complete textbook

Mathematics Preliminary Course

maths

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Margaret Grove

Mathematics Preliminary Course

maths

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Text © 2010 Grove and Associates Pty Ltd Illustrations and design © 2010 McGraw-Hill Australia Pty Ltd Additional owners of copyright are acknowledged in on-page credits

Every effort has been made to trace and acknowledge copyrighted material. The authors and publishers tender their apologies should any infringement have occurred.

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 institution (or the body that administers it) has sent a Statutory Educational notice to Copyright Agency Limited (CAL) and been granted a licence. For details of statutory educational and other copyright licences contact: Copyright Agency Limited, Level 15, 233 Castlereagh Street, Sydney NSW 2000. Telephone: (02) 9394 7600. Website: www.copyright.com.au

Reproduction and communication for other purposes Apart from any fair dealing for the purposes of study, research, criticism or review, as permitted under the Act, no part of this publication may be reproduced, distributed or transmitted in any form or by any means, or stored in a database or retrieval system, without the written permission of McGraw-Hill Australia including, but not limited to, any network or other electronic storage.

Enquiries should be made to the publisher via www.mcgraw-hill.com.au

National Library of Australia Cataloguing-in-Publication DataAuthor: Grove, Margaret.Title: Maths in focus: mathematics preliminary course/Margaret Grove.Edition: 2nd ed.ISBN: 9780070278561 (pbk.)Target Audience: For secondary school age.Subjects: Mathematics–Problems, exercises, etc. Mathematics–Textbooks.Dewey Number: 510.76

Published in Australia by McGraw-Hill Australia Pty Ltd Level 2, 82 Waterloo Road, North Ryde NSW 2113 Publisher: Eiko Bron Managing Editor: Kathryn Fairfax Production Editor: Natalie Crouch Editorial Assistant: Ivy Chung Art Director: Astred Hicks Cover and Internal Design: Simon Rattray, Squirt Creative Cover Image: Corbis Proofreader: Ron Buck CD-ROM Preparation: Nicole McKenzie Typeset in ITC Stone serif, 10/14 by diacriTech Printed in China on 80 gsm matt art by iBook

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v

PREFACE viii

ACKNOWLEDGEMENTS viii

CREDITS viii

FEATURES OF THIS BOOK viii

SYLLABUS MATRIX ix

STUDY SKILLS ix

Chapter 1: Basic Arithmetic 2

INTRODUCTION 3REAL NUMBERS 3DIRECTED NUMBERS 9FRACTIONS, DECIMALS AND PERCENTAGES 12POWERS AND ROOTS 19ABSOLUTE VALUE 37TEST YOURSELF 1 41CHALLENGE EXERCISE 1 43

Chapter 2: Algebra and Surds 44

INTRODUCTION 45SIMPLIFYING EXPRESSIONS 45BINOMIAL PRODUCTS 51FACTORISATION 55COMPLETING THE SQUARE 69ALGEBRAIC FRACTIONS 71SUBSTITUTION 73SURDS 76TEST YOURSELF 2 90CHALLENGE EXERCISE 2 93

Chapter 3: Equations 94

INTRODUCTION 95SIMPLE EQUATIONS 95SUBSTITUTION 100INEQUATIONS 103EQUATIONS AND INEQUATIONS INVOLVING ABSOLUTE VALUES 107EXPONENTIAL EQUATIONS 114QUADRATIC EQUATIONS 118QUADRATIC INEQUATIONS 125SIMULTANEOUS EQUATIONS 127TEST YOURSELF 3 133CHALLENGE EXERCISE 3 134

Contents

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vi

Chapter 4: Geometry 1 136

INTRODUCTION 137NOTATION 137TYPES OF ANGLES 138PARALLEL LINES 145TYPES OF TRIANGLES 149CONGRUENT TRIANGLES 155SIMILAR TRIANGLES 159PYTHAGORAS’ THEOREM 167TYPES OF QUADRILATERALS 173POLYGONS 180AREAS 184TEST YOURSELF 4 191CHALLENGE EXERCISE 4 193

Practice Assessment Task Set 1 195

Chapter 5: Functions and Graphs 200

INTRODUCTION 201FUNCTIONS 201GRAPHING TECHNIQUES 212LINEAR FUNCTION 220QUADRATIC FUNCTION 224ABSOLUTE VALUE FUNCTION 230THE HYPERBOLA 238CIRCLES AND SEMI-CIRCLES 242OTHER GRAPHS 250LIMITS AND CONTINUITY 256REGIONS 260TEST YOURSELF 5 270CHALLENGE EXERCISE 5 271

Chapter 6: Trigonometry 274

INTRODUCTION 275TRIGONOMETRIC RATIOS 275RIGHT-ANGLED TRIANGLE PROBLEMS 283APPLICATIONS 292EXACT RATIOS 302ANGLES OF ANY MAGNITUDE 306TRIGONOMETRIC EQUATIONS 320TRIGONOMETRIC IDENTITIES 326NON-RIGHT-ANGLED TRIANGLE RESULTS 331APPLICATIONS 342AREA 346TEST YOURSELF 6 349CHALLENGE EXERCISE 6 350

Chapter 7: Linear Functions 352

INTRODUCTION 353DISTANCE 353MIDPOINT 358

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vii

GRADIENT 360EQUATION OF A STRAIGHT LINE 370PARALLEL AND PERPENDICULAR LINES 374INTERSECTION OF LINES 379PERPENDICULAR DISTANCE 384TEST YOURSELF 7 389CHALLENGE EXERCISE 7 390

Chapter 8: Introduction to Calculus 392

INTRODUCTION 393GRADIENT 394DIFFERENTIATION FROM FIRST PRINCIPLES 403SHORT METHODS OF DIFFERENTIATION 419TANGENTS AND NORMALS 425FURTHER DIFFERENTIATION AND INDICES 430COMPOSITE FUNCTION RULE 432PRODUCT RULE 436QUOTIENT RULE 439TEST YOURSELF 8 442CHALLENGE EXERCISE 8 443


Chapter 9: The Quadratic Function 450

INTRODUCTION 451GRAPH OF A QUADRATIC FUNCTION 451QUADRATIC INEQUALITIES 457THE DISCRIMINANT 461QUADRATIC IDENTITIES 468SUM AND PRODUCT OF ROOTS 472EQUATIONS REDUCIBLE TO QUADRATICS 477TEST YOURSELF 9 481CHALLENGE EXERCISE 9 482

Chapter 10: Locus and the Parabola 484

INTRODUCTION 485LOCUS 485CIRCLE AS A LOCUS 493PARABOLA AS A LOCUS 497GENERAL PARABOLA 516TANGENTS AND NORMALS 531TEST YOURSELF 10 534CHALLENGE EXERCISE 10 535


Answers 540

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viii

PREFACE This book covers the Preliminary syllabus for Mathematics. The syllabus is available through the NSW Board of Studies website on www.boardofstudies.nsw.edu.au. You can also access resources, study techniques, examination technique, sample and past examination papers through other websites such as www.math.nsw.edu.au and www.csu.edu.au. Searching the Internet generally will pick up many websites supporting the work in this course.

Each chapter has comprehensive fully worked examples and explanations as well as ample sets of graded exercises. The theory follows a logical order, although some topics may be learned in any order. Each chapter contains Test Yourself and Challenge exercises, and there are several practice assessment tasks throughout the book.

If you have trouble doing the Test Yourself exercises at the end of a chapter, you will need to go back into the chapter and revise it before trying them again. Don’t attempt to do the Challenge exercises until you are confi dent that you can do the Test Yourself exercises, as these are more diffi cult and are designed to test the more able students who understand the topic really well.

ACKNOWLEDGEMENTS Thanks go to my family, especially my husband Geoff, for supporting me in writing this book.

CREDITSFairfax Photos: p 311Istockphoto: p 101, p 167Margaret Grove: p 18, p 37, p 159, p 202, p 242, p 256, p 275, p 292 (bottom), p 294, p 295, p 297, p 300, p 353, p 497Photolibrary: p 201Shutterstock: p 74, p 160, p 225, p 292 (top), p 486

FEATURES OF THIS BOOK This second edition retains all the features of previous Maths in Focus books while adding in new improvements.

The main feature of Maths in Focus is in its readability, its plentiful worked examples and straightforward language so that students can understand it and use it in self-paced learning. The logical progression of topics, the comprehensive fully worked examples and graded exercises are still major features.

A wide variety of questions is maintained, with more comprehensive and more diffi cult questions included in each topic. At the end of each chapter is a consolidation set of exercises (Test yourself) in no particular order that will test whether the student has grasped the concepts contained in the chapter. There is also a challenge set for the more able students.

The three practice assessment tasks provide a comprehensive variety of mixed questions from various chapters. These have been extended to contain questions in the form of sample examination questions, including short answer, free response and multiple-choice questions that students may encounter in assessments.

The second edition also features a short summary of general study skills that students will fi nd useful, both in the classroom and when doing assessment tasks and examinations. These study skills are also repeated in the HSC book.

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ix

A syllabus matrix is included to show where each syllabus topic fi ts into the book. Topics are generally arranged in a logical order. For example, arithmetic and algebra are needed in most, if not all other topics, so these are treated at the beginning of the book.

Some teachers like to introduce particular topics before others, e.g. linear functions before more general functions. However, part of the work on gradient requires some knowledge of trigonometry and the topic of angles of any magnitude in trigonometry needs some knowledge of functions. So the order of most chapters in the book have been carefully thought out. Some chapters, however, could be covered in a different order, such as geometry which is covered in Chapter 4, and quadratic functions and locus, which are near the end of the book.

SYLLABUS MATRIX This matrix shows how the syllabus is organised in the chapters of this book.

Mathematics (2 Unit)

Basic arithmetic and algebra (1.1 – 1.4) Chapter 1: Basic arithmeticChapter 2: Algebra and surdsChapter 3: Equations

Real functions (4.1 – 4.4) Chapter 5: Functions and graphs

Trigonometric ratios (5.1 – 5.5) Chapter 6: Trigonometry

Linear functions (6.1 – 6.5, 6.7) Chapter 7: Linear functions

The quadratic polynomial and the parabola (9.1 – 9.5) Chapter 9: The quadratic functionChapter 10: Locus and the parabola

Plane geometry (2.1 – 2.4) Chapter 4: Geometry 1

Tangent to a curve and derivative of a function (8.1 – 8.9) Chapter 8: Introduction to calculus

STUDY SKILLS You may have coasted through previous stages without needing to rely on regular study, but in this course many of the topics are new and you will need to systematically revise in order to build up your skills and to remember them.

The Preliminary course introduces the basics of topics such as calculus that are then applied in the HSC course. You will struggle in the HSC if you don’t set yourself up to revise the preliminary topics as you learn new HSC topics.

Your teachers will be able to help you build up and manage good study habits. Here are a few hints to get you started.

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x

There is no right or wrong way to learn. Different styles of learning suit different people. There is also no magical number of hours a week that you should study, as this will be different for every student. But just listening in class and taking notes is not enough, especially when learning material that is totally new.

You wouldn’t go for your driver’s licence after just one trip in the car, or enter a dance competition after learning a dance routine once. These skills take a lot of practice. Studying mathematics is just the same.

If a skill is not practised within the fi rst 24 hours, up to 50% can be forgotten. If it is not practised within 72 hours, up to 85–90% can be forgotten! So it is really important that whatever your study timetable, new work must be looked at soon after it is presented to you.

With a continual succession of new work to learn and retain, this is a challenge. But the good news is that you don’t have to study for hours on end!

In the classroom

In order to remember, fi rst you need to focus on what is being said and done. According to an ancient proverb:

‘I hear and I forget

I see and I remember

I do and I understand’

If you chat to friends and just take notes without really paying attention, you aren’t giving yourself a chance to remember anything and will have to study harder at home.

If you have just had a fi ght with a friend, have been chatting about weekend activities or myriad other conversations outside the classroom, it helps if you can check these at the door and don’t keep chatting about them once the lesson starts.

If you are unsure of something that the teacher has said, the chances are that others are also not sure. Asking questions and clarifying things will ultimately help you gain better results, especially in a subject like mathematics where much of the knowledge and skills depends on being able to understand the basics.

Learning is all about knowing what you know and what you don’t know. Many students feel like they don’t know anything, but it’s surprising just how much they know already. Picking up the main concepts in class and not worrying too much about other less important parts can really help. The teacher can guide you on this.

Here are some pointers to get the best out of classroom learning:

Take control and be responsible for your own learning ■

Clear your head of other issues in the classroom ■

Active, not passive, learning is more memorable ■

Ask questions if you don’t understand something ■

Listen for cues from the teacher ■

Look out for what are the main concepts ■

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xi

Note taking varies from class to class, but there are some general guidelines that will help when you come to read over your notes later on at home:

Write legibly ■

Use different colours to highlight important points or formulae ■

Make notes in textbooks (using pencil if you don’t own the textbook) ■

Use highlighter pens to point out important points ■

Summarise the main points ■

If notes are scribbled, rewrite them at home ■

At home

You are responsible for your own learning and nobody else can tell you how best to study. Some people need more revision time than others, some study better in the mornings while others do better at night, and some can work at home while others prefer a library.

There are some general guidelines for studying at home:

Revise both new and older topics regularly ■

Have a realistic timetable and be fl exible ■

Summarise the main points ■

Revise when you are fresh and energetic ■

Divide study time into smaller rather than longer chunks ■

Study in a quiet environment ■

Have a balanced life and don’t forget to have fun! ■

If you are given exercises out of a textbook to do for homework, consider asking the teacher if you can leave some of them till later and use these for revision. It is not necessary to do every exercise at one sitting, and you learn better if you can spread these over time.

People use different learning styles to help them study. The more variety the better, and you will fi nd some that help you more than others. Some people (around 35%) learn best visually, some (25%) learn best by hearing and others (40%) learn by doing.

Here are some ideas to give you a variety of ways to study:

Summarise on cue cards or in a small notebook ■

Use colourful posters ■

Use mindmaps and diagrams ■

Discuss work with a group of friends ■

Read notes out aloud ■

Make up songs and rhymes ■

Do exercises regularly ■

Role play teaching someone else ■

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xii

Assessment tasks and exams

Many of the assessment tasks for maths are closed book examinations. You will cope better in exams if you have practised doing sample exams under exam conditions.

Regular revision will give you confi dence and if you feel well prepared, this will help get rid of nerves in the exam. You will also cope better if you have had a reasonable night’s sleep before the exam.

One of the biggest problems students have with exams is in timing. Make sure you don’t spend too much time on questions you’re unsure about, but work through and fi nd questions you can do fi rst.

Divide the time up into smaller chunks for each question and allow some extra time to go back to questions you couldn’t do or fi nish. For example, in a 2 hour exam with 6 questions, allow around 15 minutes for each question. This will give an extra half hour at the end to tidy up and fi nish off questions.

Here are some general guidelines for doing exams:

Read through and ensure you know how many questions there are ■

Divide your time between questions with extra time at the end ■

Don’t spend too much time on one question ■

Read each question carefully, underlining key words ■

Show all working out, including diagrams and formulae ■

Cross out mistakes with a single line so it can still be read ■

Write legibly ■

And fi nally…

Study involves knowing what you don’t know, and putting in a lot of time into concentrating on these areas. This is a positive way to learn. Rather than just saying, ‘I can’t do this’, say instead, ‘I can’t do this yet’, and use your teachers, friends, textbooks and other ways of fi nding out.

With the parts of the course that you do know, make sure you can remember these easily under exam pressure by putting in lots of practice.

Remember to look at new work

today ■

tomorrow ■

in a week ■

in a month ■

Some people hardly ever fi nd time to study while others give up their outside lives to devote their time to study. The ideal situation is to balance study with other aspects of your life, including going out with friends, working and keeping up with sport and other activities that you enjoy.

Good luck with your studies!

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TERMINOLOGYTERMINOLOGY TERMINOLOGY

1 Basic Arithmetic

Absolute value: The distance of a number from zero on the number line. Hence it is the magnitude or value of a number without the sign

Directed numbers: The set of integers or whole numbers 3, 2, 1, 0, 1, 2, 3,f f- - -

Exponent: Power or index of a number. For example 23 has a base number of 2 and an exponent of 3

Index: The power of a base number showing how many times this number is multiplied by itself e.g. 2 2 2 2.3

# #= The index is 3

Indices: More than one index (plural)

Recurring decimal: A repeating decimal that does not terminate e.g. 0.777777 … is a recurring decimal that can be written as a fraction. More than one digit can recur e.g. 0.14141414 ...

Scientifi c notation: Sometimes called standard notation. A standard form to write very large or very small numbers as a product of a number between 1 and 10 and a power of 10 e.g. 765 000 000 is 7.65 108

# in scientifi c notation

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3Chapter 1 Basic Arithmetic

INTRODUCTION

THIS CHAPTER GIVES A review of basic arithmetic skills, including knowing the correct order of operations, rounding off, and working with fractions, decimals and percentages. Work on signifi cant fi gures, scientifi c notation and indices is also included, as are the concepts of absolute values. Basic calculator skills are also covered in this chapter.

Real Numbers

Types of numbers

Irrationalnumbers

Unreal or imaginarynumbers

Integers

Rationalnumbers

Real numbers

Integers are whole numbers that may be positive, negative or zero. e.g. , , ,4 7 0 11- - Rational numbers can be written in the form of a fraction

ba

where a and b are integers, .b 0! e.g. , . , . ,143 3 7 0 5 5

•-

Irrational numbers cannot be written in the form of a fraction ba

(that is, they are not rational) e.g. ,2 r

EXAMPLE

Which of these numbers are rational and which are irrational?

, . , , , , .3 1 353 9

42 65

• r-

Solution

34

and r are irrational as they cannot be written as fractions (r is irrational).

. , .1 3 131 9

13 2 65 2

2013and

•= = - = - so they are all rational.

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4 Maths In Focus Mathematics Preliminary Course

Order of operations

1. Brackets: do calculations inside grouping symbols fi rst. (For example, a fraction line, square root sign or absolute value sign can act as a grouping symbol.) 2. Multiply or divide from left to right. 3. Add or subtract from left to right.

EXAMPLE

Evaluate .40 3 5 4- +] g

Solution

40 3(5 4) 40 3 9

40 27

13

#- + = -

= -

=

PROBLEM

What is wrong with this calculation?

Evaluate 1 219 4

+

-

- +Press19 4 1 2 19 4 1 2'+- ='17

What is the correct answer?

BRACKETS KEYS

Use ( and ) to open and close brackets. Always use them in pairs. For example, to evaluate 40 5 43- +] g press 40 3 ( 5 4 )

31#

=

- + =

To evaluate 1.69 2.775.67 3.49

+

- correct to 1 decimal place

press ( ( 5.67 3.49 ) ( 1.69 2.77 ) )': - + =

0.7

correct to 1decimal place

=

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Rounding off

Rounding off is often done in everyday life. A quick look at a newspaper will give plenty of examples. For example in the sports section, a newspaper may report that 50 000 fans attended a football match.

An accurate number is not always necessary. There may have been exactly 49 976 people at the football game, but 50 000 gives an idea of the size of the crowd.

EXAMPLES

1. Round off 24 629 to the nearest thousand.

Solution

This number is between 24 000 and 25 000, but it is closer to 25 000.

24 629 25 000` = to the nearest thousand

CONTINUED

MEMORY KEYS

Use STO to store a number in memory. There are several memories that you can use at the same time—any letter from A to F, or X, Y and M on the keypad.

To store the number 50 in, say, A press 50 STO A

To recall this number, press ALPHA A =

To clear all memories press SHIFT CLR

X -1 KEY

Use this key to fi nd the reciprocal of x . For example, to evaluate

7.6 2.1

1#-

0.063= -

press ( ( ) 7.6 2.1 ) x 1#- =

-

(correct to 3 decimal places)

Different calculators use different keys so check

the instructions for your calculator.

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2. Write 850 to the nearest hundred. Solution

This number is exactly halfway between 800 and 900. When a number is halfway, we round it off to the larger number. 850 900` = to the nearest hundred

In this course you will need to round off decimals, especially when using trigonometry or logarithms.

To round a number off to a certain number of decimal places, look at the next digit to the right. If this digit is 5 or more, add 1 to the digit before it and drop all the other digits after it. If the digit to the right is less than 5, leave the digit before it and drop all the digits to the right.

EXAMPLES

1. Round off 0.6825371 correct to 1 decimal place.

Solution

.

. .0 6825371

0 6825371 0 7 correct to 1 decimal place` =#

2. Round off 0.6825371 correct to 2 decimal places. Solution

.

. .0 6825371

0 6825371 0 68 correct to 2 decimal places` =#

3. Evaluate . .3 56 2 1' correct to 2 decimal places. Solution

. . . 5

.

3 56 2 1 1 69 238095

1 70 correct to 2 decimal places

' =

=#

Drop off the 2 and all digits to the right as 2 is smaller than 5.

Add 1 to the 6 as the 8 is greater than 5.

Check this on your calculator. Add 1 to the 69 as 5 is too large to just drop off.

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While using a fi xed number of decimal places on the display, the calculator still keeps track internally of the full number of decimal places.

EXAMPLE

Calculate . . . .3 25 1 72 5 97 7 32#' + correct to 2 decimal places.

Solution

. . . . . . .

. .

.

3 25 1 72 5 97 7 32 1 889534884 5 97 7 32

11 28052326 7 32

18 60052326

18.60 correct to 2 decimal places

' # #+ = +

= +

=

=

If the FIX key is set to 2 decimal places, then the display will show 2 decimal places at each step.

3.25 1.72 5.97 7.32 1.89 5.97 7.32

. .

.

11 28 7 32

18 60

' # #+ = +

= +

=

If you then set the calculator back to normal, the display will show the full answer of 18.60052326.

Don’t round off at each step of a series of

calculations.

The calculator does not round off at each step. If it did, the answer might not be as accurate. This is an important point, since some students round off each step in calculations and then wonder why they do not get the same answer as other students and the textbook.

1.1 Exercises

FIX KEY

Use MODE or SET UP to fi x the number of decimal places (see the instructions for your calculator). This will cause all answers to have a fi xed number of decimal places until the calculator is turned off or switched back to normal.

1. State which numbers are rational and which are irrational.

(a) 169

0.546 (b)

(c) 17-

(d) 3r

(e) .0 34•

(f) 218

(g) 2 2

(h) 271

17.4% (i)

(j) 5

1

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2. Evaluate (a) 20 8 4'-

(b) 3 7 2 5# #-

(c) 4 27 3 6# ' '] g (d) 17 3 2#+ -

(e) . .1 9 2 3 1#-

(f) 1 3

14 7'- +

(g) 253

51

32

#-

(h)

65

143

81

-

(i)

41

81

85

65

'

+

(j) 1

41

21

351

107

-

-

3. Evaluate correct to 2 decimal places.

(a) 2.36 4.2 0.3'+ (b) . . .2 36 4 2 0 3'+] g (c) 12.7 3.95 5.7# ' (d) 8.2 0.4 4.1 0.54' #+ (e) . . . .3 2 6 5 1 3 2 7#- +] ]g g (f) 4.7 1.3

1+

(g) 4.51 3.28

1+

(h) 5.2 3.60.9 1.4

-

+

(i) 1.23 3.155.33 2.87

-

+

(j) 1.7 8.9 3.942 2 2+ -

4. Round off 1289 to the nearest hundred.

5. Write 947 to the nearest ten.

6. Round off 3200 to the nearest thousand.

7. A crowd of 10 739 spectators attended a tennis match. Write this fi gure to the nearest thousand.

8. A school has 623 students. What is this to the nearest hundred?

9. A bank made loans to the value of $7 635 718 last year. Round this off to the nearest million.

10. A company made a profi t of $34 562 991.39 last year. Write this to the nearest hundred thousand.

11. The distance between two cities is 843.72 km. What is this to the nearest kilometre?

12. Write 0.72548 correct to 2 decimal places.

13. Round off 32.569148 to the nearest unit.

14. Round off 3.24819 to 3 decimal places.

15. Evaluate 2.45 1.72# correct to 2 decimal places.

16. Evaluate 8.7 5' correct to 1 decimal place.

17. If pies are on special at 3 for $2.38, fi nd the cost of each pie.

18. Evaluate 7.48 correct to 2 decimal places.

19. Evaluate 8

6.4 2.3+ correct to

1 decimal place.

20. Find the length of each piece of material, to 1 decimal place, if 25 m of material is cut into 7 equal pieces.

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DID YOU KNOW?

In building, engineering and other industries where accurate measurements are used, the number of decimal places used indicates how accurate the measurements are.

For example, if a 2.431 m length of timber is cut into 8 equal parts, according to the calculator each part should be 0.303875 m. However, a machine could not cut this accurately. A length of 2.431 m shows that the measurement of the timber is only accurate to the nearest mm (2.431 m is 2431 mm). The cut pieces can also only be accurate to the nearest mm (0.304 m or 304 mm).

The error in measurement is related to rounding off, as the error is half the smallest measurement. In the above example, the measurement error is half a millimetre. The length of timber could be anywhere between 2430.5 mm and 2431.5 mm.

Directed Numbers

Many students use the calculator with work on directed numbers (numbers that can be positive or negative). Directed numbers occur in algebra and other topics, where you will need to remember how to use them. A good understanding of directed numbers will make your algebra skills much better.

-^ h KEY

Use this key to enter negative numbers. For example,

press ( ) 3- =

21. How much will 7.5 m2 of tiles cost, at $37.59 per m 2 ?

22. Divide 12.9 grams of salt into 7 equal portions, to 1 decimal place.

23. The cost of 9 peaches is $5.72. How much would 5 peaches cost?


(a) 17.3 4.33 2.16#-

(b) . . . .8 72 5 68 4 9 3 98# #-

(c) 5.6 4.35

3.5 9.8+

+

(d) 7.63 5.12

15.9 6.3 7.8-

+ -

(e) 6.87 3.21

1-

25. Evaluate .. ..

5 399 68 5 479 91

2

-- ] g

correct to 1 decimal place.

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Adding and subtracting

To add: move to the right along the number line To subtract: move to the left along the number line

AddSubtract

-4 -3 -2 -1 0 1 2 3 4

Same signs

Different signs

= +

+ + = +

- =

= -

+ - = -

- + = -

- +

EXAMPLES

Evaluate

1. 4 3- + Solution

Start at 4- and move 3 places to the right.

-4 -3 -2 -1 0 1 2 3 4

4 3 1- + = -

2. 1 2- - Solution

Start at 1- and move 2 places to the left.

-4 -3 -2 -1 0 1 2 3 4

1 2 3- - = -

Multiplying and dividing

To multiply or divide, follow these rules. This rule also works if there are two signs together without a number in between e.g. 32 - -

You can also do these on a calculator, or you may have a different way of working these out.

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EXAMPLES

Evaluate 1. 2 7#- Solution

Different signs ( 2 7and- + ) give a negative answer. 2 7 14#- = -

2. 12 4'- - Solution

Same signs ( 12 4and- - ) give a positive answer. 12 4 3'- - =

3. 1 3- - - Solution

The signs together are the same (both negative) so give a positive answer.

1 3

2= - +

=

1 3- - -

1. 2 3- +

2. 7 4- -

3. 8 7# -

4. 37 -- ] g 5. 28 7' -

6. . .4 9 3 7- +

7. . .2 14 5 37- -

8. . .4 8 7 4# -

9. . .1 7 4 87- -] g 10.

53 1

32

- -

11. 5 3 4#-

12. 2 7 3#- + -

13. 4 3 2#- -

14. 1 2- - -

15. 7 2+ -

16. 2 1- -] g 17. 2 15 5'- +

18. 2 6 5# #- -

19. 28 7 5#'- - -

20. 3 2-] g

1.2 Exercises

Evaluate

Start at 1- and move 3 places to the right.

ch1.indd 11 7/16/09 1:12:23 PM


Fractions, Decimals and Percentages

EXAMPLES

1. Write 0.45 as a fraction in its simplest form. Solution

.0 45

10045

55

209

'=

=

2. Convert 83 to a decimal.

Solution

..

.

8 3 0000 375

83 0 375So =

g

3. Change 35.5% to a fraction. Solution

. % .35 5

10035 5

22

20071

#=

=

4. Write 0.436 as a percentage. Solution

. . %

. %

0 436 0 436 100

43 6

#=

=

5. Write 20 g as a fraction of 1 kg in its simplest form. Solution

1 1000kg g=

1

201000

20

501

kg

gg

g=

=

Multiply by 100% to change a fraction or decimal to a percentage.

Conversions

You can do all these conversions on your calculator using the

acb

or S D+ key.

8

3 means 3 8.'

ch1.indd 12 7/16/09 1:12:24 PM


Sometimes decimals repeat, or recur. Example

. 0.31 0 33333333 3

•f= =

There are different methods that can be used to change a recurring decimal into a fraction. Here is one way of doing it. Later you will discover another method when studying series. (See HSC Course book, Chapter 8.)

EXAMPLES

1. Write .0 4• as a rational number.

Solution

. ( )

. ( )

( ) ( ):

n

n

n

n

0 44444 1

10 4 44444 2

2 1 9 4

94

Let

Then

f

f

=

=

- =

=

2. Change .1 329• •

to a fraction. Solution

. ( )

. ( )

( ) ( ): .

.

n

n

n

n

1 3292929 1

100 132 9292929 2

2 1 99 131 6

99131 6

1010

9901316

1495163

Let

Then

#

f

f

=

=

- =

=

=

=

A rational number is any number that can be

written as a fraction.

Check this on your calculator by dividing

4 by 9.

Try multiplying n by 10. Why doesn’t this work?

