Mathematics Analysis and Approaches for IB Diploma Program SL 2 Worked Solutions 9781925489835

Citation preview

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HAESE

MATHEMATICS

Analysis and Approaches SL

S ~ =i v

L 3

e

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for use with

IB Diploma Programme Bradley Steventon Charlotie Frost

Joseph Small Michael Mampusti

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MATHEMATICS: ANALYSIS AND APPROACHES Bradley Steventon Charlotte Frost Joseph Small Michael Mampusti

SL WORKED

SOLUTIONS

B.Ma.Sc. B.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: Web:

[email protected] www.haesemathematics.com

National Library of Australia Card Number & ISBN

978-1-925489-83-5

© 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 Deanne Gallasch. 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 1n 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 it 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) in each chapter of our textbook Mathematics: Analysis and Approaches 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. We have a list of errata for our books on our website. Please contact us 1f you notice any errors in this book.

BS

e-mail: web:

[email protected] www.haesemathematics.com

CF

JS

MM

TABLE OF CONTENTS

Chapter 1

THE BINOMIAL THEOREM

Chapter 2

QUADRATIC FUNCTIONS

25

Chapter 3

FUNCTIONS

98

Chapter 4

TRANSFORMATIONS

Chapter 5

EXPONENTIAL FUNCTIONS

187

Chapter 6

LOGARITHMS

226

Chapter 7

THE UNIT CIRCLE AND RADIAN MEASURE

274

Chapter 8

TRIGONOMETRIC

FUNCTIONS

313

Chapter 9

TRIGONOMETRIC

EQUATIONS AND IDENTITIES

356

Chapter 10

REASONING AND PROOF

401

Chapter 11

INTRODUCTION

417

Chapter 12

RULES OF DIFFERENTIATION

449

Chapter 13

PROPERTIES OF CURVES

520

Chapter 14

APPLICATIONS OF DIFFERENTIATION

617

Chapter 15

INTRODUCTION TO INTEGRATION

655

Chapter 16

TECHNIQUES

673

Chapter 17

DEFINITE INTEGRALS

709

Chapter 18

KINEMATICS

770

Chapter 19

BIVARIATE STATISTICS

306

Chapter 20

DISCRETE RANDOM

348

Chapter 21

THE NORMAL DISTRIBUTION

OF FUNCTIONS

TO DIFFERENTIAL CALCULUS

FOR INTEGRATION

VARIABLES

152

381

Chapter 1

THE BINOMIAL THEOREM EXERGSE1A 1

a

21=2x1

b

31=3x2x1

¢

4'=4x3x2x1

d

Hl=5x4x3x2x1

e

6l=0x5x4x3x2x1

f

I00=10x9x8x7Txb6xHx4x3x2x1

— 720 2

= 3628 800

a

4x3x2x1=4l

b

"TXx6Xx5Hx4x3x2x1="T

.

6>1

n+2)!

nl

(n+2)x(n+1)xnal

al

=n+2)(n+1),

n=>0

(n+1)!:(n+1)> — 27x° + 92 — 1

f (2z+5)° = (22)% + 3(22)%(5) + 3(22)(5)* + (5)* — 823 + 60x* + 150z + 125 g

(2a—1b)® =(2a)° + 3(2a)*(—b) + 3(2a)(—b)* + (=b)> — 8a® — 12a%b + 6ab* — b°

h

3z — 1) = (32)® + 3(32)%(—%) + 3(3z)(—2) + (—1)°

=272° —92° + 7 — & (233 + 1)3 = (2z)° + 3(2z)° (l) + 3(2x) (1)2 + (1)3 I

I

I

£

:8$3+12&3+2—|—$—13 (Ve —1)° = (Vz)’ + 3(Vx)*(-1) + 3(vVz)(—-1)* + (-1)° =z

k

— 3z +3v/xr -1

(22 +2)° = (22)® 4+ 3(2?)?(2) + 3(z2)(2)? + (2)° + 122° + 8 =% + 62*

| (3;2 3 i)3 — (22)? + 3(22)? (—mig) 4 3(x2)(—i)2 4 (—i)g 72 G

2

Exercise 1B

9

3

1

(a+b)* = a* + 4a®b + 6a*b* + 4ab® + b*

a (1+x2)*=1*4+41)°z+6(1)%z* +4(1)z> + 2* — 1 +4x + 62> + 423 + z*

b (p—q)* =p* +4p°(—q) + 6p*(—q)* + 4p(—q)° + (—q)* = p* —4p°q + 6p°q® — 4pg” + ¢° ¢ (z—2)*=z*+42°(-2) + 62%(—2)% + 4z(-2)° + (-2)* — 2% — 823+ 2472°% — 322 + 16

8

Chapter 1 (The binomial theorem)

d

Exercise 1B

(3—2)"=(3)*+4(3)°(—2z) + 6(3)*(—x)* + 4(3)(—2)° + (—2)* — 81 — 108« + 54x% — 1223 + z*

e (1+22)*=(1)*+4(1)°2z)+6(1)°(2x)* +4(1)(2z)* + (2z)* = 1+ 8z + 24z* + 3223 + 16z*

f (22 —3)* = (22)* +4(22)*(=3) + 6(22)*(—3)* + 4(22)(-3)* + (=3)* — 16x* — 12 x 823 + 54 x 42° — 108 x 2x + 81 = 16x2* — 9622 + 21622 — 216z + 81

g

(2z+b)* = (22)* +4(22)%b + 6(22)%b* + 4(22)b® + b? = 16z* + 322°b + 242°b* + 8xb® + b* 4

h (:Hl) i

:m4+4$3(1)+6x2(1) €I

W

2

3

+4a;(1) +(1) H

4

HI

:x4+4x2+6+ifi+i4

i (23; - i)d — (22)* + 4(22)> (—i) +6(22)? (—i)z +4(22) (—i)g 4 (—if = 16m4—32m2+24—$%+m_{1 3

a

i

(a—0b)°=a’+3a°(-b)+3a(-b)*+ (-b)* = a’ — 3a®b + 3ab® — b°

ii

(a—b)* =a+4a°(—b) + 6a*(—b)* + 4a(—b)° + (—b)* = a* — 4a°b + 6a°b* — 4ab® + b*

b

The terms are the same, except for their signs. The signs in the expansions of

(a+b)*

are all positive, whereas the signs in the expansions of (a —b)®

with a positive term and then alternate 4

a

1 I

b

4 5

6 10

4 10

5

1 0,

b > 0).

(a + b)®

and (a — b)* start

the 4th row the 5th row

(a+b)°=a®+ 5a*b + 10a>b? + 10a?b3 + 5ab* + b°

¢

i

(z+2)°=2"4+52*(2)+102°(2)* +102%(2)° + 52(2)* + (2)° = 2° + 10z* 4 40z° + 802° + 80x + 32

(1 —2)” = (1) +5(1)*(~2) +10(1)*(—2)* +10(1)*(~z)* + 5(1)(~z)* + (-=)" =1— 5z + 10z — 10x® + 52* — 2°

(14 22)° = (1)° + 5(1)*(2x) + 10(1)%(2z)? + 10(1)*(2z)% + 5(1)(2z)* + (2z)° = 1+ 10z + 402° + 80x° + 80z* + 32x°

iv

(z—2y)° =2° + 52 (—2y) + 102°(—2y)? + 102*(—2y)> + 5x(—2y)* + (—2y)° = z° — 102ty + 4023y? — 80z2y> + 80xy* — 32¢°

v (22 +1)° = (2%)° +5(%)*(1) + 10(=?)%(1)? + 10(=*)?(1)° + 5(=?)(1)* + (1)® = 210 + 52% + 102° + 10z* + 5% + 1

and

Chapter 1 (The binomial theorem)

