

67

75

83

91

99

score About

2.15% + 0.13%

=

2.28%

of

students would be expected to have scored less than 59.

51

59

67

75

83

91

99

score About

34.13% + 34.13% + 13.59% = 81.85% of students would be expected to have scored between 67 and 91.

51

0.13%

59

67

75

83

91

99

34.13%;

2.15%

.core

34.13%;

13.59% U—30

a

pU—20

P(value between ~ (0.3413 + 0.3413 ~ (.6826

i — o

2.15%

0.13%

. 13.59% h—0

L4

and u+ o)

[+o

b

p+20

L+30

P(value > u + 20) ~ 0.0215 4 0.0013 ~ 0.0228

588 5

Chapter 15 (The normal distribution) a

|

Exercise 15B.1

About 34.13% of female students have a height between 162 ¢cm and 170 cm.

34.13%

146 154 162 170 178 186 194 X (om) i

About 34.13% + 13.59% = 47.72% of female students have a height between 170 cm and 186 cm.

34.13% 13.59%

146 154 162 170 178 186 194 X (com) b

i About

2.15% 4+ 0.13%

=

2.28%

of

female students have a height less than 154 cm.

=i P(height

is less than 154 cm)

0.13% 2.15% '

R il

About

146 154 162 170 178 186 194 x (em) 3413%

+

50%

—

8413%

of

34.13%

female students have a height greater than 162 cm.

P(height is greater than 162 c¢m) e

~ (0.8413

¢

d

About 13.59% of the female students have a height between 178 cm and 186 cm. So, we would expect about 13.59% of 500 ~ 68 students to have a height between 178 cm and 186 cm.

146 154 162 170 178 186 194 X (com)

one standard deviation above the mean. k 1s about o above the mean p ~

a

13.59%

Approximately 16% of data lies more than

k~

6

146 154 162 170 178 186 194 X (cm)

170+ & :

178

170

About

3413%

+

50%

female

frilled

sharks

—

8413%

measure

of

more

adult

&k

:

X (cm)

34_13%

than

1.25 m long.

0.95 1.1 1.25 1.4 155 1.7 1.85 X (m)

Chapter 15 (The normal distribution) About

13.59% + 34.13% + 34.13%

— 81.85% of adult female frilled sharks

measure between 1.1 m and 1.55 m long.

Exercise 15B.1

589

34.13%

5 13.59%

0.95 1.1 1.25 1.4 1.55 1.7 1.85 x (m) About 50% + 34.13% = 84.13% born weighed less than 3.2 kg.

of babies

34.13%

So, about 84.13% x 545 ~ 459 babies born weighed less than 3.2 kg.

24 About

34.13% + 34.13% + 13.59%

= 81.85%

2.6 28

3.0 3.2 34

34.13%

34.13%

of babies born weighed

13.59%

between 2.8 kg and 3.4 kg.

So, about 81.85% x 545 =~ 446 babies born weighed between 2.8 kg and 3.4 kg. Under

18000

bottles

are

filled

on

3.6 X(kg)

24

2.6

2.8

3.0

32

34

3.6 X(kg)

14

16

18

20

22

24

26

18

20

22

24

26

about

0.13%+2.15%+13.59% = 15.87% of days. So, under 18000

bottles are filled on about

15.87% x 260 ~ 41 days.

Over 16000 bottles are filled on about 13.59% +34.13% + 50% = 97.72% of days.

34.13%

So,

13.59%

over

16000

bottles are filled on about

X (x1000 bottles)

97.72% x 260 =~ 254 days.

14

Between 18000 and 24 000 bottles are filled

on about So,

filled on about

About

34.13%

34.13% + 34.13% + 13.59% = 81.85% of days.

between

18000

and

24 000

competitors longer than So, about completed

11 minutes.

X (x1000 bottles)

34.13% 13.59%

bottles are

81.85% x 260 ~ 213 days.

2.15% + 0.13% = 2.28%

16

14

16

18

20

22

:

:

:

:

24

26

X (x1000 bottles)

of

completed the race 1n a time 11 minutes. 2.28% x 200 =~ 5 competitors the race in a time longer than

93

10

10%

105

103

11|

11 1 ¥ (min)

590

Chapter 15 (The normal distribution)

b

Exercise 15B.1

About 0.13%+2.15%+ 13.59% = 15.87% of competitors completed the race in a time less than 10 minutes 15 seconds. So, about 15.87% x 200 ~ 32 competitors completed the race in a time less than 10 minutes 15 seconds.

¢

10 minutes 15 seconds and 10 minutes 45 seconds. a

10

10§ 105 103

About 34.13% + 34.13% = 68.26% of competitors completed the race in a time between 10 minutes 15 seconds and 10 minutes 45 seconds.

So, about 68.26% x 200 ~ 137 competitors completed the race in a time between

10

o}

Approximately

84%

of data is more

11 113 v

o

34.13%

310

107

105

103

152

p

200

11

113 X (min)

than

one standard deviation below the mean, and

34.13% + 50% = 84%. 152 grams 1s about o below the mean u u~152+4+o0 ... (1) Approximately 16% of data one standard deviation above 13.59% + 2.15% + 0.13% =~ 200 grams 1s o above the nw~~200—-0 .. (2)

is more than the mean, and 16%. mean u

Equating (1) and (2) gives:

152+ 0 ~ 200 — o

X ()

20 ~ 48 .

and

ox=24

p~ 200 — 24 ~ 176

So, b

p~

176 grams

and

{using (2)}

o ~ 24 grams.

About

34.13% + 34.13% + 13.59% — 81.85% of the oranges weigh between 152 grams and 224 grams.

34.13%

34.13% 13.59%

104 128 152 176 200 224 248 X (g)

11

a

|

About 50% + 34.13% = 84.13% of radishes grown without fertiliser will have weights less than 50 grams.

34.13%

10

20

30

40

50

60 E

70

X(g) -

Chapter 15 (The normal distribution)

il

b

About

0.13% + 2.15%

= 2.28%

Exercise 15B.1

591

of

radishes grown with fertiliser will have weights less than 60 grams.

0.13% 9.15% PRV

100

140

180

220

260

-

20

60

X (g)

20

60 100 140 180 220 260 X (g)

i About 2.15% of radishes grown with fertiliser will have weights between 20 g and 60 g.

P(radish grown with fertiliser weighs between 20 g and 60 g) ~ 0.0215 ii

AbOUt

34.13%

13.59% + 34.13% + 34.13% + 13.59%

= 95.44% of radishes grown without fertiliser will have weights between 20 g and 60 g.

34.13%

13.59%

=

10

20

13.59%

30

40

50

60

70

X(g)

P(radish grown without fertiliser weighs between 20 g and 60 g) ~ 0.9544 ¢

About 13.59% + 34.13% + 50% = 97.72% of radishes grown with fertiliser will have weights more than 60 g.

20

60 100 140 180 220 260 X (g)

P(radish grown with fertiliser weighs more than 60 g) ~ 0.9772

About

2.15% + 0.13% = 2.28%

grown without fertiliser will more than 60 g.

of radishes

have

weights

10

20

30

40

P(radish grown without fertiliser weighs more than 60 g) ~ 0.0228 P(both radishes weigh more than 60 g) = P(radish grown with fertiliser weighs more than 60 g) x P(radish grown without fertiliser weighs more than 60 g) ~ 0.9772 x 0.0228 ~ (0.0223

50

60

70 X (g

592

Chapter 15 (The normal distribution)

Exercise 15B.2

EXERCISE 15B.2 1

X ~ N(60, 5%) a

To find

P(60 < X < 65),

El Deglormi) [b/ 64), we use a very high value such as 10 to represent the upper bound. El

DeglHorm1] [ab/c][Real

Normal Data Lower Upper

C.D :Variable ' 684 :%E+99

G L

: : 60

E]

DeglNorml] [ab/c]Real

Normal C.D D =0.21185539 Z:Low=0.8 Z:Up =2E+98

P

[(None |INEYD

60 64

X

P(X > 64) ~ 0.212 d

To find

P(X < 68),

El DeglMorm1) [sb7cReal Normal C.D Data :Variable Lower :-1E+99 Upper 68 c : 9 L ' 60

P(X < 68) ~ 0.945

we use a very low value such as —10% to represent the lower bound. E] Deg)Mormi] [sb7c]Real Normal C.D D =0.9452007 Z:Low=-2E+98 Z:Up =1.6

Chapter 15 (The normal distribution) e

To find

P(X < 61),

Sl

L

60

o

593

we use a very low value such as —10°? to represent the lower bound.

B DegFormi) [zb/c)Real Normal C.D Data :Variable Lower :-1+99 Upper

Exercise 15B.2

g8

BegFormi) [zb/c)Real

Normal

C.D

P

=0.5792597

z:Up

=0.2

Z:Low=-2E+98

)

60 61

X

P(X < 61) ~ 0.579 f

To find g

P(57.5 < X 4) ~ 0.841 34 ~ 0.841

about 84.1% of patients show a drop of more than 4 units. Let Y be the number of patients with a drop of more than

|g

Y ~ B(8, 0.84134)

: : 1CD{B, : BDI"“"”E 8. 8 Ans? 0. 8798100088

A

unit

P(Y > 6) ~ 0.880

11

Let X cm be the length of a randomly selected red snapper. X ~ N(58,

182)

NormCD(4,1x1099,1.9, 5>

2t Deg) Fora)

NormCD(-199,38,18, 5>

-

0.1332602629

P(X < 38)~0.13326 ~ 0.133

Let Y be the number of red snapper that are long enough to keep.

[

From a, the probability that a single snapper can be kept 1s about 1 —0.13326 ~ 0.86674.

1-Ans

Y ~ B(14, 0.866 74) P(Y > 10) ~ 0.970

HathlDegNormi (d7c)Real

NormCD(-199,38,18, 5>

0.1332602629 0.8667397371

BinomialCD(10,14,14,>

0

0.9703362804

Chapter 15 (The normal distribution)

2]

Radormi) [sb/c)Real

Inverse Data

Normal :Variable

Area o

SR 3

Tail L

2

RadForml) [eb/c)(Real

Inverse

Normal

xInv=18.4267985

:Left : 20

[None |IHEID

If

P(Xa)=1-P(X =1

x)=0.7

P(X

60) ~ 0.0766 2] Deg)Mormi) [b7c)Real Inverse Normal Data :Variable Tail :Left Area :0.77 o Sl

L

st ]

[None |INERD

If

P(X

>k)=0.23

P(X

< k)=0.77 k ~ 55.2

E

Deglorn]) (zb/c](Real

Inverse Normal xInv=55.1719279

50 k

.

618 7

Chapter 15 (The normal distribution) Let X

Review set 15B

seconds be the time a contestant holds their breath.

;X'anH15U$122) =

DeglNormi] [ab/c][Real

Inverse Data Tail Area

=

Normal :Variable :Left :0.85

o

[DeglNorm1] [ab/c][Real

Inverse Normal xInv=162.437201

r12

: 150

[None JE]

P(X >t)=0.15 P(X 80

{the population mean is greater than 80}

b T=83.1,

p,=2380,

s=129,

The value of the test statistic ¢t =

n =200 T — o = 80,

t~3.40,

and

the p-value = P(T" > t) where ~ P(T > 3.40)

n = 200,

j

T ~ t199

199

pevlie

~ (0.000407

340

3

d

Since

p-value < 0.01 = «,

we have enough evidence to reject Hy

e

We conclude that the population mean is greater than 80 at the 1% significance level.

1% significance level.