6. Find the percentage of people who prefer to drink Lemon Fuzzy, if 24 out of every 30 people prefer it. Solution

% %3024

1100 80# =

CONTINUED

ch1.indd 13 7/16/09 1:12:25 PM


1. Write each decimal as a fraction in its lowest terms.

0.64 (a) 0.051 (b) 5.05 (c) 11.8 (d)

2. Change each fraction into a decimal.

(a) 52

(b) 187

(c) 125

(d) 117

3. Convert each percentage to a fraction in its simplest form.

2% (a) 37.5% (b) 0.1% (c) 109.7% (d)

4. Write each percentage as a decimal. 27% (a) 109% (b) 0.3% (c) 6.23% (d)

5. Write each fraction as a percentage.

(a) 207

(b) 31

(c) 2154

(d) 1000

1

6. Write each decimal as a percentage.

1.24 (a) 0.7 (b) 0.405 (c) 1.2794 (d)

7. Write each percentage as a decimal and as a fraction.

52% (a) 7% (b) 16.8% (c) 109% (d) 43.4% (e)

(f) %1241

8. Write these fractions as recurring decimals.

(a) 65

(b) 799

(c) 9913

(d) 61

(e) 32

1.3 Exercises

Another method

Let .

. ( )

. ( )

( ) ( ):

n

n

n

n

n

1 3292929

10 13 2929292 1

1000 1329 292929 2

2 1 990 1316

9901316

1495163

Then

and

f

f

f

=

=

=

- =

=

=

This method avoids decimals in the fraction at the end.

ch1.indd 14 7/16/09 1:12:25 PM


Investigation

Explore patterns in recurring decimals by dividing numbers by 3, 6, 9, 11, and so on.

Can you predict what the recurring decimal will be if a fraction has 3 in the denominator? What about 9 in the denominator? What about 11?

Can you predict what fraction certain recurring decimals will be? What denominator would 1 digit recurring give? What denominator would you have for 2 digits recurring?

Operations with fractions, decimals and percentages

You will need to know how to work with fractions without using a calculator, as they occur in other areas such as algebra, trigonometry and surds.

(f) 335

(g) 71

(h) 1112

9. Express as fractions in lowest terms.

(a) .0 8•

(b) .0 2•

(c) .1 5•

(d) .3 7•

(e) .0 67• •

(f) .0 54• •

(g) .0 15•

(h) .0 216•

(i) .0 219• •

(j) .1 074• •

10. Evaluate and express as a decimal.

(a) 3 6

5+

(b) 8 3 5'-

(c) 12 34 7

+

+

(d) 19931

-

(e) 7 413 6

+

+

11. Evaluate and write as a fraction. (a) . . .7 5 4 1 7 9' +] g (b) 4.5 1.315.7 8.9

-

-

(c) 12.3 8.9 7.6

6.3 1.7- +

+

(d) . .

.11 5 9 7

4 3-

(e) 8100

64

12. Angel scored 17 out of 23 in a class test. What was her score as a percentage, to the nearest unit?

13. A survey showed that 31 out of 40 people watched the news on Monday night. What percentage of people watched the news?

14. What percentage of 2 kg is 350 g?

15. Write 25 minutes as a percentage of an hour.

ch1.indd 15 7/16/09 1:12:26 PM


DID YOU KNOW?

Some countries use a comma for the decimal point—for example, 0,45 for 0.45. This is the reason that our large numbers now have spaces instead of commas between

digits—for example, 15 000 rather than 15,000.

EXAMPLES

1. Evaluate 1 .52

43

- Solution

152

43

57

43

2028

2015

2013

- = -

= -

=

2. Evaluate 221 3' .

Solution

221 3

25

13

25

31

56

' '

#

=

=

=

3. Evaluate . .0 056 100# Solution

. .0 056 100 5 6# = Move the decimal point 2 places to the right.

The examples on fractions show how to add, subtract, multiply or divide fractions both with and without the calculator. The decimal examples will help with some simple multiplying and the percentage examples will be useful in Chapter 8 of the HSC Course book when doing compound interest.

Most students use their calculators for decimal calculations. However, it is important for you to know how to operate with decimals. Sometimes the calculator can give a wrong answer if the wrong key is pressed. If you can estimate the size of the answer, you can work out if it makes sense or not. You can also save time by doing simple calculations in your head.

ch1.indd 16 7/16/09 1:12:27 PM


4. Evaluate . . .0 02 0 3# Solution

. . .0 02 0 3 0 006# =

5. Evaluate 10

8.753 . Solution

. .8 753 10 0 8753' =

6. The price of a $75 tennis racquet increased by %.521 Find the new

price. Solution

% $ . $

$ .

5 75 0 055 75

4 13

of` #=

=

% . % $ . $

$ .

521 0 055 105

21 75 1 055 75

79 13

21

or of #= =

=

So the price increases by $4.13 to $79.13.

7. The price of a book increased by 12%. If it now costs $18.00, what did it cost before the price rise? Solution

The new price is 112% (old price 100%, plus 12%)

1%$ .

100%$ .

$16.07

11218 00

11218 00

1100

`

#

=

=

=

So the old price was $16.07.

1.4 Exercises

1. Write 18 minutes as a fraction of 2 hours in its lowest terms.

2. Write 350 mL as a fraction of 1 litre in its simplest form.

3. Evaluate

(a) 53

41

+

(b) 352 2

107

-

(c) 43 1

52

#

(d) 73 4'

(e) 153 2

32

'

Multiply the numbers and count the number

of decimal places in the question.

Move the decimal point 1 place to

the left.

ch1.indd 17 7/16/09 1:12:28 PM


4. Find 53 of $912.60.

5. Find 75 of 1 kg, in grams correct

to 1 decimal place.

6. Trinh spends 31 of her day

sleeping, 247 at work and

121

eating. What fraction of the day is left?

7. I get $150.00 a week for a casual

job. If I spend 101 on bus fares,

152 on lunches and

31 on outings,

how much money is left over for savings?

8. John grew by 20017 of his height

this year. If he was 165 cm tall last year, what is his height now, to the nearest cm?

9. Evaluate (a) 8.9 3+ (b) 9 3.7- (c) .1 9 10# (d) .0 032 100# (e) .0 7 5# (f) . .0 8 0 3# (g) . .0 02 0 009# (h) .5 72 1000#

(i) 1008.74

(j) . .3 76 0 1#

10. Find 7% of $750.

11. Find 6.5% of 845 mL.

12. What is 12.5% of 9217 g?

13. Find 3.7% of $289.45.

14. If Kaye makes a profi t of $5 by selling a bike for $85, fi nd the profi t as a percentage of the selling price.

15. Increase 350 g by 15%.

16. Decrease 45 m by %.821

17. The cost of a calculator is now $32. If it has increased by 3.5%, how much was the old cost?

18. A tree now measures 3.5 m, which is 8.3% more than its previous year’s height. How high was the tree then, to 1 decimal place?

19. This month there has been a 4.9% increase in stolen cars. If 546 cars were stolen last month, how many were stolen this month?

20. George’s computer cost $3500. If it has depreciated by 17.2%, what is the computer worth now?

ch1.indd 18 7/16/09 1:12:28 PM


Powers and Roots

A power (or index ) of a number shows how many times a number is multiplied by itself.

PROBLEM

If both the hour hand and minute hand start at the same position at 12 o’clock, when is the fi rst time, correct to a fraction of a minute, that the two hands will be together again?

EXAMPLES

1. 4 4 4 4 643# #= =

2. 2 2 2 2 2 2 325# # # #= =

In 43 the 4 is called the base number and the 3 is called

the index or power.

A root of a number is the inverse of the power.

EXAMPLES

1. 36 6= since 6 362 =

2. 8 23 = since 2 83 =

3. 64 26 = since 2 646 =

DID YOU KNOW?

Many formulae use indices (powers and roots). For example the compound interest formula that you will study in Chapter 8 of the HSC

Course book is 1A P rn

= +^ h

Geometry uses formulae involving indices, such as 34

V r3r= . Do you know what this formula is for?

In Chapter 7, the formula for the distance between 2 points on a number plane is

d x x y y( ) ( )2 1

2

2 1

2= - + -

See if you can fi nd other formulae involving indices.

ch1.indd 19 7/16/09 1:12:32 PM


Proof

( )( )( )

a aaa

a a aa a a m

na a a m n

a1

timestimes

times

m nn

m

m n

'

# # #

# # #

# # #

ff

f

=

=

=-

= -

Index laws

There are some general laws that simplify calculations with indices.

a a am n m n# = +

Proof

( ) ( )a a a a a a a a

a a a

a

m n

m n

m n

m n

times times

times

# # # # # # # #

# # #

f f

f

=

=

= +

+

1 2 34444 4444 1 2 34444 4444

1 2 34444 4444

These laws work for any m and n , including fractions and negative numbers.

a a am n m n' = -

a=( )am n mn

Proof

( ) ( )

( )

a a a a a n

a n

a

times

times

m n m m m m

m m m m

mn

# # # #f=

=

=

f+ + + +

POWER AND ROOT KEYS

Use the x2 and x3 keys for squares and cubes.

Use the xy or ^ key to fi nd powers of numbers.

Use the key for square roots.

Use the 3 key for cube roots. Use the x for other roots.

ch1.indd 20 7/16/09 1:12:33 PM


( )ab a bn n n=

Proof

( ) ( )

( ) ( )ab ab ab ab ab n

a a a b b b

a b

timesn

n n

n ntimes times

# # # #

# # # # # # #

f

f f

=

=

=

1 2 34444 4444 1 2 34444 4444

ba

ban

n

n

=c m

Proof

( )

( )( )

ba

ba

ba

ba

ba n

b b b ba a a a n

n

ba

times

timestimes

n

n

n

# # # #

# # # #

# # # #

f

ff

=

=

=

c m

EXAMPLES

Simplify

1. m m m9 7 2# '

Solution

m m m m

m

9 7 2 9 7 2

14

# ' =

=

+ -

2. 3( )y2 4

Solution

( ) ( )y y

y

y

2 2

2

8

4 3 3 4 3

3 4 3

12

=

=

=

#

CONTINUED

ch1.indd 21 7/16/09 1:12:33 PM


1. Evaluate without using a calculator.

(a) 5 23 2#

(b) 3 84 2+

(c) 41 3c m

(d) 273 (e) 164

2. Evaluate correct to 1 decimal place.

(a) 3.72 (b) 1.061.5 (c) 2.3 0.2- (d) 193 (e) . . .34 8 1 2 43 13 #-

(f) 0.99 5.61

13 +

3. Simplify (a) a a a6 9 2# #

(b) y y y3 8 5# #

- (c) a a1 3#

- -

(d) 2 2w w#1 1

(e) x x6'

(f) p p3 7'

-

(g) y

y5

11

(h) ( )x7 3 (i) (2 )x5 2 (j) (3 )y 2 4- (k) a a a3 5 7# '

(l) yx

9

2 5f p

(m) w

w w3

6 7#

(n) ( )

p

p p9

2 3 4#

(o) x

x x2

6 7'

(p) ( )

a b

a b4 9

2 2 6

#

#

(q) ( ) ( )

x y

x y1 4

2 3 3 2

#

#

-

-

4. Simplify (a) x x5 9#

(b) a a1 6#

- -

(c) mm

3

7

(d) k k k13 6 9# '

(e) a a a5 4 7# #

- -

(f) 5 5x x#2 3

(g) m nm n

4 2

5 4

#

#

1.5 Exercises

3. ( )

y

y y5

6 3 4#

-

Solution

( )

y

y y

y

y y

y

y

y

y

y

( )

5

6 3 4

5

18 4

5

18 4

5

14

9

# #=

=

=

=

- -

+ -

ch1.indd 22 7/16/09 1:12:34 PM


(h) 2 2

p

p p2

#1 1

(i) (3 )x11 2

(j) ( )

x

x3

4 6

5. Simplify (a) 5( )pq3

(b) ba 8c m

(c) 4ba4

3d n

(7 (d) a 5 b ) 2

(e) (2 )

m

m4

7 3

(f) ( )

xyxy xy3 2 4#

(g) 3

4

( )

( )

k

k

6

23

8

(h) yy

28

5 712

#_ i

(i) a

a a11

6 4 3#

-e o

(j) x y

xy58 3

9 3

#f p

6. Evaluate a 3 b 2 when 2a = and

43b = .

7. If 32x = and

91,y = fi nd the value

of xy

x y5

3 2

.

8. If 21,

31a b= = and

41,c =

evaluate c

a b4

2 3

as a fraction .

9. (a) Simplify a ba b

8 7

11 8

.

Hence evaluate (b) a ba b

8 7

11 8

when

52a = and

85b = as a fraction .

10. (a) Simplify p q r

p q r4 6 2

5 8 4

.

(b) Hence evaluate p q r

p q r4 6 2

5 8 4

as a

fraction when 87,

32p q= = and

43r = .

11. Evaluate ( )a4 3 when 6.a

32

=

1

c m

12. Evaluate b

a b4

3 6

when a21

= and

b32

= .

13. Evaluate x y

x y5 5

4 7

when x31

= and

y92

= .

14. Evaluate kk

9

5

-

-

when .k31

=

15. Evaluate ( )a b

a b3 2 2

4 6

when a43

= and

b91

= .

16. Evaluate a ba b

5 2

6 3

#

# as a fraction

when a91

= and b43

= .

17. Evaluate a ba b

3

2 7

as a fraction in

index form when a52 4

= c m and

b85 3

= c m .

18. Evaluate ( )

( )

a b c

a b c2 4 3

3 2 4

as a fraction

when ,a31

= b76

= and c97

= .

ch1.indd 23 7/16/09 1:12:35 PM


Proof

x x x

x

x xxx

x

1

1

n n n n

n nn

n

0

0

'

'

`

=

=

=

=

=

-

Negative and zero indices

Class Investigation

Explore zero and negative indices by looking at these questions.

For example simplify x x3 5' using (i) index laws and (ii) cancelling.

(i) x x x3 5 2' = - by index laws

(ii) xx

x x x x xx x x

x1

5

3

2

# # # #

# #=

=

xx1So 2

2=-

Now simplify these questions by (i) index laws and (ii) cancelling. (a) x x2 3'

(b) x x2 4'

(c) x x2 5'

(d) x x3 6'

(e) x x3 3'

(f) x x2 2'

(g) x x2'

(h) x x5 6'

(i) x x4 7'

(j) x x3'

Use your results to complete:

x

x

0

n

=

=-

x 10 =

ch1.indd 24 7/16/09 1:12:36 PM


1xx

nn=-

Proof

x x x

x

x xxx

x

xx

1

1

n n

n

nn

n

nn

0 0

00

'

'

`

=

=

=

=

=

-

-

-

EXAMPLES

1. Simplify .abcab c

4

5 0e o Solution

1abcab c

4

5 0

=e o

2. Evaluate .2 3- Solution

2

21

81

33

=

=

-

3. Write in index form.

(a) 1x2

(b) 3x5

(c) 51x

(d) x 1

1+

CONTINUED

ch1.indd 25 7/16/09 1:12:37 PM


1. Evaluate as a fraction or whole number.

(a) 3 3- (b) 4 1- (c) 7 3- (d) 10 4- (e) 2 8- 6 (f) 0 (g) 2 5- (h) 3 4- (i) 7 1- (j) 9 2- (k) 2 6- (l) 3 2- 4 (m) 0 (n) 6 2- (o) 5 3- (p) 10 5- (q) 2 7- (r) 20 (s) 8 2- (t) 4 3-

2. Evaluate (a) 20

(b) 21 4-c m

(c) 32 1-c m

(d) 65 2-c m

(e) 3

2

x y

x y 0

-

+f p

(f) 51 3-c m

(g) 43 1-c m

(h) 71 2-c m

(i) 32 3-c m

(j) 21 5-c m

(k) 73 1-c m

1.6 Exercises

Solution

(a) 1x

x2

2= -

(b) x x

x

3 3 1

3

5 5

5

#=

= -

(c) x x

x

51

51 1

51 1

#=

= -

(d) ( )x xx

11

11

1

1

1

+=

+

= + -] g

4. Write a −3 without the negative index. Solution

aa13

3=-

ch1.indd 26 7/16/09 1:12:38 PM


(l) 98 0c m

(m) 76 2-c m

(n) 109 2-c m

(o) 116 0c m

(p) 41 2

--c m

(q) 52 3

--c m

(r) 372 1

--c m

(s) 83 0

-c m

(t) 141 2

--c m

3. Change into index form.

(a) 1m3

(b) 1x

(c) 1p7

(d) 1d9

(e) 1k5

(f) 1x2

(g) 2x4

(h) 3y2

(i) 21z6

(j) 53t8

(k) 72x

(l) 2

5m6

(m) 32y7

(n) (3 4)

1x 2+

(o) ( )

1a b 8+

(p) 2

1x -

(q) ( )p5 1

13+

(r) (4 9)

2t 5-

(s) ( )x4 1

111+

(t) 9( 3 )

5a b 7+

4. Write without negative indices.

(a) t 5-

(b) x 6-

(c) y 3-

(d) n 8-

(e) w 10-

(f) x2 1-

(g) 3m 4-

(h) 5x 7-

(i) 2x 3-] g

(j) n4 1-] g

(k) x 1 6+ -] g

(l) y z8 1+ -^ h

(m) 3k 2- -] g

(n) 3 2x y 9+ -^ h

(o) 1x

5-b l

(p) y1 10-c m

(q) 2p

1-d n

(r) 1a b

2

+

-c m

(s) x yx y 1

-

+ -e o

(t) 32

x yw z 7

+

- -e o

ch1.indd 27 7/16/09 1:12:39 PM


Proof

n

n

a a

a a

a a

by index lawsn

n n

n`

=

=

=

1

1

` ^^j hh

Fractional indices

Class Investigation

Explore fractional indices by looking at these questions. For example simplify (i) 2x

21` j and (ii) .x2^ h

2( ) x x

xi by index laws

21=

=

1` ^j h

2

2

( ) x x

x x x

x x

ii

So

2

22

`

=

= =

=

1

1

^` ^

hj h

Now simplify these questions.

(a) 2x21^ h

(b) x2

(c) 3x31` j

(d) 3x31^ h

(e) x3 3^ h

(f) x33

(g) 4x41` j

(h) 4x41^ h

(i) x4 4^ h

(j) x44

Use your results to complete:

nx =1

na an=1

ch1.indd 28 7/16/09 1:12:40 PM


EXAMPLES

1. Evaluate (a) 249

1

(b) 3271

Solution

(a) 249 497

=

=

1

(b) 3

27 273

3=

=

1

2. Write x3 2- in index form. Solution

2( )x x3 2 3 2- = -1

3. Write 7( )a b+1

without fractional indices. Solution

7( )a b a b7+ = +1

Proof

n n

n n

a a

a

aa

m

n m

m

mn

=

=

a =

=

m

m

1

1

`^^

jhh

Putting the fractional and negative indices together gives this rule.

- na

a1

n=

1

Here are some further rules.

n

( )a a

a

mn

n m

=

=

m

ch1.indd 29 7/16/09 1:12:41 PM


ba

abn n

=-c bm l

EXAMPLES

1. Evaluate

(a) 384

(b) -

31251

(c) 32 3-c m

Solution

(a) 3 ( ) ( )

8 8 8216

or3 4 43

4

=

=

=

4

(b) -

3

3125

125

1

1251

51

3

=

=

=

1

1

Proof

ba

ba

ba

ba

ab

ab

ab

1

1

1

1

n

n

n

n

n

n

n

n

n

n

n

'

#

=

=

=

=

=

=

-c c

b

m m

l

ch1.indd 30 7/16/09 1:12:42 PM


(c) 32

23

827

383

3 3

=

=

=

-c cm m

2. Write in index form. (a) x5

(b) ( )x4 1

12 23 -

Solution

(a) 2x x5 =5

(b)

-

3

3

( ) ( )

( )

x x

x

4 1

1

4 1

1

4 1

2 23 2

2

-=

-

= -

2

2

3. Write -

5r3

without the negative and fractional indices. Solution

-5

5

rr

r

1

135

=

=

3

3

DID YOU KNOW?

Nicole Oresme (1323–82) was the fi rst mathematician to use fractional indices. John Wallis (1616–1703) was the fi rst person to explain the signifi cance of zero, negative

and fractional indices. He also introduced the symbol 3 for infi nity. Do an Internet search on these mathematicians and fi nd out more about their work and

backgrounds. You could use keywords such as indices and infi nity as well as their names to fi nd this information.

ch1.indd 31 7/16/09 1:12:42 PM


1. Evaluate

(a) 2811

(b) 3271

(c) 2161

(d) 381

(e) 2491

(f) 310001

(g) 4161

(h) 2641

(i) 3641

(j) 711

(k) 4811

(l) 5321

(m) 801

(n) 31251

(o) 33431

(p) 71281

(q) 42561

(r) 293

(s) -

381

(t) -

3642


(a) 4231

(b) 45.84

(c) 1.24 4.327 +

(d) 12.91

5

(e) . .. .

1 5 3 73 6 1 48

+

-

(f) . .

. .8 79 1 4

5 9 3 74 #

-

3. Write without fractional indices.

(a) 3y1

(b) 3y2

(c) 2x-

1

(d) 2( )x2 5+1

(e) -

2( )x3 1-1

(f) 3( )q r6 +1

(g) -

5( )x 7+2


(a) t

(b) y5

(c) x3

(d) 9 x3 - (e) s4 1+

(f) 2 3

1t +

(g) (5 )

1

x y 3-

(h) ( )x3 1 5+

(i) ( 2)

1

x 23 -

(j) 2 7

1y +

(k) 4

5x3 +

(l) y3 1

22 -

(m) 5 ( 2)

3

x2 34 +

5. Write in index form and simplify.

(a) x x

(b) xx

(c) x

x3

(d) x

x3

2

(e) x x4

1.7 Exercises

ch1.indd 32 7/16/09 1:12:43 PM


6. Expand and simplify, and write in index form.

(a) ( )x x 2+ (b) ( )( )a b a b3 3 3 3+ -

(c) 1pp

2

+f p

(d) ( 1 )xx

2+

(e) ( )

x

x x x3 13

2 - +

7. Write without fractional or negative indices.

(a) -

3( )a b2-1

(b) 3( )y 3--

2

(c) -

7( )a4 6 1+4

(d)

-4( )x y

3

+5

(e)

-9( )x

76 3 8+

2

Scientifi c notation (standard form)

Very large or very small numbers are usually written in scientifi c notation to make them easier to read. What could be done to make the fi gures in the box below easier to read?

DID YOU KNOW?

The Bay of Fundy, Canada, has the largest tidal changes in the world. About 100 000 000 000 tons of water are moved with each tide change.

The dinosaurs dwelt on Earth for 185 000 000 years until they died out 65 000 000 years ago. The width of one plant cell is about 0.000 06 m. In 2005, the total storage capacity of dams in Australia was 83 853 000 000 000 litres and

households in Australia used 2 108 000 000 000 litres of water.

A number in scientifi c notation is written as a number between 1 and 10 multiplied by a power of 10.

EXAMPLES

1. Write 320 000 000 in scientifi c notation.

Solution

.320 000 000 3 2 108#=

2. Write .7 1 10 5#

- as a decimal number.

Solution

. .

.7 1 10 7 1 10

0 000 071

5 5# '=

=

-

Write the number between 1 and 10

and count the decimal places moved.

Count 5 places to the left.

ch1.indd 33 7/31/09 3:40:42 PM


SIGNIFICANT FIGURES

The concept of signifi cant fi gures is related to rounding off. When we look at very large (or very small) numbers, some of the smaller digits are not signifi cant.

For example, in a football crowd of 49 976, the 6 people are not really signifi cant in terms of a crowd of about 50 000! Even the 76 people are not signifi cant.

When a company makes a profi t of $5 012 342.87, the amount of 87 cents is not exactly a signifi cant sum! Nor is the sum of $342.87.

To round off to a certain number of signifi cant fi gures, we count from the fi rst non-zero digit.

In any number, non-zero digits are always signifi cant. Zeros are not signifi cant, except between two non-zero digits or at the end of a decimal number.

Even though zeros may not be signifi cant, they are still necessary. For example 31, 310, 3100, 31 000 and 310 000 all have 2 signifi cant fi gures but are very different numbers!

Scientifi c notation uses the signifi cant fi gures in a number.

SCIENTIFIC NOTATION KEY

Use the EXP or 10x# key to put numbers in scientifi c notation.

For example, to evaluate 3.1 10 2.5 10 ,4 2# ' #

-

press 3.1 EXP 4 2.5 EXP ( ) 2

1240 000' =-

=

DID YOU KNOW?

Engineering notation is similar to scientifi c notation, except the powers of 10 are always multiples of 3. For example,

3.5 103#

15.4 10 6#

-

EXAMPLES

. ( )

. . ( )

. . ( )

12 000 1 2 10 2

0 000 043 5 4 35 10 3

0 020 7 2 07 10 3

significant figures

significant figures

significant figures

4

5

2

#

#

#

=

=

=

-

-

When rounding off to signifi cant fi gures, use the usual rules for rounding off.

ch1.indd 34 7/31/09 3:40:43 PM


EXAMPLES

1. Round off 4 592 170 to 3 signifi cant fi gures. Solution

4 592 170 4 590 000= to 3 signifi cant fi gures

2. Round off 0.248 391 to 2 signifi cant fi gures. Solution

. .0 248 391 0 25= to 2 signifi cant fi gures

3. Round off 1.396 794 to 3 signifi cant fi gures. Solution

. .1 396 794 1 40= to 3 signifi cant fi gures

1. Write in scientifi c notation . 3 800 (a) 1 230 000 (b) 61 900 (c) 12 000 000 (d) 8 670 000 000 (e) 416 000 (f) 900 (g) 13 760 (h) 20 000 000 (i) 80 000 (j)

2. Write in scientifi c notation. 0.057 (a) 0.000 055 (b) 0.004 (c) 0.000 62 (d) 0.000 002 (e) 0.000 000 08 (f) 0.000 007 6 (g) 0.23 (h) 0.008 5 (i) 0.000 000 000 07 (j)

3. Write as a decimal number. (a) .3 6 104#

(b) .2 78 107#

(c) .9 25 103#

(d) .6 33 106#

(e) 4 105#

(f) .7 23 10 2#

- (g) .9 7 10 5#

- (h) .3 8 10 8#

- (i) 7 10 6#

- (j) 5 10 4#

-

4. Round these numbers to 2 signifi cant fi gures.

235 980 (a) 9 234 605 (b) 10 742 (c) 0.364 258 (d) 1.293 542 (e) 8.973 498 011 (f) 15.694 (g) 322.78 (h) 2904.686 (i) 9.0741 (j)

1.8 Exercises

Remember to put the 0’s in!

ch1.indd 35 7/16/09 1:12:46 PM


5. Evaluate correct to 3 signifi cant fi gures.

(a) . .14 6 0 453# (b) .4 8 7' (c) 4. . .47 2 59 1 46#+

(d) . .3 47 2 7

1-

6. Evaluate . . ,4 5 10 2 9 104 5# # #

giving your answer in scientifi c notation.

7. Calculate ..

1 34 108 72 10

7

3

#

#-

and write

your answer in standard form correct to 3 signifi cant fi gures.

Investigation

A logarithm is an index. It is a way of fi nding the power (or index) to which a base number is raised. For example, when solving ,3 9x = the solution is .x 2=

The 3 is called the base number and the x is the index or power.

You will learn about logarithms in the HSC course.

If a yx = then log y xa =

The expression log 1. 7 49 means the power of 7 that gives 49. The solution is 2 since .7 492 = The expression log 2. 2 16 means the power of 2 that gives 16. The solution is 4 since .2 164 =

Can you evaluate these logarithms? log 1. 3 27 log 2. 5 25 log 3. 10 10 000 log 4. 2 64 log 5. 4 4 log 6. 7 7 log 7. 3 1 log 8. 4 2

9. 31log3

10. 41log2

The a is called the base number and the x is the index or power.

ch1.indd 36 7/16/09 1:12:47 PM


Absolute Value

Negative numbers are used in maths and science, to show opposite directions. For example, temperatures can be positive or negative.

But sometimes it is not appropriate to use negative numbers. For example, solving 9c2 = gives two solutions, c 3!= . However when solving 9,c2 = using Pythagoras’ theorem, we only use

the positive answer, 3,c = as this gives the length of the side of a triangle. The negative answer doesn’t make sense.

We don’t use negative numbers in other situations, such as speed. In science we would talk about a vehicle travelling at –60k/h going in a negative direction, but we would not commonly use this when talking about the speed of our cars!

Absolute value defi nitions

We write the absolute value of x as x

xx x

x

0 when

when x 01

$=

-)

EXAMPLES

1. Evaluate .4 Solution

4 4 04 since $=

We can also defi ne x as the distance of x from 0 on the

number line. We will use this in Chapter 3.

CONTINUED

ch1.indd 37 7/16/09 1:12:47 PM


2. Evaluate .3- Solution

3 3 3 03

since 1- = - - -

=

] g

The absolute value has some properties shown below.

Properties of absolute value

a 9= = =

| | | | | | | | | | | |

| | | |

| | | || | | | | | | |

| | | | | | | |

| | | | | | | | | | | | | | | | | |

ab a b

a

a aa a

a b b a

a b a b

2 3 2 3 6

3 3

5 5 57 7 7

2 3 3 2 1

2 3 2 3 3 4 3 4

e.g.

e.g.

e.g.e.g.

e.g.

e.g. but

2 2 2 2

2 2

# # #

1#

= - = - =

- -

= = =

- = - = =

- = - - = - =

+ + + = + - + - +

] g

EXAMPLES

1. Evaluate 2 1 3 2- - + - .

Solution

2 1 3 2 1 3

2 1 9

10

22- - + - = - +

= - +

=

2. Show that a b a b#+ + when a 2= - and 3b = . Solution

a b

2 3

11

LHS = +

= - +

=

=

LHS means Left Hand Side.

ch1.indd 38 7/16/09 1:12:49 PM


a b

2 32 3

5

RHS = +

= - +

= +

=

a b a b

1 5Since 1

#+ +

3. Write expressions for 2 4x - without the absolute value signs. Solution

1

x x xx

x

x x xx x

x

2 4 2 4 2 4 02 4

2

2 4 2 4 2 4 02 4 2 4

2

wheni.e.

wheni.e.