Exercise 1B

o (o—5) =< Haat () () 2imet0) el ) BT 5

2

£

£

3

£

1

£L

£

5 £

=5~;5—5:1::3—|—10;1:—E—l—i—i

a

1 1

b

b 6

10 15

10 20

5 15

1 + 15a%b* + 6ab® + b°

¢

i (z+2)°=2%+62°(2)+152%(2)* + 202°(2)% + 152%(2)* + 62(2)° + (2)° = 2% + 122° + 60z* + 16022 + 2402* + 192z + 64

i

(22 —1)° = (22)° +6(22)°(—1) + 15(22)*(—1)* + 20(22)*(—1)° + 15(2x)*(—1)* +6(2z)(—1)° + (-1)° — 642° — 6 x 322° + 15 x 162* — 20 x 8> + 15 x 42° — 6 x 2z + 1

= 642°% — 1922° + 240z* — 16022 + 602* — 12z + 1

il

(3: 4 i)fi 2

— 25 + 62° (l) + 1522 (1) £

i

1+ 2023 (l) €T

3

+ 1522 (l) £

4

+ 6z (1) €T

5

+ (l)

6

€I

:m6+6m4+15$2+20+1—2+%+% £

£

£L

a (1++v2)=1)°%+3(1)%2(v2)+3(1)(V2)? + (vV2)° =14+3vV2+3x2+2V2 x

=1+3V2+6+2V2

=7+5V2 b

¢

(V5+2)* = (V5)*+4(v5)3(2) +6(v5)%(2)? + 4(V5)(2)® + (2)* =25+ 8 x 5v/5 +24 x 5+ 325 + 16 = 25 + 405+ 120 + 32v/5+ 16 — 161 + 72V5 (2—v2)°

= (2)° +5(2)*(=v2) + 10(2)°(=Vv2)? + 10(2)*(—v2)* + 5(2)' (=v2)* + (-v2)° — 32 — 80vV/2+ 160 — 80v/2 + 40 — 4V/2 — 232 — 164V/2 7

a

(24+12)°=(2)°+6(2)°x+ 15(2)*z? + 20(2)°2® + 15(2)%2* + 6(2)2° + «° = 64 + 192z + 2402% 4 160x> + 60z* + 122° + 2°

9

10

Chapter 1 (The binomial theorem) b

(2.01)°

Activity

64

is obtained by letting = = 0.01

1.92

(2.01)°

0.024 0.000 16 0.0000006

= 64+ 192 x (0.01) + 240 x (0.01)* 4 160 x (0.01)° +60 x (0.01)* 412 x (0.01)° + (0.01)° = 65.944 160601 201

0.000 000001 2

+

0.000000 000001 60.944 160 601 201

8

(2x + 3)(z 4 1)

a

= (22 + 3)(z* + 42> + 62° + 4z + 1)

= 22° + 8x* + 122° + 82 4+ 2z + 32 + 122° + 182° + 122 + 3 = 22° + 11z* + 2423 4 2622 + 14z + 3 b

(z — 1)(2z + 1)3 = (z — 1) [(22)® + 3(22)*(1) 4+ 3(2z) (1)

+ (1)°]

= (z — 1)(8z° + 122° + 6z + 1)

= 8z* + 122% + 6% + x — 823 — 1222 — 6z — 1 —8x* +423 9

a

—62°

— 52—1

(3a + b)° = (3a)® + 5(3a)*b + 10(3a)3b* + .... the coefficient of a>b?

b

is

10 x 32 = 270

(2a + 3b)6 = (20)% + 6(2a)%(3b) + 15(2a)*(3b) + 20(2a)3(3b)3 + the coefficient of a3b°

is

20 x 23 x 33 = 4320

[ {apl ] 1

2

Diagonal 1: Diagonal 2: Diagonal 3: Diagonal 4: Diagonal5: Diagonal6:

1 1+ 1 =2 241=3 14+3+1=5 34+4+1=28 1+6+5+1=13

\‘B\ 1

4

6

15

1 1

20

15

6

The sequence of numbers formed by the answer to 1 1s the Fibonacci sequence.

terms in the nth shallow diagonal

(n > 3) is the sum of the terms in the (n—2)th

The sum of the

and

(n—1)th

shallow diagonals. For example, consider the term 6 in diagonal 6. It 1s obtained by adding together the 3 in diagonal 4 and the 3 in diagonal 5. We could repeat this process for the other terms in diagonal 6.

Chapter 1 (The binomial theorem)

PART 1: 1

Investigation 2

The binomial coefficient

11

COUNTING

a

| il il

b

There are 8 options for who can be listed third. There are 7 options for who can be listed fourth. There are 6 options for who can be listed fifth.

The total number of orders in which the members can be listed is:

options for first x options for second X .... X options for tenth — J0X

9O XxE8XTXbXdHx4dXxIx2x]1

= 10!

2

a

The total number of ways in which the first four members can be listed is: options for first x options for second x options for third x options for fourth = 10X 9x&XT

b

10 X9 x8XT7T=

I0 X9

X8XTXxb6Xbx4x3x2x1 bxXxdox4dx3x2x1

_10!

o6l "

The 6 people not in the team can be ordered in 6 X 5 x4 X 3 x 2 x 1 = 6! ways.

d 3

Since the order of the 6 people left out of the team 1s not important, we divided the total number of orders in which the members can be listed by 6!

Of the 4 members who are in the team: options for first x options for second x options for third x options for fourth =4 x3x2x1 — 4! There are

L&

4! = 24

ways in which the 4 members who are in the team can be ordered.

The total number of ways in which the first four members can be listed 1s

10X 9X8XT7T="26!

f{from 2}

Since the order of the four members in the team 1s not important, we divide by the number of

possible orderings 4!

{from 3}

the number of ways in which the team of four can be chosen is 5

10!

P . X

0!

The total number of ways in which » members of a team of n can be listed 1s n!

(n—7r)!

{similar to &}

Since the order of the »r members in the team is not important, we divide by the number of possible

orderings !

{similar to 4}

the number of ways i which the team of 7 can be be chosen 1s

—

X

(

n!

T

{n—r).

12

Chapter 1 (The binomial theorem)

PART 2: 1

2

Exercise 1C

THE BINOMIAL COEFFICIENT

Multiplying by (a + b) will double the amount of terms in the expression each time. there would be 2" terms.

a

If we choose b r times, then we choose a

(n —r)

times,

b

The order of each lot of r is not important, so from 5 there are

n!

(n —7r)lr!

r lots of b from n brackets.

¢

:

) 2

n!

After collecting the “like” terms, there will be

L)

)

G

(0)

o)

()

Q)

G

(o) G) G

() G G

(1)

() G

(2)

() @

ways to choose

terms of the form

(n —r)! x r!

11

(5)

)

()

T

(¢)

)

(2)

The rth number in the nth row of Pascal’s triangle 1s

1 (:’)

where

1

1 6

1

5

1 2 4 15

3

10

6 20

n,r €N,

1

3

4

10

15

a”~"b".

1

5

n>1,

1 6

1

r 2}

16(n —1) =15n 16n — 16 = 15n

n =16

11

Review set 1B

and so

k:-%:_%

(m—-2n)""=m'" + (110) m”(—2n) + (120) m°(=2n)? + .... + (—2n)' =m'" — 20m”n + 45m°(4n®) — .... + 1024n"° =m'’ — 20m’n + 180m°n* — .... + 1024n'°

k = 180

23

24

Chapter 1 (The binomial theorem)

Review set 1B

12 (24k) b T = ()@ () 8

which has constant term

(333 + i)4 3

has

)

(

r

( (

Tyy1 = (

)

( (

q*(70¢" — 6) = 70q° — 6 = 2

{qg =0 6

T =70

_ 335

q==1/5

gives a trivial solution}

Cha

pter 2

QUADRATIC FUNCTIONS a

y = 2z° — 4z + 10

is a relationship between two variables z and y which is in the form y = ax® + bx + ¢ where a, b, ¢ are constants, a # 0.

is a quadratic function.

y = 222 —4x + 10

y = 8¢ + 3

cannot be written in the form

y = 8x + 3

y = az? +bx +¢,

a # 0.