Step 1:

T

in favour of H;

on a

Let pu be the population mean weight of the bags. The factory wants to determine whether the weight has decreased, so the hypotheses to be considered are:

Hy:

po=100

{the mean weight is 100 g per bag}

Hqi:

p

Test

(d/c)Real

x2=10.3414634

|[Expected:List2 df :4 Save

Using technology, the

Since

¥2 GOF

a1

YIS

L

=N

X% ~ 10.3

(dic)Real

Step 5:

2

y

p-value =~ 0.0351.

p-value < 0.1 = a,

a 10% significance level.

we have enough evidence to reject Hy in favour of H; on

Step 6:

Since we have accepted H;, we conclude that the proportions of each ice cream flavour sold are not all the same at a 10% significance level. Brian should therefore change the amounts of each ice cream flavour that he makes.

Step 1:

Let p1, p2, p3, pa, and ps be the population proportions of people living in London who identify as being White, Asian/Asian British, Black/Black British, Mixed, and Other respectively.

The hypotheses that should be tested are:

Step 2. Step

4

Hy:

pp =0.712,

py =0.121,

p3 = 0.109,

Hy:

atleast one of p; # 0.712,

ps # 0.121,

py = 0.032, ....,

or

Since no significance level 1s specified, we assume that

3:

Ethnic group White

iohs

p5; = 0.026 p5 #* 0.026.

a = 0.05.

Jexp

4887435 | 8173941 x 0.712 =~ 5819846

_

(fobs = foxp) faxp

~ 149 384

Asian/Asian British | 1511546 |

8173941 x 0.121 ~ 989047

~ 276029

Black/Black British | 1088640 |

8173941 x 0.109 ~ 890960

~ 43 860

Mixed

405279

8173941

x 0.032 =~ 261 566

~ 78961

Other

281041

8173941

x 0.026 =~ 212522

~ 22091

Total

So,

X?alc ~ 570000

~ 570325

2

Chapter 16 (Hypothesis testing)

635

df=5—-1=4

Step4. =

SUB

([d/c](Real List 1 | List 2 | List 3 | List 4

E ¥2GOF

1 4.88E6|5.81E6

Test

x2=570323.942

Expected:List2

2| 1.51E6|

989047

df

4

281686

save

40b279|

= [Deg)Horn]] ¥2 GOF Test

Observed:Listl

3/1.08e6 sgngsn|

Using technology, the

p

=0

df=4

4

NS

AR

CNTRB:List3

Res:None

GphColor:Blue

(DY EHEPINDEES> )

p-value =~ 0.000.

Step 5:

Since p-value < 0.05 = a, we have sufficient evidence to reject Hy in favour of H; on a 5% significance level. We therefore accept H;.

Step 6:

Since we have accepted H;, we conclude on a 5% significance level that there was a significant change in London’s demographics between 2001 and 2011.

a

Band

Expected frequency

10

150 x 0.079 = 11.85

9

150 x 0.167 = 25.05

8

150 x 0.298 = 44.7

7

150 x 0.297 = 44.55

6

150 x 0.135 = 20.25

150 x 0.024 = 3.6

5 and below

b

Exercise 16D.1

Step I

Step 2. Step 3:

Let p1o, po, Ps, P7, Pg, and ps be the population proportions of Year 9 students at a particular school who are in NAPLAN bands 10, 9, 8, 7, 6, and 5 and below respectively. The hypotheses that should be tested are: HUI

P1o

—

0.079,.

Pg

—

0.167,

pPs

Hy:

at least one of pig # 0.079,

—

0.298,

D7

pg # 0.167,

The significance level is a = 0.01. Band

fnbs

fexp

(f obs

f exp)

fflxp

10

D

11.85

~ 3.9097

9

9

25.05

~ 10.2835

8

D0

44.7

~ 2.3734

i

53 | 44.55

~ 1.6027

6

23 I 20.25

~ 0.3735

5

3.6

~ (0.5444

Total

~ 19.1372

5 and below |

So, xZ;. ~ 19.1

2

—

0.297,

Pe

—

0.135,

.... , or ps # 0.024.

Ps

—

0.024

636

Chapter 16 (Hypothesis testing) Step4. B

SUB

Exercise 16D.1

df=6—-1=5 Deglornd) [d7c]Real List1 | List 2 | List 3 | List 4

1

5

3 4

66| 53|

2

2 Degformd) [d7c]Real ¥2 GOF Test

Observed:Listl

11.85

g|

df

44.7 44.55

Hy

Step 6:

¢

p

' H

SN Save

:List3 Res:None

GphColor:Blue

Using technology, the

Since

x¥2=19.1372666

Expected:List2

25.05

GRAPH] CALC J TESTJ INTR J DIST /IIENE

Step 3:

2 DegHormi) [d7c]Real x2 GOF Test

=1.8122e-03

df=5

CNTRB:List3

J

p-value ~ 0.001 81 .

p-value < 0.01 = a,

we have enough evidence to reject Hy in favour of

on a 1% significance level. We therefore accept H;.

Since we have accepted H;, we conclude on a 1% significance level that the NAPLAN results of Year 9s at this school are significantly different from the national results.

Since the expected frequency for “Band 5 and below” 1s less than 5, the sample size 1s not

large enough for y? to be distributed appropriately. By combining “Band 6” and “Band 5 and below” we can obtain more reliable results.

d

Step 1:

Let py1g, p9, ps, p7, and pg be the population proportions of Year 9 students at a particular school who are in NAPLAN bands 10, 9, 8, 7, and 6 and below

respectively. The hypotheses that should be tested are: Hy:

p1o = 0.079,

pg = 0.167,

pg = 0.298,

p7; = 0.297,

pe = 0.135 + 0.024 = 0.159 Hy:

Step 2:

at least one of pyg # 0.079,

The significance level is

Step 3:

Band

pg # 0.167,

....,

or

o = 0.01.

f obs

R

f exp

(ffl

B

SUB

9

2aills

~ 10.2835

8

5%

44,7

~ 2.3734

7

D3

44.55

~ 1.6027 ~ (.7221

g

~ 18.9414

X?alc ~ 18.9

df=5—-1=4 El

List 1 | List 2 | List 3 | List 4

3

p)

~ 3.9597

DeglNorml] [d/c][Real

1 2

fexp

fe

11.85

Total

Step4:

:

10

6 and below | 23+ 5 =28 | 20.25 + 3.6 = 23.85

S0,

pg # 0.159.

5l 9|

5|

53|

1.5 25.05

¥2 GOF

||

44.7

44.55

GRAPH) CALC J TESTJ INTR J DIST JIIEN

Using technology, the

[eg|Norm1] [d/c]Real

E]

Test

x2 GOF

Observed:Listl

p-value ~ 0.000 807.

Test

x2=18.9414828

|[Expected:List2 '4 df

Save Res:None GphColor:Blue

[Deg|(Norm1] [d/c]Real

P =8.0702x19°4 df=4

CNTRB:List3 J

Chapter 16 (Hypothesis testing)

Activity 1

637

Step 5:

Since p-value < 0.01 = «, we have enough evidence to reject Hy in favour of H, on a 1% significance level. We therefore accept H;.

Step 6:

Since we have accepted H, there is still sufficient evidence to support the claim that there 1s a substantial difference between the school’s results and the rest of the

nation at a 1% significance level.

Although we have obtained the same result both times, the result 1s more reliable now that each of the expected frequencies 1s sufficiently large.

ACTIVITY 1 1

a

Each question has 1 correct choice and 4 possible choices, so the probability of answering the

question correctly with a random guess is 1.

©

|

e

Each question can be answered either correctly (success) or incorrectly (failure). So, each trial has only two possible results.

e

The probability of answering correctly (success) is the same for each question.

e

There are 10 questions, and we assume the probability of answering each question correctly does not change based on the results from previous questions. So, there are a fixed number of independent trials.

Therefore, the number of questions the person answers correctly X is a binomial random variable.

il X ~ B(10, 1) 2

a,b

¢

T

P(X =) | Expected frequency

0 1 2

~ (0.056 31 ~ 0.18771 ~ (0.281 57

~ 8.45 ~ 28.16 ~ 42.24

9 4

~ (0.250 28 ~ (0.146 00

~ 37.94 ~ 21.90

5 or more | ~ 0.07813

~ 11.72

We would expect to see 5 correct answers about

150 x P(X = 5) ~ 8.76 times,

but expect

to see 6 or more correct answers about 11.72 — 8.76 ~ 2.96 times. This value is less than 5, which means y? may not be distributed appropriately. Therefore outcomes greater than or equal to 5 have been combined into a single “5 or more” category.

d

Step I:

Let pg, p1, p2, P3, P4, and p5 be the population proportions of people who correctly guess 0, 1, 2, 3, 4, and 5 or more answers respectively.

The hypotheses to be considered are: Hy: H;y:

po=P(X

pp =P(X

=1),

ps = P(X =4), ps = P(X > 5) atleast one of pg # P(X =0), or

Step 2.

=0),

ps

75 P(X

=

5)

The significance level is

o = 0.05.

po =P(X =2), p1 #P(X

=1),

p3 =P(X =3), ..., pys # P(X =4),

638

Chapter 16 (Hypothesis testing) Step 3:

Activity 1

Using the expected frequencies calculated in b: Number of correct answers | fobs

_ 2 (Jobs — Jexp)

Jexp

Jiexp

0 1 2

6 ~ 8.45 13 | =~ 28.16 33 | = 42.24

~ 0.7089 ~ 8.1589 ~ 2.0194

3 4

R 30 | = 21.90

~ 1.4814 ~ 2.9962

5 or more

23 || F=al LI

~ 10.8593

Total

~ 26.2241

S0, X% A 26.2 Step4. B

SUB

1

2

3

4

df=6—-1=5 [Cedforn] (dFcIRed)

g

List 1 | List 2 | List 3 | List 4

6

13|

33|

46|

s45

28.16

x2 GOF

||

x2 GOF

ok

37.54

Test

P =8.0597x1s°5

51

CNTRB

DegMornd) (d7c)Real

X2=26.227896

|Expected:List2

42.24

df=5

+List3

CNTRB:List3

Save Res:None GphColor:Blue

Using technology, the

k

E

Test

Observed:Listl

GRAPHJ CALCJ TEST [ INTR ] DIST [

3

Degforni) (d7c](Real

4

p-value ~ 0.0000806.

Step 5:

Since p-value < 0.05 = «, we have enough evidence to reject Hy in favour of H, on a 5% significance level. We therefore accept H;.

Step 6:

Since we have accepted H;, we conclude on a 5% significance level that the model in which every person randomly guesses every answer 1s not appropriate.

a

The model is not likely to be realistic. While it 1s possible that some people some answers, it 18 unlikely that every person guessed every answer, which assumes. It may be that the probability of answering correctly 1s not the same For example, there may be an option which 1s obviously wrong, and so can

b

From 2 b i, we have assumed that the probability of answering each question correctly does not change based on the results from previous questions.

a

If each person can reject exactly one out of the four answers, and will then guess one answer randomly from the remaining 3, then the probability of answering a question correctly becomes % Using the same reasoning as in 1, we conclude that

randomly guessed is what our model for each question. be rejected.

Y ~ B(10, %)

The probability distribution of Y 1s: Y

0

P(Y =) | ~0.01734

1

2

3

4

5 Or more

~ (0.086 71

~ 0.19509

~ (0.260 12

~ 0.22761

~ 0.213 13

Chapter 16 (Hypothesis testing) b

Step I:

Po

—

P(Y

—

ps =P(Y

H;:

0),,

=4),

P1

—

P(Y

ps =P(Y

—

1),

P2

atleast one of pg #P(Y =0),

—

2),

or ps # P(Y > 5).

pr #P(Y =1),

The significance level 1s

Step 3:

Using the probability distribution from a, and

P3

—

P(Y

—

..., ps #P(Y =4),

Jexp

fex, = 150 x P(Y = y): e 2 (Jobs — fexp) fexp

0 1 2

0 ~ 2.60 13 | =~ 13.01 33 | =~ 29.26

~ 4.4408 ~ (0.0000 ~ 0.4770

3

45 | ~ 39.02

~ 0.9170

4

30 | = 34.14

~ 0.5023

23 |

=~ 31.97

~ 2.5164

Total

~ 8.8039

Xfalc ~ 8.89

df=6—-1=5 [Deg)(Norm1]

List 1 | List 2 | List 3 | List 4 1

6/

2

13|

3 L

33| 45|

2.6

x2 GOF

||

13.01

20.26 39.02

D EESINMDEES> |

Using technology, the

E

Test

x2 GOF

Observed:Listl

x2=8.85947125

|[Expected:List?2

p

3

CNTRB +List3 save Res:None GphColor:Blue

Test

[d/c]Real

=0.11480237

df=5

CNTRB:List3

N}

p-value ~ 0.115.