1

1

$

$

$

- = - -

- = - - -

= - +

] g

Class Discussion

Are these statements true? If so, are there some values for which the expression is undefi ned (values of x or y that the expression cannot have)?

1. xx 1=

2. 2 2x x=

3. 2 2x x=

4. x y x y+ = +

5. x x2 2=

6. x x3 3=

7. x x1 1+ = +

8. xx

3 23 2

1-

-=

9. x

x1

2=

10. x 0$

Discuss absolute value and its defi nition in relation to these statements.

RHS means Right Hand Side.

ch1.indd 39 7/16/09 1:12:50 PM


1. Evaluate (a) 7 (b) 5- (c) 6- (d) 0 (e) 2 (f) 11- (g) 2 3- (h) 3 8- (i) 5 2-

(j) 5 3-

2. Evaluate (a) 3 2+ - (b) 3 4- - (c) 5 3- + (d) 2 7#- (e) 3 1- + - (f) 5 2 6 2

#- - (g) 2 5 1#- + - (h) 3 4- (i) 2 3 3 4- - - (j) 5 7 4 2- + -

3. Evaluate a b- if

(a) 5 2a band= = (b) 1 2a band= - = (c) 2 3a band= - = - (d) 4 7a band= = (e) .a b1 2and= - = -

4. Write an expression for

(a) a a 0when 2

(b) 0a awhen 1

(c) 0a awhen =

(d) 0a a3 when 2

(e) 0a a3 when 1

(f) 0a a3 when =

(g) a a1 1when 2+ -

(h) 1a a1 when 1+ -

(i) 2x x2 when 2-

(j) 2x x2 when 1- .

5. Show that a b a b#+ + when

(a) 2 4a band= = (b) 1 2a band= - = - (c) 2 3a band= - = (d) 4 5a band= - = (e) .a b7 3and= - = -

6. Show that x x2 = when (a) 5x = (b) x 2= - (c) x 3= - (d) 4x = (e) .x 9= -

7. Use the defi nition of absolute value to write each expression without the absolute value signs

(a) x 5+ (b) 3b - (c) 4a + (d) 2 6y - (e) 3 9x + (f) 4 x- (g) k2 1+ (h) 5 2x - (i) a b+ (j) p q-

8. Find values of x for which .x 3=

9. Simplify nn

where .n 0!

10. Simplify 22

xx-

- and state which

value x cannot be.

1.9 Exercises

ch1.indd 40 7/16/09 1:12:51 PM


1. Convert 0.45 to a fraction (a) 14% to a decimal (b)

(c) 85 to a decimal

78.5% to a fraction (d) 0.012 to a percentage (e)

(f) 1511 to a percentage

2. Evaluate as a fraction.

(a) 7 2- (b) 5 1-

(c) 29-

1

3. Evaluate correct to 3 signifi cant fi gures.

(a) . .4 5 7 62 2+

(b) 4.30.3

(c) 5.72

3

(d) ..

3 8 101 3 10

6

9

#

#

(e) -

362

4. Evaluate (a) | | | |3 2- - (b) |4 5 |- (c) 7 4 8#+ (d) [( ) ( ) ]3 2 5 1 4 8# '+ - - (e) 4 3 9- + - (f) 12- - - (g) 24 6'- -

5. Simplify

(a) x x x5 7 3# '

(b) (5 )y3 2

(c) ( )

a b

a b9

5 4 7

(d) 3

2x6 3d n

(e) a bab

5 6

4 0e o

6. Evaluate

(a) 153

87

-

(b) 76 3

32

#

(c) 943

'

(d) 52 2

101

+

(e) 1565

#

7. Evaluate (a) 4-

(b) 2361

(c) 5 2 32- -

(d) 4 3- as fraction

(e) 382

(f) 2 1- -

(g) 249-

1

as a fraction

(h) 4161

(i) 3 0-] g (j) 4 7 2 32

- - - -

8. Simplify (a) a a14 9'

(b) x y5 3 6_ i (c) p p p6 5 2# '

(d) 2b9 4^ h

(e) (2 )

x y

x y10

7 3 2


(a) n

(b) 1x5

(c) 1x y+

(d) x 14 +

Test Yourself 1

ch1.indd 41 7/16/09 1:12:52 PM


(e) a b7 +

(f) 2x

(g) 21x3

(h) x43

(i) (5 3)x 97 +

(j) 1

m34

10. Write without fractional or negative indices .

(a) a 5-

(b) 4n1

(c) 2( )x 1+1

(d) ( )x y 1- - (e) (4 7)t 4- -

(f) 5( )a b+1

(g) 3x-

1

(h) 4b3

(i) 3( )x2 3+4

(j) -

2x3

11. Show that a b a b#+ + when 5a = and 3b = - .

12. Evaluate a 2 b 4 when 259a = and 1

32b = .

13. If 31a

4

= c m and 43,b = evaluate ab3 as a

fraction.

14. Increase 650 mL by 6%.

15. Johan spends 31 of his 24-hour day

sleeping and 41 at work.

How many hours does Johan spend (a) at work?

What fraction of his day is spent at (b) work or sleeping?

If he spends 3 hours watching TV, (c) what fraction of the day is this?

What percentage of the day does he (d) spend sleeping?

16. The price of a car increased by 12%. If the car cost $34 500 previously, what is its new price?

17. Rachel scored 56 out of 80 for a maths test. What percentage did she score?

18. Evaluate ,2118 and write your answer in scientifi c notation correct to 1 decimal place.

19. Write in index form. (a) x

(b) 1y

(c) 3x6 +

(d) (2 3)

1x 11-

(e) y73

20. Write in scientifi c notation. 0.000 013 (a)

123 000 000 000 (b) 21. Convert to a fraction.

(a) .0 7•

(b) .0 124• •

22. Write without the negative index. (a) x 3-

(b) ( )a2 5 1+ -

(c) ba 5-c m

23. The number of people attending a football match increased by 4% from last week. If there were 15 080 people at the match this week, how many attended last week?

24. Show that | |a b a b#+ + when 2a = - and 5.b = -

ch1.indd 42 7/16/09 1:12:54 PM


1. Simplify 843 3

32 4 1 .

52

87

'+ -c cm m

2. Simplify .53

125

180149

307

+ + -

3. Arrange in increasing order of size: 51%,

0.502, . ,0 5•

.9951

4. Mark spends 31 of his day sleeping,

121

of the day eating and 201 of the day

watching TV. What percentage of the day is left?

5. Write -

3642

as a rational number.

6. Express . .3 2 0 01425' in scientifi c

notation correct to 3 signifi cant fi gures.

7. Vinh scored 1721 out of 20 for a maths

test, 19 out of 23 for English and 5521

out of 70 for physics. Find his average score as a percentage, to the nearest whole percentage.

8. Write .1 3274• • •


9. The distance from the Earth to the moon is .3 84 105

# km. How long would it take a rocket travelling at .2 13 10 km h4

# to reach the moon, to the nearest hour?

10. Evaluate . . .

. .0 2 5 4 1 3

8 3 4 13

'

#

+ correct to

3 signifi cant fi gures.

11. Show that ( ) ( ) .2 2 1 2 2 2 1k k k1 1- + = -+ +

12. Find the value of b c

a3 2

in index form if

., a b c52

31

53and

4 3 2

= = - =c c cm m m

13. Which of the following are rational numbers: , . , , , . , ,3 0 34 2 3 1 5 0

73•

r- ?

14. The percentage of salt in 1 L of water is 10%. If 500 mL of water is added to this mixture, what percentage of salt is there now?

15. Simplify | |

x

x

1

12 -

+ for .x 1!!

16. Evaluate 2.4 3.314.3 2.9

3 2

1.36

+

- correct to

2 decimal places.

17. Write 15 g as a percentage of 2.5 kg.

18. Evaluate . .2 3 5 7 10.1 8 2#+ - correct to


19. Evaluate ( . )

. .6 9 10

3 4 10 1 7 105 3

3 2

#

# #- +- -

and

express your answer in scientifi c notation correct to 3 signifi cant fi gures.

20. Prove | | | | | |a b a b#+ + for all real a , b .

Challenge Exercise 1

ch1.indd 43 7/16/09 1:12:56 PM

TERMINOLOGY

2 Algebra and Surds

Binomial: A mathematical expression consisting of two terms such as 3x + or x3 1-

Binomial product: The product of two binomial expressions such as ( 3) (2 4)x x+ -

Expression: A mathematical statement involving numbers, pronumerals and symbols e.g. x2 3-

Factorise: The process of writing an expression as a product of its factors. It is the reverse operation of expanding brackets i.e. take out the highest common factor in an expression and place the rest in brackets e.g. 2 8 2( 4)y y= --

Pronumeral: A letter or symbol that stands for a number

Rationalising the denominator: A process for replacing a surd in the denominator by a rational number without altering its value

Surd: From ‘absurd’. The root of a number that has an irrational value e.g. 3 . It cannot be expressed as a rational number

Term: An element of an expression containing pronumerals and/or numbers separated by an operation such as , , or# '+ - e.g. 2 , 3x -

Trinomial: An expression with three terms such as x x3 2 12

- +

ch2.indd 44 7/17/09 11:54:59 AM

45Chapter 2 Algebra and Surds

DID YOU KNOW?

Box text...

INTRODUCTION

THIS CHAPTER REVIEWS ALGEBRA skills, including simplifying expressions, removing grouping symbols, factorising, completing the square and simplifying algebraic fractions . Operations with surds , including rationalising the denominator , are also studied in this chapter .

DID YOU KNOW?

One of the earliest mathematicians to use algebra was Diophantus of Alexandria . It is not known when he lived, but it is thought this may have been around 250 AD.

In Baghdad around 700–800 AD a mathematician named Mohammed Un-Musa Al-Khowarezmi wrote books on algebra and Hindu numerals. One of his books was named Al-Jabr wa’l Migabaloh , and the word algebra comes from the fi rst word in this title.

Simplifying Expressions

Addition and subtraction

EXAMPLES

Simplify

1. x x7 - Solution

7 7 1

6

x x x x

x

- = -

=

2. x x x4 3 62 2 2- + Solution

4 3 6 6

7

x x x x x

x

2 2 2 2 2

2

- + = +

=

Here x is called a pronumeral.

CONTINUED

ch2.indd 45 7/17/09 11:55:13 AM


3. x x x3 5 43 - - + Solution

x x x x x3 5 4 8 43 3- - + = - +

4. a b a b3 4 5- - - Solution

3 4 5 3 5 4

2 5

a b a b a a b b

a b

- - - = - - -

= - -

Only add or subtract ‘like’ terms. These have the same pronumeral (for example, 3 x and 5 x ).

1. 2 5x x+

2. 9 6a a-

3. 5 4z z-

4. 5a a+

5. b b4 -

6. r r2 5-

7. y y4 3- +

8. x x2 3- -

9. 2 2a a-

10. k k4 7- +

11. 3 4 2t t t+ +

12. w w w8 3- +

13. m m m4 3 2- -

14. 3 5x x x+ -

15. 8 7h h h- -

16. b b b7 3+ -

17. 3 5 4 9b b b b- + +

18. x x x x5 3 7- + - -

19. x y y6 5- -

20. a b b a8 4 7+ - -

21. 2 3xy y xy+ +

22. 2 5 3ab ab ab2 2 2- -

23. m m m5 122 - - +

24. 7 5 6p p p2 - + -

25. 3 7 5 4x y x y+ + -

26. 2 3 8ab b ab b+ - +

27. ab bc ab ac bc+ - - +

28. a x a x7 2 15 3 5 3- + - +

29. 3 4 2x xy x y x y xy y3 2 2 2 2 3- + - + +

30. 3 4 3 5 4 6x x x x x3 2 2- - + - -

2.1 Exercises

Simplify

ch2.indd 46 7/17/09 11:55:25 AM


Multiplication

EXAMPLES

Simplify

1. x y x5 3 2# #- Solution

5 3 2 30

30

x y x xyx

x y2

# #- = -

= -

2. 3 4x y xy3 2 5#- - Solution

x y xy x y3 4 123 2 5 4 7#- - =

Use index laws to simplify this

question.

1. b5 2#

2. x y2 4#

3. p p5 2#

4. z w3 2#-

5. a b5 3#- -

6. x y z2 7# #

7. ab c8 6#

8. d d4 3#

9. a a a3 4# #

10. y3 3-^ h

11. 2x2 5^ h

12. ab a2 33 #

13. a b ab5 22 # -

14. pq p q7 32 2 2#

15. ab a b5 2 2#

16. h h4 23 7# -

17. k p p3 2#

18. t3 3 4-^ h

19. m m7 26 5# -

20. x x y xy2 3 42 3 2# #- -

2.2 Exercises

Simplify

ch2.indd 47 7/17/09 11:55:28 AM


Division

Use cancelling or index laws to simplify divisions.

EXAMPLES

Simplify

1. v y vy6 22 ' Solution

By cancelling,

v y vyvy

v y

v y

v v y

v

6 22

6

2

6

3

22

1 1

3 1 1

'

# #

# # #

=

=

=

Using index laws,

v y vy v y

v yv

6 2 3

33

2 2 1 1 1

1 0

' =

=

=

- -

2. 155

aba b

2

3

Solution

3

3

aba b a b

a b

ba

155

3

2

33 1 1 2

2 1

2

=

=

=

- -

-

1

1

1. x30 5'

2. y y2 '

3. 2

8a2

4. 8aa2

5. aa

28 2

6. x

xy

2

7. p p12 43 2'

8. 6

3ab

a b2 2

9. 1520

xyx

10. xx

39

4

7-

2.3 Exercises

Simplify

ch2.indd 48 7/17/09 11:55:29 AM


11. ab b15 5'- -

12. 62a bab2 3

13. pqs

p4

8-

14. cd c d14 212 3 3'

15. 4

2

x y z

xy z3 2

2 3

16. pq

p q

7

423

5 4

17. a b c a b c5 209 4 2 5 3 1'

- - -

18. a b

a b

4

29 2 1

5 2 4

- -

-

^^

hh

19. x y z xy z5 154 7 8 2'- -

20. a b a b9 184 1 3 1 3'- -- -^ h

Removing grouping symbols

The distributive law of numbers is given by

a b c ab ac+ = +] g

EXAMPLE

( )7 9 11 7 20

140

# #+ =

=

Using the distributive law,

( )7 9 11 7 9 7 11

63 77

140

# # #+ = +

= +

=

EXAMPLES

Expand and simplify. 1. a2 3+] g Solution

2( 3) 2 2 3

2 6

a a

a

# #+ = +

= +

This rule is used in algebra to help remove grouping symbols.

CONTINUED

ch2.indd 49 7/17/09 11:55:30 AM


2. x2 5- -] g Solution

( ) ( )x x

x

x

2 5 1 2 5

1 2 1 5

2 5

# #

- - = - -

= - - -

= - +

3. a ab c5 4 32 + -] g Solution

( )a ab c a a ab a c

a a b a c

5 4 3 5 4 5 3 5

20 15 5

2 2 2 2

2 3 2

# # #+ - = + -

= + -

4. y5 2 3- +^ h Solution

( )y y

y

y

5 2 3 5 2 2 3

5 2 6

2 1

# #- + = - -

= - -

= - -

5. b b2 5 1- - +] ]g g Solution

( ) ( )b b b b

b b

b

2 5 1 2 2 5 1 1 1

2 10 1

11

# # # #- - + = + - - -

= - - -

= -

1. x2 4-] g 2. h3 2 3+] g 3. a5 2- -] g 4. x y2 3+^ h 5. x x 2-] g 6. a a b2 3 8-] g

7. ab a b2 +] g 8. n n5 4-] g 9. x y xy y3 22 2+_ i 10. k3 4 1+ +] g 11. t2 7 3- -] g

12. y y y4 3 8+ +^ h

2.4 Exercises

Expand and simplify

ch2.indd 50 7/17/09 11:55:31 AM


13. b9 5 3- +] g

14. x3 2 5- -] g

15. m m5 3 2 7 2- + -] ]g g

16. h h2 4 3 2 9+ + -] ]g g

17. d d3 2 3 5 3- - -] ]g g

18. a a a a2 1 3 42+ - + -] ^g h

19. x x x3 4 5 1- - +] ]g g

20. ab a b a2 3 4 1- - -] ]g g

21. x x5 2 3- - -] g

22. y y8 4 2 1- + +^ h

23. a b a b+ --] ]g g

24. t t2 3 4 1 3- - + +] ]g g

25. a a4 3 5 7+ + --] ]g g

Binomial Products

A binomial expression consists of two numbers , for example 3.x + A set of two binomial expressions multiplied together is called a binomial

product. Example: x x3 2+ -] ]g g . Each term in the fi rst bracket is multiplied by each term in the second

bracket.

a b x y ax ay bx by+ + = + + +] ^g h

Proof

a b c d a c d b c d

ac ad bc bd+ + = + + +

= + + +

] ] ] ]g g g g

EXAMPLES

Expand and simplify 1. 3 4p q+ -^ ^h h

Solution

p q pq p q3 4 4 3 12+ - = - + -^ ^h h

2. 5a 2+] g

Solution

( 5)( 5)

5 5 25

10 25

a a a

a a a

a a

5 2

2

2

+ = + +

= + + +

= + +

] g

Can you see a quick way of doing this?

ch2.indd 51 7/31/09 3:43:28 PM


The rule below is not a binomial product (one expression is a trinomial), but it works the same way.

a b x y z ax ay az bx by bz+ + + = + + + + +] ^g h

EXAMPLE

Expand and simplify .x x y4 2 3 1+ - -] ^g h

Solution

( ) ( )x x y x xy x x y

x xy x y

4 2 3 1 2 3 8 12 4

2 3 7 12 4

2

2

+ - - = - - + - -

= - + - -

1. 5 2a a+ +] ]g g

2. x x3 1+ -] ]g g

3. 2 3 5y y- +^ ^h h

4. 4 2m m- -] ]g g

5. 4 3x x+ +] ]g g

6. 2 5y y+ -^ ^h h

7. 2 3 2x x- +] ]g g

8. 7 3h h- -] ]g g

9. 5 5x x+ -] ]g g

10. a a5 4 3 1- -] ]g g

11. 2 3 4 3y y+ -^ ^h h

12. 4 7x y- +] g h

13. 3 2x x2 + -^ ]h g

14. 2 2n n+ -] ]g g

15. 2 3 2 3x x+ -] ]g g

16. 4 7 4 7y y- +^ ^h h

17. 2 2a b a b+ -] ]g g

18. 3 4 3 4x y x y- +^ ^h h

19. 3 3x x+ -] ]g g

20. 6 6y y- +^ ^h h

21. a a3 1 3 1+ -] ]g g

22. 2 7 2 7z z- +] ]g g

23. 9 2 2x x y+ - +] g h

24. b a b3 2 2 1- + -] ]g g

25. 2 2 4x x x2+ - +] g h

26. 3 3 9a a a2- + +] g h

27. 9a 2+] g

28. 4k 2-] g

29. 2x 2+] g

30. 7y 2-^ h

31. 2 3x 2+] g

32. 2 1t 2-] g

2.5 Exercises

Expand and simplify

ch2.indd 52 7/31/09 3:43:29 PM


33. 3 4a b 2+] g

34. 5x y 2-^ h

35. 2a b 2+] g

36. a b a b- +] ]g g

37. a b 2+] g

38. a b 2-] g

39. a b a ab b2 2+ - +] ^g h 40. a b a ab b2 2- + +] ^g h

Some binomial products have special results and can be simplifi ed quickly using their special properties. Binomial products involving perfect squares and the difference of two squares occur in many topics in mathematics. Their expansions are given below.

Difference of 2 squares

a b a b a b2 2+ - = -] ]g g

Proof

( ) ( )a b a b a ab ab b

a b

2 2

2 2

+ - = - + -

= -

a b a ab b22 2 2+ = + +] g

Perfect squares

Proof

( ) ( )

2

a b a b a b

a ab ab b

a ab b

2

2 2

2 2

+ = + +

= + + +

= + +

] g

2a b a ab b2 2 2- = - +] g

Proof

( ) ( )

2

a b a b a b

a ab ab b

a ab b

2

2 2

2 2

- = - -

= - - +

= - +

] g

ch2.indd 53 7/17/09 11:55:35 AM


EXAMPLES

Expand and simplify 1. 2 3x 2-] g

Solution

( )x x x

x x

2 3 2 2 2 3 3

4 12 9

2 2 2

2

- = - +

= - +

] ]g g

2. 3 4 3 4y y- +^ ^h h Solution

(3 4)(3 4) 4

9 16

y y y

y

3 2 2

2

- + = -

= -

^ h

1. 4t 2+] g

2. 6z 2-] g

3. x 1 2-] g 4. 8y 2+^ h

5. 3q 2+^ h

6. 7k 2-] g

7. n 1 2+] g 8. 2 5b 2+] g

9. 3 x 2-] g

10. y3 1 2-^ h

11. x y 2+^ h

12. a b3 2-] g 13. 4 5d e 2+] g

14. 4 4t t+ -] ]g g 15. x x3 3- +] ]g g

16. p p1 1+ -^ ^h h 17. 6 6r r+ -] ]g g 18. x x10 10- +] ]g g 19. 2 3 2 3a a+ -] ]g g 20. 5 5x y x y- +^ ^h h 21. a a4 1 4 1+ -] ]g g 22. 7 3 7 3x x- +] ]g g 23. 2 2x x2 2+ -^ ^h h 24. 5x2 2

+^ h

25. 3 4 3 4ab c ab c- +] ]g g 26. 2x x

2

+b l

27. 1 1a a a a- +b bl l 28. x y x y2 2+ - - -_ _i i6 6@ @

29. a b c 2+ +] g6 @

2.6 Exercises Expand and simplify

ch2.indd 54 7/17/09 11:55:36 AM


30. x y1 2+ -] g7 A

31. a a3 32 2+ - -] ]g g 32. 16 4 4z z- - +] ]g g 33. 2 3 1 4x x 2+ + -] g

34. 2x y x y2+ - -^ ^h h 35. n n n4 3 4 3 2 52- + - +] ]g g

36. x 4 3-] g 37. x x x

1 1 22 2

- - +b bl l

38. x y x y42 2 2 2 2+ -_ i

39. 2 5a 3+] g

40. x x x2 1 2 1 2 2- + +] ] ]g g g

Expand (x 4) (x 4) .- -2

PROBLEM

Find values of all pronumerals that make this true.

i i c c b

a b c

d e

f e b

i i i h g

#

Try c 9.=

Factorisation

Simple factors

Factors are numbers that exactly divide or go into an equal or larger number, without leaving a remainder.

EXAMPLES

The numbers 1, 2, 3, 4, 6, 8, 12 and 24 are all the factors of 24. Factors of 5 x are 1, 5, x and 5 x .

To factorise an expression, we use the distributive law.

a bax bx x ++ = ] g

ch2.indd 55 7/17/09 11:55:38 AM


EXAMPLES

Factorise

1. 3 12x + Solution

The highest common factor is 3. x x3 12 3 4+ = +] g

2. 2y y2 - Solution

The highest common factor is y. y y y y2 22 - = -^ h 3. 2x x3 2- Solution

x and x 2 are both common factors. We take out the highest common factor which is x 2 . x x x x2 23 2 2- = -] g

4. x xy5 3 32+ ++] ]g g Solution

The highest common factor is 3x + . x x x yy5 3 3 3 5 22+ + + ++ =] ] ] ^g g g h 5. 8 2a b ab3 2 3- Solution

There are several common factors here. The highest common factor is 2 ab 2 . 8 2 2 4a b ab ab a b3 2 3 2 2- = -^ h

Check answers by expanding brackets.

Divide each term by 3 to fi nd the terms inside the brackets.

ch2.indd 56 7/17/09 11:55:39 AM


1. 2 6y +

2. x5 10-

3. 3 9m -

4. 8 2x +

5. y24 18-

6. 2x x2 +

7. 3m m2 -

8. 2 4y y2 +

9. 15 3a a2-

10. ab ab2 +

11. 4 2x y xy2 -

12. 3 9mn mn3 +

13. 8 2x z xz2 2-

14. 6 3 2ab a a2+ -

15. 5 2x x xy2 - +

16. 3 2q q5 2-

17. 5 15b b3 2+

18. 6 3a b a b2 3 3 2-

19. x m m5 7 5+ + +] ]g g 20. y y y2 1 1- - -^ ^h h 21. 4 7 3 7y x y+ - +^ ^h h 22. 6 2 5 2x a a- + -] ]g g 23. x t y t2 1 2 1+ - +] ]g g 24. a x b x3 2 2 3 2- + -] ]g g c x3 3 2- -] g 25. 6 9x x3 2+

26. 3 6pq q5 3-

27. 15 3a b ab4 3 +

28. 4 24x x3 2-

29. 35 25m n m n3 4 2-

30. 24 16a b ab2 5 2+

31. r rh2 22r r+

32. 3 5 3x x2- + -] ]g g 33. 4 2 4y x x2 + + +] ]g g 34. a a a1 1 2+ - +] ]g g 35. ab a a4 1 3 12 2+ - +^ ^h h

2.7 Exercises

Factorise

Grouping in pairs

If an expression has 4 terms, it may be factorised in pairs.

( ) ( )

( ) ( )

ax bx ay by x a b y a b

a b x y

+ + + = + + +

= + +

ch2.indd 57 7/17/09 11:55:40 AM


EXAMPLES

Factorise

1. 2 3 6x x x2 - + - Solution

2 3 6 ( 2) 3( 2)

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

x x

2 - + - = - + -

= - +

2. 2 4 6 3x y xy- + - Solution

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

x y xy x y x

x y x

x y

x y xy x y x

x y x

x y

2 4 6 3 2 2 3 2

2 2 3 2

2 2 3

2 4 6 3 2 2 3 2

2 2 3 2

2 2 3

or

- + - = - + -

= - - -

= - -

- + - = - - - +

= - - -

= - -

1. 2 8 4x bx b+ + +

2. 3 3ay a by b- + -

3. x x x5 2 102 + + +

4. 2 3 6m m m2 - + -

5. ad ac bd bc- + -

6. 3 3x x x3 2+ + +

7. ab b a5 3 10 6- + -

8. 2 2xy x y xy2 2- + -

9. ay a y 1+ + +

10. 5 5x x x2 + - -

11. 3 3y ay a+ + +

12. 2 4 2m y my- + -

13. x xy xy y2 10 3 152 2+ - -

14. 4 4a b ab a b2 3 2+ - -

15. x x x5 3 152- - +

16. 7 4 28x x x4 3+ - -

17. 7 21 3x xy y- - +

18. 4 12 3d de e+ - -

19. x xy y3 12 4+- -

20. a ab b2 6 3+ - -

21. x x x3 6 183 2 +- -

22. pq p q q3 32+- -

2.8 Exercises

Factorise

ch2.indd 58 7/17/09 11:55:43 AM


23. x x x3 6 5 103 2- - +

24. 4 12 3a b ac bc- + -

25. 7 4 28xy x y+ - -

26. x x x4 5 204 3- - +

27. x x x4 6 8 123 2- + -

28. 3 9 6 18a a ab b2 + + +

29. y xy x5 15 10 30+- -

30. r r r2 3 62r r+ - -

Trinomials

A trinomial is an expression with three terms, for example 4 3.x x2 - + Factorising a trinomial usually gives a binomial product.

x a b x a x bx ab2 + + ++ + =] ] ]g g g

Proof

( )

( ) ( )

( ) ( )

x a b x ab x ax bx abx x a b x a

x a x b

2 2+ + + = + + +

= + + +

= + +

EXAMPLES

Factorise

1. 5 6m m2 - + Solution

a b 5+ = - and 6ab = +

6235

+-

-

-

'

Numbers with sum 5- and product 6+ are 2- and 3.-

[ ] [ ]m m m mm m

5 6 2 32 3

2` - + = + - + -

= - -

] ]] ]

g gg g

2. 2y y2 + - Solution

1a b+ = + and 2ab = -

2211

-+

-+

'

Two numbers with sum 1+ and product 2- are 2+ and 1- . y y y y2 2 12` + - = + -^ ^h h

Guess and check by trying 2- and 3-

or 1- and .6-

Guess and check by trying 2 and 1- or

2- and 1.

ch2.indd 59 7/17/09 11:55:44 AM


The result x a b x a x bx ab2 + + + ++ =] ] ]g g g only works when the coeffi cient of x2 (the number in front of x2 ) is 1. When the coeffi cient of x2 is not 1, for example in the expression 5 2 4,x x2 - + we need to use a different method to factorise the trinomial.

There are different ways of factorising these trinomials. One method is the cross method . Another is called the PSF method . Or you can simply guess and check.