1s not a quadratic function.

y = —2x* 2 1s a relationship between two variables = and y which 1s in the form y = ax?® + bz + ¢ where a, b, c are constants, a # 0. 2

y = —2x“

18 a quadratic function.

y ==z +6—2?

canbe writtenas

y==2x+6—2* the form

y = —z* + sz +6.

is a relationship between two variables = and y which can be written in

y = ax? + bz + ¢

y ==z +6— 2

where a, b, ¢ are constants,

a # 0.

isa quadratic function.

2y +x —3 =0 cannot be written in the form vy = ax® +bx +¢, 2y +x — 3 =0 1s not a quadratic function.

y—2z% =3x —1

can be written as

a # 0.

y = 2z% + 3z — 1.

y — 22 = 3z — 1 1is a relationship between two variables = and y which can be written in the form y = axz? + bz + ¢ where a, b, ¢ are constants, a # 0.

y —2x? = 3x — 1 When

is a quadratic function.

= =1,

b

y=1°+3(1) -7

When

=

-2,

y = —2(—2)* +5(-2) + 2

=143-—-7

= —8—10+42

= -3

= —16

When

x = 3,

d

y=3(3)*—-2(3) -5

When

=

—1,

y=—3(-1)* +7(-1) — 2

=27T—6-—9

=—-3-7-2

— 16

= —12 b

y:$2—3$+1

N

21

-1120

11

D

y=2z°—z+3

y:$2+2$—5

1

2

x|

-2

—-11]

0

1

2

1 | -1

-1

yl|

—o|

6|

-0

—-2]

3

d

y=-3z°+2zx+4

26 5

Chapter 2 (Quadratic functions) a

When

Exercise 2A b

=0,

When

y=2(02+5

z =2,

y=(2)°—-3(2)+2

=9

=4—-6+42

(0, 4)

=0

does not satisfy the function

(2, 0)

y = 22° + 5.

satisfies the function

y =x° — 3z + 2. ¢

When

z = —1,

d

When

y=—(-1)+2(-1) -5

y=—-2(3-34+6

—1-2-5 = —8

=—18—3+6

(—1, —8)

(3, —15) satisfies the function y = —21% — x + 6.

= —15

satisfies the function

y = —x° + 2x — 5. e

When

f

z =2,

y = 3(2)% —4(2) + 10

When

=—-248-1 =

(2, 10) does not satisfy the function y = 3z% — 4z + 10.

a

If y=4

then

(2, 5)

b

If y=23

¢

x4z

s

e+ D (x+2)=0

If y=—4

S

+4=0

(=22 =0

£=—1or-2

e

d

then

If y=4

z* —6x+1=—4 ot

then

2 — 4z +7=3

ST+ 3r+2=0 .

satisfies the function

y=—sz°+4z — 1.

2 +3x+6=4

S

z =2,

y=—3(2)°+4(2)- 1

=12-84+10 =14

5

z =3,

(x=1)(z—-5)=0

then

2t +5xr+1=4

s

—6x+5=0

22° + 52 —3=0

2

—1D)(x+3)=0

r=1o0rd e

If y=1

If y=2

$:%DI‘—3

then

%:{:QJr%:I:—Q:l

—%$2+2$—1:2

%$2+%m—3:0

—éa:2+2:r:—3:0

. 2+ 5x—6=0 S

"

f

then

0

(x+6)(x—1)=0 r=—06orl

ozt

—4dr+6=0

which has A = (—4)% — 4(1)(6) = —-8+3 hasvertex (1, 3). The axis of symmetry is x = 1. When

=0,

b

y=(-1*+3=4

When

a > 0 so the shape 1s \/ YA

y=2(xr+2)?+1 hasvertex (-2, 1). The axis of symmetry is z = —2. z=0,

y=2(2*+1=9

a > 0 so the shape 1s \/

y=(x—1)*+3

y=2(x+2)°+1

Ay

=Y

¢

y=—2(x—1)>—3 hasvertex (1, —3). The axis of symmetry 1s = = 1.

When z =0,

y=-2(-1)"-3

d

y=2(z—3)%>+2

has vertex

(3,2).

The axis of symmetry is = = 3.

When

=0,

y=1(-3)%+2 13 Ay 2

a > 0 so the shape 1s \/ Y

y=i(z—3)*+2

sY

3

Chapter 2 (Quadratic functions)

y= —%(:1:— 1)2 +4

has vertex

The axis of symmetry 1s

=0,

(1, 4).

=z = 1.

y=—3(-1)°"+14 =3

f y=—15(x+2)*—3

has vertex

(-2, =3). The axis of symmetry i1s = = —2. When

wolbo

When

=z =0,

y= —%(2)2 — 3

= —32

a < 0 so the shape is /\

a < 0 so the shape is /\

iy;

2| 32] ;V(1,4)

AY

£ w1/

5

32

=

B

5

Y

y=—35(x+2)? -3

y=—(r+1)°+3

has vertex (—1, 3). The axis of symmetry 1s = = —1. When =0, y=—(1)*+3 = 2

y=—2(x — 3)* +2 has vertex (3, 2). The axis of symmetry 1s = = 3.

When

z =0,

y=—2(—3)*+2 = —16

a < 0 so the shape is /\

a < 0 so the shape is /\

The only graph with all of these properties is G.

The only graph with all of these properties is A.

34Y -

/\

Y |

y = x* + 2

has vertex

(0, 2).

The axis of symmetry 1s z = 0.

When

a > 0

2z =0,

y=0°+2=2

/

y=—(x—1)*+1

3

-

\

:I:

has vertex

The only graph with all of these properties is E.

The axis of symmetry 1s = = 1.

When

=0,

y=—(-1)*+1=0

The only graph with all of these properties is B.

=Y

8Y

Ay

P—

(1, 1).

a < 0 so the shape is /\

so the shape 1s \/

A

e

Exercise 2B.1

33

Chapter 2 (Quadratic functions)

e

Exercise 2B.1

y=(r—2)?—2 hasvertex (2, —2). The axis of symmetry is = = 2. When =0, y=(-2)°—2 = a >0 so the shapei1s

f

y=3(z+3)*—3 has vertex (—3, —3). The axis of symmetry 1s = = —3. When z =0, y:%(3)2—3

\/

— 0

k=0

50 Hic Shape 15 \/

is 1.

The only graph with all of these properties is €.

A

HT\

—

7

"?..».3

The only graph with all of these properties

=Y

34

Y

g

y=—x*

has vertex

(0, 0).

h

The axis of symmetry is z = 0. When z=0, y=0

e /\

a

-

y=—35(z—1)*+1

hasvertex

The axis of symmetry is = = 1. When z =0, y:—%(—l)erl _1 a < 0

so the shape is /\

The only graph with all of these properties s F. .

2_._

—-

/{}/-\

-t

1

x

=

v

Y

i

AY 34

hasvertex (-2, —1). y=2(x+2)? -1 The axis of symmetry is x = —2. y=2(2)°-1 When z=0, =7

a >0

Sotheshapeis\/

The only graph with all of these properties is H.

(1, 1).

E

=\_}7:

" v

Chapter 2 (Quadratic functions)

Exercise 2B.2

EXERCISE 2B.2 a

b

y=1x°—2x+3

y=x° +4x — 2 y=2x°+4x+2° -2 —2°

y=x°—2x+ (—=1)* +3 — (—1)° y=(x—1)°"+2 The vertex is (1, 2), y-intercept 1s 3.

y=(x+2)°—6

and the

The vertex is (—2, —6), y-intercept 1s —2.