Step 5:

Since p-value > 0.05 = «, we do not have enough evidence to reject Hy in favour of Hy on a 5% significance level. We therefore accept Hy.

Step 6:

Since we have accepted Hy, we conclude on a 5% significance level that the model in which every person rejects one incorrect answer and then randomly guesses from the remaining three answers for each question is appropriate.

The goodness of fit test could also be used to assess continuous distributions, such as the normal distribution, by considering the class intervals that the data i1s organised into as the categories.

EXERCISE 16D.2 1

3),

a = 0.05.

D Or more

So,

P(Y

> 5)

Number of correct answers | fobs

5

—

Step 2:

SUB

639

Letpo, p1, p2, P3, P4, and ps be the population proportions of people who correctly guess 0, 1, 2, 3, 4, and 5 or more answers respectively. The hypotheses to be considered are: H[}Z

Step4:

Exercise 16D.2

a

There are 5 categories so

df =5 — 1 =4,

and the significance level i1s o = 0.05.

So, from the table, xZ. = 9.49.

b

x2.=10.3>x2,,

sowe reject Hy on a 5% significance level.

640 2

Chapter 16 (Hypothesis testing) a

Exercise 16D.2

Let py, p2, p3, and ps be the population proportions of the lollies made by Chewy Chews, corresponding to the colours red, yellow, green, and blue respectively. Chewy Chews claims to make the same proportions of each colour, so the hypotheses that should be tested are: P2=%=

p4=%

P3=%:

Ho:

P1=%,

Hy:

at least one of p1, p2, p3, ps # 7. £.0)2

—

b

Colour

f obs

red

(fnbs

fexp

fe p)

12 | 60 x 0.20 = 16.25

~ 1.112

yellow |

17

16.25

~ 0.035

green

20

16.25

~ (0.865

blue

16

16.25

~ (0.004

Total

~ 2.016

So,

¢

f exp

X

~ 2.02

There are 4 categories, so

df =4 — 1 = 3,

and the significance level

a =0.1.

So, from the table, x2. = 6.25. d

x2,. 0.1 = «, so we again conclude that H level. 3

Step I:

=0.56921971

df=3

:List3

Save Res:None GphColor:Blue

HDIEPINDIED. >|

Test

x2=2.01538462

Expected:List?2 CNTRB

Using technology,

Degfornl) (G7c)Feal

should not be rejected at a 10% significance

Let pi, p2, p3, psa, and ps be the proportions of responses from the initial survey, corresponding to “very satistied”, “satisfied”, “neutral”, “dissatisfied”, and “‘very satisfied” respectively.

The hypotheses that should be tested are:

Step 2:

H(]:

D1

—

0.05,

Do

—

0.25,

Hy:

atleast one of p; # 0.05,

The significance level 1s

P3

—

0.41,

PDqg —

py # 0.25,

a = 0.01.

0.2,

...,

Ps

—

0.09

or ps #* 0.09.

Chapter 16 (Hypothesis testing) Step 3:

Response

f obs

fexp

Activity 2

Mendel’s data

641

(f obs ™ f exp)2 fim

very satisfied

25 | 233 x 0.05 = 11.65

~ 15.298

satisfied

78 | 233 x 0.2b = 58.25

~ 6.696

neutral

77 | 233 x 0.41 = 95.53

~ 3.094

36

~ 2.411

dissatisfied

233 X 0.2 = 46.6

very dissatisfied | 17 | 233 x 0.09 = 20.97 Total

~ (0.752 ~ 28.751

So, X3 ~ 28.8

Step 4.

df =5—1 =4,

Step 5:

X2, > X2,

so from the table, x%. = 13.28. S0 we have sufficient evidence to reject Hy in favour of H; at the 1%

significance level. We therefore accept H,.

Step 6:

Since we have accepted H;, we conclude at the 1% significance level that the ISP’s changes had a significant impact. The expected numbers of people who were either “dissatisfied” or “very dissatisfied” based on the initial survey were both higher than the results observed 1n the latter survey, so we conclude that the changes were effective.

ACTIVITY 2 1

Type of pea Yellow round seeds Green round seeds Yellow wrinkled seeds Green wrinkled seeds

2

Proportion )

Expected frequency _ 2|

556 x % — 312.75

_ 3 16

HH6 X % = 104.25

S

S

HH6 X % = 104.25

E

=B

HHb X % = 34.75

9+3+3+1 S 9+3+3+1 94+3+3+1

94+3+3+1

16

16

16

Let p1, p2, p3, and py be the population proportions of yellow round, green round, yellow wrinkled, and green wrinkled seeds respectively. The hypotheses to be tested are:

Hyo: p1 =+, H,:

D2===,

D3 =5,

Pa = 15

at least one of p; # 19—53 p2 # 1—%: p3 F %,

{Mendel’s model is true} or py # 1—16.

{Mendel’s model is false}

Chapter 16 (Hypothesis testing)

642

Yellow round seeds | 315 | 312.75

~ (0.0162

Green round seeds

108 | 104.25

~ (0.1349

Yellow wrinkled seeds | 101 | 104.25

~ 0.1013

Green wrinkled seeds | 32 |

34.75

~ 0.2176

Total

~ 0.4700

E

DeglNorn] [d/c](Real

SUB

1]

2 3 4

List 1 | List 2 | List 3

315/312. 108| 101 32|

75

g

INTR J DIST JIE=EE

DegMorml) [d7c)[Real ¥2 GOF Test ¥2=0.47002398 p =0.92542589 df=3 CNTRB:List3

DeglNorml] [d/c]Real

x¥2 GOF Test Observed:Listl Expected:List2 df e

SNNNs

So, X2, ~ 0.470,

Note:

List 4

2

104.25 104.25 34.75

GRAPH] CALC J TESTJ

3

Exercise 16E.1

AR

Save Res:None GphColor:Blue

¥

Ry

and using technology, the p-value =~ 0.925.

For common significance levels (o = 0.1, 0.05, or 0.01 the p-value 1s quite large. So, we would conclude that Mendel’s model 1s true.

for example),

we would retain H( since

Expected frequency tables can be found either by hand, or using the y? test functionality on your graphics calculator.

250

105 X 195 _ 69.96 | 165 250

250)

85X 58 _ 1979 | 8586 _ 9994 | B5X106 _ 5504 | g5 250 250 2

86

250

106

250

Chapter 16 (Hypothesis testing) ¢

The

2 x 3

Exercise 16E.1

contingency table is:

3

5

13

3+564+13 =21

36

17

1

36 +17+1

3+ 36 =39

13+1=14

G — R

b

In a sample of 100 students, we would expect 30 to be male and pass the Maths test.

c

_

24 | 30 | 24—-30=—-6| 26 | 20 |

(—6)>=36 |

=12

26—20=6

6% = 36

%zl.B

36 | 30 | 36—-30=6

6% = 36

e o0

(—6)% =36

30— 1.8

Total

6

14 | 20 | 14—-20=-6 | So, X% = 6. Hy:

weight and diabetes are independent

Hi:

weight and diabetes are dependent.

Step 2.

The significance level 1s

Step 3:

df=(2—-1)(3—1) =2

Step 4:

The

a = 0.05.

2 x 3 contingency table is:

=54

21454 =175

The expected frequency table is:

Step 1:

643

2

11

19

79

ROW-0P, ROW

¥2

3

68 IR

EB}

Using technology,

L

DeglHorm1] (d/c)Resl

p

=0.036735011

x2,. ~ 6.61.

Step 5:

From the screenshots above,

Step 6:

Since the p-value < 0.05 = o, we have enough evidence to reject Hy in favour of H, at the 5% significance level. We therefore accept H.

Step 7

We conclude that weight and diabetes are dependent.

a b

Test

Expected: Save Res:None GphColor:Blue Execute

69

JCOLUMNJN]N)

B

[Deg) Horm1]

I bO

ol

1

E

fi

2> D

Deglorni] [d/c](Real

Exercise 16E.1

Q.

Chapter 16 (Hypothesis testing)

644

df=(2—-1)(3—1) =2, Step I:

and the significance level « = 0.1, so from the table, x2. = 4.61.

Hy:

age of a voter and party they wish to vote for are independent

Hli

age of a voter and party they wish to vote for are dependent.

Steps 2 and 3: Step 40

the p-value ~ 0.0368.

The

Froma,

2 x 3

o =0.1

and

df = 2.

contingency table is:

Age of voter 18 to 35 | 36 to B9 | 60+

(] A

1[

2

Party A

89

95

131

Party B

168

197

I8

DeglMorml] [d/c]Real 1 2 g5 85

168

197

3

g

Observed:

13q

Expected:Mat save

Execute

ROW-0P| ROW JCOLUMNEZ]M

Using technology,

Mat | PMAT

xZ,. ~ 8.58.

O =4.61, xZ;

Step 5:

From

Step 6:

Since x2,. > X2,

Res:None

B

GphColor:Blue

e

a,

Deg/Normi] [d/c)(Real

s0

X 2 . > X 2 -

we have sufficient evidence to reject Hy in favour of H; at a

10% significance level. We therefore accept H;. Step 7:

Since we have accepted H;, we conclude at a 10% significance level that age of voter and party they wish to vote for are dependent.

Step I

H@Z

reason for travelling and rating are independent

H1:

reason for travelling and rating are dependent.

Step 2:

The significance level 1s

Step 3:

df=(2-1)4—-1)=3

a = 0.05.

Chapter 16 (Hypothesis testing)

Step 4:

The

Exercise 16E.1

645

2 x 4 contingency table 1s: Rating Poor | Fair | Good | Excellent

Reason for

Business

travelling | Holiday | =

[DeglNormi] ([d/c)(Real

A

1

1

2[

27

2

9

25

3

17

20

pa)

20

8

9

i

24

30

=

4

8

24

}

A

(DELETE](INSERT|[ ADD ]

DeglNorml] [d/c]Real

=]

¥2 Test Observed:Mat

¥2

A

Expected:Mat

Test ¥2=23.624288

P

B

Save Res:None GphColor:Blue Execute

DegNorml] [d/c]Real

=2.9923E-05

df=3

AN

Using technology,

b

o

xZ,. =~ 23.6.

Step 5:

From the screenshots above,

the p-value ~ 0.0000299.

Step 6:

Since the p-value < 0.05 = «, we have enough evidence to reject Hy in favour of H, at a 5% significance level. We therefore accept H;.

Step 7.

Since we have accepted H;, we conclude that, at a 5% significance level, reason for travelling and rating are dependent.

From the contingency table, it appears that guests travelling for a holiday are more likely to give a higher rating than those travelling for business. Step 1.

Hy:

position and injury type are independent

Hq:

position and injury type are dependent.

Step 2:

The significance level 1s

Step 3.

df=(3—-1)(4—1)=6

Step 4.

The

a = 0.1.