1. 4 3x x2 + +

2. y y7 122 + +

3. m m2 12 + +

4. t t8 162 + +

5. 6z z2 + -

6. 5 6x x2 - -

7. v v8 152 - +

8. 6 9t t2 - +

9. x x9 102 + -

10. 10 21y y2 - +

11. m m9 182 - +

12. y y9 362 + -

13. 5 24x x2 - -

14. 4 4a a2 - +

15. x x14 322 + -

16. 5 36y y2 - -

17. n n10 242 +-

18. x x10 252 +-

19. p p8 92 + -

20. k k7 102 +-

21. x x 122 + -

22. m m6 72 - -

23. 12 20q q2 + +

24. d d4 52 - -

25. l l11 182 +-

2.9 Exercises

Factorise

EXAMPLES

Factorise

1. 5 13 6y y2 - +

Solution—guess and check

For 5 y 2 , one bracket will have 5 y and the other y : .y y5^ ^h h Now look at the constant (term without y in it): .6+

ch2.indd 60 7/17/09 11:55:46 AM


The two numbers inside the brackets must multiply to give 6.+ To get a positive answer, they must both have the same signs. But there is a negative sign in front of 13 y so the numbers cannot be both positive. They must both be negative. y y5 - -^ ^h h To get a product of 6, the numbers must be 2 and 3 or 1 and 6. Guess 2 and 3 and check:

3 5 15 2 6

5 17 6

y y y y y

y y

5 2 2

2

- = - - +

= - +

-^ ^h h

This is not correct. Notice that we are mainly interested in checking the middle two terms, .y y15 2and- - Try 2 and 3 the other way around: .y y5 3 2- -^ ^h h Checking the middle terms: y y y10 3 13- - = - This is correct, so the answer is .y y5 3 2- -^ ^h h Note: If this did not check out, do the same with 1 and 6.

Solution — cross method

Factors of 5y2 are 5 y and y. Factors of 6 are 1- and 6- or 2- and .3- Possible combinations that give a middle term of y13- are

By guessing and checking, we choose the correct combination.

y13-

y y

y y

5 2 10

3 3

#

#

- = -

- = -

y y y y5 13 6 5 3 22` - + = - -^ ^h h Solution — PSF method

P: Product of fi rst and last terms 30y2 S: Sum or middle term y13- F: Factors of P that give S ,y y3 10- -

y

yyy

3031013

2 -

-

-

)

y y y y y

y y y

y y

5 13 6 5 3 10 65 3 2 5 3

5 3 2

2 2` - + = - - +

= - - -

= - -

^ ^^ ^

h hh h

5y

y 3-

2- 5y

y 2-

3- 5y

y 6-

1- 5y

y 1-

6-

5y

y 2-

3-

CONTINUED

ch2.indd 61 7/17/09 11:55:48 AM


2. 4 4 3y y2 + -

Solution—guess and check

For 4 y 2 , both brackets will have 2 y or one bracket will have 4 y and the other y . Try 2 y in each bracket: .y y2 2^ ^h h Now look at the constant: .3- The two numbers inside the brackets must multiply to give .3- To get a negative answer, they must have different signs. y y2 2 +-^ ^h h To get a product of 3, the numbers must be 1 and 3. Guess and check: y y2 3 2 1+-^ ^h h Checking the middle terms: y y y2 6 4- = - This is almost correct, as the sign is wrong but the coeffi cient is right (the number in front of y ). Swap the signs around:

4 6 2 3

4 4 3

y y y y y

y y

2 1 2 3 2

2

+ = +

= +

- - -

-

^ ^h h

This is correct, so the answer is .y y2 1 2 3- +^ ^h h

Solution — cross method

Factors of 4y2 are 4 y and y or 2 y and 2 y . Factors of 3 are 1- and 3 or 3- and 1. Trying combinations of these factors gives

2y# 3

2 1 2y y

y4

#- = -

= 6y

y y y y4 4 3 2 3 2 12` + - = + -^ ^h h

Solution — PSF method

P: Product of fi rst and last terms y12 2- S: Sum or middle term 4 y F: Factors of P that give S ,y y6 2+ -

y

yyy

12624

2-+

-

+

)

y y y y y

y y y

y y

4 4 3 4 6 2 32 2 3 1 2 3

2 3 2 1

2 2` + - = + - -

= + - +

= + -

^ ^

^ ^

h h

h h

2y

2y 1-

3

ch2.indd 62 8/1/09 6:13:20 PM


Perfect squares

You have looked at some special binomial products, including 2a b a ab b2 2 2+ = + +] g and 2 .a b a ab b2 2 2- = - +] g

When factorising, use these results the other way around.

Factorise

1. a a2 11 52 + +

2. 5 7 2y y2 + +

3. x x3 10 72 + +

4. 3 8 4x x2 + +

5. 2 5 3b b2 - +

6. 7 9 2x x2 - +

7. 3 5 2y y2 + -

8. x x2 11 122 + +

9. p p5 13 62 + -

10. x x6 13 52 + +

11. y y2 11 62 - -

12. x x10 3 12 + -

13. 8 14 3t t2 - +

14. x x6 122 - -

15. 6 47 8y y2 + -

16. n n4 11 62 +-

17. t t8 18 52 + -

18. q q12 23 102 + +

19. r r8 22 62 + -

20. x x4 4 152 - -

21. y y6 13 22 +-

22. p p6 5 62 - -

23. x x8 31 212 + +

24. b b12 43 362 +-

25. x x6 53 92 - -

26. 9 30 25x x2 + +

27. 16 24 9y y2 + +

28. k k25 20 42 +-

29. a a36 12 12 +-

30. m m49 84 362 + +

2.10 Exercises

a ab b a b

a ab b a b

2

2

2 2 2

2 2 2

+ + = +

- + = -

]]

gg

ch2.indd 63 7/17/09 11:55:52 AM


EXAMPLES

Factorise

1. 8 16x x2 - + Solution

8 16 2(4) 4x x x x

x 4

2 2 2

2

- + = - +

= -] g

2. 4 20 25a a2 + + Solution

4 20 25 2(2 )(5) 5a a a a

a

2

2 5

2 2 2

2

+ + = + +

= +

]]

gg

Factorise

1. y y2 12 - +

2. 6 9x x2 + +

3. m m10 252 + +

4. 4 4t t2 - +

5. x x12 362 - +

6. x x4 12 92 + +

7. b b16 8 12 - +

8. a a9 12 42 + +

9. x x25 40 162 - +

10. y y49 14 12 + +

11. y y9 30 252 +-

12. k k16 24 92 +-

13. 25 10 1x x2 + +

14. a a81 36 42 +-

15. 49 84 36m m2 + +

16. t t412 + +

17. x x34

942 - +

18. yy

956

2512 + +

19. xx

2 122

+ +

20. kk

25 0 4222

- +

2.11 Exercises

In a perfect square, the constant term is always a square number.

ch2.indd 64 7/17/09 11:55:54 AM


Difference of 2 squares

A special case of binomial products is a b a b a b2 2+ - = -] ]g g .

a b a ba b2 2 + -- = ] ]g g

EXAMPLES

Factorise

1. 36d2 -

Solution

d d

d d36 6

6 6

2 2 2=

= +

- -

-] ]g g

2. b9 12 -

Solution

( ) ( )

b bb b

9 1 3 13 1 3 1

2 2 2- = -

= + -

] g

3. ( ) ( )a b3 12 2+ - -

Solution

[( ) ( )] [( ) ( )]( ) ( )

( ) ( )

a b a b a ba b a b

a b a b

3 1 3 1 3 13 1 3 1

2 4

2 2+ - - = + + - + - -

= + + - + - +

= + + - +

] ]g g

Factorise

1. 4a2 -

2. 9x2 -

3. y 12 -

4. 25x2 -

5. 4 49x2 -

6. 16 9y2 -

7. z1 4 2-

8. t25 12 -

9. 9 4t2 -

10. x9 16 2-

11. 4x y2 2-

12. 36x y2 2-

2.12 Exercises

ch2.indd 65 7/31/09 3:43:29 PM


13. 4 9a b2 2-

14. x y1002 2-

15. 4 81a b2 2-

16. 2x y2 2+ -] g

17. a b1 22 2- - -] ]g g 18. z w12 2- +] g

19. x412 -

20. y

91

2

-

21. x y2 2 12 2+ - +] ^g h

22. x 14 -

23. 9 4x y6 2-

24. x y164 4-

25. 1a8 -

Sums and differences of 2 cubes

a b a ab ba b3 3 2 2+ - ++ = ] ^g h

a b a b a ab b3 3 2 2- = - + +] ^g h

Proof

( ) ( )a b a ab b a a b ab a b ab b

a b

2 2 3 2 2 2 2 3

3 3

+ - + = - + + - +

= +

Proof

( ) ( )a b a ab b a a b ab a b ab b

a b

2 2 3 2 2 2 2 3

3 3

- + + = + + - - -

= -

EXAMPLES

Factorise

1. 8 1x3 + Solution

( ) [ ( ) ( ) ]

( ) ( )

x x

x x x

x x x

8 1 2 1

2 1 2 2 1 1

2 1 4 2 1

3 3 3

2 2

2

+ = +

= + - +

= + - +

]]

gg

ch2.indd 66 7/17/09 11:55:58 AM


Factorise

1. b 83 -

2. 27x3 +

3. 1t3 +

4. 64a3 -

5. 1 x3-

6. 8 27y3+

7. 8y z3 3+

8. 125x y3 3-

9. 8 27x y3 3+

10. 1a b3 3 -

11. 1000 8t3+

12. x8

273

-

13. a b

1000 13 3

+

14. x y1 3 3+ -] g

15. x y z216125 3 3 3+

16. 2 1a a3 3- - +] ]g g

17. x127

3

-

18. 3y x3 3+ +] g

19. x y1 23 3+ + -] ^g h

20. 8 3a b3 3+ -] g

2.13 Exercises

2. 27 64a b3 3- Solution

( ) [ ( ) ( ) ]

( ) ( )

a b a b

a b a a b b

a b a ab b

27 64 3 4

3 4 3 3 4 4

3 4 9 12 16

3 3 3 3

2 2

2 2

- = -

= - + +

= - + +

] ]] ]

g gg g

Mixed factors

Sometimes more than one method of factorising is needed to completely factorise an expression.

EXAMPLE

Factorise 5 45.x2 - Solution

5 45 5( 9) (using simple factors)

5( 3)( 3) (the difference of two squares)x x

x x

2 2- = -

= + -

ch2.indd 67 7/17/09 11:56:00 AM


Factorise

1. x2 182 -

2. p p3 3 362 - -

3. y5 53 -

4. 4 8 24a b a b ab a b3 2 2 2 2+ - -

5. a a5 10 52 - +

6. x x2 11 122- + -

7. z z z3 27 603 2+ +

8. ab a b9 4 3 3-

9. x x3 -

10. x x6 8 82 + -

11. m n mn3 15 5- - +

12. x x3 42 2- - +] ]g g

13. y y y5 5162 + +-^ ^h h

14. x x x8 84 3- + -

15. x 16 -

16. x x x3 103 2- -

17. x x x3 9 273 2- - +

18. 4x y y2 3 -

19. 24 3b3-

20. 18 33 30x x2 + -

21. 3 6 3x x2 - +

22. 2 25 50x x x3 2+ - -

23. 6 9z z z3 2+ +

24. 4 13 9x x4 2- +

25. 2 2 8 8x x y x y5 2 3 3 3+ - -

26. 4 36a a3 -

27. 40 5x x4-

28. a a13 364 2 +-

29. k k k4 40 1003 2+ +

30. x x x3 9 3 93 2+ - -

2.14 Exercises

DID YOU KNOW?

Long division can be used to fi nd factors of an expression. For example, 1x - is a factor of 4 5x x+ -3 . We can fi nd the other factor by dividing 4 5x x+ -3 by 1.x -

-

5

4

5 5

5 5

0

x x

x

x xx x

x x

x

x

2

3

2

-

+ +

+

-

-

2

3

2

1x - + 4 5x -g

So the other factor of 4 5x x+ -3 is 5x x2 + + 4 5 ( 1) ( 5)x x x x x3` + - = - + +2

ch2.indd 68 7/31/09 3:43:30 PM


Completing the Square

Factorising a perfect square uses the results a ab b a b22 2 2!! + = ] g

EXAMPLES

1. Complete the square on .x x62 + Solution

Using 2 :a ab b2 2+ +

a x

ab x2 6

=

=

Substituting :a x=

xb x

b

2 6

3

=

=

To complete the square:

a ab b a b

x x x

x x x

2

2 3 3 3

6 9 3

2 2 2

2 2 2

2 2

+ + = +

+ + = +

+ + = +

]] ]

]g

g gg

2. Complete the square on .n n102 - Solution

Using :a ab b22 2+-

a n

ab x2 10

=

=

Substituting :a n=

nb n

b

2 10

5

=

=

To complete the square:

a ab b a b

n n n

n n n

2

2 5 5 5

10 25 5

2 2 2

2 2 2

2 2

- + = -

- + = -

- + = -

]] ]

]g

g gg

Notice that 3 is half of 6.

Notice that 5 is half of 10.

To complete the square on ,a pa2 + divide p by 2 and square it.

2 2

a pap

ap

22 2

+ + = +d dn n

ch2.indd 69 7/17/09 11:56:04 AM


EXAMPLES

1. Complete the square on .x x122 + Solution

Divide 12 by 2 and square it:

x x x x

x x

x

12212 12 6

12 36

6

22

2 2

2

2

+ + = + +

= + +

= +

c

]

m

g

2. Complete the square on .y y22 - Solution

Divide 2 by 2 and square it:

y y y y

y y

y

222 2 1

2 1

1

22

2 2

2

2

+ = +

= +

=

- -

-

-

c

^

m

h

Complete the square on

1. x x42 +

2. 6b b2 -

3. 10x x2 -

4. 8y y2 +

5. 14m m2 -

6. 18q q2 +

7. 2x x2 +

8. 16t t2 -

9. 20x x2 -

10. 44w w2 +

11. 32x x2 -

12. y y32 +

13. 7x x2 -

14. a a2 +

15. 9x x2 +

16. yy

2

52 -

17. k k2

112 -

18. 6x xy2 +

19. a ab42 -

20. p pq82 -

2.15 Exercises

ch2.indd 70 7/17/09 11:56:05 AM


Simplify

1. a5

5 10+

2. t3

6 3-

3. y

6

8 2+

4. d4 2

8-

5. x x

x5 22

2

-

6. y y

y

8 16

42 - +

-

7. a aab a

32 4

2

2

-

-

8. s ss s

5 62

2

2

+ +

+ -

9. bb

11

2

3

-

-

10. p

p p

6 92 7 152

-

+ -

11. a a

a2 3

12

2

+ -

-

12. x

x xy

8

2 233 -

- -+] ]g g

13. x x

x x x6 9

3 9 272

3 2

+ +

+ - -

14. p

p p

8 1

2 3 23

2

+

- -

15. 2 2ay by ax bx

ay ax by bx

- - +

- + -

2.16 Exercises

Algebraic Fractions

Simplifying fractions

EXAMPLES

Simplify 1.

24 2x +

Solution

x x

x2

4 22

2 1

2 1

2+=

+

= +

] g

2. 82 3 2

xx x

3

2

-

- -

Solution

xx x

x x x

x x

x xx

82 3 2

2 2 4

2 1 2

2 42 1

3

2

2

2

-

- -=

- + +

+ -

=+ +

+

] ^] ]

g hg g

Factorise fi rst, then cancel.

ch2.indd 71 7/17/09 11:56:07 AM


Operations with algebraic fractions

EXAMPLES

Simplify

1. x x5

14

3--

+

Solution

x x x x

x x

x

51

43

204 1 3

204 4 5 15

2019

5--

+=

- +

=- - -

=- -

-] ]g g

2. b

a b abb

a27

2 104 12

253

2 2

'+

+

+

-

Solution

ba b ab

ba

ba b ab

ab

b b b

ab aa a

b

a b bab

272 10

4 1225

272 10

254 12

3 3 9

2 55 5

4 3

5 3 98

3

2 2

3

2

2

2

2

' #

#

+

+

+

-=

+

+

-

+

=+ - +

+

+ -

+

=- - +

] ^]

] ]]

] ^g h

gg gg

g h

3. 5

22

1x x-

++

Solution

x x x xx x

x xx x

x xx

52

21

5 22 5

5 22 4 5

5 23 1

2 1-

++

=- +

+ -

=- +

+ + -

=- +

-

+] ]] ]

] ]] ]

g gg g

g gg g

Do algebraic fractions the same way as ordinary fractions.

ch2.indd 72 7/17/09 11:56:09 AM


1. Simplify

(a) 2 4

3x x+

(b) 5

1

3

2y y++

(c) 3

24

a a+-

(d) 6

32

2p p-+

+

(e) 2

53

1x x--

-

2. Simplify

(a) 2

36 3

2b a

b b2

#+ -

+

(b) 2 1

421

q q

p

p

q2

2 3

#+ +

-

+

+

(c) xyab

x y xyab a

53

212 62

2 2'

+

-

(d) x y

ax ay bx by

ab a b

x y2 2 2 2

3 3

#-

- + -

+

+

(e) x

x xx xx x

256 9

4 55 6

2

2

2

2

'-

- +

+ -

- +

3. Simplify

(a) 2 3x x+

(b) 1

1 2x x-

-

(c) 1 3a b

++

(d) 2

xx

x2

-+

(e) 1p q p q- ++

(f) 1

13

1x x+

+-

(g) 4

22

3x x2 -

-+

(h) 2 11

11

a a a2 + ++

+

(i) 2

23

11

5y y y+

-+

+-

(j) 16

212

7x x x2 2-

-- -

4. Simplify

(a) y

xx

y

yx x

4 123

6 24

9

272 82 2

3

2

# #- -

-

+

- -

(b) y y

a aya

ay

y y

4 45

43 15

52

2

2

2

2

' #- +

-

-

- - -

(c) x x

xx

x x3

39

2 84 16

32

2

#-

+-

+

-

+

(d) b

bb b

bb

b2 6

56 12

2

'+ + -

-+

(e) x x

x xx

xx

x x5 10

8 1510

92 10

5 62

2

2

2 2

' #+

- + -

-

+ +

5. Simplify

(a) 7 101

2 152

64

x x x x x x2 2 2- +-

- -+

+ -

(b) 4

52

32

2x x x2 -

--

-+

(c) 2 3p pq pq q2 2+

+-

(d) 1a b

aa b

ba b2 2+

--

+-

(e) x yx y

y xx

y x

y2 2-

++

--

-

2.17 Exercises

Substitution

Algebra is used in writing general formulae or rules. For example, the formula A lb= is used to fi nd the area of a rectangle with length l and breadth b . We can substitute any values for l and b to fi nd the area of different rectangles.

ch2.indd 73 7/17/09 11:56:11 AM


EXAMPLES

1. P l b2 2= + is the formula for fi nding the perimeter of a rectangle with length l and breadth b . Find P when .l 1 3= and . .b 3 2= Solution

. .

. .

P l b2 2

2 1 3 2 3 22 6 6 4

9

= +

= +

= +

=

] ]g g

2. V r h2r= is the formula for fi nding the volume of a cylinder with radius r and height h . Find V (correct to 1 decimal place) when 2.1r = and 8.7.h = Solution

. ( . )120.5

V r h

2 1 8 7correct to 1 decimal place

2

2

r

r

=

=

=

] g

3. If F C

59 32= + is the formula for changing degrees Celsius °C] g into

degrees Fahrenheit °F] g fi nd F when 25.C = Solution

F C5

9 32

525

32

5225 32

5225 160

5385

77

9

= +

= +

= +

=+

=

=

] g

This means that °25 C is the same as .°77 F

ch2.indd 74 7/17/09 11:56:13 AM


1. Given 3.1a = and 2.3b = - fi nd, correct to 1 decimal place.

(a) ab 3 (b) b (c) a5 2

(d) ab3

(e) a b 2+] g

(f) a b-

(g) b2-

2. T a n d1= + -] g is the formula for fi nding the term of an arithmetic series. Find T when ,a n4 18= - = and .d 3=

3. Given ,y mx b= + the equation of a straight line, fi nd y if ,m x3 2= = - and 1.b = -

4. If 100 5h t t2= - is the height of a particle at time t , fi nd h when 5.t =

5. Given vertical velocity ,v gt= - fi nd v when 9.8g = and 20.t =

6. If 2 3y x= + is the equation of a function, fi nd y when 1.3,x = correct to 1 decimal place.

7. S r r h2r= +] g is the formula for the surface area of a cylinder. Find S when 5r = and 7,h = correct to the nearest whole number.

8. A r2r= is the area of a circle with radius r . Find A when 9.5,r = correct to 3 signifi cant fi gures.

9. Given u ar 1n

n= - is the n th term of a geometric series, fi nd un if 5,a = 2r = - and 4.n =

10. Given 3V lbh= 1 is the volume

formula for a rectangular pyramid, fi nd V if . , .l b4 7 5 1= = and 6.5.h =

11. The gradient of a straight line is

given by .m x xy y

2 1

2 1=

-

- Find m

if , ,x x y3 1 21 2 1= = - = - and 5.y2 =

12. If 2A h a b= +1 ] g gives the area

of a trapezium, fi nd A when , .h a7 2 5= = and 3.9.b =

13. Find V if 3V r3r= 4 is the volume

formula for a sphere with radius r and 7.6,r = to 1 decimal place.

14. The velocity of an object at a certain time t is given by the formula .v u at= + Find v when

4 5,u a= =1 3 and 6 .t = 5

15. Given 1

,Sr

a=

- fi nd S if 5a =

and 3 .r = 2 S is the sum to infi nity of a geometric series.

16. ,c a b2 2= + according to Pythagoras’ theorem. Find the value of c if 6a = and 8.b =

17. Given 16y x2= - is the equation of a semicircle, fi nd the exact value of y when 2.x =

2.18 Exercises

ch2.indd 75 7/17/09 11:56:22 AM


18. Find the value of E in the energy equation E mc2= if 8.3m = and 1.7.c =

19. 1100

A P r n

= +c m is the formula

for fi nding compound interest. Find A when ,P r200 12= = and 5,n = correct to 2 decimal places.

20. If Srra

11

=-

-n^ h is the sum of

a geometric series, fi nd S if ,a r3 2= = and 5.n =

21. Find the value of c

a b2

3 2

if

4 3,a b2 3

= =3 2c cm m and .c21 4

= c m

Surds

An irrational number is a number that cannot be written as a ratio or fraction (rational). Surds are special types of irrational numbers, such as 2, 3 and 5 .

Some surds give rational values: for example, 9 3.= Others, like 2, do not have an exact decimal value. If a question involving surds asks for an exact answer, then leave it as a surd rather than giving a decimal approximation.

Simplifying surds

a b ab

a bb

aba

#

'

=

= =

Class Investigations

Is there an exact decimal equivalent for 1. 2 ? Can you draw a line of length exactly 2. 2 ? Do these calculations give the same results? 3.

(a) 9 4# and 9 4#

(b) 9

4 and

94

(c) 9 4+ and 9 4+

(d) 9 4- and 9 4-

Here are some basic properties of surds.

x x x2 2= =^ h

ch2.indd 76 7/17/09 11:56:25 AM


EXAMPLES

1. Express in simplest surd form 45 . Solution

45 9 5

9 5

3 5

3 5

#

#

#

=

=

=

=

2. Simplify 3 40 . Solution

3 3

3

3 2

6

40 4 10

4 10

10

10

#

# #

# #

=

=

=

=

3. Write 5 2 as a single surd. Solution

5 2 25 2

50

#=

=

54 also equals 3 15# but this will

not simplify. We look for a number that is a

perfect square.

Find a factor of 40 that is a perfect square.

1. Express these surds in simplest surd form.

(a) 12

(b) 63

(c) 24

(d) 50

(e) 72

(f) 200

(g) 48

(h) 75

(i) 32

(j) 54

(k) 112

(l) 300

(m) 128

(n) 243

(o) 245

(p) 108

(q) 99

(r) 125

2. Simplify

(a) 2 27

(b) 5 80

2.19 Exercises

ch2.indd 77 7/17/09 11:56:27 AM


(c) 4 98

(d) 2 28

(e) 8 20

(f) 4 56

(g) 8 405

(h) 15 8

(i) 7 40

(j) 8 45

3. Write as a single surd.

(a) 3 2

(b) 2 5

(c) 4 11

(d) 8 2

(e) 5 3

(f) 4 10

(g) 3 13

(h) 7 2

(i) 11 3

(j) 12 7

4. Evaluate x if

(a) 3 5x =

(b) 2 3 x=

(c) 3 7 x=

(d) 5 2 x=

(e) 2 11 x=

(f) 7 3x =

(g) 4 19 x=

(h) 6 23x =

(i) 5 31 x=

(j) 8 15x =

Addition and subtraction

Calculations with surds are similar to calculations in algebra. We can only add or subtract ‘like terms’ with algebraic expressions. This is the same with surds.

EXAMPLES

1. Simplify 3 2 4 2 .+ Solution

3 4 72 2 2+ =

2. Simplify 3 12 .- Solution

First, change into ‘like’ surds.

3 12 3 4 3

3 2 3

3

#- = -

= -

= -

3. Simplify 2 2 2 3 .- + Solution

2 2 2 3 2 3- + = +

ch2.indd 78 7/17/09 11:56:28 AM


Multiplication and division

Simplify

1. 5 2 5+

2. 3 2 2 2-

3. 3 5 3+

4. 7 3 4 3-

5. 5 4 5-

6. 4 6 6-

7. 2 8 2-

8. 5 4 5 3 5+ +

9. 2 2 2 3 2- -

10. 5 45+

11. 8 2-

12. 3 48+

13. 12 27-

14. 50 32-

15. 28 63+

16. 2 8 18-

17. 3 54 2 24+

18. 90 5 40 2 10- -

19. 4 48 3 147 5 12+ +

20. 3 2 8 12+ -

21. 2863 50--

22. 12 45 48 5-- -

23. 150 45 24+ +

24. 32 243 50 147-- +

25. 80 3 245 2 50- +

2.20 Exercises

To get a b c d ac bd ,# = multiply surds with surds and

rationals with rationals.

a b ab

a b c d ac bd

a a a a2

#

#

#

=

=

= =

EXAMPLES

Simplify 1. 2 2 5 7#- Solution

2 2 5 7 10 14#- = -

b

aba

=

CONTINUED

ch2.indd 79 7/17/09 11:56:31 AM


2. 4 2 5 18# Solution

4 2 5 18 20 36

20 6

120

#

#

=

=

=

3. 4 2

2 14

Solution

4 2

2 14

4 2

2 2 7

27

#=

=

4. 15 2

3 10

Solution

15 2

3 10

15 2

3 5 2

55

# #=

=

5. 310 2d n

Solution

33

310

3

10

310

2

2

2

=

=

= 1

d ^^n h

h

ch2.indd 80 7/17/09 11:56:35 AM


Simplify

1. 7 3#

2. 3 5#

3. 2 3 3#

4. 5 7 2 2#

5. 3 3 2 2#-

6. 5 3 2 3#

7. 4 5 3 11#-

8. 2 7 7#

9. 2 3 5 12#

10. 6 2#

11. 28 6#

12. 3 2 5 14#

13. 10 2 2#

14. 2 6 7 6#-

15. 22^ h

16. 2 72^ h

17. 3 5 2# #

18. 2 3 7 5# #-

19. 2 6 3 3# #

20. 2 5 3 2 5 5# #- -

21. 2 2

4 12

22. 3 6

12 18

23. 10 2

5 8

24. 2 12

16 2

25. 5 10

10 30

26. 6 20

2 2

27. 8 10

4 2

28. 3 15

3

29. 8

2

30. 6 10

3 15

31. 5 8

5 12

32. 10 10

15 18

33. 2 6

15

34. 32 2d n

35. 75 2d n

2.21 Exercises

Expanding brackets

The same rules for expanding brackets and binomial products that you use in algebra also apply to surds.

ch2.indd 81 7/17/09 11:56:37 AM


Simplifying surds by removing grouping symbols uses these general rules.

b c ab aca + = +^ h

Proof

a b c a b a c

ab ac

# #+ = +

= +

^ h

Binomial product:

a b c d ad bdac bc+ + = + + +^ ^h h

Proof

a b c d a c a d b c b d

ac ad bc bd

# # # #+ + = + + +

= + + +

^ ^h h

Perfect squares:

a b a ab b22

+ = + +^ h

Proof

a b a b a b

a ab ab ba ab b2

2

2 2

+ = + +

= + + +

= + +

^ ^ ^h h h

a b a ab b22

- = - +^ h

Proof

a b a b a b

a ab ab ba ab b2

2

2 2

- = - -

= - - +

= - +

^ ^ ^h h h

Difference of two squares:

a b a b a b+ - = -^ ^h h

Proof

a b a b a ab ab ba b

2 2+ - = - + -

= -

^ ^h h

ch2.indd 82 7/17/09 11:56:40 AM


EXAMPLES

Expand and simplify 1. 2 5 2+^ h Solution

( )2 5 2 2 5 2 2

10 4

10 2

# #+ = +

= +

= +

2. 3 7 2 3 3 2-^ h Solution

( )3 7 2 3 3 2 3 7 2 3 3 7 3 2

6 21 9 14

# #- = -

= -

3. 2 3 5 3 2+ -^ ^h h Solution

( ) ( )2 3 5 3 2 2 3 2 2 3 5 3 3 5 2

6 2 3 15 3 10

# # # #+ - = - + -

= - + -

4. 5 2 3 5 2 3+ -^ ^h h Solution

( 2 )( 2 ) 2 2 2 2

5 2 2 4 35 12

7

5 3 5 3 5 5 5 3 3 5 3 3

15 15

# # # #

#

+ - = - + -

= - + -

= -

= -

Another way to do this question is by using the difference of two squares.

