this edition published 2010 by the australian copyright...

This edition published 2010 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10.5/12.5pt Times First edition published 2007. John Wiley & Sons Australia, Ltd 2007, 2010 The moral rights of the contributors have been asserted. ISBN 978 174216 0573 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Illustrated by the Wiley Art Studio and Aptara Typeset in India by Aptara Printed in Singapore by Markono Print Media Pte Ltd 10 9 8 7 6 5 4 3 2 1 The publishers would like to thank the following contributors: Ross Allen Murray Anderson Bill Corbett Elena Iampolsky Andy Lezdkalns Image on page 416 Brian Hodgson

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Table of contents About eBookPlus v Chapter 1 Coordinate geometry 1

Exercise 1A Sketch graphs of y = axm + bxn + c where m = 1 or 2 and n = 1 or 2 1

Exercise 1B Reciprocal graphs 9 Exercise 1C Graphs of circles and ellipses 12 Exercise 1D Graphs of hyperbolas 17 Exercise 1E Partial fractions 22 Exercise 1F Sketch graphs using partial fractions 26 Chapter review 33 Short answer 33 Multiple choice 37 Extended response 39

Chapter 2 Circular functions 42 Exercise 2A Reciprocal trigonometric functions 42 Exercise 2B Graphs of reciprocal trigonometric

functions 44 Exercise 2C Trigonometric identities 49 Exercise 2D Compound- and double-angle

formulas 52 Exercise 2E Inverse circular functions and

their graphs 58 Chapter review 61 Short answer 61 Multiple choice 65 Extended response 68

Chapter 3 Complex numbers 71 Exercise 3A Introduction to complex numbers 71 Exercise 3B Basic operations using complex

numbers 73 Exercise 3C Conjugates and division of complex

numbers 76 Exercise 3D Complex numbers in polar form 79 Exercise 3E Basic operations on complex numbers in

polar form 83 Exercise 3F Factorisation of polynomials in C 89 Exercise 3G Solving equations in C 93 Chapter review 99 Short answer 99 Multiple choice 102 Extended response 104

Chapter 4 Relations and regions of the complex plane 107 Exercise 4A Rays and lines 107 Exercise 4B Circles and ellipses 108 Exercise 4C Combination graphs and regions 116 Exercise 4D Graphs of other simple curves 119 Chapter review 123 Short answer 123 Multiple choice 125 Extended response 126

Exam Practice 1 130 Short answer 130 Multiple choice 131 Extended response 133

Chapter 5 Differential calculus 136 Exercise 5A The derivative of tan (kx) 136

Exercise 5B Second derivatives 139 Exercise 5C Analysing the behaviour of functions

using the second derivative 142 Exercise 5D Derivatives of inverse circular functions 149 Exercise 5E Antidifferentiation involving inverse

circular functions 154 Exercise 5F Implicit differentiation 160 Chapter review 163 Short answer 163 Multiple choice 167 Extended response 169

Chapter 6 Integral calculus 173 Exercise 6A Substitution where the derivative is

present in the integrand 173 Exercise 6B Linear substitution 179 Exercise 6C Antiderivatives involving trigonometric

identities 185 Exercise 6D Antidifferentiation using

partial fractions 196 Exercise 6E Definite integrals 208 Exercise 6F Applications of integration 216 Exercise 6G Volumes of solids of revolution 222 Exercise 6H Graphs of the antiderivatives

of functions 231 Chapter review 232 Short answer 232 Multiple choice 239 Extended response 241

Chapter 7 Differential equations 245 Exercise 7A Differential equations: related rates 245 Exercise 7B Verifying solutions 249 Exercise 7C Differential equations of the

form ddyx

= f (x) 251

Exercise 7D Differential equations of the

form 2

2dd

yx

= f (x) 256

Exercise 7E Differential equations of the

form ddyx

= g(y) 260

Exercise 7F Setting up and solving differential equations 266

Exercise 7G Input/output of mixing problems 269 Exercise 7H Numerical solutions of differential

equations 272 Exercise 7I Direction field for a differential equation 275 Chapter review 278 Short answer 278 Multiple choice 280 Extended response 282


Chapter 8 Kinematics 295 Exercise 8A Differentiation and displacement,

velocity and acceleration 295 Exercise 8B Using antidifferentiation 300 Exercise 8C Motion under constant acceleration 305 Exercise 8D Velocitytime graphs 309

iv

Exercise 8E Applying differential equations to rectilinear motion 315

Chapter review 320 Short answer 320 Multiple choice 321 Extended response 323

Chapter 9 Vectors 328 Exercise 9A Vectors and scalars 328 Exercise 9B Position vectors in two and three

dimensions 330 Exercise 9C Multiplying two vectors the dot

product 334 Exercise 9D Using vectors in geometry 336 Exercise 9E Resolving vectors scalar and

vector resolutes 340 Exercise 9F Time-varying vectors 343 Chapter review 345 Short answer 345 Multiple choice 347 Extended response 348

Chapter 10 Vector calculus 352 Exercise 10A Position, velocity and acceleration 352 Exercise 10B Cartesian equations and

antidifferentiation of vectors 355 Exercise 10C Applications of vector calculus 360 Exercise 10D Projectile motion 369 Chapter review 375 Short answer 375 Multiple choice 378 Extended response 379

Chapter 11 Mechanics 385 Exercise 11A Force diagrams and the triangle

of forces 385 Exercise 11B Newtons First Law of Motion 388

Exercise 11C Newtons Second Law of Motion 390 Exercise 11D Applications of Newtons First and

Second Laws of Motion 393 Exercise 11E Applications of Newtons First and Second

Laws to connected masses 397 Exercise 11F Variable forces 399 Exercise 11G Momentum and Newtons Third

Law of Motion 402 Chapter review 403 Short answer 403 Multiple choice 405 Extended response 406


Solutions to investigations 420 Chapter 3 420 Investigation The exact values of

cos 25

and sin 25

420

Chapter 5 420 Investigation Rate of change of angle 420 Chapter 7 421 Investigation Solutions to differential equations 421 Chapter 8 422 Investigation The falling ball bearing 422 Chapter 10 423 Investigation Distance travelled along a curve 423

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 1

Exercise 1A Sketch graphs of y = axm + bxn + c where m = 1 or 2 and n = 1 or 2

1 a 2 2y xx

= +

As |x| 0, 2x

and hence

|y| Vertical asymptote: x = 0 As |x| , 2

x 0 and hence

|y| 2x Oblique asymptote: y = 2x

b 4 2y xx

= +

As |x| 0, 4x

and hence


x 0 and hence

|y| x 2 Oblique asymptote: y = x 2

c 4 2y xx

= +

As |x| 0, 4x

and hence


x 0 and hence

|y| 2x Oblique asymptote: y = 2x

d 2 23y xx

=

As |x| 0, 2x

and hence


x 0 and hence

|y| 3 x2 Curved asymptote: y = 3 x2

e 221y xx

= +

As |x| 0, 21x and hence

|y| Vertical asymptote: x = 0 As |x| , 21x 0 and hence

|y| x2 Curved asymptote: y = x2

f 24 2y xx

= +

As |x| 0, 24x and hence

|y| Vertical asymptote: x = 0 As |x| , 24x 0 and hence

|y| x 2 Oblique asymptote: y = x 2

g 21 4 3y xx

= +

As |x| 0, 1x

and hence


x 0 and hence

|y| 3 4x2 Curved asymptote: y = 3 4x2

h 8 2 1y xx

= + +

As |x| 0, 8x

and hence


x 0 and hence

|y| 2x + 1 Oblique asymptote: y = 2x + 1

2 a 6 5y xx

= +

Asymptotes: x = 0 and y = 5x The answer is B.

b 4 2 1y xx

= +

Asymptotes: x = 0 and y = 1 2x The answer is A.

c 2 64y x xx

= +

Asymptotes: x = 0 and y = x2 4x The answer is A.

d 279 4y xx

= +

Asymptotes: x = 0 and y = 9 4x The answer is E.

3 a 2 2y xx

= +

Let y1 = 2x and y2 = 2x

For y1 = 2x Straight line passing through (0, 0) Let x = 1, y1 = 2 1 = 2 (1, 2) is on the straight line.

For y2 = 2x (hyperbola)

Vertical asymptote: x = 0 Horizontal asymptote: y2 = 0 Let x = 1, y2 = 21

= 2 So (1, 2) is on the hyperbola. Let x = 1, y2 = 21

= 2 So (1, 2) is on the hyperbola.

For 2 2y xx

= +

x-intercept: y = 0

2x

+ 2x = 0

2 + 2x2 = 0 2x2 = 2 x2 = 1 x = 1 No x-intercepts. No y-intercept as the y-axis is a

vertical asymptote. Asymptotes are x = 0 and y = 2x

b y = 4x

+ x 2

Let y1 = x 2 and y2 = 4x

For y1 = x 2 Straight line with x-intercept: x 2 = 0 x = 2 y-intercept: y1 = 0 2 = 2 So (2, 0) and (0, 2) are on the

line. For y2 = 4x (hyperbola)

Vertical asymptote: x = 0 Horizontal asymptote: y = 0 Let x = 1, y2 = 41

= 4 Let x = 1, y2 = 41

= 4 So (1, 4) and (1, 4) lie on the

hyperbola.

For y = 4x

+ x 2

x-intercept: y = 0

4x

+ x 2 = 0

4+ x2 2x = 0 x2 2x + 4 = 0

22 ( 2) 4(1)(4)2 1

2 122

x = =

No x-intercepts. No y-intercepts since y-axis is a

vertical asymptote.

Chapter 1 Coordinate geometry

S M 1 2 - 1 2 C o o r d i n a t e g e o m e t r y

Asymptotes are x = 0 and 2y x=

c 4 2y xx

= +

Let 1 2y x= and 24yx

=

For 1 2y x= Straight line passing through (0, 0) Let x = 1, y1 = 2 1 = 2 (1, 2) is on the straight line.

For 24yx

= (hyperbola)

Vertical asymptote: x = 0 Horizontal asymptote: y = 0 Hyperbola passing through (1, 4)

and (1, 4)

For 4 2y xx

= +

Asymptotes are x = 0 and y = 2x x-intercept: y = 0

4 2xx

+ = 0

24 2x+ = 0 2x = 2 No x-intercepts No y-intercepts since y-axis is a

vertical asymptote.

d 2 23y xx

=

Let 21 3y x= and 22yx

=

For 21 3y x= Inverted parabola will vertex at (0, 3) Let x = 1, y1 = 3 12 = 2 (1, 2) and by symmetry (1, 2) are

on the parabola

For 22yx

=


and (1, 2)

For 2 23y xx

=

Asymptotes are x = 0 and y = 3 x2

x-intercept: y = 0

2 23 xx

= 0

33 2x x = 0 or x3 3x + 2 = 0 (x 1)(x2 + x 2) = 0 (Since P(1) = 0) (x 1)(x + 2)(x 1) = 0 x = 1 or 2 x-intercepts are (2, 0) and (1, 0) No y-intercept since x = 0 is a

vertical asymptote.

e 221y xx

= +

Let y1 = x2 and y2 = 21x

For y1 = x2 Upright parabola with vertex at (0, 0) Let x = 1, y1 = 12

= 1 (1, 1) and by symmetry (1, 1) are

on the parabola.

For y2 = 21x

(hyperbola)


= 1 (1, 1) is on the hyperbola. Let x = 1, y2 = 21( 1)

= 1 (1, 1) is on the hyperbola.

For 221y xx

= +

Asymptotes are x = 0 and y = x2

x-intercept: y = 0

221 xx

+ = 0

1 + x4 = 0 x4 = 1

x = 14( 1)

No x-intercepts. No y-intercept since y-axis is a

vertical asymptote.

f 24 2y xx

= +


For y1 = x 2, A straight line with x-intercept: x 2 = 0 x = 2 and y-intercept: y1 = 0 2 = 2 Straight line passes through (2, 0)

and (0, 2)

For y2 = 24x

(hyperbola)


= 4 (1, 4) is on the hyperbola. Let x = 1, y2 = 24( 1)

= 4 (1, 4) is on the hyperbola.