Ay -

\y=m2+4m—2

4

and the A P

/

=

y=x°>—2z+3

V(1,2) -

s

V(—2, —6)

-

x

Y

C

y =

d

% — 4z

y =2 —4r+(-2)° — (-2)7

y=(z—2)° -4

The vertex is (2, —4),

and the

y-1ntercept 1s 0.

y =2° + 3

y=x"4+3z+(3)° —(2)°

y=(z+35)" The vertex is

(—%, —%),

and the

y-intercept 1s 0. YA

=Y

y=z?+3x

e

y = x° + br — 2

y=a’+5x+(2)°—2—(2)°

y=(z+3)" - % The vertex is (—2, —=22), y-intercept 1s —2.

and the

The vertex is (3, —+),

and the

y-intercept 1s 2. YA

1\@)’:5{32"—517_2

-

‘W/E

=Y

1

V(=3,—%)

v

35

36

Chapter 2 (Quadratic functions)

Exercise 2B.2

y=1a°—6x+5

h

y=2a°+8c —2 .2 2 2 y=x"+8xr+4"—-2-4

y=2x°—6x+ (=3)°+5— (—3)° y=(x—3)°—4 The vertex is (3, —4),

y = (z+4)% — 18

and the

The vertex is (—4, —18),

y-1tercept 18 5.

y-ntercept 1s —2.

AY y=1°>—6x+5

y=x°+8x — 2

D =

AY

P

=

T

V(3. —4)

V(—4, —18)

y =2 — bz + 1 y=z"—bx+

> T

—2

=

vy i

and the

Y

AY (—2)°+1—(-2)° 21

y=(@-3)0°-% .

y:$2—5;t:-|—

-

1/

1

-

2 The vertex is (3, —%t), and the

y-intercept 1s 1.

a

I

y = 2z° +4x +5

iv

y=2z"+4z+5

o

=2[2° + 22+ 1° + 2 — 17] _=2[(z 9 +1)"2 + 3]3

V(_1.3)

y=2(x+1)*+3 The vertex is (—1, 3). The y-intercept 1s 5.

i i b

=

i

y =2x° — 8z + 3 =

Al

— 4z + (—2)* + 2 — (—2)7]

-

=2[(z - 2)* - 3]

y=2(z—2)° -5 The vertex is

i C

(2, —5).

The y-intercept 1s 3.

i i

y = 2z° — 6 + 1

= 2[z* — 3z + 3]

=20z?

— 3z + (-3 + 3 - (-3

The vertex is (3, —%). The y-intercept 1s 1.

T

Ay

v

2[z* — 4z + 2]

=2z

t—>

Y

oo

2

» _%Tl)

Iv

Y= 22 — 8x + 3 P

T

Chapter 2 (Quadratic functions)

i

y =3z

Exercise 2B.3

—6x+5

= 3[z” — 2z + 2]

= 3a? =22+ (—1)* + § - (-1)? =3[(z = 1) + 5]

y=3(x—1)*+2 il

The vertex is (1, 2).

ili

The y-intercept 1s 5. i

y=—x°+4x

42

= —[z° — 4o — 2] = —[z® — 4z + (—2)° — 2 — (—2)°]

= —[(z - 2)* - 6] y=—(x—2)°+6

il

The vertex is (2, 6).

i

The y-intercept 1s 2. i

y = —2z% — 5x + 3

= 2[(x+3)* -& - 2 =2z +2)° - & y=—2(z+2)°+2 il

The vertex is (—2, %).

i

The y-intercept is 3.

EXERCISE 2B.3 |

y=x°—4x+2 has a=1, b=—4,

b _—(-9) _,

2a

i ¢=2

2

a >0,

= = 2.

b=2,

¢= -3

When

y=2"-4(2)+2= -2

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

=z = —1,

y=(-1)*+2(-1) -3 =—4

(2, —2).

so the shape 1s \/

2(1)

The axis of symmetry 1s = = —1.

z = 2,

the vertex is

a=1,

e

The axis of symmetry 1s

il

has

2(1)

When

y=2°+2z—3

the vertex is (—1, —4). il

a >0,

so the shape 1s \/

the vertex (—1, —4) is a minimum turning point.

3

38

Chapter 2 (Quadratic functions) ¢

i

Exercise 2B.3

y=2z*+4 has

a=2,

—b_

=0

20

2(2)

d b=0,

i

c=4

y=-32°+1 has

a=-3,

20

2(-3)

Sy —

_ 0

The axis of symmetry is x = 0. When =0, y=4

e

i

y=222+8x—-7 has a =2, b=8,

il

is a minimum f

i

¢c=—7

(0, 1).

so the shape is /\

the vertex (0, 1) turning point.

is a maximum

y=—2°—42-9 has a=—-1, b= -4,

¢=—9

__b:_(_4):_2

2(2)

2a

2(—1)

The axis of symmetry is = = —2.

The axis of symmetry 1s = = —2.

When

When z = —2,

. il

a 0, so the shape is \/ the vertex (0, 4) turning point.

c=1

The axis of symmetry i1s = = 0. When =0, y=1

the vertex is (0, 4). i

b=0,

z = -2,

y=2(-2)"+8(-2) -7

y=—(—2)% —4(~2) - 9

= —15

= —4+8-9

the vertex is (—2, —15).

a >0,

— _§

so the shape is \/

the vertex

(—2, —15)

is a

il

minimum turning point.

the vertex is

a 0,

.

(2, —32).

so the shape is

the vertex

(2, —%)

\/

isa

minimum turning point.

Chapter 2 (Quadratic functions)

y:—%m2+m—5

i

i

y:%mz—fiflrfi has

-6

-1

20

a:i,

2a

The axis of symmetry 1s

= = 1.

The axis of symmetry is

z =1,

When B

(1, —3)

il

a> 0, so the shape is \_/

the vertex

isa

b 2a

has

a=1,

b=-8,

c=7.

—(=8) _ 2(1) =4 =4,

y=4°—8(4)+7

the vertex is

= —9 (4, —9).

The y-intercept 1s 7. When

y =0,

S

Since iv

The axis of symmetry 1s = = 4.

ii When

(14, —43)

is a

minimum turning point.

maximum turning point. y=2*°—-8x+7

= = 14,

the vertex is (14, —43).

(1, —%).

a < 0, so the shape is /\

the vertex

= = 14.

y=+(14)° = 7(14) + 6 = —43

b2

the vertex is

a

¢c=06

2(7)1

y=—=2(1)"+1-5

2

b=-T7,

__b:_(_T):M

2(-3)

When

Exercise 2B.3

r° —8xr+7=0

(e—=1D)(x—-T7)=0 r=1

the z-intercepts are 1 and 7.

or

7

a >0, Yy

theshapeis\/

39

40

Chapter 2 (Quadratic functions) b

y=—-2°—6z—8

has

Exercise 2B.3

a=-1,

b=

-6,

c=—8.

Since a < 0, the shape is /\ 2—5 =

2((_5’)) = -3

\)

The axis of symmetry 1s i

When

x* = —3.

= —3,

y=—(-3)" —6(~3) - 8 =1

the vertex is (—3, 1). il

The y-intercept 1s —8. When

y =0,

—x° —6x—8=0 —(z? +6x+8) =0

—(z4+2)(x+4)=0 .

x=-2

or

—4

the z-intercepts are —2 and —4.

¢

y=6x—2*>

b _

has

6

2

2(—1)

a=-1,

b=6,

¢c=0.

— 3

Since

a < 0, the shape is /\

iv

Ay

1V (3, 9)

The axis of symmetry i1s = = 3.

i When

z=3,

y=6r —x

y=6(3)—3° =9

:

the vertex is (3, 9). il

-

The y-intercept is 0.

When

y =0,

) |

6x—2°=0 (6 —z)=0 o x=0

z=3 or

the z-intercepts are 0 and 6.

6

\

>

Chapter 2 (Quadratic functions)

y=—22+3x—-2

sy

has

regl

2a

a=-1,

3

2(=1)

2

Since

a < 0,

the shape is

/\

When z =2

the vertex is i

¢=—2. iv

The axis of symmetry 1s = =

ii

b=3,

41

b|

d

Exercise 2B.3

(2, 7).