( ) ( )5 2 3 5 2 3 5 2 3

5 4 3

7

2 2

#

+ - = -

= -

= -

^ ^h h

Notice that using the difference of two

squares gives a rational answer.

ch2.indd 83 7/17/09 11:56:43 AM


1. Expand and simplify (a) 5 32 +^ h (b) 2 2 53 -^ h (c) 3 3 2 54 +^ h (d) 5 2 2 37 -^ h (e) 3 2 4 6-- ^ h (f) 3 5 11 3 7+^ h (g) 3 2 2 4 3- +^ h (h) 5 5 35 -^ h (i) 3 12 10+^ h (j) 2 3 18 3+^ h (k) 4 2 3 62- -^ h (l) 7 3 20 2 35- - +^ h (m) 10 3 2 2 12-^ h (n) 5 22 +- ^ h (o) 2 3 2 12-^ h

2. Expand and simplify (a) 2 3 5 3 3+ +^ ^h h (b) 5 2 2 7- -^ ^h h (c) 2 5 3 2 5 3 2+ -^ ^h h (d) 3 10 2 5 4 2 6 6- +^ ^h h (e) 2 5 7 2 5 3 2- -^ ^h h (f) 5 6 2 3 5 3+ -^ ^h h (g) 7 3 7 3+ -^ ^h h (h) 2 3 2 3- +^ ^h h (i) 6 3 2 6 3 2+ -^ ^h h (j) 3 5 2 3 5 2+ -^ ^h h (k) 8 5 8 5- +^ ^h h (l) 2 9 3 2 9 3+ -^ ^h h

(m) 2 11 5 2 2 11 5 2+ -^ ^h h (n) 5 2

2+^ h

(o) 2 2 32

-^ h

(p) 3 2 72

+^ h

(q) 2 3 3 52

+^ h

(r) 7 2 52

-^ h

(s) 2 8 3 52

-^ h

(t) 3 5 2 22

+^ h

3. If 3 2a = , simplify (a) a 2 2 (b) a 3 (2 (c) a ) 3 (d) 1a 2+] g (e) –a a3 3+] ]g g

4. Evaluate a and b if (a) 2 5 1 a b

2+ = +^ h

(b) 2 2 5 2 3 5- -^ ^h h a b 10= +

5. Expand and simplify (a) a a3 2 3 2+ - + +^ ^h h (b) 1p p

2- -_ i

6. Evaluate k if .k2 7 3 2 7 3- + =^ ^h h

7. Simplify .x y x y2 3+ -_ _i i

8. If 2 3 5 a b2

- = -^ h , evaluate a and b.

9. Evaluate a and b if .a b7 2 3 2

2- = +^ h

10. A rectangle has sides 5 1+ and 2 5 1- . Find its exact area.

2.22 Exercises

Rationalising the denominator

Rationalising the denominator of a fractional surd means writing it with a rational number (not a surd) in the denominator. For example, after

rationalising the denominator, 5

3 becomes 5

3 5.

ch2.indd 84 7/17/09 11:56:46 AM


Squaring a surd in the denominator will rationalise it since .x x2=^ h

DID YOU KNOW?

A major reason for rationalising the denominator used to be to make it easier to evaluate the fraction (before calculators were available). It is easier to divide by a rational number than an irrational one; for example,

5

33 2.236'=

5

3 53 2.236 5# '=

This is hard to do without a calculator.

This is easier to calculate.

b

ab

bb

a b# =

Multiplying by b

b

is the same as multiplying by 1.

Proof

ba

b

b

b

a b

ba b

2# =

=

EXAMPLES

1. Rationalise the denominator of 5

3 . Solution

5

35

55

3 5# =

2. Rationalise the denominator of 5 3

2 . Solution

5 32

3

3

5 9

2 3

5 32 3

152 3

#

#

=

=

=

Don’t multiply by

5 3

5 3 as it takes

longer to simplify.

ch2.indd 85 7/17/09 11:56:50 AM


When there is a binomial denominator, we use the difference of two squares to rationalise it, as the result is always a rational number.

To rationalise the denominator of c d

a b

+

+ , multiply by

c d

c d

-

-

Proof

c d

a b

c d

c d

c d

a b c d

c d

a b c d

c da b c d

c d

2 2

#+

+

-

-=

+ -

+ -

=-

+ -

=-

+ -

^ ^^ ^

^ ^^ ^^ ^

h hh h

h hh hh h

EXAMPLES

1. Write with a rational denominator

.2 3

5

-

Solution

2 3

5

2 3

2 3

2 3

5 2 3

2 910 3 5

710 3 5

710 3 5

2 2#

- +

+=

-

+

=-

+

=-

+

= -+

^^h

h

2. Write with a rational denominator

3 4 2

2 3 5.

+

+

Solution

3 4 2

2 3 5

3 4 2

3 4 2

3 4 2

2 3 5 3 4 2

3 16 22 3 8 6 15 4 10

2 2#

#

#

+

+

-

-=

-

+ -

=-

- + -

^ ^^ ^

h hh h

Multiply by the conjugate surd 2 3.+

ch2.indd 86 7/17/09 11:56:53 AM


296 8 6 15 4 10

296 8 6 15 4 10

=-

- + -

=- + - +

3. Evaluate a and b if .a b3 2

3 3

-= +

Solution

3 2

3 3

3 2

3 2

3 2 3 2

3 3 3 2

3 2

3 9 3 6

3 23 3 3 6

19 3 6

9 3 6

9 9 6

9 54

2 2

#

#

#

- +

+=

- +

+

=-

+

=-

+

=+

= +

= +

= +

^ ^^

^ ^h h

h

h h

.a b9 54So and= =

4. Evaluate as a fraction with rational denominator

3 2

23 2

5.

++

-

Solution

3 22

3 22

3 2

5

3 2 3 2

3 2

3 2

2 3 4 15 2 5

3 42 3 4 15 2 5

12 3 4 15 2 5

2 3 4 15 2 5

5

2 2

++

-=

+ -

- +

=-

- + +

=-

- + +

=-

- + +

= - + - -

+

^ ^^ ^

^h hh h

h

ch2.indd 87 7/17/09 11:56:56 AM


1. Express with rational denominator

(a) 7

1

(b) 2 2

3

(c) 5

2 3

(d) 5 2

6 7

(e) 3

1 2+

(f) 2

6 5-

(g) 5

5 2 2+

(h) 2 7

3 2 4-

(i) 4 5

8 3 2+

(j) 7 5

4 3 2 2-

2. Express with rational denominator

(a) 3 2

4+

(b) 2 7

3

-

(c) 5 2 6

2 3

+

(d) 3 4

3 4

+

-

(e) 3 2

2 5

-

+

(f) 2 5 3 2

3 3 2

+

+

3. Express as a single fraction with rational denominator

(a) 2 1

12 1

1+

+-

(b) 2 3

2

2 33

--

+

(c) 5 2

13 2 5

3+

+-

(d) 2 3

2 7

2 3 2

2#

+

-

+

(e) tt1

+ where t 3 2= -

(f) zz12

2- where z 1 2= +

(g) 6 3

3 2 4

6 3

2 1

6 12

-

++

+

--

-

(h) 2

2 3

31+

+

(i) 2 3

3

3

2

++

(j) 6 2

5

5 32

+-

(k) 4 3

2 7

4 3

2

+

+-

-

(l) 3 2

5 2

3 1

2 3

-

--

+

+

4. Find a and b if

(a) ba

2 53

=

(b) b

a

4 2

3 6=

(c) a b5 1

2 5+

= +

(d) a b7 4

2 77

-= +

(e) a b2 1

2 3

-

+= +

2.23 Exercises

ch2.indd 88 7/17/09 11:56:59 AM


5. Show that 2 1

2 1

24

+

-+ is

rational.

6. If x 3 2= + , simplify

(a) 1x x+

(b) 1xx

22

+

(c) 1x x

2

+b l

7. Write 5 2

25 2

1+

+-

-

3

5 1+ as a single fraction with

rational denominator .

8. Show that 3 2 2

22

8+

+ is

rational .

9. If x2 1 3+ = , where ,x 0!

fi nd x as a surd with rational denominator .

10. Rationalise the denominator of

2b

b 2

-

+ b 4!] g

ch2.indd 89 7/17/09 11:57:03 AM


1. Simplify (a) y y5 7-

(b) a3

3 12+

(c) k k2 33 2#-

(d) x y

3 5+

(e) 4 3 5a b a b- - - (f) 8 32+ (g) 3 5 20 45- +

2. Factorise (a) 36x2 - (b) 2 3a a2 + - (c) 4 8ab ab2 - (d) y xy x5 15 3- + - (e) 4 2 6n p- + (f) 8 x3-

3. Expand and simplify (a) bb 23 -+ ] g (b) x x2 1 3- +] ]g g (c) m m5 3 2+ --] ]g g (d) 4 3x 2-] g (e) 5 5p p- +^ ^h h (f) a a47 2 5+- -] g (g) 2 2 53 -^ h (h) 3 7 3 2+ -^ ^h h

4. Simplify

(a) b

aa

b5

4 1227

103 3#

-

-

(b) m m

mm

m2

5 103 3

42

2

'- -

+

+

-

5. The volume of a cube is .V s3= Evaluate V when 5.4.s =

6. (a) Expand and simplify .2 5 3 2 5 3+ -^ ^h h

Rationalise the denominator of (b)

.2 5 3

3 3

+

7. Simplify .x x x x2

33

16

22-

++

-+ -

8. If ,a b4 3= = - and ,c 2= - fi nd the value of

(a) ab2 (b) a bc- (c) a (d) bc 3] g (e) c a b2 3+] g

9. Simplify

(a) 6 15

3 12

(b) 2 2

4 32

10. The formula for the distance an object falls is given by 5 .d t2= Find d when 1.5.t =


(a) 5 3

2

(b) 2

1 3+

12. Expand and simplify (a) 3 2 4 3 2- -^ ^h h (b) 7 2

2+^ h

13. Factorise fully (a) 3 27x2 - (b) x x6 12 182 - - (c) 5 40y3 +

Test Yourself 2

ch2.indd 90 7/17/09 5:04:47 PM


14. Simplify

(a) 9

3

xy

x y5

4

(b) 15 5

5x -

15. Simplify

(a) 3 112^ h

(b) 2 33^ h

16. Expand and simplify (a) a b a b+ -] ]g g (b) a b 2+] g (c) a b 2-] g

17. Factorise (a) 2a ab b2 2- + (b) a b3 3-

18. If 3 1,x = + simplify 1x x+ and give your answer with a rational denominator.

19. Simplify

(a) 4 3a b+

(b) 2

35

2x x--

-

20. Simplify 5 2

32 2 1

2,

+-

- writing

your answer with a rational denominator.

21. Simplify (a) 3 8 (b) 2 2 4 3#- (c) 108 48-

(d) 2 18

8 6

(e) 5 3 2a b a# #- -

(f) 62m nm n

2 5

3

(g) 3 2x y x y- - -

22. Expand and simplify (a) 2 3 22 +^ h (b) 5 7 3 5 2 2 3- -^ ^h h (c) 3 2 3 2+ -^ ^h h (d) 4 3 5 4 3 5- +^ ^h h (e) 3 7 2

2-^ h


(a) 7

3

(b) 5 3

2

(c) 5 1

2-

(d) 3 2 3

2 2

+

(e) 4 5 3 3

5 2

-

+

24. Simplify

(a) x x53

22

--

(b) a a7

23

2 3++

-

(c) x x1

11

22 -

-+

(d) 2 34

31

k k k2 + -+

+

(e) 2 5

3

3 25

+-

-

25. Evaluate n if (a) 108 12 n- =

(b) 112 7 n+ =

(c) 2 8 200 n+ =

(d) 4 147 3 75 n+ =

(e) n2 2452180

+ =

ch2.indd 91 7/17/09 11:57:20 AM


26. Evaluate xx12

2+ if x

1 2 3

1 2 3=

-

+

27. Rationalise the denominator of 2 7

3

(there may be more than one answer).

(a) 2821

(b) 28

2 21

(c) 1421

(d) 721

28. Simplify .x x5

34

1--

+

(a) x20

7+- ] g

(b) 20

7x +

(c) x20

17+

(d) x20

17- +] g

29. Factorise 4 4x x x3 2- - + (there may be more than one answer).

(a) x x1 42 - -^ ]h g (b) x x1 42 + -^ ]h g (c) x x 42 -] g (d) x x x4 1 1- + -] ] ]g g g

30. Simplify .3 2 2 98+ (a) 5 2 (b) 5 10 (c) 17 2 (d) 10 2

31. Simplify .x x x4

32

22

12 -

+-

-+

(a) 2 2

5x x

x+ -

+] ]g g (b)

2 21

x xx

+ -

+] ]g g (c)

2 29

x xx

+ -

+] ]g g (d)

2 23

x xx

+ -

-] ]g g 32. Simplify .ab a ab a5 2 7 32 2- - -

(a) 2ab a2+ (b) ab a2 5 2- - (c) a b13 3- (d) 2 5ab a2- +

33. Simplify .2780

(a) 3 3

4 5

(b) 9 3

4 5

(c) 9 3

8 5

(d) 3 3

8 5

34. Expand and simplify .x y3 2 2-^ h (a) x xy y3 12 22 2- - (b) x xy y9 12 42 2- - (c) 3 6 2x xy y2 2- + (d) x xy y9 12 42 2- +

35. Complete the square on .a a162 - (a) a a a16 16 42 2- + = -^ h (b) a a a16 64 82 2- + = -^ h (c) a a a16 8 42 2- + = -^ h (d) a a a16 4 22 2- + = -^ h

ch2.indd 92 7/17/09 11:57:32 AM


1. Expand and simplify (a) ab a b b aa4 2 32 2- --] ]g g (b) 2 2y y2 2- +_ _i i (c) x2 5 3-] g

2. Find the value of x y+ with rational denominator if x 3 1= + and

2 5 3

1 .y =-

3. Simplify .7 6 54

2 3

-

4. Complete the square on .x ab x2 +

5. Factorise (a) ( ) ( )x x4 5 42+ + + (b) 6x x y y4 2 2- - (c) 125 343x3 + (d) 2 4 8a b a b2 2- - +

6. Complete the square on .x x4 122 +

7. Simplify .x x

xy x y

4 16 12

2 2 6 62 - +

+ - -

8. | |

da b

ax by c2 2

1 1=

+

+ + is the formula for

the perpendicular distance from a point to a line. Find the exact value of d with a rational denominator if , , ,a b c x2 1 3 41= = - = = - and 5.y1 =

9. Simplify 1

.a

a

13

3

+

+^ h

10. Factorise .x b

a42 2

2

-

11. Simplify .x

x y

x

x y

x x

x y

3

2

3 6

3 22-

++

+

--

+ -

+

12. (a) Expand .x2 1 3-^ h

Simplify (b) .x x x

x x8 12 6 1

6 5 43 2

2

- + -

+ -

13. Expand and simplify 3x x1 2- -] ^g h .

14. Simplify and express with rational

denominator .3 4

2 5

2 1

5 3

+

+-

-

15. Complete the square on 3 .x x2 + 2

16. If ,xk l

lx kx1 2=

+

+ fi nd the value of x when

, ,k l x3 2 51= = - = and 4.x2 =

17. Find the exact value with rational

denominator of x x x2 3 12 - + if 2 5 .x =

18. Find the exact value of

(a) xx12

2+ if x

1 2 3

1 2 3=

-

+

(b) a and b if a b2 3 3

3 43

+

-= +

19. 2A r2i= 1 is the area of a sector of a

circle. Find the value of i when A 12= and 4.r =

20. If V r h2r= is the volume of a cylinder, fi nd the exact value of r when 9V = and 16.h =

21. If 2 ,s u at2= + 1 fi nd the exact value of s

when ,u a2 3= = and 2 3 .t =


ch2.indd 93 7/17/09 11:57:46 AM

TERMINOLOGY

3 Equations

Absolute value: the distance of a number from zero on a number line

Equation: A mathematical statement that has a pronumeral or unknown number and an equal sign. An equation can be solved to fi nd the value of the unknown number e.g. 2 3 5x - =

Exponential equation: Equation where the unknown pronumeral is the power or index e.g. 2 8x

=

Inequation: A mathematical statement involving an inequality sign, , , or1 2 # $ that has an unknown

pronumeral that is solved to fi nd values that make the statement true e.g. 2 3 4x 2-

Quadratic equation: An equation involving x2 as the highest power of x that may have two, one or no solutions

Simultaneous equations: Two or more independent equations that can be solved together to produce a solution that makes each equation true at the same time. The number of equations required is the same as the number of unknowns

ch3.indd 94 7/31/09 6:58:40 PM

95Chapter 3 Equations

DID YOU KNOW?

Box text...

INTRODUCTION

EQUATIONS ARE FOUND IN most branches of mathematics. They are also important in many other fi elds, such as science, economics, statistics and engineering. In this chapter you will revise basic equations and inequations. Equations involving absolute values, exponential equations, quadratic equations and simultaneous equations are also covered here.

DID YOU KNOW?

Algebra was known in ancient civilisations. Many equations were known in Babylonia, although general solutions were diffi cult because symbols were not used in those times.

Diophantus , around 250 AD, fi rst used algebraic notation and symbols (e.g. the minus sign). He wrote a treatise on algebra in his Arithmetica , comprising 13 books. Only six of these books survived. About 400 AD, Hypatia of Alexandria wrote a commentary on them.

Hypatia was the daughter of Theon, a mathematician who ensured that she had the best education. She was the fi rst female mathematician on record, and was a philosopher and teacher. She was murdered for her philosophical views by a fanatical Christian sect.

In 1799 Carl Friedrich Gauss proved the Fundamental Theorem of Algebra: that every algebraic equation has a solution.

PROBLEM

The age of Diophantus at his death can be calculated from this epitaph:

Diophantus passed one-sixth of his life in childhood, one-twelfth in youth, and one-seventh more as a bachelor; fi ve years after his marriage a son was born who died four years before his father at half his father’s fi nal age. How old was Diophantus?

Simple Equations

Here are the four rules for changing numbers or pronumerals from one side of an equation to the other.

If a number is • added, subtract it from both sides If a number is • subtracted, add it to both sides If a number is • multiplied, divide both sides by the number If a number is • divided, multiply both sides by the number

Do the opposite operation to take a number to the

other side of an equation.

ch3.indd 95 8/7/09 11:48:39 AM


EXAMPLES

Solve 1. x3 5 17+ =

Solution

x

x

x

x

x

3 5 17

3 5 17

3 12

3 12

4

5 5

3 3

+ =

+ =

=

=

=

- -

You can check the solution by substituting the value into the equation.

3 5

( )

x

3 4 5

12 5

17

LHS

RHS

= +

= +

= +

=

=

Since , 4xLHS RHS= = is the correct solution.

2. 4 3 8 21y y- = +

Solution

y y

y y

y

y

y

y

y

y

y y

4 3 8 21

4 3 8 21

3 4 21

3 4 21

24 4

24 4

6

6

4 4

21 21

4 4

`

- = +

- = +

- = +

- = +

- =

-=

- =

= -

- -

- -

3. x x2 3 7 6 1+ = - -] ]g g Solution

( ) ( )x x

x x

x

x x

x

x x

2 3 7 6 1

6 14 6 1

7

6 14 7

7 14 7

+ = - -

+ = - +

= -

+ = -

+ =

+ +

Check these solutions by substituting them into the equation.

ch3.indd 96 7/17/09 2:01:27 PM


x

x

x

x

7 14 7

7 7

7 7

1

14 14

7 7

+ =

= -

=-

= -

- -

1. t 4 1+ = -

2. . .z 1 7 3 9+ = -

3. y 3 2- = -

4. . .w 2 6 4 1- =

5. x5 7= -

6. 1.5 6x =

7. 35y = 1

8. 7

5b=

9. n28

- =

10. 6 3

2r=

11. y2 1 19+ =

12. 33 4 9k= +

13. d7 2 12- =

14. x2 5 27- = -

15. y

34 9+ =

16. x2

3 7- =

17. m5

7 11+ =

18. x3 5 17+ =

19. a4 7 21+ = -

20. y7 1 20- =

21. b8 4 36- = -

22. 3( 2) 15x + =

23. ( )a2 3 1 8- + =

24. t t7 4 3 12+ = -

25. 3 6 9x x- = -

26. 2( 2) 4 3a a- = -

27. 5 2 3( 1)b b+ = - -

28. 3( 7) 2(2 9)t t+ = -

29. ( ) ( )p p p2 5 1 5 2+ - = - -

30. . . . .x x3 7 1 2 5 4 6 3+ = -

3.1 Exercises

Solve

A STARTLING FACT!

Half full half empty

full empty`

=

=

ch3.indd 97 7/17/09 2:01:28 PM


Equations involving fractions

There are different ways to solve this type of equation. One way is to multiply both sides of the equation by the common denominator of the fractions.

EXAMPLES

Solve

1. m3

421

- =

Solution

2

2

( )

m

m

m

m

m

m

m

34

21

34

2 24 3

2 24 3

2 27

2 27

227

13

6 6 6

24 24

2 2

- =

- =

- =

- =

=

=

=

=

+ +

1

1

c cm m

2. x x3

14

5++ =

Solution

(5)

( )

12 12 12

x x

x x

x x

x x

x

x

x

x

x

31

45

31

44 1 3 60

4 4 3 60

7 4 60

7 4 60

7 56

7 56

8

4 4

7 7

++ =

++ =

+ + =

+ + =

+ =

+ =

=

=

=

- -

c cm m

Multiply by the common denominator, 6.

The common denominator of 3 and 4 is 12.

ch3.indd 98 7/17/09 2:01:29 PM


3. y y

51

3

2

65+

--

=

Solution

4

( ) ( )

y y

y y

y y

y y

y

y

y

y

y

51

3

2

65

51

3

2

65

6 1 10 2 25

6 6 10 20 25

4 26 25

4 26 25

4 1

4 1

30 30 30

26 26

4 4

+-

-=

+-

-=

+ - - =

+ - + =

- + =

- + =

- = -

-=

-

=

- -

- -

1

e e co o m

When there is a fraction on either side of the equation, multiplying by the common denominator is the same as cross multiplying.

The common denominator of 5, 3

and 6 is 30.

EXAMPLES

1. Solve ( )x x538 0!=

Solution

=

xx

x

x

538

8 15

8 15

187

8 8

=

=

=

2. Solve 0n

n58

23

!= ^ h Solution

nn

n

n

58

23

16 15

16 15

1615

16 16

=

=

=

=

ch3.indd 99 7/17/09 2:01:29 PM


1. b5 3

2=

2. ( )x x751 0!=

3. ( )y y4109 0!=

4. 45

711x

=

5. ( )k

k54

29 0!=

6. x3

4 8- =

7. 45

43t

=

8. 7

572x+

=

9. y

2 53

= -

10. x9 3

2 7- =

11. 2

3 5w -=

12. t t52

32- =

13. x4 2

1 4+ =

14. x x5 2 10

3- =

15. 3

42

1x x++ =

16. p p

23

32

2-

+ =

17. t t7

33

1 4++

-=

18. x x9

55

2 1+-

+=

19. q q

31

42

2-

--

=

20. x x5

3 22

7++ =

+

21. b b43

51

2- =

22. a3 4

385

+ =

23. x x2

5 3+

= ,x 0 2! -^ h

24. y y1

13 1

1+

=-

,y 131

! -c m

25. t t3

24

1 0-

++

= ,t 3 4! -^ h

3.2 Exercises

Solve

Substitution

Sometimes substituting values into a formula involves solving an equation.

Investigation

Body mass index (BMI) is a formula that is used to measure body fatness and is used by health professionals to screen for weight categories that may lead to health problems.

ch3.indd 100 7/17/09 2:01:30 PM


This is not the only measure that is used when looking for health problems, however. For example, there are other factors in cardiac (heart) disease. Research these to fi nd out what other things doctors look for.

The BMI is used in a different way with children and teens, and is taken in relation to the child’s age.

The formula for BMI is hwBMI

2= where w is weight in kg and h is height

in metres.

For adults over 20, a BMI under 18.5 means that the person is underweight and over 25 is overweight. Over 30 is obese.

The BMI may not always be reliable in measuring body fat. Can you think of some reasons?

Is it important where the body fat is stored? Does it make a difference if it is on the hips or the stomach?

Research these questions and fi nd out more about BMI generally.

EXAMPLES

1. The formula for the surface area of a rectangular prism is given by ( ) .S lb bh lh2= + + Find the value of b when 180, 9 6.S l hand= = =

Solution

( )

( )

( )

.

S lb bh lh

b b

b

b

b

b

b

b

2

180 2 9 6 9 6

2 15 54

30 108

30 108

72 30

72 30

2 4

108 108

30 30

180

#

= + +

= + +

= +

= +

= +

=

=

=

- -

CONTINUED

Another way of doing this would be to change

the subject of the formula fi rst.

ch3.indd 101 7/17/09 2:01:31 PM


2. The volume of a cylinder is given by V r h2r= . Evaluate the radius r, correct to 2 decimal places, when 350V = and . .h 6 5=

Solution

( . )

( . )

.

.

..

. .

V r h

r

r

r

r

r

r

350 6 5

350 6 5

6 5350

6 5350

6 5350

4 14

6 5 6 5

2

2

2

2

2

r

r

r

r

r

r

r r

=

=

=

=

=

=

=

1. Given that u atv = + is the formula for the velocity of a particle at time t, fi nd the value of t when . ,u 17 3= . . .v a100 6 9 8and= =

2. The sum of an arithmetic series is

given by ( ) .S n a l2

= + Find l if

3, 26 1625.a n Sand= = =

3. The formula for fi nding the area

of a triangle is 2 .A bh= 1 Find b

when 36 9.A hand= =

4. The area of a trapezium is given

by 2 ( ) .A h a b= +1 Find

the value of a when 120,A =

.h b5 7and= =

5. Find the value of y when 3,x = given the straight line equation 5 2 7 0.x y- - =

6. The area of a circle is given by .A r2r= Find r correct to 3 signifi cant fi gures if 140.A =

7. The area of a rhombus is given by

the formula 2A xy= 1 where x and

y are its diagonals. Find the value of x correct to 2 decimal places when 7.8 25.1.y Aand= =

8. The simple interest formula is

.PrnI100

= Find n if 14.5,r =

150 326.25.P Iand= =

9. The gradient of a straight

line is given by .x xy y

m2 1

2 1

-

-=

Find y1 when 6,m = - 5

, .y x x7 3 1and2 2 1= = - =

10. The surface area of a cylinder is given by the formula .S r r h2r= +] g Evaluate h correct to 1 decimal place if 232 4.5.S rand= =

3.3 Exercises

ch3.indd 102 7/17/09 2:01:33 PM


11. The formula for body mass index

is hwBMI

2= . Evaluate

the (a) BMI when 65w = and 1.6h =

(b) w when 21.5BMI = and 1.8h =

(c) h when 19.7BMI = and . .w 73 8=

12. A formula for depreciation is .D P r1 n= -] g Find r if ,D P12 000 15 000= = and 3.n =

13. The x -value of the midpoint is

given by x x

x2

1 2+= . Find x 1

when 2x = - and .x 52 =

14. Given the height of a particle at time t is ,h t5 2= evaluate t when .h 23=

15. If ,y x 12= + evaluate x when .y 5=

16. If the surface area of a sphere is ,S r4 2r= evaluate r to 3 signifi cant fi gures when . .S 56 3=

17. The area of a sector of a circle

is .A r2

21

i= Evaluate r when

24.6A = and . .0 45i=

18. If x

y1

23 -

= , fi nd the value of x

when .y 3=

19. Given 2 5,y x= + evaluate x when .y 4=

20. The volume of a sphere is

V r3

34r= . Evaluate r to 1 decimal

place when .V 150=

There are two solutions to this question.

Inequations

• 2 means greater than • 1 means less than • $ means greater than or equal to • # means less than or equal to

In order to solve inequations, we need to see what effect one operation applied to both sides has on the inequality sign.

If a b2 then a c b c2+ + for all c

For example, 3 22 and 3 1 2 12+ + are both true.

If a b2 then a c b c2- - for all c

For example, 3 22 and 3 1 2 12- - are both true.

ch3.indd 103 7/17/09 2:01:34 PM


If a b2 then ac bc2 for all c 02

For example, 3 22 and 3 2 2 2# #2 are both true.

If a b2 then ac bc1 for all c 01

On the number plane, we graph inequalities using arrows and circles (open for greater than and less than and closed in for greater than or equal to and less than or equal to)

1

2

#

$

If a b2 then a c b c' '2 for all 0c 2

If a b2 then a c b c' '1 for all c 01

If a b2 then a b1 11 for all positive numbers a and b

The inequality sign reverses when: multiplying by a negative • dividing by a negative • taking the reciprocal of both sides •

For example, 3 22 but .3 2 2 2# #1- -

For example, 6 42 and 6 2 4 2' '2 are both true.

For example, 6 42 but .6 2 4 2' '1- -

For example, 3 22 but .31

21

1

ch3.indd 104 7/17/09 2:01:35 PM


EXAMPLES

Solve and show the solutions on a number line 1. x5 7 17$+

Solution

x

x

x

x

x

5 7 17

5 7 17

5 10

5 10

2

7 7

5 5

$

$

$

$

$

+

+ - -

2. t t3 2 5 42- +

Solution

t t

t t

t

t

t

t

t

t t

t t

t

t

t

t

t

t t

t t

3 2 5 4

3 2 5 4

2 2 4

2 2 4

6 2

6 2

3

3 2 5 4

3 2 5 4

2 2 4

2 2 4

2 6

2 6

3

3 3

4 4

2 2

5 5

2 2

2 2

or

2

2

2

2

2

2

2

2

2

2

2

2

2

1

- +

- +

- +

- +

-

-

-

- +

- +

- -

- -

-

-

-

- -

- -

- -

+ +

- -

CONTINUED

Remember to change the inequality sign when

dividing by -2.