For 24 2y xx

= +

Asymptotes are x = 0 and y = x 2

No y-intercept since the y-axis is a vertical asymptote.

g 21 4 3y xx

= +

Let y1 = 3 4x2 and y2 = 1x

For y1 = 3 4x2 Inverted parabola with vertex (0, 3) Let x = 1, y1 = 3 4(1)2

= 1 (1, 1) and by symmetry (1, 1)

lie on the parabola.


Vertical asymptote: x = 0 Horizontal asymptote: y = 0 Hyperbola passes through (1, 1)

and (1, 1)

For 21 4 3y xx

= +

Asymptotes are x = 0 and y = 3 4x2

x-intercept: 21 4 3xx

+ = 0

1 4x3 + 3x = 0 or 4x3 3x 1 = 0 (x 1)(4x2 + 4x + 1) = 0 (Since P(1) = 0) (x 1)(2x + 1)2 = 0 x = 1 or x = 1

2

x-intercepts are (1, 0) and ( 12

, 0)

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 3 No y-intercept since y-axis is a vertical asymptote.

h 8 2 1y xx

= + +

Let y1 = 2x + 1 and y2 = 8x

For y1 = 2x + 1 x-intercept: 2x + 1 = 0 x = 1

2

y-intercept: y = 2(0) + 1 = 1 ( 1

2, 0) and (0, 1) lie on the straight line.


Vertical asymptote: x = 0 Horizontal asymptote: y = 0 Hyperbola passing through (1, 8) and (1, 8)

For 8 2 1y xx

= + +

Asymptotes are x = 0 and y = 2x + 1 x-intercept: y = 0

8 2 1xx

+ + = 0

8 + 2x2 + x = 0 2x2 + x + 8 = 0

x = 21 1 4(2)(8)2 2

= 1 634

No x-intercepts. No y-intercept since y-axis is a vertical asymptote.

4 a y = 2 2xx

+

y = 2x1 + 2x

ddyx

= 2x2 + 2

For ddyx

= 0,

2x2 + 2 = 0 2 + 2x2 = 0 2x2 = 2 x2 = 1 x = 1 x = 1 or 1 For x = 1, y = 2

1 + 2 1x

= 2 + 2 = 4

(1, 4) is a stationary point. For x = 1, y = 2

1 + 2 1

= 2 + 2 = 4 (1, 4) is a stationary point.

x < 1 (= 2) 1 > 1( 1

2)

< 1 ( 1

2) 1

> 1 (2)

ddyx

1 12

0 6 6 0 1 12

Slope / \ \ /

Therefore (1, 4) is a local maximum and (1, 4) is a local minimum.

b y = 4x

+ x 2

= 4x1 + x 2

ddyx

= 4x2 + 1

For ddyx

= 0,

4x2 + 1 = 0 4 + x2 = 0 x2 = 4 x = 4 x = 2 or 2 For x = 2, y = 4

2 + (2) 2

= 6 (2, 6) is a stationary point. For x = 2, y = 4

2 + (2) 2

= 2 (2, 2) is a stationary point.

x < 2(3) = 2> 2(1)

< 2 (1) = 2

> 2 (3)

ddyx

59

0 3 3 0 59

Slope / \ \ /

(2, 6) is a local maximum and (2, 2) is a local minimum.

c y = 4x

+ 2x

= 4x1 + 2x

ddyx

= 4x2 + 2

For ddyx

= 0,

4x2 + 2 = 0 4 + 2x2 = 0 2x2 = 4 x2 = 2 x = 2 x = 2 or 2 For x = 2, y = 4

2 + 2( 2 )

= 2 2 2 2 = 4 2 ( 2, 4 2) is a stationary point.

For x = 2 , y = 42

+ 2 2

= 2 2 2 2+ = 4 2


( 2 , 4 2 ) is a stationary point.

x < 2 (2)

= 2 > 2 (1)

< 2 (1)

= 2 > 2(2)

ddyx

1 0 2 2 0 1

Slope / \ \ /

( 2 , 4 2 ) is a local maximum and ( 2 , 4 2 ) is a local minimum.

d y = 3 x2 2x

= 3 x2 2x1

ddyx

= 2x + 2x2

For ddyx

= 0,

2x + 2x2 = 0 2x3 + 2 = 0 2x3 = 2 x3 = 1 x = 1 For x = 1, y = 3 12 2

1

= 3 1 2 = 0 (1, 0) is a stationary point.

x < 1 ( 1

2) = 1

> 1 (2)

ddyx

7 0 3 12

Slope / \ (1, 0) is a local maximum.

e y = 21x

+ x2

= x 2 + x2

ddyx

= 2x3 + 2x

For ddyx

= 0,

2x3 + 2x = 0 2 + 2x4 = 0 2x4 = 2 x4 = 1 x = 1 or 1 For x = 1, y = 211 + 1

2

= 2 For x = 1, y = 21( 1) + (1)

2

= 2 (1, 2) and (1, 2) are the stationary points.

x < 1 (2) = 1 > 1 ( 1

2)

< 1 ( 1

2) = 1

> 1 (2)

ddyx

3 34

0 15 15 0 3 34

Slope \ / \ __ / (1, 2) is a local minimum and

(1, 2) is a local minimum.

f y = x + 24x

2

= x + 4x2 2

ddyx

= 1 8x3

For ddyx

= 0,

1 8x3 = 0 x3 8 = 0 x3 = 8 x = 2 For x = 2, y = 2 + 242 2

= 2 + 1 2 = 1 (2, 1) is a stationary point.

x < 2 (1) = 2 > 2 (3)

ddyx

7 0 1927

Slope \ /


g y = 1x

4x2 + 3

= x1 4x2 + 3

ddyx

= x2 8x

For ddyx

= 0,

x2 8x = 0 1 8x3 = 0 8x3 = 1 x3 = 1

8

x = 12

For x = 12

, y = 12

1

421

2

+ 3

= 2 1 + 3 = 0 ( 1

2, 0) is a stationary point.

x < 12

(1) = 1

2

> 12

14

ddyx

7 0 14

Slope / \

( 12

, 0) is a local maximum.

h y = 8x

+ 2x + 1

= 8x 1 + 2x + 1

ddyx

= 8x2 + 2

For ddyx

= 0,

8x2 + 2 = 0 8 + 2x2 = 0 2x2 = 8 x2 = 4 x = 2 or 2

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 5 For x = 2, y = 8

2 + 2(2) + 1

= 7 For x = 2, y = 8

2 + 2(2) + 1

= 9 (2, 7) and (2, 9) are stationary

points.

x < 2 (3) = 2 > 2 (1)

< 2 (1) = 2

> 2 (3)

ddyx

1 19

0 6 6 0 1 19

Slope / \ \ /

(2, 7) is a local maximum and (2, 9) is a local minimum.

5 a 6 3y xx

=

Asymptotes are x = 0 and y = 3x.

x-intercepts: 6x

3x = 0

6 3x2 = 0 3x2 = 6 x2 = 2 x = 2 or 2 The answer is A.

b y = 3 25x

Asymptotes are x = 0 and y = 3

x-intercepts: 3 25x

= 0

3x2 5 = 0 3x2 = 5 x2 = 5

3

x = 53

or 53

The answer is A.

c y = x2 22x


x-intercepts: x2 22x

= 0

x4 2 = 0 x4 = 2

x =142 or (

142 )

The answer is B.

d y = 3x 1 + 5x

Asymptotes are x = 0 and y = 3x 1

x-intercepts: 3x 1 + 5x

= 0

3x2 x + 5 = 0

x = 21 ( 1) 4(3)(5)

2 3

= 1 596

No x-intercepts.

ddyx

= 3 5x2

For ddyx

= 0,

3 5x2 = 0 3x2 5 = 0 3x2 = 5 x2 = 5

3

x = 53

Turning points where x = 53

and

x = 53

The answer is C.

6 a 2216y xx

= +

Let y1 = x2 and y2 = 216x

For y1 = x2 An upright parabola with vertex

(0, 0). Let x = 1, y1 = 12 = 1 (1, 1) and by symmetry (1, 1) are

on the parabola.

For y2 = 216x


and (1, 16)

For 2216y xx

= +


x-intercepts: y = 0

2216 xx

+ = 0

16 + x4 = 0 x4 = 16 No x-intercepts No y-intercept since y-axis is a

vertical asymptote.

ddyx

= 32x3 + 2x

For ddyx

= 0,

32x3 + 2x = 0 32 + 2x4 = 0 2x4 = 32 x4 = 16 x = 2 or 2 For x = 2, y = 16 2( 2)

+ 2( 2)

= 4 + 4 = 8 For x = 2, y = 1622

+ 22

= 4 + 4 = 8 Stationary points are (2, 8) and

(2, 8).

x < 2(3) = 2> 2(1)

< 2 (1) = 2

> 2(3)

ddyx

4 2227 0 30 30 0 4

2227

Slope \ / \ /

(2, 8) and (2, 8) are local minimums.

b 9y xx

=

Let y1 = x and y2 = 9x

For y1 = x, A straight line passing through

(0, 0) and (1, 1)

For y2 = 9x

Asymptotes: x = 0 and y = 0 A hyperbola passing through (1, 9)

and (1, 9)

For 9y xx

=

Asymptotes: x = 0 and y = x

x-intercept: 9 xx

= 0

9 x2 = 0 x2 = 9 x = 3 (3, 0) and (3, 0) are the

x-intercepts. No y-intercept since x = 0 is a

vertical asymptote.

ddyx

= 9x2 1

For ddyx

= 0,

9x2 1 = 0 9 x2 = 0 x2 = 9 No turning points.

c 2 24 1 4 1x xyx x x

+= = +

1 4y xx

= +

Let y1 = 4x and y2 = 1x

For y1 = 4x A straight line passing through

(0, 0) and (1, 4)

For y2 = 1x

Asymptotes: x = 0 and y = 0


A hyperbola passing through (1, 1) and (1, 1)

For 1 4y xx

= +

Asymptotes: x = 0 and y = 4x

x-intercepts: 1 4xx

+ = 0

1 + 4x2 = 0 4x2 = 1 x2 = 1

4

No x-intercepts No y-intercept since y-axis is a

vertical asymptote.

ddyx

= x2 + 4

For ddyx

= 0,

x2 + 4 = 0 1 + 4x2 = 0 4x2 = 1 x2 = 1

4

x = 12

or 12

For x = 12, y = 11 2

2

1 4

+

= 2 2 = 4

For x = 12, y = 11 2

2

1 4

+

= 2 + 2 = 4 ( 1

2, 4) and ( 1

2, 4) are stationary

points.

x < 12

(1) = 1

2 > 1

2

14

< 12

14

= 1

2 > 1

2

(1)

ddyx

3 0 12 12 0 3

Slope / \ \ /

( 12

, 4) is a local maximum and

( 12

, 4) is a local minimum.

d 2 3 2x xy

x +=

23y xx

= +


For y1 = x 3

A straight line with intercepts (3, 0) and (0, 3)

For y2 = 2x

Asymptotes: x = 0 and y = 0 A hyperbola passing through (1, 2)

and (1, 2)

For 23y xx

= +

Asymptotes: x = 0 and y = x 3

x-intercepts: 23xx

+ = 0

x2 3x + 2 = 0 (x 2)(x 1) = 0 x = 1 or 2 x-intercepts are (1, 0) and (2, 0) No y-intercept since y-axis is a

vertical asymptote.

ddyx

= 1 2x2

For ddyx

= 0,

1 2x2 = 0 x2 2 = 0 x2 = 2 x = 2 or 2 For x = 2 , y = 2 3 + 2

( 2 )

= 2 3 2 = 3 2 2 For x = 2 , y = 2 3 + 2

2

= 2 3 2 + = 2 2 3 ( 2 , 3 2 2 ) and

( 2 , 2 2 3 ) are stationary points.

x < 2 (2)

= 2 > 2 (1)

< 2 (1)

= 2 > 2 (2)

ddyx

12

0 1 1 0 12

Slope / \ \ /

( 2 , 3 2 2 ) is a local maximum and ( 2 , 2 2 3 ) is a local minimum.

e 21 8y xx

= +

Let y1 = 8x2 and y2 = 1x

For y1 = 8x2

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 7 An upright parabola with vertex (0, 0) and passing

through (1, 8) and (1, 8).