3 x 4 contingency table is: Position

Forward | Midfielder | Defender | Goalkeeper Inju

A

23

18

24

7

Mild injury — Serious injury

14

34

23

11

10

16

S

7

SeNE

JUrY type

E]

No injury

DegNorm1] [dic]Real

1

2

3

1

23

14

10

2

18

34

16

3

24

23

4

13 K|

=

DegHormi) [d7c)(Real

x2

Test

alllObserved

1{] 7

ROW-0P] ROW JCOLUMNHZDIAM]

:Mat

Expected:Mat

|Save

Res:None

=]

A

B

GphColor:Blue

x2

DegMorni) (d7c)Real

Test

X2=7.9418852 p

=0.24239194

df=6

Execute

AN

739

Using technology, x3,. ~ 7.94. Step 5.

From the screenshots above,

the p-value ~ 0.242.

Step 6:

Since p-value > 0.1 = a, we do not have enough evidence to reject Hy in favour of H, at a 10% significance level. We therefore accept Hy.

646

Chapter 16 (Hypothesis testing) Step 7:

7

a

The

Exercise 16E.1

Since we have accepted Hj, we conclude at a 10% significance level that position and injury type are independent. 4 x 2

contingency table is: Own a pet?

The expected frequency table is: Own a pet?

8x235118

402 |

54 x 118

~ 971

EXMU7 239 54 x 117

235

~ 50.9

100 x 117

LT

96.9

~ 49&

2

0.05 = a,

we do not have sufficient evidence to reject Hy in

favour of H, at a 5% significance level. We therefore accept H.

Since we have accepted Hy, we conclude that there is not a link between age and owning a pet.

8

a

The

3 x4

contingency table is:

Intelligence level

Smoking habits

The expected frequency table is:

Intelligence level

‘

Smoking

habits

763 x 502

763 x 734

763 x 218

763 X 9

1463

1463

1463

1463

~ 202

== 383

~ 114

~ 4.69

387 x 502

387 x 734

387 x 218

387 x 9

~ 133

~ 194

313 x 502

313 x 734

313 x 218

313 x 9

1463

1463

1463

1463

~ 46.6

~ 1.93

1463

=

1463

==

1463

~577 |

1463

~2.38

648

Chapter 16 (Hypothesis testing) b

Step I:

Exercise 16E.1

Hy:

smoking habits and intelligence level are independent

Hq:

smoking habits and intelligence level are not independent.

Step 2:

The significance level 1s a = 0.01.

Step 3:

df=(3—-1)(4—1)=6

Step 4: = A

1 2 3

1

279 123 100

(d/c)Real 2 386 201 147

3

96 58 s

4

2 5 .| .

E x2

Test

(dic]Real

2 X2

:

DeglHorn]] Test x2=16.9174413 p =9.5915E-03 df=6

Xpected: : |Save Res:None

GphColor:Blue Execute

Using technology and the tables from a, x2, ~ 16.9.

¢

Step 5.

From the screenshots above,

the p-value =~ 0.009 59.

Step 6:

Since the p-value < 0.01 = «, we have sufficient evidence to reject Hy in favour of Hy at a 1% significance level. We therefore accept H;.

Step 7.

Since we have accepted H;, we conclude at a 1% significance level that there is a link between smoking habits and intelligence level.

Each expected frequency in the very high intelligence level 1s less than 5, so we should combine it with the high intelligence level, which gives: The new

3 X 3 contingency table: Intelligence level

Smoking habits

96 + 2 = 98 123

201

o8 + 9o = 63

387

100

147

64 4+ 2 = 66

313

502

734

98 + 63 + 66 = 227 | 1463

The new expected frequency table: Intelligence level 763 X 227 1463

Smoking habits

132.8 |

194.2

1074 |

157.0

387 x 227 1463 313 x 227

SlaciN

1463

~~ 118.4

60.0

Yol

Chapter 16 (Hypothesis testing) d

Step I:

Exercise 16E.2

Hy:

smoking habits and intelligence level are independent

Hy:

smoking habits and intelligence level are not independent.

Step 2:

The significance level 1s

Step 3:

df=(3-1)(3—-1) =4

649

a = 0.01.

Step 4. E

A

Degl{form1] [d7c]Real

1

1I[ 2 3

279 123 100

2

3

386 201

127

=

98 63 G

Deglformi] [d7c)Real

:

Save

Res:None

=]

x2 :

Test x2=13.1846438 p =0.01040795 df=4

GphColor:Blue

o

[d7c)Real

Execute

ROW-0F[ ROW JCOLUNNINVNM Using technology and the tables from ¢, 2, Step 5:

From the screenshots above,

Step 6:

Since

= 13.2.

the p-value ~ 0.0104.

the p-value > 0.01 = a,

we do not have sufficient evidence to reject Hy in

favour of H, at a 1% significance level. We therefore accept H,.

Step 7.

a

The

Since we have accepted H, we conclude that there 1s not a significant link between smoking habits and intelligence level, which 1s different from the conclusion we arrived at in b.

2 x 2

contingency table is: Result

Guess

Heads

Tails

Sum

Heads

54

50

54 4+ 50 = 104

Tails

41

55

41 4 55 = 96

Sum | 94441 =95 | 50+ 55 =105 | 104 + 96 = 200 The expected frequency table 1s: Result Heads Heads Guess

M

200

= 494

Tails 20X

200

—sad.0

Tails | 22292 _ 456 | 2910 _ 594 200 200

Chapter 16 (Hypothesis testing)

650

b

Exercise 16E.2

We find x?,. using Yates’ continuity correction:

54 1494 50 | 54.6 41 | 45.6

4.6 —4.6 —4.6

4.1 el 4.1

16.81 16.81 16.81

~ 0.3403 ~ 0.3079 ~ (.3686

Gy [y (A

4.6

4.1

16.81

= [(r3335

Total

~ 1.3503

So, X2, ~ 1.35 Hy:

Horace’s guess and the result of the toss are independent

Hy:

Horace’s guess and the result of the toss are not independent.

Since

a = 0.05,

x2. = 3.84,

and

xZ, ~ 1.35,

then

X2, < x%,.

This means that we

do not have enough evidence to reject Hy in favour of H; at a 5% significance level. conclude that Horace’s guess and the result of the toss are independent.

d

According to the test performed in €, we conclude that Horace’s claim is not valid. The

2 x 2

contingency table is: Result

56 176

Country

29 48

56 + 29 = 85 176 + 48 = 224

56 + 176 =232 | 29448 =77 | 85+ 224 = 309 The expected frequency table is: Result

85X 232 _ oo | 85XTT o1, 309 309 224 X 232 _ 1p0 o | 224X 7T oo 309 SO

i

At a 10% significance level with df = (2—-1)(2—-1)=1,

56 | = 63.8 |

x2. =2.71.

~ —T7.8

~ 7.3

~ 53.29

~ (0.835

20 | ~21.2

~ 7.8

~ 7.3

~ 03.29

~ 2.514

176 | =~ 168.2|

=~ 17.8

~ 7.3

~ 53.29

~ (0.317

48 | ~55.8 |

~ —7.8

~ 7.3

~ 53.29

~ 0.955

Total

~ 4.621

So,

xZ,. ~4.62.

We

Chapter 16 (Hypothesis testing)

d

Hj,: Hy:

Review set 16A

651

the country where a motorbike test took place and the result are independent the country where a motorbike test took place and the result are not independent.

From b and ¢ we know that

2,

> x2.,

and so we have sufficient evidence to reject Hy

at the 10% level of significance. We therefore conclude that the country in which a motorbike test took place is not independent of the result at a 10% level of significance.

REVIEW SET 16A 1

Let p be the population mean number of minutes after 9:45 am that the bus arrives. The hypotheses that should be considered are:

Hy: Hy: 2

a

=0 >0

{the buses arrive on time} {the buses arrive late}

Thereis a =~ 7.94% chance of observing this result if the null hypothesis is true. For a 10% significance level, we reject H result.

¢ 3

As the p-value < 0.1 = «,

if there is less than a 10% chance of observing this

there 1s enough evidence to reject Hy in favour of H;.

We are asked to test a hypothesis about a single population mean, so we conduct Student’s ¢-test for a population mean. Step 1:

Let u be the population mean number of shaves. The hypotheses to be tested are:

Hy: Hq:

p =13 p# 13

{the manufacturer’s blades last 13 shaves} {the manufacturer’s blades do not last 13 shaves}

Step 2:

The significance level is

Step 3:

T =128,

py=13,

o = 0.05.

s=1.6,

n =30

Coe

The value of the test statistic is

12.8 — 13

¢t = ———

~ —0.685.

/30 Step 4.

Since

Hy:

p # 13

and

n = 30,

the p-value

=2

PO where

el Form1

5T T ~ tog

~ 2 x P(T > 0.685)

~ (0.499

p-value > 0.05 = a,

Student-t

C.D

Data

:Variable

g%per

§%5+99

Lower

TN

g

Student-t

:0.685

I

phColor:Blue

(dZc)Real

C.D

p

=0.24939144

t:Up

=1E+99

t:Low=0.685

T

Step 5:

Since

we do not have enough evidence to reject Hy in favour of

Step 6:

Since we have accepted Hy, we conclude that the population mean number of shaves 1s 13. The manufacturer’s claim is valid.

H, at a 5% significance level. We therefore accept H.

652 4

Chapter 16 (Hypothesis testing)

Review set 16A

We need to test a hypothesis about a single population mean, so we conduct Student’s ¢-test for a population mean. a

b

Let u be the population mean of Rosario’s apricots this year. The hypotheses to be tested are: Hy:

p =90

{the mean weight of the harvest is the same as last year}

Hyi:

p )| |1

Using technology, Since

t ~ —3.59

the p-value < 0.01 = «,

ave

ISP

[d/c]Real tTest :List 0.05 = «,

at the 5% significance level.

¢ =~ 0.558,

=]

P

DeglHorml] [d/c]Real

2—-Sample tTest i >U2 t =(0.55834287 p =(0.29112441 df =22 X1 =10.9

X2

and the

=10.25

y

p-value ~ 0.291.

we do not have enough evidence to reject Hy in favour of H,

We therefore conclude that the two friends catch the same number of fish on average. So Joe’s claim 1is not justified. 6

Stepl:

Step 2:

Let pu, and pug be the population mean amount of time spent shopping supermarkets A and B respectively. The hypotheses to be considered are:

Hy:

pa = pup

{customers spend the same amount of time at each supermarket}

Hi:

p, # pg

{customers spend different amounts of time at each supermarket}

The significance level 1s

a = 0.1.

in

Chapter 16 (Hypothesis testing)

Review set 16A

653

Step 3: =

=

List 1 | List 2 | List 3 | List 4

SUB

1

12

2 3

I

28 13

4

||

35 32

7

21

[Deg){Norm1)

=]

2—-Sample tTest Data ‘List

¢l

2—Sample "l FU2

P ER2

List( '

Deg)(Norm1]

'Listl :List2

t

=-1.8040534

x1

=13.9

P df

'

tTest

X2

=0.08630669 =20

=21.1666667

J

Using technology, the value of the test statistic 1s ¢ ~ —1.80. Step 4.

From the screenshots above, the

Step 5:

Since the p-value < 0.1 = o, we have enough evidence to reject Hy in favour of H, at the 10% significance level. We therefore accept H.

Step 6:

Since we have accepted H; at the 10% significance level, we therefore conclude that there is a significant difference between the mean amount of time spent at supermarket A and supermarket B.

a

I

p-value ~ 0.0863.

Let p1, p2, p3, ps, and ps be the population proportions of sizes Small, Medium, Large,

X-Large, and XX-Large respectively.

The null hypothesis 1s therefore: Hy: p1 =0.1, pa =0.2, p3 =0.35, There are 5 categories, so b

Size

i

fexp

4

70x01=7

Medium

7

70x02=14

Large

22 | 70 x 0.35 = 24.5

X-Large

24 | 70 x 0.25 =17.5 13

0x01=7

[d7]Real

SUB 1

2 3

4

eRT

List1 | List 2 | List 3 | List 4 4

7 22| 24|

7

14 24.5

¢

[d7c] Real

¥2 GOF Test Observed:Listl Expected:List2

'y CNTRB

17.5

INTR J DIST /I

Using technology, the

ps =0.1

df =5 —1 = 4.