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 105 7/17/09 2:01:35 PM


3. Solve .z1 2 7 111 #+

Solution

Method 1: Separate into two separate questions. (i) z

z

z

z

z

1 2 7

1 2 7

6 2

6 2

3

7 7

2 2

1

1

1

1

1

+

+

-

-

-

- -

(ii) z

z

z

z

z

2 7 11

2 7 11

2 4

2 4

2

7 7

2 2

#

#

#

#

#

+

+ - -

Putting these together gives the solution .z3 21 #-

Method 2: Do as a single question.

z

z

z

z

z

1 2 7 11

1 2 7 11

6 2 4

6 2 4

3 2

7 7 7

2 2 2

1

1

1

1

1

#

#

#

#

#

+

+

-

-

-

- - -

Solving this inequation as a single question is quicker than splitting it into two parts.

Notice that the circle is not fi lled in for 1 and fi lled in for #.

1. Solve and plot the solution on a number line

(a) x 4 72+ (b) y 3 1#-

2. Solve (a) t5 352 (b) x3 7 2$- (c) 2( 5) 8p 2+ (d) 4 ( 1) 7x #- - (e) y y3 5 2 42+ - (f) a a2 6 5 3#- - (g) 3 4 2(1 )y y$+ - -

(h) 2 9 1 4 ( 1)x x1+ - +

(i) a2

3# -

(j) y

83

22

(k) b2

5 41+ -

(l) x3

4 62-

(m) x41

51#+

(n) m4

332

2-

3.4 Exercises

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 106 7/17/09 2:11:55 PM


(o) b52

21 6$-

(p) r2

3 6#-

-

(q) z9

1 2 32+

+

(r) w w6 3

2 5 41++

(s) x x2

13

2 7$+

--

(t) t t7

22

3 2#+

-+

(u) q q

32

243

1-

+

(v) x x32

21

92

2--

(w) b b8

2 5 312

6#

-+

+

3. Solve and plot the solutions on a number line

(a) 3 2 9x 11 + (b) p4 2 101#- (c) 2 3 1 11x1 1- (d) y6 5 9 34# #- + (e) 2 3 (2 1) 7y1 1- -

PROBLEM

Find a solution for this sum. Is it a unique solution?

D A N G E R

C R O S S

R O A D S+

Equations and Inequations Involving Absolute Values

On a number line, x means the distance of x from zero in either direction.

EXAMPLES

Plot on a number line and evaluate x 1. x 2=

Solution

x 2= means the distance of x from zero is 2 (in either direction).

x 2!=

CONTINUED

2 2

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 107 7/17/09 2:14:40 PM


Class Discussion

What does a b- mean as a distance along the number line? Select different values of a and b to help with this discussion.

The solution of | x | 21 would be

2 x 2.1 1-

The solution of | x | 2$ would be x 2, x 2.# $-

x a= means x a!=

x a1 means a x a1 1-

x a2 means ,x a x a2 1 -

We use absolute value as a distance on a number line to solve equations and inequations involving absolute values.

2. x 2#

Solution

x 2# means the distance of x from zero is less than or equal to 2 (in either direction).

Notice that there is one region on the number line. We can write this as the single statement .x2 2# #-

3. 2x 2

Solution

2x 2 means the distance of x from zero is greater than 2 (in either direction).

There are two regions on the number line, so we write two separate inequalities , .x x2 21 2-

-4 -3 -2 -1 0 1 2 3 4

2 2

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 108 7/17/09 2:29:40 PM


EXAMPLES

Solve

1. 4x 7+ =

Solution

This means that the distance from 4x + to zero is 7 in either direction. So .x 4 7!+ = 4x 7+ =

7-

7 4

=

- -=

x

x

x

4

4

11

4

+

+

= -

-

x

x

x

4 7

4 7

3

4 4

or+ =

+ =

=

- -

2. 2 1y 51-

Solution

This means that the distance from 2 1y - to zero is less than 5 in either direction. So it means .y5 2 1 51 1- -

y

y

y

y

5 2 1 5

5 2 1 5

4 2 6

2 3

1 1 1

2 2 2

1 1

1 1

1 1

1 1

- -

- -

-

-

+ + +

3. 5 7b 3$-

Solution

b5 7 3$- means that the distance from 5 7b - to zero is greater than or equal to 3 in either direction.

.

b b

b b

b b

b b

5 7 3 5 7 3

5 4 5 10

5 4 5 10

54 2

54

5 5 5 5

# $

# $

# $

# $

- - -

+ + + +

,

b b

b b

5 7 3 5 7 3

2

7 7 7 7

So

# $

# $

- - -

You could solve these as two separate inequations.

These must be solved and written as two

separate inequations.

ch3.indd 109 7/17/09 2:16:48 PM


While it is always a good habit to check solutions to equations and inequations by substituting in values, in these next examples it is essential to check, as some of the solutions are impossible!

EXAMPLES

Solve

1. x x2 1 3 2+ = -

Solution

x x2 1 3 2+ = - means that x2 1+ is at a distance of x3 2- from zero.

x x2 1 3 2!+ = -] g This question is impossible if x3 2- is negative. Can you see why? If x2 1+ is equal to a negative number, this is impossible as the absolute value is always positive. Case (i)

x x

x

x

x xx x

x

2 1 3 2

1 2

3

2 2

2 2

2 1 3 2

1 2

+ = -

= -

=

- -

+ +

+ = -

= -

Check solution is possible: Substitute x 3= into .x x2 1 3 2+ = -

3 3 2

2 3 1

77

9 2

7

LHS

RHS

#

#

= +

=

= -

= -

=

=

Since , 3xLHS RHS= = is a solution. Case (ii)

( )x x

x

x x

x

x

x

x

x

x x

2 1 3 2

3 2

2 1 3 2

5 1 2

5 1 2

5 1

5 1

51

3 3

1 1

5 5

+ = - -

= - +

+ = - +

+ =

+ =

=

=

=

+ +

- -

ch3.indd 110 7/17/09 2:16:49 PM


Check: Substitute x

51

= into .x x2 1 3 2+ = -

3 2

251 1

152

152

51

53 2

52

LHS

RHS

#

#

= +

=

= -

= -

1

=

= -

Since , x51LHS RHS! = is not a solution.

So the only solution is .x 3=

2. x x2 3 1 9- + + =

Solution

In this question it is diffi cult to use distances on the number line, so we use the defi nition of absolute value.

( )

( )

x x xx x

x x xx x

2 3 2 3 2 3 02 3 2 3 0

1 1 1 01 1 0

whenwhenwhenwhen

1

1

$

$

- =- -

- - -

++ +

- + +=

''

This gives 4 cases: (i) ( ) ( )x x2 3 1 9- + + = (ii) ( ) ( )x x2 3 1 9- - + = (iii) ( ) ( )x x2 3 1 9- - + + = (iv) ( ) ( )x x2 3 1 9- - - + =

Case (i)

( ) ( )x x

x x

x

x

x

x

x

2 3 1 9

2 3 1 9

3 2 9

3 2 9

3 11

3 11

332

2 2

3 3

- + + =

- + + =

- =

- =

=

=

=

+ +

Check by substituting x 332

= into .x x2 3 1 9- + + =

CONTINUED

It is often easier to solve these harder equations graphically. You will do

this in Chapter 5.

ch3.indd 111 7/17/09 2:16:50 PM


2 332 3 3

32 1

431 4

32

431 4

32

9

LHS #= - + +

= +

=

RHS

= +

=

So x 332

= is a solution.

Case (ii)

( ) ( )x x

x x

x

x

x

2 3 1 9

2 3 1 9

4 9

4 9

13

4 4

- - + =

- - - =

- =

- =

=

+ +

Check by substituting x 13= into .x x2 3 1 9- + + =

2 13 3 13 1

23 1423 14

37

LHS

RHS

#

!

= - + +

= +

= +

=

So x 13= is not a solution. Case (iii)

( ) ( )x x

x x

x

x

x

x

x

2 3 1 9

2 3 1 9

4 9

4 9

5

5

5

4 4

1 1

- - + + =

- + + + =

- + =

- + =

- =

-=

= -

- -

- -

Check by substituting x 5= - into .x x2 3 1 9- + + =

2 5 3 5 1

13 413 4

17

LHS #= - - + - +

- -

= +

=

RHS!

= +

So x 5= - is not a solution. Case (iv)

( ) ( )x x

x x

x

x

x

2 3 1 9

2 3 1 9

3 2 9

3 2 9

3 7

2 2

- - - + =

- + - - =

- + =

- + =

- =

- -

ch3.indd 112 7/17/09 2:16:51 PM


x

x

3 7

231

3 3-

=

= -

- -

Check by substituting x 231

= - into .x x2 3 1 9- + + =

2 231 3 2

31 1

732 1

31

732 1

31

9

LHS #= - - + - +

- -

= +

=

RHS

= +

=

So x 231

= - is a solution.

So solutions are , .x 332 2

31

= -

While you should always check solutions, you can see that there are some cases where this is really important.

You will learn how to solve equations involving

absolute values graphically in Chapter 5. With

graphical solutions it is easy to see how many

solutions there are.

1. Solve (a) x 5=

(b) y 8=

(c) a 41

(d) k 1$

(e) x 62

(f) p 10#

(g) x 0=

(h) a 142

(i) y 121

(j) b 20$

2. Solve (a) x 2 7+ =

(b) n 1 3- =

(c) 2a 42

(d) 5x 1#-

(e) x9 2 3= +

(f) x7 1 34- =

(g) y4 3 111+

(h) x2 3 15$-

(i) x3

4=

(j) a2

3 2#-

3. Solve (a) x x2 5 3+ = - (b) a a2 1 2- = + (c) b b3 2 4- = - (d) k k3 2 4- = - (e) y y6 23 7+ = - (f) x x4 3 5 4+ = - (g) m m2 5- = (h) d d3 1 6+ = +

(i) y y5 4 1- = + (j) t t2 7 3- = -

4. Solve (a) x x3 3 1+ = -

(b) y y2 5 2- = -

(c) a a3 1 2 9+ = -

(d) x x2 5 17+ + =

(e) d d3 2 4 18- + + =

5. (a) Solve .t t4 3 1 11- + - = By plotting the solutions on (b)

a number line and looking at values in between the solutions, solve .t t4 3 1 111- + -

3.5 Exercises Remember to check solutions

in questions 3, 4 and 5.

ch3.indd 113 7/17/09 2:16:51 PM


To solve equations, use inverse operations: For squares, take the square root For cubes, take the cube root For square roots, take the square For cube roots, take the cube

You have previously used these rules when substituting into formulae involving squares and cubes.

EXAMPLES

Solve

1. x 92 =

Solution

x

xx

9

93

2

2 !

` !

=

=

=

2. n5 403 =

Solution

n

n

n

nn

5 40

5 40

8

82

5 5

3

3

3

33 3

=

=

=

=

=

There are two possible solutions for x – one positive and one negative since 3 92

= and ( 3) 9.2

- =

There is only one answer for this question since 2 83

= but ( 2) 8.3

- = -

Exponential Equations

An exponential equation involves an unknown index or power e.g. .2 8x = We can also solve other equations involving indices. In order to solve

these, you need to understand their relationship. For example, squares and square roots are the reverse of each other (we call them inverse operations). Similarly cubes and cube roots are inverses, and this extends to all indices.

ch3.indd 114 7/31/09 6:58:44 PM


Investigation

Investigate equations of the type x kn = where k is a constant, for example, .x 9n =

Look at these questions:

What is the solution when 1. ?n 0= What is the solution when 2. ?n 1= How many solutions are there when 3. ?n 2= How many solutions are there when 4. ?n 3= How many solutions are there when 5. n is even? How many solutions are there when 6. n is odd?

3. 3a 4=2

Solution

We use the fact that 32

23

.a a a= =2

33

2` `j j

3

3

2

22

a

a

a

a

4

4

4

428

3

3

`

=

=

=

=

=

=

2

2

3

33

^

`

h

j

In other types of equations, the pronumeral (or unknown variable) is in the index. We call these exponential equations , and we use the fact that if the base numbers are equal, then the powers (or indices or exponents) must be equal.

EXAMPLES

Solve 1. 3 81x =

Solution

3 81x = Equating indices:

x

3 34

x 4

`

=

=

CONTINUED

ch3.indd 115 7/17/09 2:16:53 PM


2. 5 25k2 1 =-

Solution

k

k

k

k

k

5 25

5 52 1 2

2 1 2

2 3

2 3

121

1 1

2 2

k

k

2 1

2 1 2

`

=

=

- =

- =

=

=

=

+ +

-

-

3. 8 4n =

Solution

It is hard to write 8 as a power of 4 or 4 as a power of 8, but both can be written as powers of 2.

( )

n

n

n

8 4

2 2

2 23 2

3 2

32

3 3

n

n

n

3 2

3 2

`

=

=

=

=

=

=

We can check this solution

by substituting k 12

1= into

the equation 5 25.2k 1=

-

1. Solve (a) x 273 = (b) y 642 = (c) n 164 = (d) ( )x 20 give the exact answer2 = (e) p 10003 = (f) x2 502 = (g) y6 4864 = (h) w 7 153 + =

(i) n6 4 922 - = (j) q3 20 43 + = -

2. Solve and give the answer correct to 2 decimal places.

(a) p 452 = (b) x 1003 = (c) n 2405 = (d) x2 702 = (e) y4 7 343 + =

(f) d3

144

=

(g) k2

3 72

- =

(h) x5

1 23 -

=

(i) y2 9 202 - = (j) y7 9 2003 + =

3.6 Exercises

ch3.indd 116 8/1/09 6:15:50 PM


3. Solve

(a) 3n 9=2

(b) 4t 8=3

(c) 5x 4=2

(d) 3t 16=4

(e) 5p 27=3

(f) 4m2 250=3

(g) 3b 3 39+ =2

(h) 3y5 405=4

(i) 7a3 2 10- =2

(j) 4t

39=

3

4. Solve (all pronumerals ! 0) (a) x 51 =- (b) a 83 =- (c) y 325 =- (d) x 1 502 + =- (e) n2 31 =-

(f) a813 =-

(g) x412 =-

(h) b911 =-

(i) x 2412 =-

(j) b81164 =-

5. Solve (all pronumerals ! 0)

(a) 3x 8=-

1

(b) -

2x125

8=

3

(c) 4a 3=-

1

(d) -

4k 125=3

(e) 3x3 12=-

2

(f) 2x81

=-

3

(g) -

3y41

=2

(h) 5n94

=-

2

(i) 3b321

=-

5

(j) -

3m4936

=2

6. Solve (a) 2 16n = (b) 3 243y = (c) 2 512m = (d) 10 100 000x = (e) 6 1m = (f) 4 64x = (g) 4 3 19x + = (h) )(5 3 45x =

(i) 4 4x =

(j) 26 18

k

=

7. Solve (a) 3 81x2 = (b) 2 16x5 1 =- (c) 4 4x 3 =+ (d) 3 1n 2 =- (e) 7 7x2 1 =+ (f) 3 27x 3 =- (g) 5 125y3 2 =+ (h) 7 49x3 4 =-

(i) 2 256x4 = (j) 9 9a3 1 =+

8. Solve (a) 4 2m = (b) 27 3x = (c) 125 5x =

(d) 491 7

k

=c m

(e) 1000

1 100k

=c m

(f) 16 8n = (g) 25 125x = (h) 64 16n =

(i) 41 2

k3

=c m

(j) 8 4x 1 =-

9. Solve (a) 2 8x x14 =+ (b) 3 9x x5 2= - (c) 7 7k k2 3 1=+ - (d) 4 8n n3 3= + (e) 6 216x x5 =- (f) 16 4x x2 1 4=- - (g) 27 3x x3=+

(h) 21

641x x2 3

=+c cm m

ch3.indd 117 7/17/09 2:16:55 PM


PUZZLE

Test your logical thinking and that of your friends. How many months have 28 days? 1. If I have 128 sheep and take away all but 10, how many 2. do I have left? A bottle and its cork cost $1.10 to make. If the bottle costs $1 more 3. than the cork, how much does each cost? What do you get if you add 1 to 15 four times? 4. On what day of the week does Good Friday fall in 2016? 5.

Quadratic Equations

A quadratic equation is an equation involving a square. For example, .x 4 02 - =

Solving by factorisation

When solving quadratic equations by factorising, we use a property of zero.

For any real numbers a and b , if ab 0= then a 0= or b 0=

EXAMPLES

Solve 1. x x 6 02 + - =

Solution

( ) ( )

x xx x

6 03 2 0

2 + - =

+ - =

(i) 43

6427x x2 3

=-c cm m

(j) 5251x

x 9

=--] cg m

10. Solve (a) 4 2m =

(b) 259

53k 3

=+c m

(c) 2

1 4 x2 5= -

(d) 3 3 3k =

(e) 271

813n3 1

=+c m

(f) 52

25n n3 1

=+ -c cm m

(g) 32161x =-

(h) 9 3 3b b2 5 =+

(i) 81 3x x1 =+

(j) 2551m

m3 5

=--c m

ch3.indd 118 7/17/09 2:16:56 PM


0

0

2

=

=

=

x x

x x

x

3 0 2

3 0 2

3

3 3 2 2

or

or

` + = -

+ = -

= -

- - + +

x

So the solution is .x 3 2or= -

2. y y7 02 - =

Solution

( )y y

y y

y y

7 07 0

0 7 0or

2

`

- =

- =

= - =

y

y

7 0

7

7 7- =

=

+ +

So the solution is y 0= or 7.

3. a a3 14 82 - = -

Solution

( ) ( )

a a

a a

a aa a

a a

a a

a a

a

a

3 14 8

3 14 8

3 14 8 03 2 4 0

3 2 0 4 0

3 2 0 4 0

3 2 4

3 2

32

8 8

2 4 4

3 3

or

or

2

2

2

`

- = -

- = -

- + =

- - =

- = - =

- = - =

= =

=

=

+ +

+ + +

So the solution is .a 4or32

=

Solve

3.7 Exercises

1. y y 02 + =

2. b b 2 02 - - =

3. p p2 15 02 + - =

4. t t5 02 - =

5. x x9 14 02 + + =

6. q 9 02 - =

ch3.indd 119 7/17/09 2:16:57 PM


7. x 1 02 - =

8. a a3 02 + =

9. x x2 8 02 + =

10. x4 1 02 - =

11. x x3 7 4 02 + + =

12. y y2 3 02 + - =

13. b b8 10 3 02 - + =

14. x x3 102 - =

15. x x3 22 =

16. x x2 7 52 = -

17. x x5 02- =

18. y y 22 = +

19. n n8 152= +

20. x x12 7 2= -

21. m m6 52 = -

22. ( ) ( )x x x1 2 0+ + =

23. ( ) ( ) ( )y y y1 5 2 0- + + =

24. ( ) ( )x x3 1 32+ - =

25. ( ) ( )m m3 4 20- - =

Application

A formula for displacement s at time t is given by 21

s ut at2= + where u is the

initial velocity and a is the acceleration. Find the time when the displacement will be zero, given 12u = - and 10.a =

2

20 12 (10)

12 5

( 12 5 )

0 or 12 5 0

s ut at

t t

t t

t t

t t

2

2

2

`

= +

= - +

= - +

= - +

= - + =

1

1

12 5 0

5 12

5 12

2.4

12 12

5 5

t

t

t

t

+ =+ +-

=

=

=

So displacement will be zero when 0t = or 2.4.

Solving by completing the square

Not all trinomials will factorise, so other methods need to be used to solve quadratic equations.

ch3.indd 120 7/17/09 2:16:58 PM


EXAMPLES

Solve 1. x 72 =

Solution

.

x

x

7

72 6

2

!

!

=

=

=

2. x 3 112+ =] g

Solution

. , .

x

x

x

x

3 11

3 11

3 11

11 30 3 6 3

3 3

2

!

!

!

+ =

+ =

+ =

= -

= -

- -

] g

3. y 2 72- =^ h

Solution

. , .

y

y

y

y

2 7

2 7

2 7

7 24 6 0 6

2 2

2

!

!

!

- =

- =

- =

= +

= -

+ +

^ h

Take the square root of both sides.

To solve a quadratic equation like ,x x6 3 02 - + = which will not factorise, we can use the method of completing the square.

You learnt how to complete the square in

Chapter 2.

EXAMPLES

Solve by completing the square 1. x x6 3 02 - + = (give exact answer)

Solution

2

x x

x x

6 3 0

6 3 3 9

2

22

2

- + =

- = - = =6c m Halve 6, square it and

add to both sides of the equation.

CONTINUED

ch3.indd 121 7/17/09 2:16:59 PM


x x

x

x

x

x

6 3

3 6

3 6

3 6

6 3

9 9

3 3

2

2

` !

!

!

- = -

- =

- =

- =

= +

+ +

+ +

] g

2. y y2 7 02 + - = (correct to 3 signifi cant fi gures)

Solution

2

. .

y y

y y

y y

y

y

y

y

y

2 7 0

2 7 1 1

2 7

1 8

1 8

1 8

8 1

2 2 11 83 3 83

1 1

1 1

or

2

22

2

2

2

` !

!

!

!

+ - =

+ = = =

+ =

+ =

+ =

+ =

= -

= -

= -

+ +

- -

2c

^

m

h

1. Solve by completing the square, giving exact answers in simplest surd form

(a) x x4 1 02 + - = (b) a a6 2 02 - + = (c) y y8 7 02 - - = (d) x x2 12 02 + - = (e) p p14 5 02 + + = (f) x x10 3 02 - - = (g) y y20 12 02 + + = (h) x x2 1 02 - - =

(i) n n24 7 02 + + = (j) y y3 1 02 - + =

2. Solve by completing the square and write your answers correct to 3 signifi cant fi gures

(a) x x2 5 02 - - = (b) x x12 34 02 + + = (c) q q18 1 02 + - = (d) x x4 2 02 - - = (e) b b16 50 02 + + = (f) x x24 112 02 - + = (g) r r22 7 02 - - = (h) x x8 5 02 + + =

(i) a a6 012 + - = (j) y y40 3 02 - - =

3.8 Exercises

Solving by formula

Completing the square is diffi cult with harder quadratic equations, for example .x x2 5 02 - - = Completing the square on a general quadratic equation gives the following formula.

ch3.indd 122 7/17/09 2:16:59 PM


For the equation ax bx c 02 + + =

xa

b b ac2

42!=

- -

Proof

Solve ax b c 02 + + = by completing the square.

ax bx c

ax bx c

x abx

ac

x abx

ac

x abx

ac

ab

ab

ab

x abx

ac

xa

bac

ab

aac b

xab

aac b

ab ac

xa

ba

b ac

xab

ab ac

ab b ac

a a a a

ac

ac

ab

ab

ab

ab

0

0

0

0

22 4

2 4

44

2 44

24

2 24

2 24

24

4 4

2 2

2

2

2

2

22 2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

'

!

!

!

!

!

+ + =

+ + =

+ + =

+ + =

+ = - = =

+ = -

+ = - +

=- +

+ =- +

=-

+ =-

=- -

=- -

- -

+ +

- -

b c

c

l m

m

EXAMPLES

1. Solve x x 2 02 - - = by using the quadratic formula.

Solution

, ,

( )( ) ( ) ( ) ( )

a b c

xa

b b ac

1 1 2

24

2 11 1 4 1 2

21 1 8

2

2

!

!

!

= = - = -

=- -

=- - - - -

=+

CONTINUED

ch3.indd 123 7/17/09 2:17:00 PM


21 9

21 3

2 1or

!

!

=

=

= -

2. Solve y y2 9 3 02 - + = by formula and give your answer correct to 2 decimal places.

Solution

, ,

. .

a b c

xa

b b ac

y

2 9 3

24

2 29 9 4 2 3

49 81 24

49 57

4 14 0 36or

2

2

!

!

!

!

Z

= = - =

=- -

=- - - -

=-

=

]] ] ] ]

gg g g g

x2

1 3!= gives two

separate solutions, 2

1 3+

and .2

1 3-

These solutions are irrational.

1. Solve by formula, correct to 3 signifi cant fi gures where necessary

(a) y y6 2 02 + + = (b) x x2 5 3 02 - + = (c) b b 9 02 - - = (d) x x2 1 02 - - = (e) x x8 3 02- + + = (f) n n8 2 02 + - = (g) m m7 10 02 + + = (h) x x7 02 - =

(i) x x5 62 + = (j) y y3 12 = -

2. Solve by formula, leaving the answer in simplest surd form

(a) x x 4 02 + - = (b) x x3 5 1 02 - + = (c) q q4 3 02 - - = (d) h h4 12 1 02 + + = (e) s s3 8 2 02 - + = (f) x x11 3 02 + - = (g) d d6 5 2 02 + - = (h) x x2 72 - =

(i) t t 12 = + (j) x x2 1 72 + =

3.9 Exercises

Class Investigation

Here is a proof that .1 2= Can you see the fault in the proof?

( ) ( ) ( )x x x x

x x x x x x x

x x x

x x2

1 2

2 2 2 2

`

- = -

- = + -

= +

=

=

ch3.indd 124 7/17/09 2:17:01 PM


In Chapter 9 you will look at how to use the number plane to solve these quadratic inequations. Here are some examples of solving quadratic inequations using the number line.

Quadratic Inequations

Solving quadratic inequations is similar to solving quadratic equations, but you need to do this in two stages. The fi rst is to solve the equation and then the second step is to look at either the number line or the number plane for the inequality.

To solve a quadratic inequation: Factorise and solve the quadratic equation 1. Test values in the inequality 2.

EXAMPLES

Solve 1. x x 6 02 2+ -

Solution

( ) ( )x x

x x

x

6 02 3 0

2 3

First solve

or

2

`

+ - =

- + =

= -

Now look at the number line.

Choose a number between 3- and 2, say .x 0= Substitute x 0= into the inequation.

x x 6 0

0 0 6 06 0 (false)

2

2

2

2

2

+ -

+ -

-

So the solution is not between 3- and 2. the solution lies either side of 3- and 2. Check by choosing a number on either side of the two numbers. Choose a number on the RHS of 2, say .x 3=

Be careful: x x 6 02 2+ - does not mean x 2 02-

and x 3 0.2+

CONTINUED

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 125 7/31/09 6:58:45 PM


Substitute x 3= into the inequation.

6 0

3 3 6 0(true)

2 2

2

+ -

So the solution is on the RHS of 2. Choose a number on the LHS of ,3- say x 4= - Substitute x 4= - into the inequation

( ) ( )4 4 6 0

6 0 (true)

2 2

2

- + - -

So the solution is on the LHS of .3-

This gives the solution , .x x3 21 2-

2. x9 02 $-

Solution

( ) ( )x

x x

x

9 03 3 0

3

First solve 2

` !

- =

- + =

=

Choose a number between 3- and 3, say .x 0= Substitute x 0= into the inequation.

x9 0

9 0 09 0 (true)

2

2

$

$

$

-

-

So the solution is between 3- and 3, that is x3 3# #- . On the number line:

Check numbers on the RHS and LHS to verify this.

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

-4 -3 -2 -1 0 1 2 3 4

ch3.indd 126 7/31/09 7:00:59 PM


Solve

1. x x3 02 1+

2. y y4 02 1-

3. n n 02 $-

4. x 4 02 $-

5. n1 02 1-

6. n n2 15 02 #+ -

7. c c 2 02 2- -

8. x x6 8 02 #+ +

9. x x9 20 02 1- +

10. b b4 10 4 02 $+ +

11. a a1 2 3 02 1- -

12. y y2 6 02 2- -

13. x x3 5 2 02 $- +

14. b b6 13 5 02 1- -

15. x x6 11 3 02 #+ +

16. y y 122 #+

17. x 162 2

18. a 12 #

19. x x 62 1 +

20. x x2 32 $ +

21. x x22 1

22. a a2 5 3 02 #- +

23. y y5 6 82 $+

24. m m6 152 2 -

25. x x3 7 42 # -

3.10 Exercises

Simultaneous Equations

Two equations, each with two unknown pronumerals, can be solved together to fi nd one solution that satisfi es both equations.

There are different ways of solving simultaneous equations. The elimination method adds or subtracts the equations. The substitution method substitutes one equation into the other.

Linear equations

These equations can be solved by either method. Many students prefer the elimination method.

ch3.indd 127 7/31/09 7:01:27 PM


EXAMPLES

Solve simultaneously 1. a b3 2 5+ = and a b2 6- = -

Solution

( )

( )

: ( )

( ): ( )

a b

a b

a b

a ba

a

3 2 5 1

2 6 2

2 2 4 2 12 3

1 3 3 2 5 17 7

1

#

+ =

- = -

- = -

+ + =

= -

= -

]

]

g

g

( )a 1 1Substitute in= -

( ) b

b

b

b

3 1 2 5

3 2 5

2 8

4

- + =

- + =

=

=

,a b1 4solution is` = - =

2. x y5 3 19- = and x y2 4 16- =

Solution

( )

( )

( ) : ( )

( ) : ( )

( ) ( ):

x y

x y

x y

x y

x

x

5 3 19 1

2 4 16 2

1 4 20 12 76 3

2 3 6 12 48 4

3 4 14 28

2

#

#

- =

- =

- =

- =

- =

=

( )x 2 2Substitute in=

( ) y

y

y

y

2 2 4 16

4 4 16

4 12

3

- =

- =

- =

= -

ch3.indd 128 7/31/09 7:01:38 PM


Solve simultaneously

1. a b a b2 4and- = - + =

2. x y x y5 2 12 3 2 4and+ = - =

3. p q p q4 3 11 5 3 7and- = + =

4. y x y x3 1 2 5and= - = +

5. x y x y2 3 14 3 4and+ = - + = -

6. t v t v7 22 4 13and+ = + =

7. x y4 5 2 0 and+ + = x y4 10 0+ + =

8. x y x y2 4 28 2 3 11and- = - = -

9. x y x y5 19 2 5 14and- = + = -

10. m n m n5 4 22 5 13and+ = - = -

11. w w w w4 3 11 3 2and1 2 1 2+ = + =

12. a b a b3 4 16 2 3 12and- = - + =

13. p q5 2 18 0 and+ + =

p q2 3 11 0- + =

14. x x x x7 3 4 3 5 2and1 2 1 2+ = + = -

15. x y x y9 2 1 7 4 9and- = - - =

16. s t5 3 13 0 and- - = s t3 7 13 0- - =

17. a b a b3 2 6 3 2and- = - - = -

18. k h3 2 14 and- = - k h2 5 13- = -

19. v v2 5 16 0 and1 2+ - = v v7 2 6 01 2+ + =

20. . . .x y1 5 3 4 7 8 and+ = . . .x y2 1 1 7 1 8- =

3.11 Exercises

PROBLEM

A group of 39 people went to see a play. There were both adults and children in the group. The total cost of the tickets was $939, with children paying $17 each and adults paying $29 each. How many in the group were adults and how many were children? (Hint: let x be the number of adults and y the number of children.)