For y2 = 1x

Asymptotes: x = 0 and y = 0 A hyperbola passing through (1, 1) and (1, 1).

For y = 1x

+ 8x2

Asymptotes: x = 0 and y = 8x2

x-intercept: 1x

+ 8x2 = 0

1 + 8x3 = 0 8x3 = 1 x3 = 1

8

x = 12

( 12

, 0) is the x-intercept.

No y-intercept.

ddyx

= x2 + 16x

For ddyx

= 0,

x2 + 16x = 0 1 + 16x3 = 0 16x3 = 1 x3 = 1

16

x = 11316

= 132 2

For x = 132 2, y = 1

32 2

1

+ 821

32 2

= 2 3 2 + 8 13 24 2

= 2 3 2 + 23 4

= 2 3 2 + 834

= 2 3 2 + 3 2 = 3 3 2 ( 132 2

, 3 3 2 ) is a stationary point.

x < 132 2

( 14

)

132 2

> 132 2

(1)

ddyx

12 0 15

Slope \ /

( 132 2, 3 3 2 ) is a local minimum.

f 3 2

23 4x xyx

+ =

y = x 24x

+ 3

Let y1 = x + 3 and y2 = 24x

For y1 = x + 3 A straight line with intercepts (3, 0) and (0, 3)

For y2 = 24x

Asymptotes: x = 0 and y = 0 A hyperbola passing through (1, 4) and (1, 4).

For y = x 24x

+ 3

Asymptotes: x = 0 and y = x + 3

x-intercept: x 24x

+ 3 = 0

x3 4 + 3x2 = 0 x3 + 3x2 4 = 0 (x 1)(x2 + 4x + 4) = 0 (Since P(1) = 0) (x 1)(x + 2)(x + 2) = 0 x = 1 or x = 2 The x-intercepts are (1, 0) and (2, 0) No y-intercept.

ddyx

= 1 + 8x3

For ddyx

= 0,

1 + 8x3 = 0 x3 + 8 = 0 x3 = 8 x = 2 For x = 2, y = 2 4 2( 2)

+ 3

= 0 (2, 0) is a stationary point.

x < 2 (3) = 2 > 2 (1)

ddyx

1927

0 7

Slope / \

(2, 0) is a local maximum.

g y = 21x

+ x 2


For y1 = x 2 A straight line with intercepts

(2, 0) and (0, 2)

For y2 = 21x

Asymptotes: x = 0 and y = 0


A hyperbola passing through (1, 1) and (1, 1)

For y = 21x

+ x 2

Asymptotes: x = 0 and y = x 2

x-intercepts: 21x

+ x 2 = 0

1 + x3 2x2 = 0 x3 2x2 + 1 = 0 (x 1)(x2 x 1) = 0 (Since P(1) = 0) x = 1 or x2 x 1 = 0

x = 21 ( 1) 4(1)( 1)

2 1

= 1 52

= 1 52

+ or 1 52

1.62 or 0.62 The x-intercepts are (1, 0),

(1.62, 0) and (0.62, 0) No y-intercept.

ddyx

= 2x3 + 1

For ddyx

= 0,

2x3 + 1 = 0 2 + x3 = 0 x3 = 2 x = 3 2 1.26 For x = 1.26, y = 1 2(1.26)

+ 1.26 2

0.11 (1.26, 0.11) is a stationary point.

x < 1.26 (1) = 1.26 > 1.26

(2) ddyx

1 0 34

Slope \ /

(1.26, 0.11) is a local minimum.

h y = 2 x 1x

Let y1 = 2 x and y2 = 1x

For y1 = 2 x A straight line with intercepts

(2, 0) and (0, 2)

For y2 = 1x

Asymptotes: x = 0 and y = 0 A hyperbola passing through

(1, 1) and (1, 1)

For y = 2 x 1x

Asymptotes are x = 0 and y = 2 x

x-intercepts: 2 x 1x

= 0

2x x2 1 = 0 or x2 2x + 1 = 0 (x 1)2 = 0 x = 1 The x-intercept is (1, 0)

No y-intercepts

ddyx

= 1 + x2

For ddyx

= 0,

1 + x2 = 0 x2 + 1 = 0 x2 = 1 x = 1 or 1 For x = 1, y = 2 (1) 1( 1)

= 4 For x = 1, y = 2 1 1

1

= 0 Stationary points are (1, 4) and

(1, 0).

x < 1(2) = 1> 1

12

< 1 12

= 1 > 1(2)

ddyx

34

0 3 3 0 34

Slope \ / / \

(1, 4) is a local minimum and (1, 0) is a local maximum.

7 a V = r2h Where V = 128 128 = r2h r2h = 128

h = 2128r

A = 2 r (r + h)

Substitute h = 2128r

A = 2 r 2128rr

+

A = 2 r2 + 256 ,r

r > 0

b Asymptotes: r = 0, A = 2 r2

ddAr

= 4 r 256 r2

For ddAr

= 0,

4 r 256 r 2 = 0 4 r3 256 = 0 r3 64 = 0 r3 = 64 r = 4

For r = 4, A = 2 42 + 2564

= 32 + 64 = 96 The stationary point is (4, 96).

r < 4 (3) = 4 > 4 (5)

ddAr

16.44 0 9.76

Slope \ /


c The minimum total surface area is

96 cm2. 8 a V = 25xy When V = 400 400 = 25xy

y = 40025x

y = 16x

A = 2(25x + 25y + xy)

Substitute y = 16x

A = 2(25x + 25 16x

+ x 16x

)

= 2(25x + 400x

+ 16)

A = 50x + 800x

+ 32, x > 0

b Asymptotes: x = 0 and A = 50x + 32

ddAr

= 50 800x2

For ddAr

= 0,

50 800x2 = 0 50x2 800 = 0 50x2 = 800 x2 = 16 x = 4 or 4 Discard x = 4 since x > 0. For x = 4, A = 50 4 + 800

4 + 32

= 200 + 200 + 32 = 432 (4, 432) is a stationary point.

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 9 x < 4 (2) = 4

> 4 (5)

ddAx

150 0 18

Slope \ /


c From part b, the minimum total

surface area is 432 cm2 when x = 4.

y = 16x

= 164

= 4 The dimensions for minimum total

surface area are x = y = 4 cm. (or length = width = 4 cm).

Exercise 1B Reciprocal graphs

1 a y = 21

4x , x 2.

Let y1 = x2 4 x-intercepts of y1 = x2 4 are x = 2. Vertical asymptotes of

y = 21

4x are x = 2.

orizontal asymptote is y = 0 y-intercept (0, 1

4)

b y = 12x +

Let y1 = x + 2 x-intercept of y1 = x + 2 is x = 2

Vertical asymptote of y = 12x +

is

x = 2 orizontal asymptote is y = 0 y-intercept (0, 1

2)

c y = 21 ,

2x x+ x 0, 2.

Let y1 = x2 + 2x x-intercepts of y1 = x2 + 2x are

x = 0 and x = 2

Vertical asymptotes of y = 21

2x x+

are x = 0 and x = 2. Horizontal asymptote is y = 0.

No y-intercept but turning point is (1, 1)

d y = 21 ,4 3x x +

x 3, 1

Let y1 = x2 4x + 3. x-intercepts of y1 = x2 4x + 3 are

x = 1, x = 3. Vertical asymptotes of

y = 21 ,4 3x x +

are x = 1, x = 3.

Horizontal asymptote is y = 0 y-intercept is (0, 1

3)

Turning point is (2, 1)

e y = 1 ,2 5x +

x 52

.

Let y1 = 2x + 5 x-intercept of y1 = 2x + 5 is

x = 2 12

.

Vertical asymptote of y = 12 5x +

is x = 2 12

Horizontal asymptote is y = 0. y-intercept is (0, 1

5)

f y = 21 ,

2x x + x 0, 2.

Let y1 = x2 + 2x x-intercepts of y1 = x2 + 2x are

x = 0 and x = 2

Vertical asymptotes of

y = 21

2x x + are x = 0 and x = 2

Horizontal asymptote is y = 0 No y-intercept Turning point is (1, 1)

g y = 21 ,

4 9x x 3

2

Let y1 = 4x2 9 x-intercepts of y1 = 4x2 9 are

x = 32

Vertical asymptotes of y = 21

4 9x

are x = 32

Horizontal asymptote is y = 0 y-intercept is (0, 1

9)

Turning point is (0, 19

)

h y = 21 ,

2 5 3x x + x 3

2, 1.

Let y1 = 2x2 + 5x 3 x-intercepts of y1 = 2x2 + 5x 3

are x = 1 and x = 32

Vertical asymptotes of

y = 21

2 5 3x x + are x = 1 and

x = 32

Horizontal asymptote is y = 0. y-intercept is (0, 1

3)

Turning point is ( 54

, 8)

2 a For f (x) = x 4 x-intercept: x 4 = 0 x = 4 y-intercept: f (0) = 4 The x-intercept is (4, 0)


y-intercept is (0, 4)

For g(x) = 14x

Vertical asymptote is x = 4. Horizontal asymptote is g(x) = 0. y-intercept is (0, 1

4)

b For f (x) = x2 4x x-intercepts: x2 4x = 0 x(x 4) = 0 x = 0 or x = 4 x-intercepts are (0, 0) and (4, 0) y-intercept is (0, 0) Turning point: x = 0 4

2+

= 2 f (2) = 22 4(2) = 4 Turning point is (2, 4)

For g(x) = 21

4x x

Vertical asymptotes are x = 0 and x = 4

Horizontal asymptote is g(x) = 0 No y-intercept Reciprocal of turning point (2, 4)

is (2, 14

)

c For f (x) = 3 x x-intercept: 3 x = 0 x = 3 y-intercept: f (0) = 3 x-intercept is (3, 0) and y-intercept is (0, 3)

For g(x) = 13 x

Vertical asymptote is x = 3 Horizontal asymptote is g(x) = 0 y-intercept is (0, 1

3)

d For f (x) = x2 + 3x + 2 x-intercepts: x2 + 3x + 2 = 0 (x + 2)(x + 1) = 0 x = 2 or x = 1

x-intercepts are (2, 0) and (1, 0) y-intercept: f (0) = 2 y-intercept is (0, 2) Turning point: x = 2 1

2 +

= 32

f 32

= 23

2

+ 3 32

+ 2

= 14


, 14

)

For g(x) = 213 2x x+ +


Horizontal asymptote is g(x) = 0 y-intercept is (0, 1

2)

Reciprocal of turning point ( 3

2, 1

4) is ( 3

2, 4)

e For f (x) = 3x + x2

x-intercepts: 3x + x2 = 0 x(3 + x) = 0 x = 0 or x = 3 x-intercepts are (0, 0) and (3, 0) y-intercept is (0, 0) Turning point: x = 0 3

2+

= 32

f 32

= 3 32

+ 23

2

= 94


, 94

)

For g(x) = 21

3x x+


Horizontal asymptote is g(x) = 0 Reciprocal of turning point

( 32

, 94

) is ( 32

, 49

)

f For f (x) = 3x2 8x 3 x-intercepts: 3x2 8x 3 = 0 (3x + 1)(x 3) = 0 x = 1

3 or x = 3

x-intercepts are ( 13

, 0) and (3, 0)

y-intercept: f (0) = 3 y-intercept is (0, 3)