The y-intercept i1s —2. When

y =0,

—z4+3x—-2=0

s

;.

—(z* -3z +2)=0 —(z—=1)(x—-2)=0 r=1

or

2

the x-intercepts are 1 and 2.

e y=222+4z-24 P o 2a

has a=2, b=4, c=-24

Since a >0, the shapeis \_/

2(2)

y=2a?+ 4 — 24

The axis of symmetry 1s = = —1. i

When

=z = —1,

y=2(—1)% +4(-1) — 24 = —26

the vertex is il

> i

(—1, —26).

The y-intercept 1s —24. When

y =0,

2r% +4r —24 =0

+2x —12=0 L2+ 2x =12 2 + 2z + 12 =12 + 12 (x+1)* =13 ;1:—|—1=i\/fi

r=—14++13

the x-intercepts are

—1 4 +/13.

=il

V(—1, —26):

—1++/13

42

Chapter 2 (Quadratic functions) f

y=-32°+4x—1

Since |

a < 0,

b

—

2a

=

has

Exercise 2B.3 a=-3,

b=4,

¢c=—1.

the shape i1s /\

-4

2

= —

iv

3

2(—3

AY

V(2,1 33

ii

When

x

Lol

The axis of symmetry is z = 2 z = 2,

o

_ 2 _— — y=—3z"+4r—1

5i

!

i

y=-3(3)"+4(3) -1

=3

2

'

MR =) pAm |=—3+35—1

ks 3

the vertex is (%, %). il

The y-intercept 1s —1. When

y =0,

—3z° +4r—1=0

—(3z° —4z+1) =0

—Bz—1)(x—1)=0 .

x==x 3

or

1

the z-intercepts are % and 1. g

y=2z*—-5x+2

has

a=2,

b=-5

. —b_ —(=5) _5 Do

ii When

z =2

Since iv

=

S,

c=2.

a >0, AY

E

y=22"— 5z +2

y=2(3)>-5(3)+2 2


—8c—5

has a=4, b=-8, c=—5.

e el

iv

20

2(4)

The axis of symmetry 1s

i

When

=1,

Since a> 0, the shapeis \_/

= = 1.

y=4(1)>—-8(1)—-5

= —9

the vertex is (1, —9). iii

The y-intercept i1s —5.

When

y =0, 4z° —8x—5=0 o 2z +1)(2x—-5) =0

LT = —%

or

%

the x-intercepts are —% and % y=—%m2—|—2$—3 Since

a < 0, —b

az—%,

b=2,

c=-3.

the shape is /\ —2

20

has

2(—7)

=4

iv

The axis of symmetry 1s

i When z=4,

= = 4.

fdl Wb

i

y=-1(4)>+2(4)-3 =t

the vertex is (4, 1). il

The y-intercept is —3. When

y = 0,

—%$2—|—23:—3=0 coxt—8x+12=0

S

(x—=2)(x—6)=0 r=2

the x-intercepts are 2 and 6.

or

6

Exercise 2B.3

43

44

1

Chapter 2 (Quadratic functions)

a

Exercise 2C

b

y=z2*+z-2 has

a=1,

b=1,

c=

-2

y=2%—-4z+1 has

A = b* — dac

a=1,

= (—4)* —4(1)(1)

=9

— 12

Since A > 0, the graph cuts the x-axis twice. Since a > 0, the graph is concave up.

Since A > 0, the graph cuts the x-axis twice. Since a > 0, the graph 1s concave up.

y=—z%2-3

y=a°+Te—2

AL 7

has

a=-1,

b=0,

c=

-3

AL \V

has

A = b* — dac

a=1,

-+

~

a=1,

b=8,

¢=16

A = b — 4dac

=Y

Since A > 0, the graph cuts the z-axis twice. Since a > 0, the graph 1s concave up.

AL \ has

a=-2,

Since A = 0, the graph touches the x-axis. Since a > 0, the graph 1s concave up.

b=3,

c=1

A = b — dac

— 32— a(-2)(1)

— 8% — 4(1)(16) = 0

A

—2

y=—22°+ 3z + 1

y=x°+8c+16 has

c=

= 72— 4(1)(~2)

= —12

Since A < 0, the graph does not cut the xr-axis. Since a < 0, the graph is concave down. The graph 1s negative definite.

b=7,

A = b? — 4ac

= 0% — 4(~1)(-3)

e

c=1

A = b? — 4ac

= 1% — 4(1)(-2)

¢

b=—-4,

= 17

Since

A > 0,

the graph cuts the z-axis

twice. Since a < 0, the graph 1s concave down.

7\

-I/\Fm

Chapter 2 (Quadratic functions) g

h y=—2*+2+6

y=6x*+5xr—4 has

a=6,

Exercise 2C

b=5,

¢=—4

has

a=—-1,

b=1,

¢=6

A = b° — dac

A = b* — 4ac

= 1% —4(=1)(6)

= 5% — 4(6)(—4) =121

— 25

Since A > 0, the graph cuts the x-axis twice. Since a > 0, the graph i1s concave up.

Since twice.

A > 0,

the graph cuts the x-axis

Since a < 0, the graph is concave down.

AL \

y = 922 + 6x + 1 c=1

b=6,

a=9,

has

A = b* — dac

= 6= — 4(9)(1) =0

Since A = 0, the graph touches the r-axIs. Since a > 0, the graph is concave up. 2

y=22*—-5x+1

has

a

Since

a > 0,

b

A=0%—4ac

a=2,

b=-5,

c=1

the graph is concave up.

= (=5)" —4(2)(1) = 17

A > 0,

the graph cuts the x-axis twice.

When

y=0,

222 —-5x+1=0

L

—bEVA -

T =

2a

—(—5) + V17 2(2) 4

r~ 228

-

4

or

3

v

(.22

the z-intercepts are

d

The y-intercept is 1.

a

y:—:r2—|—43:—7

has

~ 2.28

and

a=-1,

~ 0.22.

b=4,

¢c= -7

A = b* — 4dac

=42 — 4(-1)(-T7) = —12

Since

A < 0,

\~0.22

{:J\/

the graph does not cut the x-axis.

[~2.28

y=2z°-5x+1

=Y

¢

Since

45

46

Chapter 2 (Quadratic functions) b

Exercise 2C

The graph 1s negative definite, since a < 0 and A < 0. This means that it lies entirely below the z-axis.

2

When

2(—1)

z=2,

> /\

—d

y=-2°+4(2)-7

8 £

V(2,-3)

= —3

the vertex is (2, —3).

Py

The y-intercept is —7.

4

a

22 —4x+7

has

y=—x24+4x— T

a=2,

b=-4,

c=T

A = b® — dac

= (—4)* — 4(2)(7) — —40

Since Since

A < 0, the graph does not cut the z-axis. a > 0, the graph i1s concave up.

The graph 1s positive definite, which means that 1t lies entirely above the x-axis. b

—22°4+3x—4

has

a=-2,

b=3,

\/ > -«

c=—-4

A = b% — 4ac

— 82— 4(=2)(—4) = —23

Since

A < 0, the graph does not cut the x-axis.

Since

a < 0,

the graph is concave down.

The graph is negative definite, which means that 1t lies entirely below the x-axis. ¢c

z°—3x+6

has

a=1,

b=-3,

-«

/\

c=6

A = b® — 4dac

= (=3)* — 4(1)(6) = —15 Since Since

A < 0, the graph does not cut the z-axis. a > 0, the graph 1s concave up.

The graph 1s positive definite, which means that it lies entirely above the z-axis. r* —3x+6

>0

forall z.