Non-linear equations

In questions involving non-linear equations there may be more than one set of solutions. In some of these, the elimination method cannot be used. Here are some examples using the substitution method.

ch3.indd 129 7/31/09 7:02:05 PM


EXAMPLES

Solve simultaneously 1. xy 6= and x y 5+ =

Solution

( )

( )

( ): ( )

xy

x y

y x

6 1

5 2

2 5 3From

=

+ =

= -

Substitute (3) in (1)

( )

( ) ( )

x x

x x

x xx x

x x

x x

5 6

5 6

0 5 60 2 3

2 0 3 0

2 3

or

or

2

2

`

- =

- =

= - +

= - -

- = - =

= =

Substitute x 2= in (3) y 5 2 3= - = Substitute x 3= in (3) y 5 3 2= - = solutions are ,x y2 3= = and ,x y3 2= =

2. x y 162 2+ = and x y3 4 20 0- - =

Solution

( )( )

:

( )

x yx y

x yx y

16 13 4 20 0 2

2 3 20 4

43 20 3

From

2 2+ =

- - =

- =

-=

] g

Substitute (3) into (1)

( )

.

x x

x x x

x x x

x x

xx

x

x

43 20 16

169 120 400 16

16 9 120 400 256

25 120 144 0

5 12 05 12 0

5 12

2 4

22

22

2 2

2

2

`

+-

=

+- +

=

+ - + =

- + =

- =

- =

=

=

c

d

m

n

.( . )

.

. , . .

x

y

x y

2 4 3

43 2 4 20

3 2

2 4 3 2

Substitute into

So the solution is

=

=-

= -

= = -

] g

ch3.indd 130 7/31/09 7:02:19 PM


Equations with 3 unknown variables

Three equations can be solved simultaneously to fi nd 3 unknown pronumerals.

Solve the simultaneous equations.

1. y x2= and y x=

2. y x2= and x y2 0+ =

3. x y 92 2+ = and x y 3+ =

4. x y 7- = and xy 12= -

5. y x x42= + and x y2 1 0- - =

6. y x2= and x y6 9 0- - =

7. x t2= and x t 2 0+ - =

8. m n 162 2+ = and m n 4 0+ + =

9. xy 2= and y x2=

10. y x3= and y x2=

11. y x 1= - and y x 32= -

12. y x 12= + and y x1 2= -

13. y x x3 72= - + and y x2 3= +

14. xy 1= and x y4 3 0- + =

15. h t2= and h t 1 2= +] g

16. x y 2+ = and x xy y2 82 2+ - =

17. y x3= and y x x62= +

18. | |y x= and y x2=

19. y x x7 62= - + and x y24 4 23 0+ - =

20. x y 12 2+ = and x y5 12 13 0+ + =

3.12 Exercises

Four unknowns need 4 equations, and so on.

EXAMPLE

Solve simultaneously ,a b c a b c7 2 4- + = + - = - and .a b c3 3- - =

Solution

b2

b

b

b

b

b

=

( )

( )

( )

( ) ( ):

( )

( ) ( ):

( )

( ) ( ):

4 8

a c

a b c

a c

a

a b c

a c

a

or a

a

a

7 1

2 4 2

3 3 3

1 2 7

2 3 4

1 3 7

3

4 10

2 5 5

4 5 2

2

- + =

+ - = -

- - =

+ =

+

+ - + =

- -

- =

- =

+ +

=

=

b

a

ba c

3

- +

=

=

a b c2

3

+ - 4= -

ch3.indd 131 7/31/09 7:02:30 PM


Substitute a 2= in (4)

( ) b

b

b

2 2 3

4 3

1

+ =

+ =

= -

Substitute a 2= and b 1= - in (1)

( ) c

c

c

c

2 1 7

2 1 7

3 7

4

- - + =

+ + =

+ =

=

solution is , ,a b c2 1 4= = - =

Solve the simultaneous equations.

1. ,x x y2 2 4= - - = and x y z6 0- + =

2. ,a a b2 2 3 1= - - = - and a b c5 9- + =

3. ,a b c a b2 1 2+ + = + = - and c 7=

4. ,a b c a b c0 4+ + = - + = - and a b c2 3 1- - = -

5. ,x y z x y z7 2 1+ - = + + = and x y z3 2 19+ - =

6. ,x y z x y z1 2 9- - = + - = - and x y z2 3 2 7- - =

7. ,p q r2 5 25+ - = p q r2 2 24- - = - and

p q r3 5 4- + =

8. ,x y z2 3 9- + = x y z3 2 2+ - = - and x y z3 5 14- + =

9. ,h j k3 3+ - = - h j k2 3+ + = - and h j k5 3 2 13- - = -

10. ,a b c2 7 3 7- + = a b c3 2 4+ + = - and a b c4 5 9+ - =

3.13 Exercises

You will solve 3 simultaneous equations in later topics (for example, in Chapter 9).

ch3.indd 132 7/31/09 7:02:42 PM


Test Yourself 3 1. Solve

(a) b8 3 22= -

(b) a a4 3

2 9-+

=

(c) ( )x x4 3 1 11 3+ = -

(d) p p3 1 9#+ +

2. The compound interest formula is

.A P r1100

n

= +c m Find correct to 2

decimal places. (a) A when ,P r1000 6= = and n 4= (b) P when , .A r12 450 5 5= = and n 7=

3. Complete the square on (a) x x82 - (b) k k42 +

4. Solve these simultaneous equations. (a) x y 7 0- + = and x y3 4 26 0- + = (b) xy 4= and x y2 7 0- - =

5. Solve (a) 3 81x 2 =+ (b) 16 2y =

6. Solve (a) b3 1 5- =

(b) g g5 3 3 1- = +

(c) x2 7 1$-

7. The area of a trapezium is given by

2 ( )A h a b= +1 . Find

(a) A when ,h a6 5= = and b 7= (b) b when ,A h40 5= = and .a 4=

8. Solve x x2 3 1 02 - + = by factorisation (a) quadratic formula. (b)

9. Solve ,y2 3 1 101 #- + and plot your solution on a number line.

10. Solve correct to 3 signifi cant fi gures (a) x x7 2 02 + + = (b) y y2 9 02 - - = (c) n n3 2 4 02 + - =

11. The surface area of a sphere is given by .A r4 2r= Evaluate to 1 decimal place

(a) A when .r 7 8= (b) r when .A 102 9=

12. Solve .x7

343 92

--

13. Solve .x x11 18 02 2- +

14. Solve the simultaneous equations x y 162 2+ = and .x y3 4 20 0+ - =

15. The volume of a sphere is .V r34 3r=

Evaluate to 2 signifi cant fi gures (a) V when r 8= (b) r when V 250=

16. Which of the following equations has (i) 2 solutions (ii) 1 solution

(iii) no solutions?

(a) x x6 9 02 - + =

(b) x2 3 7- =

(c) x x2 7- = -

(d) x x 4 02 - + =

(e) x x2 1 2+ = -

17. Solve simultaneously , , .a b a b c a b c5 2 4 5+ = + + = - - =

18. Solve ,n3 5 52+ and plot the solution on a number line.

19. Solve , .x x x

13 4 0 1!+

= -^ h

ch3.indd 133 7/31/09 7:02:53 PM


20. Solve .9 27x x2 1 =+

21. Solve (a) y y2 3 5 52- +^ h (b) n n3 02 #+ (c) 3 27x2 1 =- (d) x5 1 393 - = (e) x5 4 11- = (f) t2 1 3$+ (g) x x2 8 02 #+ -

(h) 8 4x x1 =+ (i) y 4 02 2- (j) x1 02 #- (k) 27 9x2 1 =- (l) b4 3 5#-

(m) x x3 2 2 3+ = - (n) t t4 5 2- = + (o) x x2 32 1 + (p) m m 62 $+

1. Find the value of y if .aa1y3 5

2=-

2. Solve .x a2 22

3. The solutions of x x6 3 02 - - = are in the form .a b 3+ Find the values of a and b .

4. Solve x x1

21

1 1- +

=- correct to 3

signifi cant fi gures. ( )x 1!!

5. Factorise .x x x9 8 725 3 2- - + Hence solve .x x x9 8 72 05 3 2- - + =

6. Solve simultaneous equations y x x3 2= + and .y x 1= +

7. Find the value of b if x x b82 2- + is a perfect square. Hence solve x x8 1 02 - - = by completing the square.

8. Considering the defi nition of absolute

value, solve ,x

xx

3

3

-

-= where .x 3!

9. Solve .t t2 3 1 51+ + -

10. Solve .x x4 1 28#- -] ]g g

11. Solve 2 .x81

=3

12. The volume of a sphere is given by

.V r34 3r= Find the value of r when

.V 51 8= (correct to three signifi cant fi gures).

13. Solve .x x x3 4 2- + + = -

14. Find the solutions of x ax b2 02 - - = by completing the square.

15. Given ,A P r1100

n

= +c m fi nd P

correct to 2 decimal places when . , .A r3281 69 1 27= = and .n 30=

16. Solve ( )x x3 8 2 12 = - and write the solution in the simplest surd form.

17. Solve .y y3 1 2 3 52- + +


ch3.indd 134 8/1/09 6:21:39 PM

ch3.indd 135 7/31/09 7:03:11 PM

TERMINOLOGY

4 Geometry 1

Altitude: Height. Any line segment from a vertex to the opposite side of a polygon that is perpendicular to that side

Congruent triangles: Identical triangles that are the same shape and size. Corresponding sides and angles are equal. The symbol is /

Interval: Part of a line including the endpoints

Median: A line segment that joins a vertex to the opposite side of a triangle that bisects that side

Perpendicular: A line that is at right angles to another line. The symbol is =

Polygon: General term for a many sided plane fi gure. A closed plane (two dimensional) fi gure with straight sides

Quadrilateral: A four-sided closed fi gure such as a square, rectangle, trapezium etc.

Similar triangles: Triangles that are the same shape but different sizes. The symbol is yz

Vertex: The point where three planes meet. The corner of a fi gure

Vertically opposite angles: Angles that are formed opposite each other when two lines intersect

ch4.indd 136 7/17/09 6:07:29 PM

137Chapter 4 Geometry 1

INTRODUCTION

GEOMETRY IS USED IN many areas, including surveying, building and graphics. These fi elds all require a knowledge of angles, parallel lines and so on, and how to measure them. In this chapter, you will study angles, parallel lines, triangles, types of quadrilaterals and general polygons.

Many exercises in this chapter on geometry need you to prove something or give reasons for your answers. The solutions to geometry proofs only give one method , but other methods are also acceptable .

DID YOU KNOW?

Geometry means measurement of the earth and comes from Greek. Geometry was used in ancient civilisations such as Babylonia. However, it was the Greeks who formalised the study of geometry, in the period between 500 BC and AD 300.

Notation

In order to show reasons for exercises, you must know how to name fi gures correctly.

• B The point is called B .

The interval (part of a line) is called AB or BA .

If AB and CD are parallel lines, we write .AB CD<

This angle is named BAC+ or .CAB+ It can sometimes be named .A+

Angles can also be written as BAC^ or BAC

This triangle is named .ABC3

ch4.indd 137 8/7/09 5:18:27 PM


This quadrilateral is called ABCD .

Line AB is produced to C .

DB bisects .ABC+

AM is a median of .ABCD

AP is an altitude of .ABCD

Types of Angles

Acute angle

0 90xc c c1 1

To name a quadrilateral, go around it: for example, BCDA is correct, but ACBD is not.

Producing a line is the same as extending it.

ABD+ and DBC+ are equal.

ch4.indd 138 7/17/09 6:07:35 PM


Right angle

A right angle is .90c Complementary angles are angles whose sum is .90c

Obtuse angle

x90 180c c c1 1

Straight angle

A straight angle is .180c Supplementary angles are angles whose sum is .180c

Refl ex angle

x180 360c c c1 1

Angle of revolution

An angle of revolution is .360c

Vertically opposite angles

AEC+ and DEB+ are called vertically opposite angles . AED+ and CEB+ are also vertically opposite angles.

ch4.indd 139 7/17/09 6:07:37 PM


Proof

( )

( ) ( )

( )

AEC x

AED x CED

DEB x AEB

x

CEB x CED

AEC DEB AED CEB

180 180

180 180 180

180 180

Let

Then straight angle,

Now straight angle,

Also straight angle,

and `

c

c c c

c c c c

c

c c c

+

+ +

+ +

+ +

+ + + +

=

= -

= - -

=

= -

= =

EXAMPLES

Find the values of all pronumerals, giving reasons.

1.

Solution

( )x ABC

x

x

154 180 180

154 180

26

154 154

is a straight angle,

`

c++ =

+ =

=

- -

2.

Solution

( )x

x

x

x

x

x

2 142 90 360 360

2 232 360

2 232 360

2 128

2 128

64

232 232

2 2

angle of revolution, c+ + =

+ =

+ =

=

=

=

- -

Vertically opposite angles are equal.

That is, AEC DEB+ += and .AED CEB+ +=

ch4.indd 140 7/17/09 6:07:38 PM


3.

Solution

( )y y

y

y

y

y

y

2 30 90 90

3 30 90

3 30 90

3 60

3 60

20

30 30

3 3

right angle, c+ + =

+ =

+ =

=

=

=

- -

4.

Solution

(

( )

(

x WZX YZV

x

x

y XZY

w WZY XZV

50 165

50 165

115

180 165 180

15

15

50 50

and vertically opposite)

straight angle,

and vertically opposite)

c

+ +

+

+ +

+ =

+ =

=

= -

=

=

- -

5.

CONTINUED

ch4.indd 141 7/17/09 6:07:39 PM


Solution

( )

( )

( )

( )

a

b

b

b

b

d

c

90

53 90 180 180

143 180

143 180

37

37

53

143 143

vertically opposite angles

straight angle,


similarly

c

=

+ + =

+ =

+ =

=

=

=

- -

6. Find the supplement of .57 12c l

Solution

Supplementary angles add up to .180c So the supplement of 57 12c l is .180 57 12 1 2 482c c c- =l l

7. Prove that AB and CD are straight lines.

Solution

x x x xx

x

x

x

6 10 30 5 30 2 10 36014 80 360

14 280

14 280

20

80 80

14 14

angle of revolution+ + + + + + + =

+ =

=

=

=

- -

^ h

( )

( )

AEC

DEB

20 30

50

2 20 10

50

#

c

c

c

c

+

+

= +

=

= +

=

These are equal vertically opposite angles . AB and CD are straight lines

C

DA

B

E(2x22 +10)c

(6x+10)c

(5x+30)c

(x+30)c

ch4.indd 142 7/17/09 6:07:41 PM


4.1 Exercises

1. Find values of all pronumerals, giving reasons.

yc 133c

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

2. Find the supplement of (a) 59c (b) 107 31c l (c) 45 12c l

3. Find the complement of (a) 48c (b) 34 23c l (c) 16 57c l

4. Find the (i) complement and (ii) supplement of

(a) 43c 81c(b) 27c(c) (d) 55c (e) 38c (f) 74 53c l (g) 42 24c l (h) 17 39c l (i) 63 49c l (j) 51 9c l

5. (a) Evaluate x . Find the complement of (b) x . Find the supplement of (c) x.

(2x+30)c

142c

ch4.indd 143 7/17/09 6:07:45 PM


6. Find the values of all pronumerals, giving reasons for each step of your working.

(a)

(b)

(c)

(d)

(e)

(f)

7.

Prove that AC and DE are straight lines.

8.

Prove that CD bisects .AFE+

9. Prove that AC is a straight line.

A

B

C

D

(110-3x)c

(3x+70)c

10. Show that + AED is a right angle.

A B

C

DE

(50-8y)c

(5y-20)c

(3y+60)c

ch4.indd 144 7/17/09 6:07:46 PM


Parallel Lines

When a transversal cuts two lines, it forms pairs of angles. When the two lines are parallel, these pairs of angles have special properties.

Alternate angles

Alternate angles form a Z shape. Can you fi nd another set of

alternate angles?

Corresponding angles form an F shape. There are 4 pairs

of corresponding angles. Can you fi nd them?

If the lines are parallel, then alternate angles are equal.

Corresponding angles

If the lines are parallel, then corresponding angles are equal.

ch4.indd 145 7/17/09 6:07:47 PM


Cointerior angles

Cointerior angles form a U shape. Can you fi nd another pair?

If AEF EFD,+ += then AB CD.<

If BEF DFG,+ += then AB CD.<

If BEF DFE 180 ,c+ ++ = then AB CD.<

If the lines are parallel, cointerior angles are supplementary (i.e. their sum is 180c ).

Tests for parallel lines

If alternate angles are equal, then the lines are parallel.

If corresponding angles are equal, then the lines are parallel.

If cointerior angles are supplementary, then the lines are parallel.

ch4.indd 146 7/17/09 6:21:59 PM


EXAMPLES

1. Find the value of y , giving reasons for each step of your working.

Solution

( )

55 ( , )

AGF FGH

y AGF CFE AB CD

180 125

55

is a straight angle

corresponding angles,`

c c

c

c

+ +

+ + <

= -

=

=

2. Prove .EF GH<

Solution

( )CBF ABC

CBF HCD

180 120

60

60

is a straight angle

`

c c

c

c

+ +

+ +

= -

=

= =

But CBF+ and HCD+ are corresponding angles EF GH` <

Can you prove this in a different way?

If 2 lines are both parallel to a third line, then the 3 lines are parallel to each other. That is, if AB CD< and ,EF CD< then .AB EF<

ch4.indd 147 7/17/09 6:24:18 PM


1. Find values of all pronumerals. (a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

2. Prove .AB CD< (a)

(b)

A

B C

D

E104c76c

(c)

4.2 Exercises Think about the reasons for each step of your calculations.

ch4.indd 148 7/17/09 6:24:25 PM


Types of Triangles

Names of triangles

A scalene triangle has no two sides or angles equal.

A right (or right-angled) triangle contains a right angle.

The side opposite the right angle (the longest side) is called the hypotenuse.

An isosceles triangle has two equal sides.

A

B

C

D

E

F

52c

128c

(d) AB

C

DE F

G

H

138c

115c23c

(e)

The angles (called the base angles) opposite the equal sides in an isosceles triangle are equal.

An equilateral triangle has three equal sides and angles.

ch4.indd 149 7/17/09 6:24:33 PM


All the angles are acute in an acute-angled triangle.

An obtuse-angled triangle contains an obtuse angle.

Angle sum of a triangle

The sum of the interior angles in any triangle is 180c ,that is, a b c 180+ + =

Proof

, YXZ a XYZ b YZX cLet andc c c+ + += = =

( , , )( )

( )

AB YZ

BXZ c BXZ XZY AB YZAXY b

YXZ AXY BXZ AXB

a b c

180

180

Draw line

Then alternate anglessimilarly

is a straight angle

`

c

c

c

+ + +

+

+ + + +

<

<=

=

+ + =

+ + =

ch4.indd 150 7/17/09 6:24:38 PM


Exterior angle of a triangle

Class Investigation

Could you prove the base angles in an isosceles triangle are equal? 1. Can there be more than one obtuse angle in a triangle? 2. Could you prove that each angle in an equilateral triangle is 3. ?60c Can a right-angled triangle be an obtuse-angled triangle? 4. Can you fi nd an isosceles triangle with a right angle in it? 5.

The exterior angle in any triangle is equal to the sum of the two opposite interior angles. That is,

x y z+ =

Proof

,

ABC x BAC y ACD z

CE AB

Let and

Draw line

c c c+ + +

<

= = =

( , , )

( , , )

z ACE ECD

ECD x ECD ABC AB CE

ACE y ACE BAC AB CEz x y

corresponding angles

alternate angles`

c

c

c

+ +

+ + +

+ + +

<

<

= +

=

=

= +

EXAMPLES

Find the values of all pronumerals, giving reasons for each step. 1.

CONTINUED

ch4.indd 151 7/17/09 6:24:42 PM


Solution

( )x

x

x

x

53 82 180 180135 180

135 180

45

135 135

angle sum of cD+ + =

+ =

+ =

=

- -

2.

Solution

( )A C x base angles of isosceles+ + D= =

( )x x

x

x

x

x

x

48 180 1802 48 180

2 48 180

2 132

2 132

66

48 48

2 2

angle sum in a cD+ + =

+ =

+ =

=

=

=

- -

3.

Solution

)y

y

y

35 14135 141

106

35 35(exterior angle of

`

D+ =

+ =

=

- -

This example can be done using the interior sum of angles.

( )

( )

BCA BCD

y

y

y

y

180 141 180

39

39 35 180 18074 180

74 180

106

74 74

is a straight angle

angle sum of

`

c c c

c

c

+ +

D

= -

=

+ + =

+ =

+ =

=

- -

ch4.indd 152 7/17/09 6:24:47 PM


1. Find the values of all pronumerals.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

2. Show that each angle in an equilateral triangle is .60c

3. Find ACB+ in terms of x .

4.3 Exercises Think of the reasons for each step of your

calculations.

ch4.indd 153 7/17/09 6:24:53 PM


4. Prove .AB ED<

5. Show ABCD is isosceles.

6. Line CE bisects .BCD+ Find the value of y , giving reasons.

7. Evaluate all pronumerals, giving reasons for your working. (a)

(b)

(c)

(d)

8. Prove IJLD is equilateral and JKLD is isosceles.

9. In triangle BCD below, .BC BD= Prove AB ED .

A

B

C

D

E

88c

46c

10. Prove that .MN QP

P

N

M

O

Q

32c

75c

73c

ch4.indd 154 7/17/09 6:25:02 PM


Congruent Triangles

Two triangles are congruent if they are the same shape and size. All pairs of corresponding sides and angles are equal.

For example:

We write .ABC XYZ/D D

Tests

To prove that two triangles are congruent, we only need to prove that certain combinations of sides or angles are equal.

Two triangles are congruent if • SSS : all three pairs of corresponding sides are equal • SAS : two pairs of corresponding sides and their included angles are

equal • AAS : two pairs of angles and one pair of corresponding sides are equal • RHS : both have a right angle, their hypotenuses are equal and one

other pair of corresponding sides are equal

EXAMPLES

1. Prove that OTS OQP/D D where O is the centre of the circle.

CONTINUED

The included angle is the angle between the 2 sides.

ch4.indd 155 7/17/09 6:25:09 PM


Solution

:

:

:

,

S

A

S

OS OQ

TOS QOP

OT OP

OTS OQP

(equal radii)

(vertically opposite angles)

(equal radii)

by SAS`

+ +

/D D

=

=

=

2. Which two triangles are congruent?

Solution

To fi nd corresponding sides, look at each side in relation to the angles. For example, one set of corresponding sides is AB , DF , GH and JL . ABC JKL A(by S S)/D D

3. Show that triangles ABC and DEC are congruent. Hence prove that .AB ED=

Solution

: ( ): ( )

: ( )

( )

AA

S

BAC CDE AB EDABC CED

AC CD

ABC DEC

AB ED

alternate angles,similarly

given

by AAS,

corresponding sides in congruent s

`

`

+ +

+ +

<

/D D

D

=

=

=

=

ch4.indd 156 7/17/09 6:25:12 PM


1. Are these triangles congruent? If they are, prove that they are congruent. (a)

(b)

X

Z

Y

B

C

A

4.7 m2.3 m

2.3 m

4.7 m110c 110c

(c)

(d)

(e)

(e)

2. Prove that these triangles are congruent. (a)

(b)

(c)

(d)

(e)

4.4 Exercises

ch4.indd 157 7/17/09 6:25:20 PM


3. Prove that (a) Δ ABD is congruent to Δ ACD

(b) AB bisects BC , given ABCD is isosceles with .AB AC=

4. Prove that triangles ABD and CDB are congruent. Hence prove that .AD BC=

5. In the circle below, O is the centre of the circle.

O

A

B

D

C

Prove that (a) OABT and OCDT are congruent.

Show that (b) .AB CD=

6. In the kite ABCD, AB AD= and .BC DC=

A

B D

C

Prove that (a) ABCT and ADCT are congruent.

Show that (b) .ABC ADC+ +=

7. The centre of a circle is O and AC is perpendicular to OB .

O

A

B

C

Show that (a) OABT and OBCT are congruent.

Prove that (b) .ABC 90c+ =

8. ABCF is a trapezium with AF BC= and .FE CD= AE and BD are perpendicular to FC.

D

A B

CFE

Show that (a) AFET and BCDT are congruent.

Prove that (b) .AFE BCD+ +=

ch4.indd 158 8/7/09 12:54:47 PM


9. The circle below has centre O and OB bisects chord AC .

O

A

B

C

Prove that (a) OABT is congruent to .OBCT

Prove that (b) OB is perpendicular to AC.

10. ABCD is a rectangle as shown below.

D

A B

C

Prove that (a) ADCT is congruent to BCDT .

Show that diagonals (b) AC and BD are equal .

Investigation

The triangle is used in many structures, for example trestle tables, stepladders and roofs.

Find out how many different ways the triangle is used in the building industry. Visit a building site, or interview a carpenter. Write a report on what you fi nd.

Similar Triangles

Triangles, for example ABC and XYZ , are similar if they are the same shape but different sizes .

As in the example, all three pairs of corresponding angles are equal. All three pairs of corresponding sides are in proportion (in the same ratio).

ch4.indd 159 7/17/09 6:25:36 PM


Application

Similar fi gures are used in many areas, including maps, scale drawings, models and enlargements.

EXAMPLE

1. Find the values of x and y in similar triangles CBA and XYZ .

Solution

First check which sides correspond to one another (by looking at their relationships to the angles). YZ and BA , XZ and CA , and XY and CB are corresponding sides.

. .

.

. . .

CAXZ

CBXY

y

y4 9 3 6

5 4

3 6 4 9 5 4

`

#

=

=

=

We write: XYZ; DABC <D XYZD is three times larger than .ABCD

ABXY

ACXZ

BCYZ

ABXY

ACXZ

BCYZ

26 3

412 3

515 3

`

= =

= =

= =

= =

This shows that all 3 pairs of sides are in proportion.

ch4.indd 160 7/17/09 6:25:53 PM


.. .

.

. ..

. . .

.. .

.

y

BAYZ

CBXY

x

x

x

3 64 9 5 4

7 35

2 3 3 65 4

3 6 2 3 5 4

3 62 3 5 4

3 45

#

#

#

=

=

=

=

=

=

=

Two triangles are similar if: three pairs of • corresponding angles are equal three pairs of • corresponding sides are in proportion two pairs of • sides are in proportion and their included angles are equal

If 2 pairs of angles are equal then the third

pair must also be equal.

EXAMPLES

1. Prove that triangles (a) ABC and ADE are similar. Hence fi nd the value of (b) y , to 1 decimal place.

Solution

(a) A+ is common

ADE; D

( )( )( )

ABC ADE BC DEACB AED

ABC

corresponding angles,similarly3 pairs of angles equal`

+ ++ +

<

<D

=

=

(b)

CONTINUED

Tests

There are three tests for similar triangles.

ch4.indd 161 7/17/09 6:26:08 PM


. .

.

. .. . .

.. .

.

.

AE

BCDE

ACAE

y

y

y

2 4 1 9

4 3

3 7 2 42 4 3 7 4 3

2 43 7 4 3

6 6

4 3

#

#

= +

=

=

=

=

=

=

2. Prove .WVZD;XYZ <D

Solution

( )

ZVXZ

ZWYZ

ZVXZ

ZWYZ

XZY WZV

3515

73

146

73


`

+ +

= =

= =

=

=

since two pairs of sides are in proportion and their included angles are equal the triangles are similar

Ratio of intercepts

The following result comes from similar triangles.

When two (or more) transversals cut a series of parallel lines, the ratios of their intercepts are equal.

: :AB BC DE EF

BCAB

EFDE

That is,

or

=

=

ch4.indd 162 7/17/09 6:26:12 PM


Proof

Draw DG and EH parallel to AC .

`

EHFD;

`

`

( )

( )

( , )

( , )( )

( )

DG AB

EH BC

BCAB

EHDG

GDE HEF DG EH

DEG EFH BE CFDGE EHF

DGE

EHDG

EFDE

BCAB

EFDE

1

2

Then opposite sides of a parallelogram

Also (similarly)

corresponding s

corresponding sangle sum of s

So

From (1) and (2):

+ + +

+ + +

+ +

<

<

<

D

D

=

=

=

=

=

=

=

=

EXAMPLES

1. Find the value of x , to 3 signifi cant fi gures.

Solution

. ..

. . .

.. .

.

x

x

x

8 9 9 31 5

9 3 8 9 1 5

9 38 9 1 5

1 44

ratios of intercepts on parallel lines

#

#

=

=

=

=

^ h

CONTINUED

ch4.indd 163 7/17/09 6:26:16 PM


2. Evaluate x and y , to 1 decimal place.

Solution

Use either similar triangles or ratios of intercepts to fi nd x . You must use similar triangles to fi nd y .