Turning point: x = 132

3 +

= 43

f 43

= 32

43

8 43

3

= 253


, 253

)

For g(x) = 21

3 8 3x x

Vertical asymptotes: x = 13, x = 3

Horizontal asymptote is g(x) = 0 y-intercept is (0, 1

3)

Reciprocal of turning point ( 4

3, 25

3) is ( 4

3, 3

25)

g For f (x) = (x 4)2

x-intercept: (x 4)2 = 0 x 4 = 0 x = 4 x-intercept is (4, 0) y-intercept: f (0) = (0 4)2

= 16 y-intercept is (0, 16) Turning point is (4, 0)

For g(x) = 21

( 4)x


16)

h For f (x) = (x + 3)2

x-intercept: (x + 3)2 = 0 x + 3 = 0 x = 3 x-intercept is ( 3, 0) y-intercept: f (0) = (0 + 3)2

= 9 y-intercept is (0, 9) Turning point is (3, 0)

For g(x) = 21

( 3)x +

Vertical asymptote is x = 3. Horizontal asymptote is g(x) = 0

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 11 y-intercept is (0, 1

9)

i For f (x) = x2 + 4x 4 x-intercept: x2 + 4x 4 = 0 (x2 4x + 4) = 0 (x 2)2 = 0 x 2 = 0 x = 2 x-intercept is (2, 0) y-intercept: f (0) = 4 y-intercept is (0, 4) Turning point: f (x) = (x 2)2 Turning point is (2, 0)

For g(x) = 214 4x x +


4)

j For f (x) = x2 + x + 1

4

= (x + 12

)2

x-intercept: (x + 12

)2 = 0

x + 12

= 0

x = 12

x-intercept is ( 12 , 0)

y-intercept: f (0) = 14

y-intercept is (0, 14

)


, 0)

For g(x) = 124

1x x+ +

Vertical asymptote is x = 12

Horizontal asymptote is g(x) = 0 y-intercept is (0, 4)

k For f (x) = x2 + 2 f (x) = 2x For f (x) = 0

2x = 0 x = 0 f (0) = 02 + 2 = 2 Turning point is (0, 2) No x-intercept as parabola is

upright with T.P. (0, 2). y-intercept is (0, 2)

For g(x) = 21

2x +

No vertical asymptotes Horizontal asymptote is g(x) = 0 Vertex is (0, 1

2)


)

l For f (x) = x2 + 2x + 4 f (x) = 2x + 2 For f (x) = 0 2x + 2 = 0 2x = 2 x = 1 f (1) = (1)2 + 2(1) + 4 = 3 Turning point is (1, 3) No x-intercept as parabola is

upright with T.P. (1, 3). y-intercept is (0, 4)

For g(x) = 212 4x x+ +

No vertical asymptotes Horizontal asymptote is g(x) = 0 Vertex is (1, 1

3)


)

3 21( )4 3

f xx x

=+ +

a Vertical asymptotes: x2 + 4x + 3 = 0 (x + 3)(x + 1) = 0 x = 3 or x = 1 Horizontal asymptote: y = 0 The answer is B. b Turning point of g(x) = x2 + 4x + 3: g (x) = 2x + 4 For g (x) = 0 2x + 4 = 0 2x = 4 x = 2

g(2) = (2)2 + 4(2) + 3 = 1 Turning point of g(x) is (2, 1) Turning point of f (x) is (2, 1)

also. y-intercept: f (0) = 1

3


)

The answer is D. c Using the information from parts a

and b, The answer is E. 4 a A = L W A = x (4 x) A = 4x x2 b V = L W H = Ah if V = 1; A = 4x x2 1 = (4x x2)h

h = 21

4x x

c x = 3.95, h = 1 24(3.95) (3.95)

= 10.1975

h 5.06 cm

d h = 21

4x x

Consider g(x) = 4x x2

x-intercepts: 4x x2 = 0 x (4 x) = 0 x = 0 or 4 x-intercepts are ( 0, 0) and (4, 0) y-intercept is (0, 0) Turning point: g (x) = 4 2x If g (x) = 0 4 2x = 0 4 = 2x x = 2 g(2) = 4(2) 22 = 4 Turning point of g(x) is (2, 4)

For h = 21

4x x, x > 0 and x < 4

Vertical asymptotes are x = 0, x = 4 Horizontal asymptote is y = 0 No y-intercept. Vertex is (2, 1

4)

e From the graph, the minimum

height is 14

or 0.25 cm when length

(= x) is 2 cm and width (4 x) is also 2 cm.

5 Since the asymptotes are x = 3 and x = 5 then, the denominator must be ( 3)( 5),x x expanding this gives

2 8 15x x + Therefore 8 and 15b c= = The turning point is (4, 1)


Therefore 1(4 3)(4 5) 1

a a = =

1a =

Exercise 1C Graphs of circles and ellipses 1 a Centre (3, 1) and radius 2 (x 3)2 + (y 1)2 = 22

(x 3)2 + (y 1)2 = 4 b Centre ( 4, 3) and radius 6 (x 4)2 + (y 3)2 = 62

(x + 4)2 + (y + 3)2 = 36 c Centre (2, 0) and radius 3 (x 2)2 + (y 0)2 = 32

(x + 2)2 + y2 = 9 d Centre (0, 1) and radius 3 (x 0)2 + (y 1)2 = 2( 3) x2 + (y 1)2 = 3

2 a x2 + y2 + 6x 4y + 12 = 0 x2 + 6x + y2 4y + 12 = 0 (x + 3)2 9 + (y 2)2 4 + 12 = 0 (x + 3)2 + (y 2)2 = 1 Centre (3, 2) and radius 1.

b x2 + y2 4x 2y + 1 = 0 x2 4x + y2 2y + 1 = 0 (x 2)2 4 + (y 1)2 1 + 1 = 0 (x 2)2 + (y 1)2 = 4

Centre (2, 1) and radius 2.

c x2 + y2 5x = 0 x2 5x + y2 = 0 (x 5

2)2 + y2 = 25

4

Centre ( 52

, 0) and radius 52

.

d x2 + y2 + 10x 4y + 25 = 0 x2 + 10x + y2 4y + 25 = 0 (x + 5)2 25 + (y 2)2 4 + 25 = 0 (x + 5)2 + (y 2)2 = 4 Centre (5, 2) and radius 2.

e 2x2 + 2y2 + 8x 6y = 0 x2 + 4x + y2 3y = 0 (x + 2)2 4 + (y 3

2)2 9

4 = 0

(x + 2)2 + (y 32

)2 = 254

Centre (2, 32

) and radius 52

.

f 3x2 + 3y2 + 9x 12y = 12 x2 + 3x + y2 4y = 4 (x + 3

2)2 9

4 + (y 2)2 4 = 4

(x + 32

)2 + (y 2)2 = 94

Centre ( 32

, 2) and radius 32

.

3 a x = 2 cos(t), y = 2 sin(t), t [0, 2]

2x = cos(t)

2y = sin(t)

2

4x +

2

4y = 1

x2 + y2 = 4 Domain is range of x = 2 cos(t) = [2, 2] Range is range of y = 2 sin(t) = [2, 2] b x = 3 cos(2t), y = 3 sin(2t), t [0, ]

cos(2t) = 3x sin(2t) =

3y

2

9x +

2

9y = 1

x2 + y2 = 9 Domain is range of x = 3 cos(2t) = [3, 3] Range is range of y = 3 sin(2t) = [3, 3] c x = 2 4 cos(t), y = 1 + 4 sin(t),

t [0, ]

cos(t) = 24

x sin(t) = 1

4y

2(2 )

16x +

2( 1)16

y = 1

(2 x)2 + (y 1)2 = 16 Domain is range of x 4 cos(t) = [2 + 4, 2 + 4] = [2, 6] Range is range of y = 1 + 4 sin(t),

t [0, ] = [1 + 0, 1 + 4] = [1, 5]

d x = 2 sin(t) 3, y = 2 + 2 cos(t), t R

32

x + = sin(t) 22

y = cos(t)

2( 3)

2x + +

2( 2)2

y = 1

(x + 3)2 + (y 2)2 = 2 Domain is range of x = 2 sin(t) 3 = [ 2 3, 2 3] Range is range of y = 2 + 2 cos(t) = [ 2 + 2, 2 + 2]

4 a 2 2( 4) ( 9) 1

81 64x y ++ =

h = 4, k = 9 a2 = 81, b2 = 64 a = 9, b = 8 The answer is D.

b 2 2( 2) 1

36 144x y + =

h = 0, k = 2 a2 = 36, b2 = 144 a = 6, b = 12 The answer is C.

c 2 2( 3) ( 2) 1

81 4x y+ ++ =

h = 3, k = 2 a2 = 81, b2 = 4 a = 9, b = 2 The answer is A.

d 2 2( 12) 1

16 4x y + =

h = 12, k = 0 a2 = 16, b2 = 4 a = 4, b = 2 The answer is E.

5 a 2 2( 2) ( 3) 1

9 4x y + =

h = 2, k = 3 and so the centre is (2, 3)

a2 = 9, b2 = 4 a = 3, b = 2 Vertices are: (2 3, 3) (2 + 3, 3) = (1, 3) = (5, 3) and (2, 3 2) (2, 3 + 2) = (2, 1) = (2, 5)

b 2 2( 2) ( 4) 1

9 4x y ++ =


a2 = 9, b2 = 4 a = 3, b = 2 Vertices are:

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 13 (2 3, 4) (2 + 3, 4) = (1, 4) = (5, 4) and (2, 4 2) (2, 4 + 2) = (2, 6) = (2, 2)

c 2 2( 1) 1

9 5x y+ + =

h = 1, k = 0 and so the centre is (1, 0) a2 = 9, b2 = 5 a = 3, b = 5 Vertices are: (1 3, 0) (1 + 3, 0) = (4, 0) = (2, 0) and (1, 0 5) (1, 0 + 5)

= (1, 5) = (1, 5)

d 2 2( 4) 1

9 4x y + =



e 2 2( 3) ( 3) 1

25 4x y + =



f 2 2( 2) ( 3) 1

9 4x y ++ =

h = 2, k = 3 and so the centre is (2, 3) a2 = 9, b2 = 4 a = 3, b = 2 Vertices are: (2 3, 3) (2 + 3, 3) = (1, 3) = (5, 3) and (2, 3 2) (2, 3 + 2) = (2, 5) = (2, 1)

g 2 2( 1) ( 2) 1

16 5x y+ + =

h = 1, k = 2 and so the centre is (1, 2) a2 = 16, b2 = 5 a = 4, b = 5 Vertices are: (1 4, 2) (1 + 4, 2) = (5, 2) = (3, 2) and (1, 2 5 ) (1, 2 + 5 )

h 2 2( 2) ( 1) 1

9 7x y+ ++ =

h = 2, k = 1 and so the centre is (2, 1) a2 = 9, b2 = 7 a = 3, b = 7 Vertices are: (2 3, 1) (2 + 3, 1) = (5, 1) = (1, 1) and (2, 1 7 ) (2, 1 + 7 )