Small

XX-Large |

pg =0.25,

! :List3

Save Res:None GphColor:Blue

[d7c)Real

x2 GOF Test x2=12.5979592 p =0.01341683

df=4 CNTRB:List3

p-value ~ 0.0134.

Using significance level a = 0.05, the p-value < o, and so we reject the null hypothesis at a 5% significance level. We therefore conclude that the proportions of shirts sold in the first week are significantly different from the initial proportions, and so the store should change the distribution of shirt sizes it stocks.

654

Chapter 16 (Hypothesis testing)

Review set 16A

super rare | 200 X 0.05 = 12.5

250 x 0.1 =25

rarc

uncommon |

250 x 0.25 = 62.5

200 x 0.6 = 150

common

b

Step I

Step 2:

Let p1, p2, p3, and pg be the population proportions of super rare, rare, uncommon,

and common items respectively. The hypotheses to be tested are: HDZ

P1

—

005,

P2

—

01,

Hy:

at least one of p; # 0.05,

The significance level 1s

P3

—

025,

P4

—

py # 0.1,

0.6

p3 # 0.25,

or py # 0.6.

a = 0.01.

Step 3: super rare

5

12.5

4.5

rare

17

29

2.956

uncommon | 76 | 62.5 common 152 | 150

2.916 ~ (0.0267

Total

~ 10.0027

So, Step 4.

xZ. ~ 10.0.

df =4—-1=3,

and

o = 0.01,

so, using the table on page 403 of the book,

X2 = 11.34. Step 5:

Since

x7;. < X4,

We do not have sufficient evidence to reject Hy in favour of

H, with a 1% level of significance. We therefore accept Hy.

Step 6:

9

Step I.

Since we have accepted Hp,

we

conclude that Emmanuel’s

justified at a 1% level of significance.

Hy:

age of a driver and opinion are independent

H1:

age of a driver and opinion are not independent.

Step 2:

The significance level iIs a = 0.1.

Step 3.

df=(2-1)(3—-1)=2

Step 40

The

2 x 3

contingency table is: Age of driver

suspicions are not

Chapter 16 (Hypothesis testing) =]

Deg|Norml] [dic)Real

A

1

1[

2

2

234

156

3

169

191

=]

x2

1311]

DegNorml] [d/c]Real

Test

Observed:Mat

=]

x2

A

Expected:Mat B Save Res:None

IRV

Test

¥2=42.05871659

=7.3689e-10

EDIT

Using technology, Step 5:

655

[Degl(Norml] [d/c]Reall

p d

GphColor:Blue Execute MY

Review set 16B

x2,. =~ 42.1.

From the screenshots above, the

p-value ~ 7.37 x 101,

Step 6:

Since p-value < 0.05 = a, we have enough evidence to reject Hy in favour of H; at a 10% significance level. We therefore accept H;.

Step 7:

Since we have accepted H;, we conclude that there 1s a significant association between the age of a driver and their opinion on the speed limit.

REVIEW SET 16B 1

Let u be the population mean minimum weight of Quickchick’s chickens. The hypotheses that should be considered are:

Hy: Hq: 2

p=1.2 p < 1.2

{the advertised weight is correct} {the chickens are less than the advertised weight}

a

Using the table on page 403 of the book, if df =6

b

x2,. ~ 571

< 12.59 = xZ.,

and

« = 0.05,

then

x2. = 12.59.

so there is insufficient evidence to reject the null hypothesis

at a 5% significance level. 3

We need to test a hypothesis about a single population mean, so we conduct Student’s t-test for a population mean. Step 1:

Let u be the population mean systolic blood pressure of the employees.

The hypotheses to be tested are:

Step 2: Step 3:

Hy:

p =140

{the employees’ average blood pressure is not too high}

Hy:

p > 140

{the employee’s average blood pressure is too high}

The significance level 1s T = 143.7 mm Hg,

a = 0.05.

p, = 140 mm Hg, .

The value of the test statistic i1s

¢t =

s = 11.2 mm Hg,

143.7 — 14

i ;71_2

n = 35 employees.

i ~ 1.95.

V35 Step 4.

Since

Hy:

p > 140

the p-value — P(T >

=P(T"

1" ~ t34

~P(T > 1.95) ~ (0.0297

Step 5:

n = 35,

B

> t)

where

and

Student-t

C.D

Data

:Variable

ggper

:%Emg

Lower

B

Student-t

:1.95

GphColor:Blue

Dediomt

C.D

b

=0.02973047

t:Up

=1E+99

t:Low=1.95

J

Since p-value < 0.05 = o, we have sufficient evidence to reject Hy in favour of Hy at a 5% significance level. We therefore accept H.

656

Chapter 16 (Hypothesis testing) Step 6:

L

Review set 16B

Since we have accepted H;, we conclude that the company’s concerns are justified at a 5% significance level.

We need to test hypotheses about a single population mean, so we conduct Student’s t-test for a population mean. Step I

Let 1 be the population mean distance that Arthur can hit a golf ball. The hypotheses to be tested are:

Step 2.

Hy:

=115

{the professional did not help Arthur’s drive distance}

Hy:

p> 115

{the professional improved Arthur’s drive distance}

The significance level 1s

a = 0.05.

Steps 3 and 4. E

Degforn) [d7c)(Real List1 | List 2 | List 3 | List 4

SUB

11

100

2

126

3

0

93

4

Using technology,

3

Since

= Begorni) (d7c)Real 1-Sample tTest 14 >11 t =1.40194855

ERINIE S

P

Vm-m_

171

D MMPHESPINDDED > |

Step 5:

& Degormi) (d7c)Real 1-Sample tTest Data tList L :>p0

¢t ~ 1.40

p-value > 0.05 = «,

Freq save

3 | Res:None

and the

X

sx n

J

=0.083577201

=122.566667

=29.56195562 =30

p-value ~ 0.0858.

we do not have sufficient evidence to reject Hy

of H, at a 5% significance level. We therefore accept Hy.

in favour

Step 6:

Since we have accepted H,, we conclude that there is insufficient evidence at the 5% significance level to claim that Arthur has improved.

Step I

Let 4y and p, be the population mean points totals of people living in Maple Grove and Berkton respectively. The hypotheses to be tested are:

Step 2:

Hy:

pqy = py

{there is no difference in average points between the two suburbs}

Hi:

pq # py

{there is a difference in average points between the two suburbs}

The significance level 1s a = 0.1.

Steps 3 and 4. El

DegNorml] [d/c](Real

SUB

List1 | List 2 | List3 | List4

1

2 3 4 R

14

11 13 13

L

0

10 12 14

NSPINBDES

Using technology,

> ||

¢ ~ 1.18

g

DeglMorml] [dic]Real

2-Sample

El

tTest

DegNorml] [dicReal

2—-Sample

tTest

Data

:List

nl

2

List(l)

:Listl

p df x1 Ix2

=0.24184195 =80 =12.275 =11.9047619

|||zl

Freq(l) Freq(2) (ISP

P ER2 :1 :1

t

Y]

=1.17912849

¥

and the p-value =~ 0.242.

Step 5:

Since p-value > 0.1 = a, we do not have enough evidence to reject Hy in favour of H, at the 10% significance level. We therefore accept Hy,.

Step 6:

Since we have accepted H, we conclude that there 1s not a significant difference between the points totals of the two suburbs.

Chapter 16 (Hypothesis testing)

6

a

Review set 16B

657

Revision course: E SUB

[Begformd (d7c)Resl List 1 | List 2 | List 3 | List 4

1

32|

4

35

2 3

=] DeglFormi (d7c]Real 1-Variable X =34.6428571

2X

39 31

=485

2. X2 cX

=16905 =2.7152254

n

—

SX

=%481772256

(Y MIPRSPINEIDEP >

Using technology, the sample mean 1s 7| ~ 34.6,

and

s; =~ 2.82.

No revision course: E

DegNormi] [d/c)Real List1 | List 2 | List 3 | List 4 1 2

E Deglforni] [d/c]Real 1 —-Variable

31

3 4

30 23

GRAPH/ CALC [ TEST/

28 INTR / DIST /HIEE

X 2x2

= =10485

n

=

OX SX

=4 ,25323406 =?fi48330235

Using technology, the sample mean 1s T ~ 32.1,

Step I: Step 2:

and

s9 ~ 4.48.

Hy:

py = po,

{the revision course had no effect}

Hy:

pq > p,

{the revision course improved examination scores}

The significance level 1s

a = 0.1.

Step 3: =

DeglMorml] [d/c]Real

2—Sample

Data nl x1 sx1 nl

=]

tTest

DeglNorml] [d/c]Real

2—-Sample

:Variable POU2 :34.6 :2.82 114

sxl1 nl X2 SX2

=

tTest

2—-Sample

:2.82 114 :32.1 14 .48 :10

nl t D df x1

n

X2

Using technology, the value of the test statistic 1s

Step 4.

From the screenshots above, the

[DeglNorm1] [d/c][Real

tTest

P2 =1.68050288 =0.05350242 =22 =34.6

=32.1

v

¢ ~ 1.68.

p-value ~ 0.0535.

Step 5:

Since the p-value < 0.1 = a, we have enough evidence to reject Hy in favour of H; at the 10% significance level. We therefore accept H;.

Step 6:

Since we have accepted H,, we conclude that there i1s sufficient evidence at the 10% significance level to say that the revision course was effective.

Step I:

Let p1, p2, p3, and p4 be the population proportions of glass, agate, alabaster, and onyx marbles respectively. The hypotheses to be tested are: 4 ‘ 0-

H:

Step 2.

P1

A+242+1

E

at least one of pl;é%,,

The significance level is

P2

pg%é,

a = 0.05.

Tk

p3

Qg2

pg;é%,,

P4

0

or

1)47&%.

Chapter 16 (Hypothesis testing)

Review set 16B (fnbs - fflfip)z

lype

Jfobs

glass

19

H0 X

2009

agate

16

H0 X

10

2

=

o

X

%

~ 0.4672 =,

o

o

J exp

~ 0.3211

1 _N050

ol

onyx

~ 2.2756

Total

=y )

W

13

o

alabaster |

f exp

S

Step 3:

So, x%,. ~ 5.22 Step 4.

There are 4 categories, so E

df =4 —1=3

DeglfNorn1] [d/c](Real

&

List 1 | List 2 | List 3 | List 4

1o/ 22.222| I

WK -

SUB

@

16)11.111 13/ 11.111 2| 5.bbbb

B

653

Y

|

DeglNorml) [(d/c)Real

¥2 GOF Test Observed:Listl Expected:List2

df

AN Ehg

save

Using technology, the

&> |

Res:None

GphColor:Blue

DeglNorml) [(d/c)Real

¥2 GOF Test ¥2=0D,215 p =0.15671398 df=3 CNTRB:List3

: 3

NS

HSINPPRE

E

J

p-value =~ 0.157.

Step 5:

Since p-value > 0.05 = a, we do not have sufficient evidence to reject Hy in favour of H; at a 5% significance level. We therefore accept Hy.

Step 6:

Since we have accepted Hy, we conclude that there i1s not enough evidence to suggest

the company’s claim is invalid at the 5% significance level.

Step I:

Hy:

P and () are independent.

Hi:

P and @ are not independent.

The significance level in a1s

Step 2:

a; = 0.05.

The significance level in B is as = 0.01. Step 3:

df = (3—1)(4—1) =6 The

Step 4.