\/

L

Chapter 2 (Quadratic functions) d

4r—22—-6

has

a=—-1,

b=4,

Exercise 2C

47

¢c= —6

A = b* — 4dac

= 42— 4(-1)(-6) = —8

Since Since

A < 0, the graph does not cut the x-axis. a < 0, the graph is concave down. ——

The graph is negative definite, which means that 1t lies entirely below the z-axis. dr — x> —6

5

Consider

/\

< 0 for all z.

y = ax? + bz + c.

Consider

y = dz* + ex + f.

The graph 1s concave up, so a > 0. The y-intercept 1s positive, so ¢ > 0. The axis of symmetry is to the right of the

The graph 1s concave down, so d < 0. The y-intercept 1s 0, so f = 0. The axis of symmetry 1s to the right of the

y-axl1s, SO

y-axi1s, SO

=

2a

. The A

> ()

graph

b0}

2d

>0

soe>0

not cut the x-axis,

so

{d < 0}

The graph cuts the z-axis twice, so

< 0.

Ao

> 0.

Constant | a | b | c | d | e | f | A1 | Ay Sign 6

ab

T

aw=111

e

b =3,

e

| ()

==

[ ==

"ei="k

A = b% — dac

= (3)* —4(1)(k) — 9 — 4k

I

The graph cuts the r-axis twice if A > 0. 9—4k

>0

4k

0 4k < 16

k4

48

Chapter 2 (Quadratic functions) c

Exercise 2D

a=k+1, b=-2k, c=k—-14 A = b? — 4ac — (—=2k)% — 4(k + 1)(k — 4)

Also, if & = —1 then the function is the line y = 2x — 5, 1n which case it cuts the x-axis only once at

x = %

= 4k — 4(k* — 3k — 4)

— 4k% — 4k* + 12k+ 16 = 12k + 16 I

The graph cuts the x-axis twice if A > 0.

12k +16 > 0 12k > —16

3z°+kr—1 Since

has

a > 0,

a=3,

The graph touches the z-axis if A = 0.

o

k>—%,

7

Il

s

o

k=-%

12E+16 0 for all real values of k. the graph cuts the z-axis twice for all k.

3x? 4+ kx — 1 is never positive definite for any value of k.

8 y=32°+(k—2)z+k*+4

has

a=3,

b=k—2,

c=k*+14

A = b* — dac

= (k—2)* —4(3)(k* +4) =k*—4k+4—2k* -8 —k* — 4k — 4 —(k* + 4k + 4) = —(k+2)°

a >0,

and

A 0.

y=4

L d=ua(-1)(—2) C.oa=2 The quadraticis

y =2(z — 1)(z — 2).

-

& Y

=Y

So,

Chapter 2 (Quadratic functions)

The graph touches the z-axis at z =2, The graph is concave up, so a > 0. When

=0,

so

y = a(x — 2)?. 12

12 = a(—2)? -

P

The quadratic is y = 3(z — 2)2.

v

Since the x-intercepts are 1 and 3, y = a(x — 1)(z — 3). The graph 1s concave up, so a > 0.

AY

=0,

49

AY

y=12

Loa=3

When

Exercise 2D

y=3

5

3

3=a(—-1)(-3) Loa=1 The quadraticis y = (z — 1)(xz — 3).

-

b v

Since the z-intercepts are —1 and 3, y = a(x + 1)(z — 3). The graph 1s concave down, so When

z=0,

e

AY

a < 0.

9

y=3

3 =a(l)(—3)

;ooa=-—1

f&

The quadratic is y = —(x + 1)(x — 3). The graph touches the z-axis at z =1, The graph 1s concave down, so When

o '

z=0,

&

Y

so y = a(x — 1)2.

a < 0.

i

-

AY -

y= -3

P

:

>

—3

—3=a(-1)" Cooa=—3

The quadratic is y = —3(z — 1)~

Y

Since the z-intercepts are —2 and 3, y = a(x + 2)(x — 3). The graph 1s concave down, so When

=0,

a < 0.

y=12


(0

3 \

v

The axis of symmetry =z = 3 lies midway between the x-intercepts. the other z-intercept 1s 4. the quadratic has the form

y = a(z —2)(z —4)

4

v

50

Chapter 2 (Quadratic functions) b

Exercise 2D

The axis of symmetry x = —1 the other z-intercept 1s 2.

lies midway between the x-intercepts.

the quadratic has the form

y=a(x+4)(x—2) But when

=0,

where

a 0. When

2 =-6,

AY (—6,9)

y=29

co9=a(—6+4)*+3

s 9=a(-2)%+3 C.

. .

9=4a+3

6 =4a H’ZE 3

The quadratic is y = 3(z +4)% + 3.

V(—4,3) Y

53

54

Chapter 2 (Quadratic functions)

Investigation 3

i Since the vertex is (3, —3), the form

When

the quadratic has

y = a(z — 5)* — 3, where

z =32,

Finding quadratics

AY

a > 0.

(2.1)

y=1


—8

¢

The graphs do not meet if A dc+8 de c

A =0

de +8 =0

>0 c>

The graphs touch if

—2

0 for any value of m the linear function

14

y=(r—2)°

and

y = max + 3

will always meet the curve

y = —2%+ bx + ¢ touch when

(3—2)°=—(3)*+b(3)+c

y = 2z —z — 2

twice.

z =3

S, 1=-943b+c 3b+c =10 c=10—-3b ... (1)

Now, consider

(z — 2)? = —z% + bz + ¢

cort—drx+4=—-2*+bzx+c

—22°+(b+4)z+c—4=0

s

This quadratic has

A = 0

(b+4)? —4(-2)(c—4)=0

b +8b+ 16+ 8 — 32 =0

b* +8b — 16 +8(10 —3b) =0 . b +8b—16+80—24b =0 b* — 16b + 64 = 0 (b—8)* =0 ‘. b=28 Substituting into (1) gives b=8,

c=—14

since the graphs touch.

{using (1)}

¢ = 10 — 3(8)

= —14

y=—x’+bx +c

62

Chapter 2 (Quadratic functions)

1

Exercise 2F

Let the smaller of the integers be x. The other integer is .

the sum of their squares is

(z + 12).

z° + (z + 12)* = 74

2 + 22 + 24z + 144 = 74 2% + 24z + 70 = 0 . 2° 4+ 120435 =0 (x+7)(x+5)=0 .

So, the integers are

2

7 and —5,

or

—7 and 5.

Let the number be z, so its reciprocal 1s They have sum

«x +

L

b

x=—70r

1

—. L

26 5

m2+1:%$

:32—2—56:fl+1:0 5z

— 26

+5 =0

5z — 1)(x — 5) =0 =

Rk

5

orH

So, the number 1s either 5 or %

3

Let the number be x.

The sum of the number and its square 1s 210. x4 x? =210

L4+

.

x—210=0

(x+15)(x—14) =0 r=—15or

14

However, the number 1s a natural number. the number is 14.

&

Suppose the numbers are z and Then x(z + 2) = 360

(x + 2).

s, 22 +2r—360=0

.

(x+20)(x—18) =0 r = —20 or 18 the numbers are

5

18 and 20,

Suppose the numbers are x and

szt

Then

or

—20 and —18&.

(z + 2).

z(x + 2) =255

+ 22 —255=0

(x —15)(x+17) the numbers are

=0 r=15or

—17

15 and 17,

or

—17 and —15.

—9

Chapter 2 (Quadratic functions)

6

If the polygon has 90 diagonals, then

g(n —3) =90

in® —3n =90 n®—3n—180=0

s .

Exercise 2F

(n—=15)(n+12) =0 S.oon=—12o0r

15

We reject the negative solution, as a polygon must have a positive number of sides.

the polygon has 15 sides.

If the width of the rectangle is w cm, then its length is (w + 4) cm.