6 a 2 2( 2) ( 3) 1

4 9x y ++ =




b 2 2( 3) 1

9 16x y ++ =

h = 0, k = 3 and so the centre is (0, 3) a2 = 9, b2 = 16 a = 3, b = 4 Vertices are: (0 3, 3) (0 + 3, 3) = (3, 3) = (3, 3) and (0, 3 4) (0, 3 + 4) = (0, 7) = (0, 1)

c 2 2

19 15x y+ =


a2 = 9, b2 = 15 a = 3, b = 15 Vertices are: (0 3, 0) (0 + 3, 0) = (3, 0) = (3, 0) and (0, 0 15 ) (0, 0 + 15 ) = (0, 15 ) = (0, 15 )

d 2 2( 4) 1

9 25x y + =

h = 0, k = 4 Centre is (0, 4) a2 = 9, b2 = 25 a = 3, b = 5 Vertices are: (0 3, 4) (0 + 3, 4) = (3, 4) = (3, 4)

and (0, 4 5) (0, 4 + 5) = (0, 1) = (0, 9)

e 2 2

125 36x y+ =

h = 0, k = 0 Centre is (0, 0) a2 = 25, b2 = 36 a = 5, b = 6 Vertices are: (0 5, 0) (0 + 5, 0) = (5, 0) = (5, 0) and (0, 0 6) (0, 0 + 6) = (0, 6) = (0, 6)

f 2 2( 2) ( 1) 1

8 9x y ++ =

h = 2, k = 1 Centre is (2, 1) a2 = 8, b2 = 9 a = 8 , b = 3 = 2 2 Vertices are: (2 2 2 , 1) (2 + 2 2 , 1) and (2, 1 3) (2, 1 + 3) = (2, 4) = (2, 2)

g 2 2( 3) ( 2) 1

9 36x y + =


C o o r d i n a t e g e o m e t r y S M 1 2 - 1 15

h 2 2( 3) 1

4 12x y+ + =

h = 3, k = 0 Centre is (3, 0) a2 = 4, b2 = 12 a = 2, b = 12 = 2 3 Vertices are: (3 2, 0) (3 + 2, 0) = (5, 0) = (1, 0) and (3, 0 2 3 ) (3, 0 + 2 3 ) = (3, 2 3 ) = (3, 2 3 )

7 2 2( 3) ( 2) 1

25 16x y+ + =

a h = 3, k = 2 Centre is (3, 2) The answer is B. b a2 = 25, b2 = 16 a = 5, b = 4 Maximum and minimum vertices are: (3, 2 + 4) (3, 2 4) = (3, 6) = (3, 2) The answer is C.

8 a 9(x 5)2 + 16(y + 1)2 = 144

2 29( 5) 16( 1) 144

144 144 144x y ++ =

2 2( 5) ( 1) 1

16 9x y ++ =


b 16x2 + 25y2 = 400

2 216 25 400

400 400 400x y+ =

2 2

25 16x y+ = 1


c 16(x 2)2 + 25y2 = 400

2 216( 2) 25

400 400x y + = 400

400

2 2( 2)

25 16x y + = 1


d 4(x 1)2 + 16(y + 3)2 = 64

2 24( 1) 16( 3)

64 64x y ++ = 64

64

2 2( 1) ( 3)

16 4x y ++ = 1



e 16(x 2)2 + 9(y + 5)2 = 144

2 216( 2) 9( 5)

144 144x y ++ = 144

144

2 2( 2) ( 5)

9 16x y ++ = 1


f 25(x 3)2 + 9(y + 2)2 = 225

2 225( 3) 9( 2)

225 25x y ++ = 225

225

2 2( 3) ( 2)

9 25x y ++ = 1


9 a 9x2 72x + y2 4y + 112 = 0 9(x2 8x) + y2 4y + 112 = 0 9((x 4)2 16) + (y 2)2 4 + 112 = 0 9(x 4)2 144 + (y 2)2 + 108 = 0 9(x 4)2 + (y 2)2 = 36

2 2( 4) ( 2)

4 36x y + = 1

Centre (4, 2) Vertices (2, 2) (6, 2) (4, 4) (4, 8)

b 9x2 + 16y2 + 32y 128 = 0 9x2 + 16(y2 + 2y) 128 = 0 9x2 + 16((y + 1)2 1) 128 = 0 9x2 + 16(y + 1)2 16 128 = 0 9x2 + 16(y + 1)2 = 144

2 2( 1)

16 9x y ++ = 1

Centre (0, 1) Vertices (0, 4) (0, 2) ( 4, 1) (+4, 1)

c 9x2 + 54x + 7y2 + 18 = 0 9(x2 + 6x) + 7y2 + 18 = 0 9((x + 3)2 9) + 7y2 + 18 = 0 9(x + 3)2 81 + 7y2 + 18 = 0 9(x + 3)2 + 7y2 = 63

2 2( 3)

7 9x y+ + = 1

Centre (3, 0) Vertices (3 7, 0) (3 + 7 , 0) (3, 3) (3, 3)

d 4x2 32x + 9y2 54y + 109 = 0 4(x2 8x) + 9(y2 6y) + 109 = 0 4((x 4)2 16) + 9((y 3)2 9) + 109 = 0 4(x 4)2 64 + 9(y 3)2 81 + 109 = 0 4(x 4)2 + 9(y 3)2 = 36

2 2( 4) ( 3)

9 4x y + = 1

Centre (4, 3) Vertices (1, 3), (7, 3), (4, 1), (4, 5)

10 a x = 2 sin(t) y = 3 cos(t)

sin(t) = 2x cos(t) =

3y

2 2

14 9x y+ =

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 17 Ellipse centre (0, 0) Domain [2, 2] Range [3, 3] b x = 1 + 3 cos(t) y = 2 4 sin(t)

cos(t) = 13

x sin(t) = 24

y +

= ( 2)4

y

2 2( 1) ( 2) 1

9 16x y + =

Ellipse centre (1, 2) Domain [2, 4] Range [2, 6] c x = cos(2t) 2 y = 3 + 2 sin(2t)

cos(2t) = x + 2 sin(2t) = 32

y

2

2 ( 3)( 2) 14

yx + + =

Ellipse centre (2, 3) top half only Domain [3, 1] Range [3, 5]

11 a 2 2 9x y+ =

2 2

19 9x y+ =

x = 3 cos(t), y = 3 sin(t)

b 2 2

14 9x y+ =

x = 2 cos(t), y = 3 sin(t) c (x 3)2 + (y + 2)2 = 16

2 2( 3) ( 2)

16 16x y ++ = 1

x = 3 + 4 cos(t), y = 4 sin(t) 2

d 2 2( 3) ( 1)

25 9x y+ + = 1

2 2

2 2( 3) ( 1)

5 3x y+ + = 1

x = 3 + 5 cos(t), y = 1 + 3 sin(t)

Exercise 1D Graphs of hyperbolas

1 a 2 2

81 64x y = 1

h = 0, k = 0 so no translations. a2 = 81, b2 = 64 a = 9, b = 8 Centre (0, 0) Vertices (9, 0) (9, 0) Asymptotes y = 8

9x and y = 8

9x

The answer is A.

b 2 2( 3) 1

36 144x y =

h = 0, k = 3 a2 = 36, b2 = 144 a = 6, b = 12 Centre (0, 3) Vertices are (0 6, 3) and (0 + 6, 3) or ( 6, 3) and (6, 3) Asymptotes: y 3 = 12

6(x) and y 3 = 12

6(x)

= 2x = 2x y = 3 2x and y = 3 + 2x The answer is B.

c 2 2( 3) 1

9 4x y+ =

h = 3, k = 0 a2 = 9, b2 = 4 a = 3, b = 2 Centre (3, 0) Vertices are (3 3, 0) and (3 + 3, 0) or ( 6, 0) and (0, 0) Asymptotes: y 0 = 2

3( 3)x +

or y 0 = 23

( 3)x +

3y = 2(x + 3) or 3y = 2(x + 3) 3y = 2x 6 or 3y = 2x + 6 3y + 2x + 6 = 0 or 3y 2x 6 = 0 The answer is E.

d 2 2( 2) ( 1) 1

16 4x y =

h = 2, k = 1 a2 = 16, b2 = 4 a = 4, b = 2 Centre (2, 1) Vertices are (2 4, 1) and (2 + 4, 1) or (2, 1) and (6, 1) Asymptotes: y 1 = 2

4( 2)x and y 1 = 2

4( 2)x

= 12

( 2)x = 12

( 2)x or 2(y 1) = (x 2) and 2(y 1) = (x 2) 2y 2 = x + 2 and 2y 2 = x 2 or 2y = 4 x and 2y = x The answer is C.

2 a 2 2

19 4x y =

h = 0, k = 0 a2 = 9, b2 = 4 a = 3, b = 2 Centre (0, 0) Vertices are (0 3, 0) and (0 + 3, 0) or (3, 0) and (3, 0) Asymptotes: y = 2

3x and y = 2

3x

b 2 2

14 9x y =

h = 0, k = 0 a2 = 4, b2 = 9 a = 2, b = 3 Centre (0, 0) Vertices (2, 0) and (2, 0) Asymptotes: y = 3

2x and y = 3

2x


c 2 2

19 9x y =


3x and y = 3

3x

or y = x and y = x

d 2 2

19 16x y =


3x and y = 4

3x

e 2 2

14 4x y =


2x and y = 2

2x

or y = x and y = x

f 2 2

116 4x y =


4x and y = 2

4x

or y = 12

x and y = 12

x

g 2 2( 2) ( 3) 1

9 4x y =

h = 2, k = 3 a2 = 9, b2 = 4 a = 3, b = 2 Centre (2, 3) Vertices (2 3, 3) and (2 + 3, 3) or (1, 3) and (5, 3) Asymptotes: y 3 = 2

3( 2)x and y 3 = 2

3( 2)x

3(y 3) = 2(x 2) and 3(y 3) = 2(x 2) 3y 9 = 2x + 4 and 3y 9 = 2x 4 3y + 2x = 13 3y 2x = 5 For 3y + 2x = 13 For 3y 2x = 5 x = 0, 3y = 13 x = 0, 3y = 5 y = 13

3 y = 5

3

(0, 133

) (0, 53

)

y = 0, 2x = 13 y = 0, 2x = 5 x = 13

2 x = 5

2

( 132

, 0) ( 52

, 0)

h 2 2( 2) ( 3) 1

25 4x y =


5( 2)x and y 3 = 2

5( 2)x

5(y 3) = 2(x 2) 5(y 3) = 2(x 2) 5y 15 = 2x + 4 5y 15 = 2x 4 5y + 2x = 19 5y 2x = 11 x = 0, 5y = 19 x = 0, 5y = 11 y = 19

5 y = 11

5

(0, 195

) (0, 115

)

y = 0, 2x = 19 y = 0, 2x = 11 x = 19

2 x = 11

2

( 192

, 0) ( 112

, 0)

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 19 i

2 2( 2) ( 2) 19 25

x y =


3( 2)x and y 2 = 5

3( 2)x


3 y = 4

3

(0, 163

) (0, 43

)

y = 0, 5x = 16 y = 0, 5x = 4 x = 16

5 x = 4

5

( 165

, 0) ( 45

, 0)

j 2 2( 1) 1

9 25x y =

h = 0, k = 1 a2 = 9, b2 = 25 a = 3, b = 5 Centre (0, 1) Vertices (3, 1) and (3, 1) Asymptotes: y 1 = 5

3x and y 1 = 5

3x

3(y 1) = 5x 3(y 1) = 5x 3y 3 = 5x 3y 3 = 5x 3y + 5x = 3 3y 5x = 3 x = 0, 3y = 3 x = 0, 3y = 3 y = 1 y = 1 (0, 1) (0, 1) y = 0, 5x = 3 y = 0, 5x = 3 x = 3

5 x = 3

5

( 35

, 0) ( 35

, 0)

k 2 2( 3) ( 3) 1

25 4x y =

h = 3, k = 3 a2 = 25, b2 = 4 a = 5, b = 2 Centre (3, 3) Vertices (3 5, 3) and (3 + 5, 3) or (2, 3) and (8, 3)

Asymptotes: y 3 = 2

5( 3)x and y 3 = 2

5( 3)x

5(y 3) = 2(x 3) 5(y 3) = 2(x 3) 5y 15 = 2x + 6 5y 15 = 2x 6 5y + 2x = 21 5y 2x = 9 x = 0, 5 y = 21 x = 0, 5y = 9 y = 21

5 y = 9

5

(0, 215

) (0, 95

)

y = 0, 2x = 21 y = 0, 2x = 9 x = 21

2 x = 9

2

( 212

, 0) ( 92

, 0)

l 2 2( 1) ( 1) 1

25 9x y + =

h = 1, k = 1 a2 = 25, b2 = 9 a = 5, b = 3 Centre (1, 1) Vertices (1 5, 1) and (1 + 5, 1) or ( 4, 1) and (6, 1) Asymptotes: y + 1 = 3

5( 1)x and y + 1 = 3

5( 1)x

5(y + 1) = 3(x 1) 5(y + 1) = 3(x 1) 5y + 5 = 3x + 3 5y + 5 = 3x 3 5y + 3x = 2 5y 3x = 8 x = 0, 5y = 2 x = 0, 5y = 8 y = 2

5 y = 8

5

(0, 25

) (0, 85

)

y = 0, 3x = 2 y = 0, 3x = 8 x = 2

3 x = 8

3

( 23

, 0) ( 83

, 0)

3 Centre (0, 0) Vertices (2, 0) and (2, 0) Asymptotes: y = 1

2x and y = 1

2x

h = 0, k = 0 a = 2, b = 1

Rule is 2 2

2 22 1x y = 1

or 2 2

4 1x y = 1

The answer is D.