3 x 4

contingency table is:

Qi | Q | Q3 | Qs P

|19

23|

27 | 39

Pl 11 |2 | 27| P | 26|39 (21| =]

A

1 2 3

Deglorml]

[d/c][Real

19 11 26

23 20 39

1

2

3

35 30 &

4

27 39 27 35 21 ELR 30

Using technology, Step 5:

¥2

DeglNorml] [d/c]Real

Test

Observed:

Expected:Mat B save Res:None GphColor:Blue Execute

x2,. ~ 13.0.

From the screenshots above, the

p-value ~ 0.0433.

E

¥2

DeglNorml) (dic]Real

Test ¥2=12.9825587 p =0.04331375 df=6

Chapter 16 (Hypothesis testing) a

Since the

p-value

< 0.05 = oy,

Review set 16B

we have sufficient evidence to reject Hy

659

in favour of H;

at

the 5% significance level. We therefore conclude that P and () are not independent. b 9

Since the p-value > 0.01 = as, we do not have sufficient evidence to reject Hy in favour of H, at the 1% significance level. We therefore conclude that P and () are independent.

We want to test whether two categories are independent, so we conduct a y? test for independence.

Step 1:

Hy:

Business success and education level are independent.

H1:

Business success and education level are dependent.

Step 2:

The significance level is

Step 3:

df=4—-1)(4—-1)=9

Step 4:

The

4 x 4

a = 0.01.

contingency table 1s: Education level

.

Graduate | Undereraduate

High school

. certificate

'S degree

Postgraduate

No success

35

30

41

25

Business | Low success

28

41

26

29

Success

515

24

41

Ho

High success

52

38

63

72

success

=]

A

1

2

3 4

1

2 28

35 52

30

41

24 38

3

41

26

a1 63

4

=]

26|

29 ||

¥2

Test

|Observed:Mat

|[Expected:Mat

56 || |[Save Res:None 721] |GphColor :Blue

=

A

B

x¥2

Test

x2=25.5557891 p

=2.4141e-03

df=9

-

CALC

I

Using technology, x2,. =~ 25.6. Step 5

xZ,. =~ 25.6 > xo. = 21.67,

so we have enough evidence to reject Hy in favour of H;

at a 1% significance level. We therefore accept H;. Step 6.

Since we have accepted Hy, we conclude at a 1% significance level that there is a link between education level and business success.

Chapter 17 VORONOI

1

a

DIAGRAMS

The diagram contains:

I

3cells

il

3 edges

i1

AY

\

3

=3

vertex.

°B

3

Y

A?

T~—_

T

~

_S.C

Y

b

1

(—1,2)

liesin cell B, so P is closest to site B.

il A has coordinates (—4, —1), so

PA=./(—4—(—1))2+ (=1 —2)2

= V(=3 + (37

— /18 = 3v/2 ~ 4.24 units

B has coordinates (2, 3), so

PB= /(2 —(—1))2 + (3 —2)2 = /3% + 12 = /10 =~ 3.16 units

C has coordinates

(0, —3),

so

PC = +/(0— (=1))2 4+ (=3 — 2)2

= V12 +(-5)? — /26 ~ 5.10 units So, ¢

i

PB
+ (-1 -2)*

— /13 units

= /22 + (-3)? = v/ 13 units

=QA ¢

d

V

(=3, 0)

is equally closest to sites A and D, and so lies on the edge adjacent to cells A and D.

(—3, 2) edge.

1s equally close to sites A and D, but is closest to site B, and so does not lie on an

Cell D 1s a triangle with base 8 units and height 8 units. area of cell D = £ x 8 x 8 — 32 units?

662 L

Chapter 17 (Voronoi diagrams) a

Exercise 17A

The statement “Py 1s closer to A than to any other site” 1s true, as P, lies in cell A.

b

The

statement “Pg

O

)

:

B

A

1s closer to B than to C” 1s

true, as Pg lies 1n cell B.

o

¢

The statement “A 1s closer to P

Pg” 1s not necessarily true.

than to

d

The statement “B 1s closer to Pg than to

C” 1s not necessarily true.

For example:

5

C

For example:

If the circle passes through another site, then P 1s equidistant from that site and X, and therefore 1s not within the interior of cell X. If another site lies within the circle, then P 1s closer to that site than to X, and therefore is not within

the interior of cell X. Since P lies within the interior of cell X, this circle cannot contain any other sites.

i

6

(0, 0)

lies within cell E, so the holding

}y (km)

post office 1s post office E.

2 \

ii (4, 1) lies within cell B, so the holding

A,

post office 1s post office B.

i

iv

(=2, 0)

lies within cell A, so the holding

2

post office is post office A. (3, —2)

//

\

/

/N

7

&

post office 1s post office C.

D

SN

o

lies within cell C, so the holding

T

E

BN

T—

| N

/4,1,/

m(kn:-)-

mm

Zd

>|/,/ =

IC

/v b

One building lies on the edge adjacent to cells D and E, and the other lies on the edge adjacent to cells A and E. Consider a building located at (—1, —2) which lies on the edge adjacent to cells D and E. A building 2.5 km north of (—1, —2) is located at (—1, —2+2.5), whichis (—1,0.5) or

(-1, 3).

Now

(—1, 4)

lies on the edge adjacent to cells A and E.

So the two buildings are located at

(—1, —2)

and

(-1, 3).

Chapter 17 (Voronoi diagrams) a

i

(—3,

—2)

lies in cell C, so the nearest

\ 4y (km)_

public school 1s school C.

e

\\ 5

(6, 2) lies in cell A, so the nearest public school 1s school A.

i

(—2,

i

4)

A

? |

1=

lies in cell D, so the nearest

T

B(4,3)

The midpoint of [AB] is

b

oB

G|

/

|, Y

y=3J3xr+1

y=3(-2)+1=-5 is a point on the line y = 3z + 1. lies on the Voronoi edge.

The distance from A(—2,5)

to (=2, =5)

is

and the distance from B(4, 3) to (=2, —5) is

/(-2 —(—2))? + (=5 —5)? = 10 units /(=2 —4)2 + (=5 — 3)2 = /36 + 64 — 10 units

d

|1

(0,0)

liesin cell B, so it is closest to site B.

il

(3, 6)

lies in cell B, so it is closest to site B.

il

(—4, —5)

lies in cell A, so it is closest to site A.

v

Chapter 17 (Voronoi diagrams)

Exercise 17B

665

3

—_PB(C, D)

H

a

A(4,7),

B(8,3),

C(0,-5)

The midpoint of [AB] 1s The gradient of [AB] is

So,

PB(A, B)

(#, S 8—4

%)

or

(6, 5).

—1.

4

has gradient 1 and passes through

The midpoint of [AC] is The gradient of [AC] is

(%

(6, 5).

. +2‘5) or (2, 1).

T

eS

0—4

—4

So, PB(A, C) has gradient —% and passes through The midpoint of [BC] 1s

The gradient of [BC] is

(%:

T

0—8

iy _5)

L

r (4, —1).

—8

So,

PB(B, C)

We

plot sites A, B, and C on a set of

axes. We draw and PB(B, C)

(2, 1).

has gradient —1 and passes through

PB(A, B), PB(A, C), as dashed lines, then

(4, —1).

i

make solid only the parts which form the Voronoi edges.

)

AY

A

4

2

PB(A, C)

T2

o

a

—

2 E

PB(A, B

d

810 PB(B, C)

=

666

Chapter 17 (Voronoi diagrams)

b A(-1,4),

B(52),

Exercise 17B

C(=5, —4)

The midpoint of [AB] is (_1; A The gradient of [AB] is So,

PB(A, B)

SN

5_—1

T

2) or (2,3).

6

3

has gradient 3 and passes through

The midpoint of [AC] is (‘1 The gradient of [AC] is

(2, 3).

= +2‘4) or (=3, 0).

_54 — 41 = _—i = 2

So, PB(A, C) has gradient —% and passes through

(-3, 0).

The midpoint of [BC] is (5 +2‘5, - +2‘4) or (0, —1).

has gradient —% and passes through

plot sites A, B, and

make solid only the parts which form the Voronoi edges.

-y

PB(A, B), PB(A, C), as dashed lines, then

.

axes. We draw and PB(B, C)

C on a set of

PB(A, C)

o

-

S

V=

We

(0, —1).

2

PB(B, C)

L/

So,

———=2 = —° — 2 e —10 H

"y

The gradient of [BC] is

C

5

a A(—10,9),

B(10,13),

The midpoint of [AB] is

o

C(-2, —7) (wj

9_;13)

or

(0, 11).

has gradient —5 and passes through

(0, 11).

The gradient of [AB] is —o—20 — 2 10 — —10

So,

PB(A, B)

20

The midpoint of [AC] is (‘10;_2., ’”2_7) or (—6, 1) The gradient of [AC] 1s So,

PB(A, C)

—7-9 16 —2—--10 8

has gradient 5 and passes through

The midpoint of [BC] is (

104+ —9r= 135 =7 ) 2 p.

(—6, 1). or

(4, 3).

B(B, (

Chapter 17 (Voronoi diagrams)

—7-13

The gradient of [BC] i1s

—2-10

So,

PB(B, C)

We

plot sites A, B, and

axes.

—12

PB(B, C)

C

PB(A, B),

3’

PB(A, ,

C),

Al _1&1? 9)

as dashed lines, then

“

has gradient —5 and passes through

(—6, 1).

x — 2y = 1(—6) — 2(1)

which1s

2y =x + 8

or

PB(B, C)

(0, 11).

y=—ox+11

has gradient 4 and passes through

its equation is

y:%$+4

has gradient —% and passes through

its equation is

—i0 f/

Sz +y = 5(0) + 1(11) or

PB(A, C)

”

PB(A, -

its equation is

(4, 3).

on a set of

make solid only the parts which form the Vorono1 edges.

PB(A, B)

667

5

has gradient —2 and passes through

We draw

and

—20

Exercise 17B

(4, 3).

3z + 5y = 3(4) + 5(3)

which 1s

b5y = —3x + 27

or

y = —%3; + %

The vertex is the point at which all 3 edges intersect, so we equate the equations found in b.

PB(A, B) and PB(A, C) intersect where

—5z+ 11 = sz +4 10r — 22 = —x — 8

11z =14 __ 14 :1::%,

y=—5(%)+11 ot

o

When

=T il

1 1

So, the vertex is V(%, %)

Check:

Using the equation of PB(B, C),

VA= /(-10- )"+ (0 - ) =/ (-2 B

+ (8)°

17126180 ~ 12.1 units

when

x = %,

y=—2 (%) +E

=2

+ (13- ) vB = \/(#)?*10= /(%+)

=

(2)’

1?318” ~ 12.1 units

668

Chapter 17 (Voronoi diagrams)

Exercise 17B

ve=y/(2- 1)+ (71— )] =V B+ () 17 680

So, the vertex V is equidistant from A, B, and C.

(—2,8)

il

(5,95)

lii

(2, —3)

a A(53), .

liesin cell A, so it is closest to site A. lies in cell B, so it is closest to site B.

lies in cell C, so it is closest to site C.

B4, —2),

C(—4, —6)

.

.