. the areais

w(w +4) = 26

Cw’ + 4w —26=0

whichhas

b=4,

¢= —26

—4 + /42 — 4(1)(—26)

v W

a=1,

2(1)

=

—4 4+ /120 2

w=—24+30 But

w >0,

so

w=—2++/30 ~ 3.477

So, the width is approximately 3.48 cm.

a

The base has sides of length = cm, so the areas of the top and

bottom surfaces are both z? c¢cm?. The box has height

(x + 1) cm,

side faces is z(x + 1) cm?. . the total surface area is .

by

so the area of each of the

A = 2z° + 4z(z + 1)

= 22° + 42° + 4

A =6x*+4r

6x° + 4z = 240 322 + 22 — 120 = 0 (3z +20)(x — 6) =0 .

but

x >0,

so

20 113:—?01’6

x =6

the box 1s 6 cm by 6 cm by 7 cm.

cm?

63

64

9

Chapter 2 (Quadratic functions)

Suppose the tinplate was When

3 cm X 3 cm

Exercise 2F

x cm X x cm.

A

squares are cut from the corners, the

base of the open box formedis (x —6) cm x (z —6) cm. The open box has height 3 cm, so its volume is

3x (x—6)x (x—6)=2380

:

T cm

3(z? — 12z + 36) = 80

which has

L

a =3,

E

|

. 322 — 362 + 108 = 80 s 3z? —36x+28 =0 .

T

T

v e

b= —-36,

i

!

X

3 om

I Nl —>

4+—

¢ = 28

E

—(=36)+ \/(—36)2 — 4(3)(28)

B

2(3)

.

ifi‘ gakd

and since

x > 6,

;;:'::lESJr—'i{)fiU ~ 11.16

the original piece of tinplate was about 11.2 ¢cm square. 10

Suppose one side of the rectangle has length has length y cm.

The perimeter is (2x 4+ 2y) cm,

so

cm and the other

2z + 2y = 20

JEm

2y = 20 — 2z Loy=10—2x

T cm

The area A = z(10 — z) cm?. If the area is 30 cm?, then

(10 — z) = 30 s

10z — 2% = 30

z° — 10z + 30 =0 Now

A = (—10)% —4(1)(30) = 100 — 120 — —20

which is < 0

There are no real solutions, indicating that this situation i1s impossible. 11

The smaller rectangle is similar to the original rectangle.

ABAD _ BYBC

Suppose

AB = z units,

and

1

r(r—1)=1 ot

1

z-—1

—r—1=0

which has

a=1,

b= -1,

:

Y

B

X

C

AD = BC = 1 unit

T _

A

¢=

—1

5

Chapter 2 (Quadratic functions)

_ ()2 : 2(1) -

(D

1+/1+4

z =

1

:

+2\/3

AB

ot

{since N

x > 0}

s

.

which 1s the golden ratio w1

the golden ratio 1s

12

+2\/5.

Let the speed of the normal train be x kmh~1. the speed of the express train is distance

Time =

(z + 10) kmh—1,

g so the normal train takes

%

hours, and the express train takes

e

to travel 160 km. The express train takes half an hour less time than the normal train to travel 160 km.

160

160

1

T

x + 10

2

160(z + 10) — 160z = sz(z + 10)

160z + 1600 — 160z = s2° + 5z 2 + 5r — 1600 = 0 . x~b5l.8 But

z >0,

so

or

—61.8

{using technology}

=z~ 51.8

the speed of the express train is x + 10 ~ 51.8 + 10 ~ 61.8 kmh™1. 13

65

2 1445 el B

= But

Exercise 2F

Let the number of elderly citizens in the original group be . The amount paid by each elderly citizen was originally

i:

dollars.

Now, 8 elderly citizens fell 1ll, so the amount paid by each elderly citizen 1s These elderly citizens had to pay $1 more than before. 160 r—38

160 - z

+1

160z = 160(x — 8) 4+ x(x — 8)

160z = 160z — 1280 + 2% — 8x . x? — 8z — 1280 =0 (x +32)(x —40) =0 .o But

£ >0,

so

r—8

= —32 or 40 =z =40

=40 — 8 = 32

elderly citizens went on the trip.

mlfiDB

dollars.

hours

66

Chapter 2 (Quadratic functions)

14

a

The parabola has vertex

Exercise 2F

(0, 8), so it has equation

y =a(x —0)* +8 oy =azx®+8 When

=3,

y=0,

so

0=a(3%) +8 9a = —8 a=—8 9

-

the equation of the parabola1s b

y = —%3’:2 + 8.

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

When

z =42,

y=—2(2)*+38 32 = —322 4+ 8 .

For heights greater than 4.44 m, the tunnel 1s less than 4 m wide. But the truck is 5 m high. the truck will not fit through the tunnel. 15

a

The position of the stone above sea level 1s plotted on the graph as shown, where the maximum height reached 1s 80 m, when ¢t = 2 s. So, the vertex is (2, 80), and the h-intercept is 60, as the height is 60 m when ¢ = 0. The quadratic has the form

h = a(t —2)* + 80, where When

t =0,

A h (m)

a

B

t (seconds)

)

The quadratic is h = —5(t — 2)* + 80, which gives the stone’s height above sea level for any

b

When

t=3,

h=-5(3—-2)°+80

= —5(1) + 80

=75

¢ > 0.

¢

The stone will hit the water when

c,

h = 0.

=5(t—2)°+80=0

5. b(t—2)* =80

The stone 1s 75 m above sea level after

c.

3 seconds.

-

(t—2)* =16 32—

+/16

S.t=2+4 but

t>0,

..

t=2+4=6

It will take 6 seconds for the stone to hit

the water.

Chapter 2 (Quadratic functions)

a

y=zx°-22x has a=1, b=-2,

Since

¢=0.

a > 0, the shape 1s \j

The minimum

value occurs when

y=17—2x — x° has a=-1, b=-2,

Since

So, the minimum value of y = 2° — 2z is —1, occurring when = = 1.

¢

y=2_8+2z— 3z? has

a=-3,

Since

b=2,

¢c=28.

a < 0, the shape is /\

The maximum

value occurs when

value occurs when

y=7-2(—1)—(=1)*=38

So, the maximum

y = 7 —2x — x*

value of

r = —1.

y=2z’4+zc—1 has a=2, b=1,

Since

c=17T.

a < 0, the shape is /\

The maximum

and

Exercise 2G

is 8, occurring when

¢c= —1.

a > 0, the shape 1s \_/

The minimum

value occurs when

So, the maximum value of

y =8+ 2z — 3z? when

e

is 8%, occurring

x = %

y=4z -z +5 has a=4, b=—-1,

Since

y =Tz — 2x° ¢c= 0.

a > 0, the shape is \_/

The minimum

value occurs when

So, the minimum value of

A2 y=4z® when

—x+95 1 o = E.

So, the minimum value of y=2r*+z—1 is —1%, occurring 1 when x = — 7

15 occurring 18. 44,

has

Since

a=

-2,

b=7,

c¢=0.

a < 0, the shape is /\

The maximum

value occurs when

So, the maximum value of

y =Tz — 2% when

fi

x = o

is 63, occurring

67

68 2

Chapter 2 (Quadratic functions) a

P=-32"+240x

—800

Exercise 2G has

a=

—3,

b=240,

¢ = —800.

Since a < 0, the shape is /\ The maximum profit occurs when

-b 2a

x =

240 2(—3)

40

So, 40 refrigerators should be made each day to maximise the total profit.

b

When

z =40,

P = —3(40)% + 240(40) — 800 = 4000

So, the maximum profit is €4000. 3

a

Let the other side be ¥ m long.

b

yym

A = 100z — z? with

a = —1,

Since 1 xm

and

2

2(—1)

— 50

y = 100 — 50 = 50

when z =y = 50, which is when the

oz +y =100

rectangle is a square.

oy=100—x

A = xy A=z(100 — x)

A =100z — z°

Let the dimensions of the paddock be If 1000 m of fence 1s available, then

2 + y = 1000 y = 1000 — 22

* m x y m.

Drain

{perimeter} .... (1)

The area of the enclosure

Since

_ —b _ —100

So, the area of the rectangle 1s maximised

2z+ 2y = 200

4

¢ =0.