4 Centre (5, 3) a2 = 9, b2 = 4 a = 3, b = 2 Vertices (5 3, 3) and (5 + 3, 3) or (2, 3) and (8, 3) The answer is A. 5 a 25x2 16y2 = 400

2 225 16

400 400x y = 400

400

2 2

16 25x y = 1


4x and y = 5

4x

b 9x2 16y2 = 144

2 29 16

144 144x y = 144

144

2 2

16 9x y = 1

h = 0, k = 0 a2 = 16, b2 = 9 a = 4, b = 3 Centre (0, 0) Vertices ( 4, 0) and (4, 0) Asymptotes: y = 3

4x and y = 3

4x

c x2 y2 = 25

2 2

25 25x y = 25

25

2 2

25 25x y = 1

h = 0, k = 0 Centre (0, 0) a2 = 25, b2 = 25 a = 5, b = 5 Vertices (5, 0) and (5, 0) Asymptotes: y = 5

5x and y = 5

5x

or y = x and y = x

d 9x2 25y2 = 225

2 29 25

225 225x y = 225

225

2 2

25 9x y = 1


5x and y = 3

5x

e 16(x 2)2 9(y + 5)2 = 144

2 216( 2) 9( 5)

144 144x y + = 144

144

2 2( 2) ( 5)

9 16x y + = 1

h = 2, k = 5 a2 = 9, b2 = 16 a = 3, b = 4 Centre (2, 5) Vertices (2 3, 5) and (2 + 3, 5) or (1, 5) and (5, 5) Asymptotes: y + 5 = 4

3(x 2) and y + 5 = 4

3(x 2)

3(y + 5) = 4(x 2) 3(y + 5) = 4(x 2) 3y + 15 = 4x + 8 3y + 15 = 4x 8 3y + 4x = 7 3y 4x = 23 x = 0, 3y = 7 x = 0, 3y = 23 y = 7

3 y = 23

3

(0, 73

) (0, 233

)

y = 0, 4x = 7 y = 0, 4x = 23 x = 7

4 x = 23

4

( 74

, 0) ( 234

, 0)

f 25(x 3)2 9(y + 2)2 = 225

2 225( 3) 9( 2)

225 225x y + = 225

225

2 2( 3) ( 2)

9 25x y + = 1

h = 3, k = 2 a2 = 9, b2 = 25 a = 3, b = 5 Centre (3, 2)

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 21 Vertices (3 3, 2) and (3 + 3, 2) or (0, 2) and (6, 2) Asymptotes: y + 2 = 5

3(x 3) and y + 2 = 5

3(x 3)

3(y + 2) = 5(x 3) 3(y + 2) = 5(x 3) 3y + 6 = 5x + 15 3y + 6 = 5x 15 3y + 5x = 9 3y 5x = 21 x = 0, 3y = 9 x = 0, 3y = 21 y = 3 y = 7 (0, 3) (0, 7) y = 0, 5x = 9 y = 0, 5x = 21 x = 9

5 x = 21

5

( 95

, 0) ( 215

, 0)

g 36(x 4)2 4(y 2)2 = 144

2 236( 4) 4( 2)

144 144x y = 144

144

2 2( 4) ( 2)

4 36x y = 1


2(x 4) and y 2 = 6

2(x 4)

y 2 = 3(x 4) y 2 = 3(x 4) = 3x + 12 = 3x 12 y + 3x = 14 y 3x = 10 x = 0, y = 14 x = 0, y = 10 (0, 14) (0, 10) y = 0, 3x = 14 y = 0, 3x = 10 x = 14

3 x = 10

3

( 143

, 0) ( 103

, 0)

h 9x2 16(y + 1)2 = 144

2 29 16( 1)

144 9x y + = 144

144

2 2( 1)

16 9x y + = 1

h = 0, k = 1 a2 = 16, b2 = 9 a = 4, b = 3 Centre (0, 1) Vertices ( 4, 1) and (4, 1)

Asymptotes: y + 1 = 3

4(x 0) and y + 1 = 3

4(x 0)

4(y + 1) = 3x 4(y + 1) = 3x 4y + 4 = 3x 4y + 4 = 3x 4y + 3x = 4 4y 3x = 4 x = 0, 4y = 4 x = 0, 4y = 4 y = 1 y = 1 (0, 1) (0, 1) y = 0, 3x = 4 y = 0, 3x = 4 x = 4

3 x = 4

3

( 43

, 0) ( 43

, 0)

i 9(x + 3)2 7y2 = 63

2 29( 3) 7

63 63x y+ = 63

63

2 2( 3)

7 9x y+ = 1

h = 3, k = 0 a2 = 7, b2 = 9 a = 7 , b = 3 Centre (3, 0) Vertices (3 7 , 0) and (3 + 7 , 0) Asymptotes: y 0 = 3

7(x + 3) and y 0 = 3

7(x + 3)

7y = 3(x + 3) 7y = 3(x + 3) = 3x 9 = 3x + 9 7 y + 3x = 9 7 y 3x = 9

x = 0, 7y = 9 x = 0, 7y = 9

y = 97

y = 97

(0, 97

) (0, 97

)

y = 0, 3x = 9 y = 0, 3x = 9 x = 3 x = 3 (3, 0) (3, 0)

j 4(x 4)2 9(y 3)2 = 36

2 24( 4) 9( 3)

36 36x y = 36

36

2 2( 4) ( 3)

9 4x y = 1

h = 4, k = 3 a2 = 9, b2 = 4 a = 3, b = 2 Centre (4, 3) Vertices (4 3, 3) and (4 + 3, 3) or (1, 3) and (7, 3)


Asymptotes: y 3 = 2

3(x 4) and y 3 = 2

3(x 4)


3 y = 1

3

(0, 173

) (0, 13

)

y = 0, 2x = 17 y = 0, 2x = 1 x = 17

2 x = 1

2

( 172

, 0) ( 12

, 0)

6 a x = 4 sec(t) and y = 3 tan(t), t R

sec(t) =4x tan(t) =

3y

sec2(t) = 2

16x tan2(t) =

2

9y

2 2

16 9x y = 2 2sec ( ) tan ( )t t

= 1 Hyperbola with centre (0, 0) Domain (, 4] [4, ) Range R. b x = 2 sec(2t) and y = 3 tan(2t), t R

sec(2t) = 2x tan(2t) =

3y

sec2(2t) = 2

4x tan2(2t) =

2

9y

2 2

4 9x y = 2 2sec (2 ) tan (2 )t t

= 1 Hyperbola with centre (0, 0) Domain (, 2] [2, ) Range R.

c x = 1 sec(t) and y = 2 + tan(t), 3,2 2

t

sec(t) = 1 x tan(t) = y 2 = (x 1) sec2(t) = (x 1)2 tan2(t) = (y 2)2

(x 1)2 (y 2)2 = sec2(t) tan2(t) = 1 Hyperbola with centre (1, 2). Right branch Domain [2, ) Range R.

d x = 2 + 3 sec(2t) and y = 4 5 tan(2t), t ,4 4

sec(2t) = 23

x tan(2t) = 45

y

= ( 4)5

y

2 2( 2) ( 4)

9 25x y = 2 2sec ( ) tan ( )t t

= 1

Hyperbola with centre (2, 4). Right Branch Domain [5, ) Range R. 7 a a = 2, b = 3, h = 2 and k = 1 x = 2 + 2 sec(t) y = 1 + 3 tan(t) b a = 5, b = 4, h = 3 and k = 1 x = 3 + 5 sec(t) y = 1 + 4 tan(t)

8 General equation of an hyperbola is 2 2

2 2( ) ( ) 1x h y k

a b =

Equation of asymptotes;

( )by k x ha

=

So 32

20 ( 0)y x =

43

y x=

Exercise 1E Partial fractions 1 a a(x 2) + b(x + 1) 3x Let x = 2, 3b = 6 b = 2 Let x = 1, 3a = 3 a = 1 a = 1 and b = 2 b a(x + 3) + b(x 3) x + 9 Let x = 3, 6b = 3 + 9 6b = 6 b = 1 Let x = 3, 6a = 3 + 9 6a = 12 a = 2 a = 2 and b = 1 c a(x + 2) + b 3x + 10 Let x = 2, b = 6 + 10 b = 4 Let x = 0, 2a + b = 10 2a + 4 = 10 2a = 6 a = 3 a = 3 and b = 4 d a(x 4) + b(2x 1) 4x + 5 Let x = 4, b(8 1) = 16 + 5 7b = 21 b = 3 Let x = 1

2, 7

2 a = 2 + 5

72

a = 7

7a = 14 a = 2 a = 2 and b = 3 e a(3x + 5) + b(4x 3) 29 Let x = 5

3, b( 20

3 3) = 29

293 b = 29

b = 3 Let x = 3

4, a( 9

4+ 5) = 29

294

a = 29

a = 4 a = 4 and b = 3

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 23 f a(2x + 3) + b(5 x) 4 6x Let x = 3

2, b(5 + 3

2) = 4 + 9

132

b = 13

b = 2 Let x = 5, a(10 + 3) = 4 30 13a = 26 a = 2 a = 2 and b = 2

2 a 3 5( 2)( 1)

xx x

++ +

2 1

a bx x

++ +

( 1) ( 2)( 2)( 1)

a x b xx x+ + ++ +

3x + 5 a(x + 1) + b(x + 2) Let x = 1, 2 = b b = 2 Let x = 2, 1 = a a = 1

3 5( 2)( 1)

xx x

++ +

1 22 1x x

++ +

b 7 3( 3)( 3)

xx x

+ +

3 3

a bx x

+ +

( 3) ( 3)( 3)( 3)

a x b xx x+ + +

7x + 3 a(x + 3) + b(x 3) Let x = 3, 18 = 6b b = 3 Let x = 3, 24 = 6a a = 4

7 3( 3)( 3)

xx x

+ +

4 33 3x x

+ +

c 5 4( 2)( 5)

xx x

+ +

2 5

a bx x

+ +

( 5) ( 2)( 2)( 5)

a x b xx x+ + +

5x + 4 a (x + 5) + b(x 2) Let x = 5, 21 = 7b b = 3 Let x = 2, 14 = 7a a = 2

5 4( 2)( 5)

xx x

+ +

2 32 5x x

+ +

d 7 14(2 1)( 4)

xx x

2 1 4

a bx x

+

( 4) (2 1)(2 1)( 4)

a x b xx x

+

7x 14 a(x 4) + b(2x 1) Let x = 4, 14 = 7b b = 2 Let x = 1

2, 7

2 14 = 7

2a

212

= 72

a

a = 3

7 14(2 1)( 4)

xx x

3 22 1 4x x

+

e 2 14

( 1)( 3)x

x x

+

1 3a b

x x+

+

( 3) ( 1)( 1)( 3)

a x b xx x + ++

2x 14 a(x 3) + b(x + 1)

Let x = 3, 8 = 4b b = 2 Let x = 1, 16 = 4a a = 4

2 14( 1)( 3)

xx x

+

4 21 3x x

+

f 22( 4)(2 5)

xx x

+

4 2 5

a bx x

+ +

(2 5) ( 4)( 4)(2 5)

a x b xx x

+ + +

x 22 a(2x + 5) + b(x 4) Let x = 5

2, 5

2 22 = (b 5

2 4)

392

= 132

b

b = 3 Let x = 4, 26 = 13a a = 2

22( 4)(2 5)

xx x

+

2 34 2 5x x

+ +

3 ( 5)ax b x+ 3 10x + Let x = 0, 5b = 10 b = 2 Let x = 5, 5a = 15 + 10 = 25 a = 5 a = 5 and b = 2 The answer is B.