The midpoint of [AB] 1s

The gradient of [AB] is So,

PB(A, B)

(

5+4

34

o

>

—2

)

or

9

e —1

4 —5

has gradient —% and passes through

The midpoint of [AC] is (5+_4 2 The gradient of [AC] is

(

3+_6) or (%, -3). 2

——> 45 — =2 9 — 1,

So, PB(A, C) has gradient —1 and passes through

The midpoint of [BC] is (4+2_4, fi) The gradient of [BC] 1s So,

PB(B, C)

We

plot sites A,

et et

—4 -4

—8

(3, —3).

r (0, —4). 1 2

has gradient —2 and passes through

axes. We draw and PB(B, C)

B, and

1

(5, 5)

] —

6

i

b3

d

C

on

a set of

PB(A, B), PB(A, C), as dashed lines, then

(0, —4). bR

A

K

make solid only the parts which form the Voronoi edges.

b

1

(3,1)

liesin cell A, so they are closest to store A.

i

(—2,2)

il

(=5, —1)

Ay (kn

lies in cell A, so they are closest to store A. lies in cell C, so they are closest to store C.

ey

=

Chapter 17 (Vorono1 diagrams)

Exercise 17B

669

I Amanda must live at the vertex of the Voronoi diagram. Using the information found in a:

PB(A, B) has equation

x+5y =1(3) +5(3)

whichi1s

by=—xz+7 or

PB(A, C)

has equation

has equation and

— 3

y=—x—1

2z + y = 2(0) —4 or

PB(A, C)

—%:1: T %

z+y=3 or

PB(B, C)

y—=

PB(B, C)

y=—2x—4

intersect where

—x —1= —2x —4 r=—3

When z=-3, y=—(-3)—1=2 So, Amanda lives at (—3, 2). Check:

il

Using the equation of PB(A, B),

when

The distance from Amanda’s house at (—3, 2)

z = -3,

y=—=(-3)+ % =2

oW/

to A(5, 3) is

V(5—(-3))2+ (3-2)2 — /2112 — /65~ 8.06 km Amanda’s house 1s equidistant from all 3 stores, so she 1s about 8.06 km from each store. 7

a

The Voronoi diagram must have an edge missing because sites A and D are in the same cell.

b

The missing edge is the perpendicular bisector of A(—5, 3)

The midpoint of [AD] is

(‘5 - LT

The gradient of [AD] is

el

So,

PB(A, D)

through

—3—-95

and

D(—3, —3).

+2_3) or (—4, 0). s

2

has gradient 4+ and passes

\

by

(—4, 0).

its equation is

E

x — 3y = 1(—4) — 3(0)

whichis

A®

\

3y =-x+4+14

— 1 4 or y=1z+14

-

]

p?

3

=

¢

=

ST

\\

I\ Y

&B

C \

670 8

Chapter 17 (Voronoi diagrams) a

Exercise 17B

Sites B(3,3) and C(1, —5) are currently in the same cell, so the missing edge must be the perpendicular bisector of [BC].

The midpoint of [BC] is (3—‘2“ g +2_5 ) or (2, —1). The gradient of [BC] is

_15 _33 s _—2 — 4

Ay

So, PB(B, C) has gradient —5 and passes through (2, —1). its equation is x + 4y = 1(2) + (—1) or x-+4y dy+2=0 -+

)

B ’:H\“*?

]

&

—3 //

A

[T

-

x

T

-317T

-

"

y

b

C

Sites A(—3,2) and C(1, 4) are currently in the same cell, so the missing edge must be the perpendicular bisector of [AC].

The midpoint of [AC] is (_32“,, 2;4) or (=1, 3). :

.

The gradient of [AC] 1s So,

PB(A,

through .

C)

4 — 2

has gradient

2

1

e

L

AY

—2

and passes

—_

(—1, 3).

Ao

its equationis 2x +y =2(—1)+3 or 20+y—1=0

/

o&

3 ‘

oD &

33

E|

\ \'\

' 9

a

>

3

‘\

\

Sites A and D are 1n the same cell, as are sites C and E.

So, there are 2 missing edges.

The missing edges are the perpendicular bisector of perpendicular bisector of C(5, —4) and E(1, 0).

The midpoint of [AD] is

(_3 . Al

The gradient of [AD] is

‘12 - 43 5 ‘?6 _ 3

So, ",

PB(A, D)

_2) or (=2, 1).

has gradient = and passes through

1ts equation 1is

r—3y=1(-2) — 3(1)

which is

or

3y=x+395

y==zr+2

A(—3,

(—2, 1).

4)

and

D(—1,

—2),

and the

Chapter 17 (Voronoi diagrams)

The midpoint of [CE] is (E2 The gradient of [CE] is

its equationis

———= — i4 — 1 — —

yh A

= —y = 1(3) — (—2)

) o

4 LN

!

-

an

T4

y=2a—5b

or

671

_4;'3) or (3, —2).

So, PB(C, E) has gradient 1 and passes through

(3, —2).

Exercise 17B

I

N 4

\\

T

DI

_,:1

/

nc

Y

¢

1

i d

(2, 3) 1

il a

liesin cell A, so it is closest to site A.

liesin cell E, so it is closest to site E.

(4, —1) and E.

lies on the edge adjacent to cells C and E, so it is equally closest to sites C

(=2, 1) lies on the vertex adjacent to cells A, D, and E, so it is equally closest to sites A, D, and E.

The blue edge has gradient 3 and passes through the point

.

the equation of the blue edgeis

or

b

Sites A(—3,4)

and

(—1, 0).

3x —y =3(—1) -0 y=3r—+3

B(4, 5) are currently in the same cell, so the missing edge must be the

5—4 4 S

The gradient of [AB] is

through

B)

has gradient —7 and passes

m—

PB(A,

(%, %)

its equationis

T

Tx+y or

‘;flx

.

=7 (%) 4+ %

—

y=—7x+38 D

¢

i i

(1,3)

B

y/(km)A|

%

G

So,

or

th‘{D o

(_3;4, %)

i

The midpoint of [AB] is

A

perpendicular bisector of [AB]. ba |

10

(—4,1)

%

/

o

liesin cell B, so Julie is nearest to campsite B.

The distance from

(1, 3)

to B(4,5)

is

So, Julie 1s about 3.61 km from campsite B.

+/(4—1)2+ (5 — 3)2

x_rll (k:mz :

~—

Simon is equally closest to campsites C and D, so he 1s on the edge adjacent to cells C and D. He 1s less than 1 km south of cell A, so we construct a line which 1s 1 km south of the edge adjacent to cells A and D. Simon 1s therefore somewhere on the line segment shown alongside in red.

y (km)4| A

\

e 3.

b

The Vorono1 diagram must have a site missing as : . : the shaded cell in the diagram alongside currently has no site.

Let the missing site be X. . . .

xr-axis because site B lies on the x-axis, and the

:

edge adjacent to sites B and X 1s perpendicular

i

3

o

PB(A, X)

Y

C

-

SA

2 —

DL, fi,.r"f

-

f"’_;

to the z-axis. ¢

yh 3l

L

X must lie on the .

‘m{

=3

/

a

3

'

B

11

B

i

d

Exercise 17B

"

Chapter 17 (Voronoi diagrams)

1y WL

672

/|7 [y

> T

°C

has gradient —1, so [AX] has gradient 1.

If we draw the line (AX), X must be the point where the line cuts the z-axis (from b).

observe that X has coordinates a

The missing site X must lie in the shaded cell, as this

cell currently has no site. Now PB(A, X) has gradient %, and PB(C, X)

ya/

AT

/

has

//‘ 3

S

gradient —2. |AX] has gradient —3, and [CX] has gradient % If we draw lines (AX) and (CX) through A and D

respectively, their intersection point must be site X. We observe that X has coordinates (—1, —4).

L ~T

_fl,__.—-'-*

&

[ 3

T \

The missing site X must lie in the shaded cell, as this cell currently has no site.

Now PB(C, X) has gradient 1.

has gradient —1, and

PB(D, X)

0

=

X"} 1\

vy b

? _

o\

AY

.

-D

3

oA

w=

|CX] has gradient 1, and [DX] has gradient —1.

If we draw lines (CX) and (DX) through C and D respectively, their intersection point must be site X.

We observe that X has coordinates

(0, —1).

=Y

12

(—3, 0).

Chapter 17 (Vorono1 diagrams)

Exercise 17B

673

Ch

13

Y

The distance from

(0, —1)

to (—=2,1)

The distance from

(0, —1)

to B(3, —1)

is

+/(=2—0)2+ (1 —(-1))2 = v/8 ~ 2.83 km. is

/(3 —0)2+ (-1 —(—1))2 =3 km.

So, if the missing vet clinic was at (—2, 1), But (0, —1) is in cell B. the missing vet clinic is not at (—2, 1).

then its cell would include

PB(A,

D)

D)

horizontal.

has gradient —1,

and

PB(C,

is

(0, —1).

y (km)4d

[AD] has gradient 1, and [CD] 1s vertical.

A

.

- 3

If we draw lines (AD) and (CD) through A and D respectively, their intersection must be site D.

We observe that the missing vet clinic D is at (—3, 1).

D | A

2

j

=t

= )

3

(

o L (km)

. B

SEEEL Y

I il

(—1,4)

lies in cell A, so it is closest to vet clinic A.

(1, —5) lies on the edge adjacent to cells B and C, so it is equally closest to vet clinics B and C.

Larissa lives equally closest to vet clinics A and D, so Larissa’s house must lie on the edge adjacent to cells A and D.

The shortest possible distance to either vet clinic is the point at which meet. From the construction lines drawn in ¢, we observe that this point is

The distance from

(—1,3)

to A(1,5)

is

+/(1—(=1))2+(5—3)?

PB(A, D)

and

[AD]

(—1, 3).

= /8~ 2.83 km.

So, the shortest distance Larissa’s house could be from vet clinics A and D 1s

2v/2 ~ 2.83 km.

674

1

Chapter 17 (Voronoi diagrams)

Exercise 17C

a

Site D lies in the existing cell C.

b

Cells A and C will be affected, as they both contain points which are closest to D.

¢

We construct PB(C, D) within cell C, which creates a new vertex at (—3, 1). Cell A is adjacent to (—3, 1), so we construct PB(A, D) from (—3, 1) through cell A.

PB(A,D)

}¥ 5

\ -

D

-3

@

$A = D

>

|

.C =.fi,‘\\| B

13 [

[= >

PB(C.D) | | ¥\ We then remove the edge segment from the original Voronoi diagram which now lies within cell D, giving us the Voronoi diagram which Includes site D.

AY

) "}_A

\ -

D

@

—3

&b

-

L~

{

oC _3\\|

B

3

>

E

A\ 2

a

Adding site D will affect all cells as cells A, B, and C each contain points which are closer to

site D than they are to sites A, B, and C respectively.

b

We construct PB(A, D) within cell creates new vertices at (—1, 0) and Cell C is adjacent to (—1, 0), so we PB(C, D) from (—1, 0) through cell This creates a new vertex at (1, —6). Cell B is adjacent to (1, —6), so we

A, which (7, 0). construct C.

Ay ©

-

construct

PB(B, D) from (1, —6) through cell B. This connects us back to

(7, 0).

A, PB(A.D) T3 D—— 3" — PB(C. o\ — e D 2

O

\\

:

PB(B D}B

Y

We then original cell D, includes

remove the segments of edges from the Voronoi diagram which now lie within giving us the Voronoi diagram which site D.

AY © =

—3

A

P

I 3

\

x

‘

c’

\ \J

\

’B

Chapter 17 (Voronoi diagrams)

Exercise 17C

675

From the diagram in b, cell D is a triangle with base 8 units and height 6 units.

area of cell D = % X 8 X 6 = 24 units?. We construct PB(A, E), PB(B, E), PB(C, E), and PB(D, E) within the original cells A, B, C, and D respectively.

AY A, PB(A.E)

|

B PB(B.E)

-

—4

o

i

D | E

-

1 | @

PB(D, E)

PR/ ) l"D'q\La. E.-}'

Al

We then original cell E, includes

remove the segments of edges from the Voronoi diagram which now lie within giving us the Voronoi diagram which site E.

A,

-

—4

y

| 3

o

D | E

C respectively.

.E

A

Lfidjfi -

—4

l

C

(B

o

Al

We construct PB(A, E), PB(B, E), and PB(C, E) within the original cells A, B, and

AY

—o

/

y

=

] | @ e

C

py [[FBBE

B

*'

~PB(C,E) \

C

o—P

4

D

\ | \y

We then original cell E, includes

remove the segments of edges from the Voronoi diagram which now lie within giving us the Voronoi diagram which site E.