The area 1s maximised when

The perimeter 1s 200 m.

.

b=100,

a < 0, the shape is /\

T

The area

is a quadratic function

y = 1000 — 22,

T m-

A = zy

A = x(1000 — 2x)

ym

= 1000z — 2z~

A = -2z 4+ 1000z A 1s a quadratic and

a < 0,

so its shape is /\

So, the area 1s maximised when

and when

x = 250,

z = i 2a

L

2(—2)

U

200

y = 1000 — 2(250) = 500

So, the paddock has a maximum

area when the dimensions are

250 m

x 500 m.

Chapter 2 (Quadratic functions)

e——y m——>»

“

b

If

92+ 8y = 1800,

then

The area of each penis 4

“

|

.

|

™T

H

I

H

I

Substituting

A=

The length of fence required for this enclosure 1s 9z + 8y. If 1800 m 1s available for this enclosure, then ’

S,

1800 8_— 9 - )

§E

e

m

so its shape is /\

x = b 2a

and when

into A we get

1800 92 — 8 8 9 92 A= —22"+225x

"

a < 0,

So, the area 1s maximised when

8

A = zy. e

(

1800 — 9x

y=

4

92 + 8y = 1800.

The area A is a quadratic with

y =

Exercise 2G

z = 100,

_2295

= 100

2(_8)

y = 1800 — 9(100) _ 1195

Hence, the area 1s maximised when the dimensions are

100 m x 112.5 m.

Let * m X y m be the dimensions of a single pen as shown below. So, the total length of fencing required 1s 6z + 6y. If there 1s 500 m of fencing available, then | e—ym—» 6x + 6y = 500

rm 4+

r+y =833

11

y=283z—z

.. (1)

The area of each pen will be ;. which 1s a quadratic with

So, at © = ;—b

8]

A = xy and substituting equation (1), we have A= x(83% — x) A=

a < 0,

—x°+ 83%:13 so its shape is /\

we have a maximum value of A.

:1:—2(_1)— M5

2

%=

andso ang

y =833= 833 —41% —

So, the dimensions that maximise the area are

=413 412 = 413 41% m

X 41% m.

Let £ m X y m be the dimensions of a single pen as shown below. So, the total length of fencing required 1s 5z + 8y. Sl If there 1s 500 m of fencing available, then

‘[

5z + 8y = 500

8y = 500 — dx

U

L 1M F

500 — 5z

l

ERiiS

y =623 — 2z

... (1)

The area of each pen will be

A = zy

and substituting equation (1), we have

A = z(625 — 2x) A= —22" 4621z

which is a quadratic with

a < 0, so its shape is /\

9

69

70

Chapter 2 (Quadratic functions) So, when

= = B —62= 1

(1

Exercise 2G

we have a maximum value of A.

=50,

and substituting

= = 50

into

y = 625 — 2z, 3

we have

y_glz 1

So, the dimensions that maximise the area are 50 m x 31+ m.

P Y

a

=Y

-

y=0.0082° — 0.8z + 50

has y-intercept when

z = 0

.y = 0.008(0)* — 0.8(0) + 50 = 50 So, the height of the platforms 1s 50 m.

©

In order to find the coordinates of L, we need to find the distance between the two platforms.

When

y =50,

50 = 0.0082% — 0.8z + 50

-, 0.008z° — 0.8z =0

.

x(0.008x — 0.8) =0 . x=0 & i—))

or SO

0.008r—-08=0 xr = 100

So, the distance between the two platforms 1s 100 m. L lies halfway between the two platforms. L 1s 50 m from each platform.

When

z =50,

vy = 0.008(50)* — 0.8(50) + 50 = 30

So, L 1s 30 m above ground level. 8

a The graphs of y =2° — 3z meet where

and y = 2z — z° 2 2 x° —3x =2x — x

co 202 — b =0 . (26 —-5)=0 B

b

m:(lorQ%

The vertical separation between the curves is given by

S = (2z — %) — (z* — 3z) S=2z—z°—2°+ 3z

S = —2z° + 5z

{y =2z —z*

isabove

y=2z"—3z

for 0

i? A >

X .

2a

Y

equation

Yy = md (:1: — E) + 0 2a

2

=57~ 3)2

Y~ % ¢

2

The parabola and (MP) meet where

4da

1

4a

1

—1° 4a

Lo,

B

X

X2

— =T+

—

2a

:

which 1s a quadratic in z with

X

==

(a: — —)

2a

o

—

X

da

—

2

X

e

2a

X2— da

= X

A = (——) 2a

-

—

1

(_) da

X 2

i

da

. XA 4a2

4a?

So, the graphs fouch at one point. (MP) is a tangent to the parabola.

d

| i

o

MNP =460

{corresponding angles}

Triangle FPM is isosceles

MFP = MNP {base angles of an isosceles A}

=0 iii iv

FPB = MFP =0

{alternate angles}

angle of incidence = angle of reflection {from the Opening Problem| angle of reflection = 6 But the angle between the normal and the focus always reflected to the focus.

v

Y

FPB = 6,

so a vertical ray of light 1s

Misha needs to use a parabolic mirror, and place his cup at the focus of the parabola.

Chapter 2 (Quadratic functions)

Exercise 2H.1

a

(r+4)(x—2)

B

|4

—4 When

« =3

=Y

has zeros —4 and 2.

|

b

(z+1)(x —5)

has zeros —1 and 5.

_I_

2 we have

2Y

sY

8Y

pol= f—--

=Y

AY

8Y

74

|

_|_

I

>

|

—1

(7)(1) > 0,

When

« =6

4

5 we have

|

>

(7)(1) > 0,

so we put a + sign here.

so we put a + sign here.

As the factors are single, the signs

As the factors are single, the signs alternate.

alternate.

_|_

—4

¢

x(x —3)

-

2

has zeros 0 and 3.

d

—1

z(r+2)

D

has zeros 0 and —2.

_I_

-

I

|

0 When

x =4

4

3 we have

|

|

> T

(4)(1) > 0,

|

29 When

so we put a + sign here.

z =1

;

0 we have

|

(1)(3) >0,

so we put a + sign here.

As the factors are single, the signs

As the factors are single, the signs

alternate.

alternate.

T

Chapter 2 (Quadratic functions)

e

(2z+1)(xz—4)

has zeros —% and 4.

f

—(z+1)(x—3)

Exercise 2H.1

has zeros —1 and 3.

_I_

-

| _% When

| 4

x =5

we have

t

_

s

-

(11)(1) > 0,

| —1 When

so we put a + sign here.

T

|

—

i

|

g

—(3z—2)(z+1)

When

|

4

2

ot

z =1

we have

>

U

-

3

h

:

>

—

i

(5—x)(1—2z)

T

=6

—(5)(1) < 0,

—(1)(2) < 0,

+

|

3

1

When

=4

-

T

has zeros 3 and 3.

|

2

_

‘

we have

|

3

4

-

,

(7)(—1) 0,

so we put a + sign here.

As the factors are single, the signs alternate. +

-

3

a

1

=

5

(xz+2)°

1

+

5

-

T

has zero —2.

b

R _9

-

When

= =0

we have

>

(z—3)? -4

(2)? > 0,

|

+

z =4

we have

S

>

L

(1)% > 0,

so we put a -+ sign here.

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

+

3 When

so we put a + sign here.

-

has zero 3.

i

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

+

|

+

>

U

75

76

Chapter 2 (Quadratic functions)

¢

—(x—4)*

has zero 4.


T

-

—(1)%

-

z°—-9=(z+3)(z—3)

b

has zeros —3 and 3.

—

|

_3

—

2

> T

4—2=2+2)2-1) has zeros —2 and 2.

+

-

|

|

-3 When

& =4

i

3 we have

-

»

T

When

(7)(1) > 0,

so we put a + sign here.

e BN

¢

—3