4 15( 3)( 1)

xx x

++

3 1

a bx x

++

( 1) ( 3)( 3)( 1)

a x b xx x + ++

x + 15 ( 1) ( 3)a x b x + + Let x = 1, 16 = 4b b = 4 Let x = 3, 12 = 4a a = 3 a = 3 and b = 4 The answer is E.

5 a 22 5

( 1)x

x+

+ 21 ( 1)

a bx x

++ +

2( 1)( 1)

a x bx+ ++

2 5x + ( 1)a x b+ + Let x = 1, 3 = b Let x = 0, 5 = a + 3 a = 2

22 5

( 1)x

x+

+ 2

2 31 ( 1)x x

++ +

b 24 9

( 2)x

x

22 ( 2)

a bx x

+

2( 2)( 2)

a x bx +

4 9x ( 2)a x b + Let x = 2, 1 = b Let x = 0, 9 = 2a 1 2a = 8 a = 4

24 9

( 2)x

x

2

4 12 ( 2)x x


c 23 7

( 3)x

x

+ 23 ( 3)

a bx x

++ +

2( 3)( 3)

a x bx+ ++

3 7x ( 3)a x b+ + Let x = 3, 2 = b Let x = 0, 7 = 3a + 2 3a = 9 a = 3

23 7

( 3)x

x

+ 2

3 23 ( 3)x x

++ +

d 210 9(2 1)

xx

22 1 (2 1)a b

x x+

2(2 1)(2 1)

a x bx +

10 9x (2 1)a x b +

Let x = 12

, 5 9 = b

b = 4 Let x = 0, 9 = a 4 a = 5 a = 5

210 9(2 1)

xx

25 4

2 1 (2 1)x x

e 22 5

(4 )x

x

24 (4 )

a bx x

+

2(4 )(4 )

a x bx

+

2 5x (4 )a x b + Let x = 4, 3 = b Let x = 0, 5 = 4a + 3 4a = 8 a = 2

22 5

(4 )x

x

2

2 34 (4 )x x +

6 a 23 4

6x

x x

= 3 4

( 3)( 2)x

x x

3 2

a bx x

+ +

2( 2) ( 3)

6a x b x

x x+ +

3 4x ( 2) ( 3)a x b x+ + Let x = 2, 10 = 5b b = 2 Let x = 3, 5 = 5a a = 1

23 4

6x

x x

1 2

3 2x x+

+

b 217

6 7x

x x

+ = 17

( 7)( 1)x

x x

+

7 1

a bx x

++

2( 1) ( 7)

6 7a x b x

x x + +

+

17x ( 1) ( 7)a x b x + + Let x = 1, 16 = 8b b = 2

Let x = 7, 24 = 8a a = 3

217

6 7x

x x

+ 3 2

7 1x x

+

c 27 10

2 8x

x x

= 7 10

( 4)( 2)x

x x

+

4 2

a bx x

+ +

2( 2) ( 4)

2 8a x b x

x x+ +

7 10x ( 2) ( 4)a x b x+ + Let x = 2, 24 = 6b b = 4 Let x = 4, 18 = 6a a = 3

27 10

2 8x

x x

3 4

4 2x x+

+

d 22525

xx

= 25

( 5)( 5)x

x x +

5 5

a bx x

+ +

2( 5) ( 5)

25a x b x

x+ +

x 25 a(x + 5) + b(x 5) Let x = 5, 20 = 10b b = 2 Let x = 5, 30 = 10a a = 3

22525

xx

3 2

5 5x x + +

e 23 19

2 5 3x

x x

= 3 19

(2 1)( 3)x

x x +

2 1 3

a bx x

++

2( 3) (2 1)

2 5 3a x b x

x x + +

3x 19 a(x 3) + b(2x + 1) Let x = 3, 28 = 7b b = 4 Let x = 1

2, 17 1

2 = (3 1

2)a

a = 5

23 19

2 5 3x

x x

5 4

2 1 3x x

+

f 240

8 2x

x x

+ = 40

(2 )(4 )x

x x

+

2 4

a bx x

++

2(4 ) (2 )

8 2a x b x

x x + +

+

40 x a(4 x) + b(2 + x) Let x = 4, 36 = 6b b = 6 Let x = 2, 42 = 6a a = 7

240

8 2x

x x

+ 7 6

2 4x x+

+

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 25 7 a

22 6 5( 2)( 1)

x xx x

+ +

= 2

22 6 5

2x xx x

+ +

2 2

2

22 2 6 5

2 2 44 1

x x x x

x xx

+ +

+

22 6 5

( 2)( 1)x x

x x+

+ 4 12

( 2)( 1)x

x x+

+

4 1( 2)( 1)

xx x

+

2 1

a bx x

++

( 1) ( 2)( 2)( 1)

a x b xx x + ++

4 1x ( 1) ( 2)a x b x + + Let x = 1, 3 = 3b b = 1 Let x = 2, 9 = 3a a = 3

4 1( 2)( 1)

xx x

+

3 12 1x x

++

22 6 5

( 2)( 1)x x

x x+

+ 3 12

2 1x x+ +

+

b 23 4 11

( 3)( 4)x xx x

+ +

= 2

23 4 11

12x xx x

+ +

2 2

2

312 3 4 11

3 3 3625

x x x x

x xx

+ +

+ +

23 4 11

( 3)( 4)x xx x

+ +

253( 3)( 4)

xx x

++ +

25( 3)( 4)

xx x

+ +

3 4

a bx x

+ +

( 4) ( 3)( 3)( 4)

a x b xx x+ + +

25x + ( 4) ( 3)a x b x+ + Let x = 4, 21 = 7b b = 3 Let x = 3, 28 = 7a a = 4

25( 3)( 4)

xx x

+ +

4 33 4x x

+

23 4 11

( 3)( 4)x xx x

+ +

4 333 4x x

+ +

c 3 24 4 18 30( 2)(2 5)

x x xx x+

+ =

3 2

24 4 18 30

2 10x x x

x x+

+

2 3

3 2

2

2

2 12 10 4 4 18 30

4 2 20

2 2 30

2 1020

xx x x x x

x x x

x x

x xx

2+

+ +

+

+

+

3 24 4 18 30( 2)(2 5)

x x xx x+

+ 202 1

( 2)(2 5)xx

x x+ +

+

20( 2)(2 5)

xx x

+

2 2 5

a bx x

+ +

(2 5) ( 2)( 2)(2 5)

a x b xx x

+ + +

20x (2 5) ( 2)a x b x+ +

Let x = 52

, 52

20 = b( 52

2)

452

= 92

b

b = 5 Let x = 2, 18 = 9a a = 2

3 24 4 18 30( 2)(2 5)

x x xx x+

+ 2 52 1

2 2 5x

x x+ +

+

d 2

24 54

12x xx x

+

= 24 54

( 4)( 3)x x

x x+

+

2 2

2

412 4 54

4 4 485 6

x x x x

x xx

+

24 54

( 4)( 3)x x

x x+

+ 5 64

( 4)( 3)x

x x+

+

5 6( 4)( 3)

xx x

+

4 3

a bx x

+ +

( 3) ( 4)( 4)( 3)

a x b xx x+ + +

5x 6 a(x + 3) + b(x 4) Let x = 3, 21 = 7b b = 3 Let x = 4, 14 = 7a a = 2

2

24 54

12x xx x

+

2 344 3x x

+ + +

e 3 2

24 28 37 10

3 8 4x x x

x x + +

+ =

3 24 28 37 10(2 1)(3 2 )

x x xx x

+ +

2 3

3 2

2

2

54 8 3 4 28 37 10

4 8 3

20 34 10

20 40 156 5

xx x x x x

x x x

x x

x xx

2

+ +

+

+

+

3 2

24 28 37 10

3 8 4x x x

x x + +

+ x 5 + 6 5

(2 1)(3 2 )x

x x

6 5(2 1)(3 2 )

xx x

2 1 3 2

a bx x

+

(3 2 ) (2 1)(2 1)(3 2 )

a x b xx x

+

6x 5 a(3 2x) + b(2x 1) Let x = 3

2, 4 = 2b

b = 2 Let x = 1

2, 2 = 2a

a = 1

3 2

24 28 37 10

3 8 4x x x

x x + +

+ x 5 1 2

2 1 3 2x x+

f 3 236 39 5 17

212 5 3

x x x

x x

+

+ =

3 236 39 5 17(4 3)(3 1)x x x

x x+

+


2 3

3 2

2

2

3 212 5 3 36 39 5 17

36 15 9

24 4 17

24 10 66 11

xx x x x x

x x x

x x

x xx

2+

+ +

+

+

+

3 2

236 39 5 17

12 5 3x x x

x x+

+ 3x + 2 + 6 11

(4 3)(3 1)x

x x +

6 11(4 3)(3 1)

xx x +

4 3 3 1

a bx x

++

(3 1) (4 3)(4 3)(3 1)

a x b xx x + ++

6x 11 a(3x 1) + b(4x + 3) Let x = 1

3, 13 = 13

3b

b = 3 Let x = 3

4, 13

2 = 13

4a

a = 2

3 2

236 39 5 17

12 5 3x x x

x x+

+ 3x + 2 + 2 3

4 3 3 1x x

+

8 a 2 3

3 2

2

2

43 10 17 51

3 10

4 7 51

4 12 405 11

xx x x x x

x x x

x x

x xx

2

+ +

+

+

++

3 2

217 51

3 10x x x

x x +

+ x 4 + 2

5 113 10

xx x

++

The answer is D.

b 25 11

3 10x

x x+

+ = 5 11

( 5)( 2)x

x x+

+

5 2

a bx x

++

( 2) ( 5)( 5)( 2)

a x b xx x + ++

5x + 11 a(x 2) + b(x + 5) Let x = 2, 21 = 7b b = 3 Let x = 5, 14 = 7a a = 2

3 2

217 51

3 10x x x

x x +

+ x 4 + 2 3

5 2x x+

+

The answer is A.