Note:

We do not need to construct PB(D, E) as there are no points in cell D which are now closest to site E.

A,

R

12 !

D

pYy

E

I/l [

\ cl

Y

\l \ oy

B

1T

>

|z

676 5

Chapter 17 (Voronoi diagrams) a

A(-3,1),

B(1,3),

Exercise 17C

C(1,-1),

D(0, —2)

We draw PB(A, B), PB(A, C), and PB(B, C),

then remove segments which do not form part of the Voronoi diagram.

AY

31 o83 S

[

-

Y

Given this Voronoi diagram, we now construct

PB(A, D)

and

PB(C, D)

within the original

ol

cells A and C respectively. Then we remove the segment of the edge from the original Voronoi diagram which now lies within cell D, giving us the Voronoi diagram for A, B,

.

C, and D.

AY

b

D¢

B

o

-

Y A(_S:

3)3

B(_3!

_3)5

C(la

1):«

D(Sa

_1)

We draw PB(A, B), PB(A, C), and PB(B, C), then remove segments which do not form part of the Voronoi diagram.

yv

/3

5 ¢

WL

E

Y

Given this Voronoi diagram, we now construct

PB(B, D)

and

PB(C, D)

within the original

cells B and C respectively. Then we remove the segment of the edge from the original Voronoi diagram which now lies within cell D, giving us the Voronoi diagram for A, B, C, and D.

yy

/. b

o

>

Chapter 17 (Voronoi diagrams)

I

i

(—2, —1)

(5, 2) I

Exercise 17C

677

lies in cell A, so it is closest to ATM A.

lies in cell B, so it is closest to ATM B.

We construct

PB(A, E),

PB(B, E),

PB(C, E),

\ 4 y(krn)'

and PB(D, E) within the original cells A, B, C, and D respectively.

Ay

3

PB(B, E)

PB(A,E)

-3

—

B

\

—

,

.

N

B(D,E)\

s

PB(C.E)

\™

;D

T (km)

C

\v

We then remove the segments of edges which now lie within cell E, giving us the Voronoi diagram which includes site E.

\ 4 y (km) \ °B ]

A

-

3/

~——

4

)i

\

3

\

o

],

T (km)

E

_3

ID

[

C

\v

il

Yes, there are residents whose nearest ATM has changed from D to E, as part of the original cell D now lies in cell E.

il

If Morris is equally closest to ATMs B, C, and E, then he is at the vertex adjacent to cells B, C, and E.

We observe from the Voronoi diagram in i that this is the point Morris is therefore located at

(5, 1).

I

(=3, 2)

lies in cell A, so polling booth A is the closest.

il

(1, —4)

lies in cell D, so polling booth D is the closest.

i

We construct PB(A, (0, 2)), PB(B, (0, 2)), and PB(D, (0, 2)) within the original cells A, B, and D respectively.

(5, 1).

A y (km)

A, PRIA-(0-9))

VNS4
(K

T

R D'

|

.C

\

We see from the Voronoi diagram in | that cell C will be unaffected, as it does not contain any points which are closer to site (0, 2) than to site C.

From the diagram in i, the new cell is a triangle with base 5 units and height 5 units. area of new cell = = x 5 x 5 =12.5 km?.

1 | Location | Temperature (°C) 28.4 25.6 27.3

¢

(5, —2) 1s closest to A, so we estimate a temperature of 28.4°C at 3 pm at (5, —2).

~—

c

\ =44

(-3, —1) 1is closest to B, so we estimate a temperature of 25.6°C at 3 pm at (—3, —1).

\

|=

b

Ce

YA

o

(1, 0) 1s closest to C, so we estimate a temperature of 27.3°C at 3 pm at (1, 0).

T

a

/‘F/

A B C

Chapter 17 (Voronoi diagrams)

Exercise 17D

679

Location | Elevation (m)

a

A(—4, 4)

57

B(—2, 0)

48

C(2, 4)

55

D(2, —6)

36

We draw PB(A, B), PB(A, C), PB(B, C), PB(B, D), and PB(C, D), then remove segments which do not form part of the Voronoi diagram.

A 1 (km) o

O

.

e)

|

C

P

-

rj

x (km)

>

e

P

D Y

b

1

(0,1)

is closestto B, so we estimate an elevation of 48 m at (0, 1).

il

(—4, 2) is closest to A, so we estimate an elevation of 57 m at (—4, 2).

il

(3, —4)

is closest to D, so we estimate an elevation of 36 m at (3, —4).

Location | Snowfall (inches)

12.2

D

9.3

a

|1

(—1, —4) is closest to D, so we estimate that 9.3 inches of snow were received at (—1, —4).

i

(2, 1) s closest to B, so we estimate that 5.5 inches of snow were received at (2, 1).

i

(1, —4)

is equally closest to C and D, so we

estimate that

12.2 4 9.3

— 10.75

were received at (1, —4).

inches of snow

Ay (km)

&

C

e

7 5.5

T WL/

A B

680

Chapter 17 (Voronoi diagrams) b

i

Exercise 17E

We construct PB(C, E) and PB(D, within the original cells C and D respectively.

E)

\

\ 5

Ao

Ay Ekm) B

—

-«

]

D

—3

3 —

o

z (km)

7/

PB(D, E)

PB(C, o

E We then remove the edge segment from the

E) CI

fy

\

original Voronoi diagram which now lies within cell E, giving us the Voronoi diagram which includes location E.

B

Ay fkm)

B

A \

3

—

-

3

—

—3

.

D

E ii

Both (—1, —4) location E.

and

(1, —4)

7

—

/

_

O

e ‘;

a

vy

from a i and a iii respectively are now closest to

The Voronoi diagram has vertex V(1, 1).

Y

V 1s equidistant from A, B, and C.

A has coordinates

(—2, 3),

so

AL

VA = /(-2 —-1)2 4+ (3 -1)2

'\\

B

3 \

‘

:'

\

A

= /(=3)? + 22 =

v/13 units

ol

So the largest empty circle has centre and radius /13

units.

x (km)

C.

So, we would now estimate 10.6 inches of snow at both locations.

1

o

V(1, 1)

_:

'_

3 Tk ’C

Chapter 17 (Voronoi diagrams)

b

The Voronoi diagram has vertices

Vi(1, 2)

and

Exercise 17E

631

V,(0, —1).

V1 1s equidistant from A, B, and C, and V5 1s equidistant from A, C, and D.

A has coordinates

(—2, 3).

Ay

VIA=/(-2-1)2+(3-2)?

At

= /(—3)2 + 12 —

Vi

/10 units

-

.-": —3

HEEP

= /(~2) + 42

BN

D-"'n.,__r___,,..-'

3

|

T

B

/20 = 2v/5 units

So, the largest empty circle has centre a

n“‘*..C

{JT\r.rfi

‘I'I

VoA =/(=2-0)* + (3 - (-1))? —

B

The Voronoi diagram has vertices

V3(0, —1)

V1(20, 30),

and radius 2+/5 units.

V2(20, 10),

and

V3(8, 2).

V; 1s equidistant from A, B, and E, V5 is equidistant from B, C, and E, and V3 is equidistant from C, D, and E.

E has coordinates

(—8, 26).

y (km) 4

ViE = /(-8 — 20)2 + (26 — 30)2 B

—

2

B

2

r;"_-

= \/(=28)2 + (—4) = v/800 = 20v/2 km

E(-8,26)

et

,

*a

o

. B(48,26) ‘~

VoE = /(=8 — 20)2 + (26 — 10)2 » 1 (km)

= /(—28)2 + 162

=

/1040 = 4/65 km

V3E = /(=8 — 8)2 + (26 — 2)2 = /(—16)2 + 242

=

/832 = 8/13 km

So, the largest empty circle has centre

V3(20, 10),

and radius 4v/65 km.

the rubbish dump should be established at (20, 10). ®

From a, the largest empty circle has radius 44/65 km. the dump is 44/65 =~ 32.2 km

¢

V3(20, 10)

from the nearest town.

is adjacent to cells B, C, and E.

So towns B, C, and E are closest to the rubbish dump.

d

To show that the location found in a is preferable, we must show that there is at least one town

which is less than ~ 32.2 km from The distance from

(25, 15)

(25, 15).

to B(48, 26)

is

/(48 — 25)2 + (26 — 15)2

— /232 4+ 112 =

So, B is closer to

(25, 15)

V650 =~ 25.5 km

than the answer found in a is to any other town.

682 3

4

Chapter 17 (Voronoi diagrams)

Exercise 17E

Let P lie in cell X. The largest empty circle centred at P would touch X. As P lies within cell X, the largest empty circle does not touch any other site, meaning that a larger empty circle could be created with centre closer to the edge of the cell. the circle centred at P 1s not the largest empty circle. the largest empty circle cannot lie within a cell.

a

i A(-15,10),

D(—3, —18)

The gradient of [AD] is

s

—3 — —15

or (—9, —4).

w)

(‘15 puia

The midpoint of [AD] is

12

3

So, PB(A, D) has gradient 2 and passes through . its equation is

3z — Ty = 3(—9) — 7(—4)

which1s

7y =3z —1

yzfix—% -

or

i C(15,0), D(—3, —18) The midpoint of [CD] is

(15 .

The gradient of [CD] is So,

PB(C, D)

3-15

T8

—18

(6, —9).

x +y =6+ (—9) or

PB(A, D)

arge.

0 +2‘18) or (6, —9).

has gradient —1 and passes through

its equationis

b

(-9, —4).

and PB(C, D)

y=—x—3

intersect where

2z 7 — =7 = —x — 3 3r—1=—-Tx — 21 10x = —20 r=—2

When z=-2, y=—(-2)—-3=-1 So, V5 has coordinates (—2, —1). ¢

From b, the Voronoi diagram has vertices

Vi(3, 14)

and

Vy(—2, —1).

V; 1s adjacent to cells A, B, and C, and V5 1s adjacent to vertices A, C, and D.

V1C = /(15— 3)2 + (0 — 14)2

Ay (km)

= /122 + (—14)2

A= B(7:32)

— v 340 km

VoC = /(15 — (—2)) + (0 — (—1))2 _ —

So,

the

17

2

—I_

largest g

V1(3, 14),

1

2

empty PY

A(=15,10) -

circle

has

centre

and radius /340 km.

the optimal position for a toxic waste dump is

Vg(_2:

Vi3, 14) -

~

_1)75

D(—3,a9 1) —18)

(3, 14).

;

—©

______ HC(]_E),

U)

T

PB(C.D)

-

(km)

Chapter 17 (Voronoi diagrams)

5 A(-5, —10),

B(11, 18),

The gradient of [AB] is PB(A, B)

—

—5+11

—10+

oo

a1

11 — —o

16

or (3, 4).

18)

4

has gradient —% and passes through

its equation is

633

C(5, —12)

a The midpoint of [AB] is (

So,

Exercise 17E

(3, 4).

4z + 7y = 4(3) + 7(4)

which is 7y = —4z + 40 or

y=—sx+ 3

The midpoint of [AC] is

(_5; e

The gradient of [AC]is

—=——10 _—2__1 b ——5 10 D

So,

PB(A, C)

(0, —11).

bx —y = 5(0) — (—11) or

The midpoint of [BC] is The gradient of [BC] is

PB(B, C)

r (0, —11).

has gradient 5 and passes through

its equation is

So,

; _12)

y=ox—11

(” = S

5b—11

B +2‘12)

e T

r (8, 3)

D.

—6

has gradient —+ and passes through

its equation is

(8, 3).

x + by = 1(8) + 5(3)

which is or

5y = —z + 23 y= —%:1: - %

We plot sites A, B, and C on a set of

AY

axes. We draw PB(A, B), PB(A, C), and PB(B, C), then remove segments which do not form part of the Voronoi diagram.

~

\a\

\x‘

B

10 \\'-fll

\\""'-._