Exercise 1F Sketch graphs using partial fractions

1 a y = 12x

Asymptotes: x = 2, y = 0

b y = 34x +


c y = 21x


d y = 3 13 2

xx

1

3 2 3 13 2

1

x xx

y = 3 13 2

xx

1 + 13 2x

Asymptotes: x = 23

, y = 1

e y = 4 52 1xx

++

2

2 1 4 54 2

3

x xx

+ ++

y = 4 52 1xx

++

2 + 32 1x +

Asymptotes: x = 12

, y = 2

f y = 4 143

xx

+

4

3 4 144 12

2

x xx

+

+

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 27 y = 4 14

3x

x +

y 4 + 23x


g y = 2 73

xx

+

2

3 2 72 6

1

x xx

+

y = 2 73

xx

+

2 13x +


2 a y = 2 1

1x x

x +

2

2

1 1

1

xx x x

x x

+

y x + 11x

Asymptotes: x = 1, y = x

b y = 22 4 3

2x x

x+ ++

2

2

22 2 4 3

2 43

xx x x

x x

+ + +

+

y 2x + 32x +

Asymptotes: x = 2, y = 2x

c y = 2 4 5

1x x

x+ +

+

2

2

31 4 5

3 53 3

2

xx x x

x xxx

++ + +

+++

y x + 3 + 21x +

Asymptotes: x = 1, y = x + 3

d y = 2 7 11

3x x

x +

2

2

43 7 11

34 114 12

1

xx x x

x xxx

+

+ +

y x 4 13x

Asymptotes: x = 3, y = x 4

e y = 23 4 3

2x x

x

2

2

3 22 3 4 3

3 62 32 4

1

xx x x

x xxx

+

y 3x + 2 + 12x

Asymptotes: x = 2, y = 3x + 2

3 a y = 2 2( 2)( 4)

xx x

+ +

2 4

a bx x

+ +


( 4) ( 2)( 2)( 4)

a x b xx x+ + +

2x + 2 a(x + 4) + b(x 2) Let x = 4, 6 = 6b b = 1 Let x = 2, 6 = 6a a = 1

y 12x

+ 14x +

For y = 0: 12x

+ 14x +

= 0

1(x + 4) + 1(x 2) = 0 x + 4 + x 2 = 0 2x + 2 = 0 2x = 2 x = 1 x = 1 when y = 0 Vertical asymptotes: x = 2, x = 4

b y = 3( 2)( 1)x x

+

2 1

a bx x

++

( 1) ( 2)( 2)( 1)

a x b xx x + ++

3 a(x 1) + b(x + 2) Let x = 1, 3 = 3b b = 1 Let x = 2, 3 = 3a a = 1

y 1 12 1x x

+

For y = 0: 1 12 1x x

+

= 0

1(x 1) 1(x + 2) = 0 x 1 x 2 = 0 No x-intercept. Vertical asymptotes: x = 2, x = 1

c y = 25 2

4x

x

= 5 2( 2)( 2)

xx x

+

2 2

a bx x

+ +

( 2) ( 2)( 2)( 2)

a x b xx x+ + +

5x 2 a(x + 2) + b(x 2) Let x = 2, 12 = 4b b = 3

Let x = 2, 8 = 4a a = 2

y = 2 32 2x x

+ +

For y = 0: 2 32 2x x

+ +

= 0

2(x + 2) + 3(x 2) = 0 2x + 4 + 3x 6 = 0 5x 2 = 0 5x = 2 x = 2

5

x = 25

when y = 0


d y = 23 5

2 3x

x x

= 3 5( 3)( 1)

xx x

+

3 1

a bx x

+ +

( 1) ( 3)( 3)( 1)

a x b xx x+ + +


y 1 23 1x x

+ +

For y = 0: 1 23 1x x

+ +

= 0

1(x + 1) + 2(x 3) = 0 x + 1 + 2x 6 = 0 3x 5 = 0 3x = 5 x = 5

3

x = 53

when y = 0


e y = 22

2 7 3x

x x +

+ +

= 2(2 1)( 3)

xx x

++ +

2 1 3

a bx x

++ +

( 3) (2 1)(2 1)( 3)

a x b xx x

+ + ++ +

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 29 x + 2 a(x + 3) + b(2x + 1) Let x = 3, 5 = 5b b = 1 Let x = 1

2, 2 1

2 = (2 1

2)a

a = 1

y 1 12 1 3x x

+ +

For y = 0: 1 12 1 3x x

+ +

= 0

1(x + 3) 1(2x + 1) = 0 x + 3 2x 1 = 0 x + 2 = 0 x = 2 x = 2 when y = 0 Vertical asymptotes: 2x + 1 = 0 and x + 3 = 0 or x = 1

2, x = 3

4 a f (x) = 23 4

3 2x

x x+

+ +

= 3 4( 2)( 1)

xx x

++ +

2 1

a bx x

++ +

( 1) ( 2)( 2)( 1)

a x b xx x+ + ++ +

3x + 4 a(x + 1) + b(x + 2) Let x = 1, 1 = b Let x = 2, 2 = a a = 2

f(x) 2 12 1x x

++ +

The answer is C. b For f(x) = 0:

2 12 1x x

++ +

= 0

2(x + 1) + 1(x + 2) = 0 2x + 2 + x + 2 = 0 3x + 4 = 0 3x = 4 x = 4

3

f (x) = 0 when x = 43

Vertical asymptotes: x = 2, x = 1 The answer is D.

5 2 3

3 2

212 2 2 11 7

2 2 2413 7

xx x x x x

x x xx

2 +

+

3 2

22 2 11 7

12x x x

x x +

2x + 2

13 712

xx x

+

2x + 13 7( 4)( 3)

xx x

+ +

Asymptotes: x = 4, x = 3 The answer is C.

6 Asymptotes where x2 4 = 0 (x 2)(x + 2) = 0 x = 2 or x = 2 The answer is D.

7 a 2 2

2

21 2 2 2

2 22

x x x

xx

+

f (x) 2 + 22

1x

x

= 2 + 2( 1)( 1)

xx x +

2( 1)( 1)

xx x +

1 1

a bx x

+ +

( 1) ( 1)( 1)( 1)

a x b xx x+ + +

2x a(x + 1) + b(x 1) Let x = 1, 2 = 2b b = 1 Let x = 1, 2 = 2a a = 1

f (x) = 2 + 1 11 1x x

+ +

Asymptotes are x = 1 and x = 1

1 11 1x x

+ +

= 0

x + 1 + x 1 = 0 2x = 0 x = 0 x = 0 is where y = 2

b f (x) = 2

25

2x

x x+

2 2

2

12 5

27

x x x

x xx

+

+

f (x) 1 + 27

2x

x x+

27

2x

x x+

= 7

( 2)( 1)x

x x+

+

2 1

a bx x

+ +

( 1) ( 2)( 2)( 1)

a x b xx x+ + +

x + 7 a(x + 1) + b(x 2) Let x = 1, 6 = 3b b = 2 Let x = 2, 9 = 3a a = 3

f (x) 1 + 3 22 1x x

+

Asymptotes: x = 2 and x = 1


3 22 1x x

+

= 0

3(x + 1) 2(x 2) = 0 3x + 3 2x + 4 = 0 x + 7 = 0 x = 7 x = 7 is where y = 1

c g(x) = 2

23 6 14

6x xx x

+ +

2 2

2

36 3 6 14

3 3 183 4

x x x x

x xx

+ +

+ +

g(x) 3 + 23 4

6x

x x

23 4

6x

x x

= 3 4

( 3)( 2)x

x x

+

3 2

a bx x

+ +

( 2) ( 3)( 3)( 2)

a x b xx x+ + +


g(x) 3 + 1 23 2x x

+ +


1 23 2x x

+ +

= 0

1(x + 2) + 2(x 3) = 0 x + 2 + 2x 6 = 0 3x 4 = 0 3x = 4 x = 4

3

x = 43

is where y = 3

d g(x) = 2

230 10

6 2x

x x

2 2

2

56 2 30 10

30 5 105

x x x

x xx

g(x) 5 + 25

6 2x

x x

25

6 2x

x x = 5

(3 2)(2 1)x

x x +

3 2 2 1

a bx x

+ +

(2 1) (3 2)(3 2)(2 1)

a x b xx x+ + +

5x (2 1) (3 2)a x b x+ +

Let x = 12

, 52

= ( 72

)b

b = 57

Let x = 23

, 103

= 73

a

a = 107

g(x) 10 557(3 2) 7(2 1)x x

+ + +


2

10 57(3 2) 7(2 1)x x

+ +

= 0

2 13 2 2 1x x

+ +

= 0

2(2x + 1) +1(3x 2) = 0 4x + 2 + 3x 2 = 0 7x = 0 x = 0 x = 0 is where y = 5

e f (x) = 2

22 7 11

4 3x x

x x +

+

2 2

2

24 3 2 7 11

2 8 65

x x x x

x xx

+ +

+

f (x) 252

4 3x

x x + +

25

4 3x

x x +

= 5( 3)( 1)

xx x

3 1

a bx x

+

( 1) ( 3)( 3)( 1)

a x b xx x +

5x ( 1 ) ( 3)a x b x + Let x = 1, 6 = 2b b = 3 Let x = 3, 8 = 2a a = 4

f (x) 2 4 33 1x x

+


4 33 1x x

+

= 0

4(x 1) + 3(x 3) = 0 4x + 4 + 3x 9 = 0 x 5 = 0 x = 5

C o o r d i n a t e g e o m e t r y S M 1 2 - 1 31 x = 5 is where y = 2

8 a y = 2 2 22 3 2 3

12 1 ( 1) ( 1)x x a b

xx x x x+ += +

++ + + +

22 3

( 1)x

x+

+ = 2

( 1)( 1)

a x bx+ ++

2x + 3 = a (x + 1) + b Let x = 1; 1 = b Let x = 0; 3 = a + 1 a = 2

22 3

2 1x

x x+

+ + = 2

2 11 ( 1)x x

++ +

Asymptote: x = 1; y = 0 y-intercept: x = 0; y = 2

1 + 1

1 = 3

(0, 3)

x-intercept: 22 3

2 1x

x x+

+ + = 0

2x + 3 = 0 x = 3

2

( 32

, 0)

ddyx

= 2 32 2

( 1) ( 1)x x

+ +

ddyx

= 0; 22

( 1)x + = 3

2( 1)x +

(x + 1)3 = (x + 1)2

(x + 1)3 (x + 1)2 = 0 (x + 1)2 (x + 1 + 1) = 0 (x + 1)2 (x + 2) = 0 x = 1, x = 2 discard x = 1

When x = 2, y = 22 1

2 1 ( 2 1)+

+ +

= 2 + 1 = 1 Stationary point is (2, 1)

x 3 2 1.5

f (x) 14 0 8

Slope \ /

(2, 1) is a minimum turning point.

b y = 2 2 21 1

48 16 ( 4) ( 4)x x a b

xx x x x = +

+

21

( 4)x

x

= 2

( 4)( 4)

a x bx +

x 1 = a(x 4) + b

Let x = 4; 3 = b Let x = 0; 1 = 4a + 3 4 = 4a a = 1

y = 21 3

4 ( 4)x x+

Asymptotes: x = 4, y = 0 y-intercept: y = 1

4 + 3

16 = 1

16; (0, 1

16)

x-intercept: 21

8 16x

x x

+ = 0

x 1 = 0 x = 1 (1, 0)

ddyx

= 2 31 6

( 4) ( 4)x x

ddyx

= 0; 3( 4) 6( 4)xx

= 0

x + 4 6 = 0 x = 2 When x = 2; y = 1

6 + 3

36 = 3

36 = 1

12

Stationary point is (2, 112

)

x 3 2 1

f (x) 1343 0 1

125

Slope \ /

(2, 112

) is a minimum turning point.

c y = 3 2

22 10 1

2 15x x x

x x

2 3 2

3 2

2 15 2 10 1

2 155 1

xx x x x x

x x xx

y = x + 25 1

2 15x

x x

y = x + 5 1( 5)( 3)

xx x

+

5 1( 5)( 3)

xx x

+

5

ax

+ 3

bx +

5x 1 = a(x + 3) + b(x 5) Let x = 3; 16 = 8b b = 2 Let x = 5; 24 = 8a a = 3

y = x + 35x

+ 23x +

Asymptotes: x = 5; x = 3; y = x y-intercept: y = 1

15; (0, 1

15)

x-intercepts: (2.25, 0), (0.10, 0) (4.35, 0) (using calculator)


d y = 3 2

22 2 13 8

6x x x

x x +

2 3 2

3 2

26 2 2 13 8

2 2 128

xx x x x x

x x xx

this edition published 2010 by the australian copyright...

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