. ..

.. .

.

. .. .

.. .

.

x

x

y

y

5 8 3 42 7

3 42 7 5 8

4 6

7 1 3 42 7 3 4

3 46 1 7 1

12 7

#

#

=

=

=

=+

=

=

1. Find the value of all pronumerals, to 1 decimal place where appropriate. (a)

(b)

(c)

(d)

(e)

4.5 Exercise s

These ratios come from intercepts on parallel lines.

These ratios come from similar triangles.

Why?

ch4.indd 164 7/17/09 6:26:20 PM


(f)

14.3

a

46c

19c

115c

46c

xc

9.125.7

8.9 y

(g)

2. Evaluate a and b to 2 decimal places.

3. Show that ABCD and CDED are similar.

4. EF bisects .GFD+ Show that DEFD and FGED are similar.

5. Show that ABCD and DEFD are similar. Hence fi nd the value of y .

4.2

4.9

6.86

1.3

5.881.82

A

C

BD

E F

yc87c

52c

6. The diagram shows two concentric circles with centre O .

Prove that (a) D .OCD;OAB <D If radius (b) . OC 5 9 cm= and

radius . OB 8 3 cm,= and the length of . CD 3 7 cm,= fi nd the length of AB , correct to 2 decimal places.

7. (a) Prove that .ADED;ABC <D Find the values of (b) x and y ,

correct to 2 decimal places.

8. ABCD is a parallelogram, with CD produced to E . Prove that .CEBD;ABF <D

ch4.indd 165 7/17/09 6:26:25 PM


9. Show that .ABC; DAED <D Find the value of m .

10. Prove that ABCD and ACDD are similar. Hence evaluate x and y .

11. Find the values of all pronumerals, to 1 decimal place. (a)

(b)

(c)

(d)

(e)

12. Show that

(a) BCAB

FGAF

=

(b) ACAB

AGAF

=

(c) CEBD

EGDF

=

13. Evaluate a and b correct to 1 decimal place.

14. Find the value of y to 2 signifi cant fi gures.

15. Evaluate x and y correct to 2 decimal places.

ch4.indd 166 8/1/09 11:56:40 AM


Pythagoras’ Theorem

DID YOU KNOW?

The triangle with sides in the proportion 3:4:5 was known to be right angled as far back as ancient Egyptian times. Egyptian surveyors used to measure right angles by stretching out a rope with knots tied in it at regular intervals.

They used the rope for forming right angles while building and dividing fi elds into rectangular plots.

It was Pythagoras (572–495 BC)who actually discovered the relationship between the sides of the right-angled triangle. He was able to generalise the rule to all right-angled triangles.

Pythagoras was a Greek mathematician, philosopher and mystic. He founded the Pythagorean School, where mathematics, science and philosophy were studied. The school developed a brotherhood and performed secret rituals. He and his followers believed that the whole universe was based on numbers.

Pythagoras was murdered when he was 77, and the brotherhood was disbanded.

The square on the hypotenuse in any right-angled triangle is equal to the sum of the squares on the other two sides. c a b

c a b

That is,

or

2 2 2

2 2

= +

= +

ch4.indd 167 7/17/09 6:26:40 PM


Proof

Draw CD perpendicular to AB Let ,AD x DB y= = Then x y c+ = In ADCD and ,ABCD A+ is common

D

D

;

;

( )ABC

ABC

equal corresponding s+

ADC ACB

ADC

ABAC

ACAD

cb

bx

b xcBDC

BCDB

ABBC

ay

ca

a yc

a b yc xcc y x

c c

c

90

Similarly,

Now

2

2

2 2

2

`

c+ +

<

<

D

D

= =

=

=

=

=

=

=

+ = +

= +

=

=

^

]

h

g

EXAMPLES

1. Find the value of x , correct to 2 decimal places.

Solution

c a b

x 7 449 16

65

2 2 2

2 2 2

= +

= +

= +

=

,c a b ABCIf then must be right angled2 2 2 D= +

ch4.indd 168 7/17/09 6:27:00 PM


. x 65

8 06 to 2 decimal places=

=

2. Find the exact value of y .

Solution

c a b

y

y

y

y

8 4

64 16

48

48

16 3

4 3

2 2 2

2 2 2

2

2

`

#

= +

= +

= +

=

=

=

=

3. Find the length of the diagonal in a square with sides 6 cm. Answer to 1 decimal place.

Solution

6 cm

6 cm

.

c a b

c

6 672

728 5

2 2 2

2 2

= +

= +

=

=

=

So the length of the diagonal is 8.5 cm.

Leave the answer in surd form for the exact

answer.

CONTINUED

ch4.indd 169 7/17/09 6:27:03 PM


1. Find the value of all pronumerals, correct to 1 decimal place. (a)

(b)

(c)

(d)

2. Find the exact value of all pronumerals. (a)

(b)

(c)

(d)

4.6 Exercises

4. A triangle has sides 5.1 cm, 6.8 cm and 8.5 cm. Prove that the triangle is right angled.

Solution

6.8 cm

8.5 cm5.1 cm

Let .c 8 5= (largest side) and a and b the other two smaller sides.

. . .

. .

a b

c

c a b

5 1 6 872 25

8 572 25

2 2 2 2

2 2

2 2 2`

+ = +

=

=

=

= +

So the triangle is right angled .

ch4.indd 170 7/17/09 6:27:10 PM


3. Find the slant height s of a cone with diameter 6.8 m and perpendicular height 5.2 m, to 1 decimal place.

4. Find the length of CE , correct to 1 decimal place, in this rectangular pyramid. 8.6 AB cm= and 15.9 .CF cm=

5. Prove that ABCD is a right-angled triangle.

6. Show that XYZD is a right-angled isosceles triangle.

X

Y Z1

1 2

7. Show that .AC BC2=

8. (a) Find the length of diagonal AC in the fi gure.

Hence, or otherwise, prove (b) that AC is perpendicular to DC .

9. Find the length of side AB in terms of b .

10. Find the exact ratio of YZXY in

terms of x and y in .XYZD

ch4.indd 171 7/17/09 6:27:15 PM


11. Show that the distance squared between A and B is given by .d t t13 180 6252 2= - +

12. An 850 mm by 1200 mm gate is to have a diagonal timber brace to give it strength. To what length should the timber be cut, to the nearest mm?

13. A rectangular park has a length of 620 m and a width of 287 m. If I walk diagonally across the park, how far do I walk?

14. The triangular garden bed below is to have a border around it. How many metres of border are needed, to 1 decimal place?

15. What is the longest length of stick that will fi t into the box below, to 1 decimal place?

16. A ramp is 4.5 m long and 1.3 m high. How far along the ground does the ramp go? Answer correct to one decimal place .

4.5 m1.3 m

17. The diagonal of a television screen is 72 cm. If the screen is 58 cm high, how wide is it?

18. A property has one side 1.3 km and another 1.1 km as shown with a straight road diagonally through the middle of the property. If the road is 1.5 km long, show that the property is not rectangular.

1.3 km

1.1 km

1.5 km

19. Jodie buys a ladder 2 m long and wants to take it home in the boot of her car. If the boot is 1.2 m by 0.7 m, will the ladder fi t?

ch4.indd 172 7/17/09 6:27:24 PM


Types of Quadrilaterals

A quadrilateral is any four-sided fi gure

In any quadrilateral the sum of the interior angles is 360c

20. A chord AB in a circle with centre O and radius 6 cm has a perpendicular line OC as shown 4 cm long.

A

B

O

C

6 cm4 cm

By fi nding the lengths of (a) AC and BC , show that OC bisects the chord .

By proving congruent (b) triangles, show that OC bisects the chord .

Proof

Draw in diagonal AC

180 ( )( )

,

ADC DCA CAD

ABC BCA CAB

ADC DCA CAD ABC BCA CAB

ADC DCB CBA BAD

180

360

360

angle sum ofsimilarly

That is

`

c

c

c

c

+ + +

+ + +

+ + + + + +

+ + + +

D+ + =

+ + =

+ + + + + =

+ + + =

ch4.indd 173 7/17/09 6:27:32 PM


opposite sides• of a parallelogram are equal • opposite angles of a parallelogram are equal • diagonals in a parallelogram bisect each other each diagonal bisects the parallelogram into two • congruent triangles

A quadrilateral is a parallelogram if: both pairs of • opposite sides are equal both pairs of • opposite angles are equal one • pair of sides is both equal and parallel the • diagonals bisect each other

These properties can all be proven.

Parallelogram

A parallelogram is a quadrilateral with opposite sides parallel

EXAMPLE

Find the value of .i

Solution

120 56 90 360

266 360

94

angle sum of quadrilaterali

i

i

+ + + =

+ =

=

^ h

PROPERTIES

TESTS

ch4.indd 174 7/17/09 6:27:36 PM


Rhombus

A rectangle is a parallelogram with one angle a right angle

the same as for a parallelogram, and also • diagonals are equal •

A quadrilateral is a rectangle if its diagonals are equal

Application

Builders use the property of equal diagonals to check if a rectangle is accurate. For example, a timber frame may look rectangular, but may be slightly slanting. Checking the diagonals makes sure that a building does not end up like the

Leaning Tower of Pisa!

It can be proved that all sides are equal.

If one angle is a right angle, then you can prove all angles are

right angles.

A rhombus is a parallelogram with a pair of adjacent sides equal

the same as for parallelogram, and also • diagonals bisect at right angles • diagonals bisect the angles of the rhombus •

Rectangle

PROPERTIES

PROPERTIES

TEST

ch4.indd 175 7/31/09 4:25:28 PM


Square

A square is a rectangle with a pair of adjacent sides equal

• the same as for rectangle, and also diagonals are perpendicular • diagonals make angles of • 45c with the sides

Trapezium

A trapezium is a quadrilateral with one pair of sides parallel

Kite

A kite is a quadrilateral with two pairs of adjacent sides equal

A quadrilateral is a rhombus if: all sides are equal • diagonals bisect each other at right angles •

TESTS

PROPERTIES

ch4.indd 176 7/17/09 6:27:44 PM


EXAMPLES

1. Find the values of ,i x and y , giving reasons.

Solution

( )

. ( )

. ( )

x

y

83

6 7

2 3

opposite s in gram

cm opposite sides in gram

cm opposite sides in gram

c + <

<

<

i =

=

=

2. Find the length of AB in square ABCD as a surd in its simplest form if 6 .BD cm=

Solution

( )

( )

AB x

ABCD AB AD x

A 90

Let

Since is a square, adjacent sides equal

Also, by definitionc+

=

= =

=

By Pythagoras’ theorem:

3

c a b

x x

x

x

x

6

36 2

18

18

2 cm

2 2 2

2 2 2

2

2

`

= +

= +

=

=

=

=

CONTINUED

ch4.indd 177 7/17/09 6:27:47 PM


1. Find the value of all pronumerals, giving reasons. (a)

(b)

(c)

(d)

(e)

(f)

(g)

4.7 Exercises

3. Two equal circles have centres (a) O and P respectively. Prove that OAPB

is a rhombus. Hence, or otherwise, show that (b) AB is the perpendicular bisector

of OP .

Solution

(a) ( )

( )

OA OB

PA PB

OA OB PA PB

equal radii

similarly

Since the circles are equal,

=

=

= = =

since all sides are equal, OAPB is a rhombus The diagonals in any rhombus are perpendicular bisectors. (b)

Since OAPB is a rhombus, with diagonals AB and OP , AB is the perpendicular bisector of OP .

ch4.indd 178 7/17/09 6:27:51 PM


2. Given ,AB AE= prove CD is perpendicular to AD .

3. (a) Show that C xc+ = and ( ) .B D x180 c+ += = -

Hence show that the sum of (b) angles of ABCD is .360c

4. Find the value of a and b .

5. Find the values of all pronumerals, giving reasons.

(a)

(b)

(c)

(d)

(e)

7

y3x

x+6

(f)

6. In the fi gure, BD bisects .ADC+ Prove BD also bisects .ABC+

7. Prove that each fi gure is a parallelogram. (a)

(b)

ch4.indd 179 7/17/09 6:27:55 PM


(c)

(d)

8. Evaluate all pronumerals.

(a)

(b)

ABCD is a kite

(c)

(d)

(e)

9. The diagonals of a rhombus are 8 cm and 10 cm long. Find the length of the sides of the rhombus.

10. ABCD is a rectangle with .EBC 59c+ = Find ,ECB EDC+ + and .ADE+

11. The diagonals of a square are 8 cm long. Find the exact length of the side of the square.

12. In the rhombus, .ECB 33c+ = Find the value of x and y .

Polygons

A polygon is a closed plane fi gure with straight sides

A regular polygon has all sides and all interior angles equal

ch4.indd 180 7/17/09 6:28:01 PM


Proof

Draw any n -sided polygon and divide it into n triangles as shown. Then the total sum of angles is n 180# c or 180 .n But this sum includes all the angles at O . So the sum of interior angles is 180 360 .n c- That is, S n

n

180 360

2 180# c

= -

= -] g

EXAMPLES

4-sided (square)

3-sided (equilateral

triangle)

5-sided (pentagon)

6-sided (hexagon)

8-sided (octagon)

10-sided (decagon)

DID YOU KNOW?

Carl Gauss (1777–1855) was a famous German mathematician, physicist and astronomer. When he was 19 years old, he showed that a 17-sided polygon could be constructed using a ruler and compasses. This was a major achievement in geometry.

Gauss made a huge contribution to the study of mathematics and science, including correctly calculating where the magnetic south pole is and designing a lens to correct astigmatism.

He was the director of the Göttingen Observatory for 40 years. It is said that he did not become a professor of mathematics because he did not like teaching.

The sum of the interior angles of an n -sided polygon is given by

( 2) 180

S n

S n

180 360

or # c

= -

= -

The sum of the exterior angles of any polygon is 360c

Proof

Draw any n -sided polygon. Then the sum of both the exterior and interior angles is .n 180# c

n

n nn n

180

180 180 360180 180 360

360

Sum of exterior angles sum of interior angles# c

c

c

c

= -

= - -

= - +

=

] g

ch4.indd 181 7/17/09 6:28:08 PM


EXAMPLES

1. Find the sum of the interior angles of a regular polygon with 15 sides. How large is each angle?

Solution

( )

( )

n

S n

15

0

15 0

0

2340

2 18

2 18

13 18

#

#

#

c

c

c

c

=

= -

= -

=

=

Each angle has size .2340 15 156'c c=

2. Find the number of sides in a regular polygon whose interior angles are .140c

Solution

Let n be the number of sides Then the sum of interior angles is 140n

( )

( )

S n

n n

n

n

n

2 180

140 2 180

180 360

360 40

9

But

So

#

#

c

c

= -

= -

= -

=

=

So the polygon has 9 sides.

There are n sides and so n angles, each 140 .c

1. Find the sum of the interior angles of

a pentagon (a) a hexagon (b) an octagon (c) a decagon (d) a 12-sided polygon (e) an 18-sided polygon (f)

2. Find the size of each interior angle of a regular

pentagon (a) octagon (b) 12-sided polygon (c) 20-sided polygon (d) 15-sided polygon (e)

3. Find the size of each exterior angle of a regular

hexagon (a) decagon (b) octagon (c) 15-sided polygon (d)

4. Calculate the size of each interior angle in a regular 7-sided polygon, to the nearest minute.

5. The sum of the interior angles of a regular polygon is .1980c

How many sides has the (a) polygon?

Find the size of each interior (b) angle, to the nearest minute.

4.8 Exercises

ch4.indd 182 7/17/09 6:28:12 PM


6. Find the number of sides of a regular polygon whose interior angles are .157 30c l

7. Find the sum of the interior angles of a regular polygon whose exterior angles are .18c

8. A regular polygon has interior angles of .156c Find the sum of its interior angles.

9. Find the size of each interior angle in a regular polygon if the sum of the interior angles is .5220c

10. Show that there is no regular polygon with interior angles of .145c

11. Find the number of sides of a regular polygon with exterior angles

(a) 40c (b) 03 c (c) 45c (d) 36c (e) 12c

12. ABCDEF is a regular hexagon.

F

E D

A B

C

Show that triangles (a) AFE and BCD are congruent .

Show that (b) AE and BD are parallel .

13. A regular octagon has a quadrilateral ACEG inscribed as shown.

D

A

B

E

C

F

G

H

Show that ACEG is a square .

14. In the regular pentagon below, show that EAC is an isosceles triangle .

D

A

BE

C

15. (a) Find the size of each exterior angle in a regular polygon with side p .

Hence show that each interior (b)

angle is ( )pp180 2-

.

ch4.indd 183 7/17/09 6:28:15 PM


Areas

Most areas of plane fi gures come from the area of a rectangle.

Rectangle

A lb=

Square

A x2=

Triangle

A bh21

=

Proof

h

b

Draw rectangle ABCD , where b length= and h breadth= .

A square is a special rectangle.

The area of a triangle is half the area of a rectangle.

ch4.indd 184 7/17/09 6:28:18 PM


bharea

21

21

21

21

` =

DEF AEFD CEF EBCFArea area and area areaD D= =

CDE ABCDarea ` D =

A bhThat is, =

area

A bh=

Proof

In parallelogram ABCD , produce DC to E and draw BE perpendicular to CE . Then ABEF is a rectangle.

Area ABEF bh= In ADFD and ,BCED

( )

( )

AFD BEC

AF BE h

AD BC

ADF BCE

ADF BCE

ABCD ABEF

bh

90

opposite sides of a rectangle

opposite sides of a parallelogram

by RHS,

area areaSo area area

`

`

c+ +

/D D

D D

= =

= =

=

=

=

=

Rhombus

The area of a parallelogram is the same as the area of

two triangles.

A xy21

=

( x and y are lengths of diagonals)

Parallelogram

ch4.indd 185 8/7/09 12:57:48 PM


( )A h a b21

= +

Proof

DE x

DF x a

FC b x ab x a

Let

Then

`

=

= +

= - +

= - -

] g

Proof

Let AC x= and BD y= By properties of a rhombus,

AE EC x21

= = and DE EB y21

= =

Also AEB 90c+ =

ABC x y

xy

ADC x y

xy

xy xy

xy

Area

Area

total area of rhombus

21

21

41

21

21

41

41

41

21

:

:

`

D

D

=

=

=

=

= +

=

Trapezium

ch4.indd 186 7/17/09 6:28:22 PM


A r2r=

EXAMPLES

1. Find the area of this trapezium.

Solution

( )

( ) ( )

24

A h a b

4 7 5

2 12

m2

21

21

#

= +

= +

=

=

2. Find the area of the shaded region in this fi gure.

8.9

cm

3.7

cm

12.1 cm

4.2 cm

CONTINUED

( )

( )

( )

ADE ABFE BFC

xh ah b x a h

h x a b x a

h a b

2

Area trapezium area area rectangle area

21

21

21

21

D D= + +

= + + - -

= + + - -

= +

Circle

ch4.indd 187 7/31/09 4:25:28 PM


Solution

. .

.

. .

. . .

.

lb

lb

8 9 12 1

107 69

3 7 4 2

15 54107 69 15 54

92 15

Area large rectangle

cmArea small rectangle

cmshaded area

cm

2

2

2

#

#

`

=

=

=

=

=

=

= -

=

3 . A park with straight sides of length 126 m and width 54 m has semi-circular ends as shown. Find its area, correct to 2 decimal places.

126 m

54 m

Solution

-Area of 2 semi circles area of 1 circle=

2

( )

.

r

A r

254

27

27

2290 22 m

2

2

r

r

=

=

=

=

=

.

.

126 54

6804

2290 22 6804

9094 22

Area rectangle

Total area

m2

#=

=

= +

=

1. Find the area of each fi gure. (a)

(b)

4.9 Exercises

ch4.indd 188 7/17/09 6:28:28 PM


(c)

(d)

(e)

(f)

(g)

2. Find the area of a rhombus with diagonals 2.3 m and 4.2 m.

3. Find each shaded area .(a)

(b)

(c)

(d)

(e)

6 cm

2 cm

4. Find the area of each fi gure. (a)

(b)

ch4.indd 189 7/17/09 6:28:31 PM


(c)

(d)

(e)

5. Find the exact area of the fi gure.

6. Find the area of this fi gure, correct to 4 signifi cant fi gures. The arch is a semicircle.

7. Jenny buys tiles for the fl oor of her bathroom (shown top next column) at $45.50 per .m2 How much do they cost altogether?

8. The dimensions of a battleaxe block of land are shown below.

Find its area. (a) A house in the district where (b)

this land is can only take up 55% of the land. How large (to the nearest m2 ) can the area of the house be?

If the house is to be a (c) rectangular shape with width 8.5 m, what will its length be?

9. A rhombus has one diagonal 25 cm long and its area is 600 .cm2 Find the length of

its other diagonal and (a) its side, to the nearest cm. (b)

10. The width w of a rectangle is a quarter the size of its length. If the width is increased by 3 units while the length remains constant, fi nd the amount of increase in its area in terms of w .

ch4.indd 190 7/17/09 6:28:34 PM


Test Yourself 4

The perimeter is the distance around the outside of the fi gure.

1. Find the values of all pronumerals (a)

(b)

(c)

x(d)

(O is the centreof the circle.)

(e)

(f)

(g)

2. Prove that AB and CD are parallel lines.

3. Find the area of the fi gure, to 2 decimal places.

4. (a) Prove that triangles ABC and ADE are similar.

Evaluate (b) x and y to 1 decimal place.

5. Find the size of each interior angle in a regular 20-sided polygon.

6. Find the volume of a cylinder with radius 5.7 cm and height 10 cm, correct to 1 decimal place.

7. Find the perimeter of the triangle below.

ch4.indd 191 7/17/09 6:28:40 PM


8. (a) Prove triangles ABC and ADC are congruent in the kite below.

Prove triangle (b) AOB and COD are congruent. ( O is the centre of the circle.)

9. Find the area of the fi gure below.

10. Prove triangle ABC is right angled.

11. Prove .AGAF

ACAB

=

12. Triangle ABC is isosceles, and AD bisects BC .

Prove triangles (a) ABD and ACD are congruent.

Prove (b) AD and BC are perpendicular.

13. Triangle ABC is isosceles, with .AB AC= Show that triangle ACD is isosceles.

14. Prove that opposite sides in any parallelogram are equal.

15. A rhombus has diagonals 6 cm and 8 cm. Find the area of the rhombus. (a) Find the length of its side. (b)

16. The interior angles in a regular polygon are .140c How many sides has the polygon?

17. Prove AB and CD are parallel.

ch4.indd 192 7/17/09 6:28:51 PM



10 cm

2 cm

5 cm

6 cm

8 cm

19. Prove that z x y= + in the triangle below.

20. (a) Prove triangles ABC and DEF are similar.

Evaluate (b) x to 1 decimal place.

1. Find the value of x .

2. Evaluate x , y and z .

3. Find the sum of the interior angles of a regular 11-sided polygon. How large is each exterior angle?

4. Given ,BAD DBC+ += show that ABDD and BCDD are similar and hence fi nd d .

5. Prove that ABCD is a parallelogram. .AB DC=

6. Find the shaded area.


ch4.indd 193 7/17/09 6:29:02 PM


7. Prove that the diagonals in a square make angles of 45c with the sides.

8. Prove that the diagonals in a kite are perpendicular.

9. Prove that MN is parallel to XY .

10. Evaluate x .

11. The letter Z is painted on a billboard.

Find the area of the letter. (a) Find the exact perimeter of the letter. (b)

12. Find the values of x and y correct to 1 decimal place.

13. Find the values of x and y , correct to 2 decimal places.

14. ABCD is a square and BD is produced to

E such that .DE BD21

=

Show that (a) ABCE is a kite.

Prove that (b) DEx

22

= units when

sides of the square are x units long.

ch4.indd 194 7/17/09 6:29:12 PM

TERMINOLOGY

5

Arc of a curve: Part or a section of a curve between two points

Asymptote: A line towards which a curve approaches but never touches

Cartesian coordinates: Named after Descartes. A system of locating points (x, y) on a number plane. Point (x, y) has Cartesian coordinates x and y

Curve: Another word for arc. When a function consists of all values of x on an interval, the graph of y f x= ] g is called a curve y f x= ] g

Dependent variable: A variable is a symbol that can represent any value in a set of values. A dependent variable is a variable whose value depends on the value chosen for the independent variable

Direct relationship: Occurs when one variable varies directly with another i.e. as one variable increases, so does the other or as one variable decreases so does the other

Discrete: Separate values of a variable rather than a continuum. The values are distinct and unrelated

Domain: The set of possible values of x in a given domain for which a function is defi ned

Even function: An even function has line symmetry (refl ection) about the y-axis, and f x f x=- -] ]g g

Function: For each value of the independent variable x, there is exactly one value of y, the dependent variable. A vertical line test can be used to determine if a relationship is a function

Independent variable: A variable is independent if it may be chosen freely within the domain of the function

Odd function: An odd function has rotational symmetry about the origin (0, 0) and where f x f x=- -] ]g g

Ordered pair: A pair of variables, one independent and one dependent, that together make up a single point in the number plane, usually written in the form (x, y)

Ordinates: The vertical or y coordinates of a point are called ordinates

Range: The set of real numbers that the dependent variable y can take over the domain (sometimes called the image of the function)

Vertical line test: A vertical line will only cut the graph of a function in at most one point. If the vertical line cuts the graph in more than one point, it is not a function

Functions and Graphs

ch5.indd 200 7/19/09 12:41:35 PM

201Chapter 5 Functions and Graphs

INTRODUCTION

FUNCTIONS AND THEIR GRAPHS are used in many areas, such as mathematics, science and economics. In this chapter you will study functions, function notation and how to sketch graphs. Some of these graphs will be studied in more detail in later chapters.

DID YOU KNOW?

The number plane is called the Cartesian plane after Rene Descartes (1596–1650). He was known as one of the fi rst modern mathematicians along with Pierre de Fermat (1601–1665). Descartes used the number plane to develop analytical geometry. He discovered that any equation with two unknown variables can be represented by a line. The points in the number plane can be called Cartesian coordinates.

Descartes used letters at the beginning of the alphabet to stand for numbers that are known, and letters near the end of the alphabet for unknown numbers. This is why we still use x and y so often!

Do a search on Descartes to fi nd out more details of his life and work.

Descartes

Functions

Defi nition of a function

Many examples of functions exist both in mathematics and in real life. These occur when we compare two different quantities. These quantities are called variables since they vary or take on different values according to some pattern. We put these two variables into a grouping called an ordered pair.

ch5.indd 201 7/19/09 12:41:48 PM


EXAMPLES

1. Eye colour

Name Anne Jacquie Donna Hien Marco Russell Trang

Colour Blue Brown Grey Brown Green Brown Brown

Ordered pairs are (Anne, Blue), (Jacquie, Brown), (Donna, Grey), (Hien, Brown), (Marco, Green), (Russell, Brown) and (Trang, Brown).

2. y x 1= +

x 1 2 3 4

y 2 3 4 5 The ordered pairs are (1, 2), (2, 3), (3, 4) and (4, 5).

3. A

B

C

D

E

1

2

3

4

The ordered pairs are (A, 1), (B, 1), (C, 4), (D, 3) and (E, 2).

Notice that in all the examples, there was only one ordered pair for each variable. For example, it would not make sense for Anne to have both blue and brown eyes! (Although in rare cases some people have one eye that’s a different colour from the other.)

A relation is a set of ordered points ( x , y ) where the variables x and y are related according to some rule.

A function is a special type of relation. It is like a machine where for every INPUT there is only one OUTPUT.

INPUT PROCESS OUTPUT

The fi rst variable (INPUT) is called the independent variable and the second (OUTPUT) the dependent variable. The process is a rule or pattern.

ch5.indd 202 7/19/09 12:42:02 PM


For example, in ,y x 1= + we can use any number for x (the independent variable), say x 3= .

When x

y

3

3 1

4

=

= +

=

As this value of y depends on the number we choose for x , y is called the dependent variable.

A function is a relationship between two variables where for every independent variable, there is only one dependent variable.

This means that for every x value, there is only one y value.

While we often call the independent variable x and the dependent

variable y, there are other pronumerals we could

use. You will meet some of these in this course.

Investigation

When we graph functions in mathematics, the independent variable (usually the x -value) is on the horizontal axis while the dependent variable (usually the y -value) is on the vertical axis.

In other areas, the dependent variable goes on the horizontal axis. Find out in which subjects this happens at school by surveying teachers or students in different subjects. Research different types of graphs on the Internet to fi nd some examples.

Here is an example of a relationship that is NOT a function. Can you see the difference between this example and the previous ones?

A

B

C

D

E

1

2

3

4

In this example the ordered pairs are (A, 1), (A, 2), (B, 1), (C, 4), (D, 3) and (E, 2).

Notice that A has two dependent variables, 1 and 2. This means that it is NOT a function.

ch5.indd 203 7/19/09 12:42:18 PM


Here are two examples of graphs on a number plane.

1.

x

y

2.

x

y

There is a very simple test to see if these graphs are functions. Notice that in the fi rst example, there are two values of y when x 0= . The y -axis passes through both these points.

x

y

ch5.indd 204 7/19/09 12:42:27 PM


If a vertical line cuts a graph only once anywhere along the graph, the graph is a function.

y

x

If a vertical line cuts a graph in more than one place anywhere along the graph, the graph is not a function.

x

y

There are also other x values that give two y values around the curve. If we drew a vertical line anywhere along the curve, it would cross the curve in two places everywhere except one point. Can you see where this is?

In the second graph, a vertical line would only ever cross the curve in one place.

So when a vertical line cuts a graph in more than one place, it shows that it is not a function.

ch5.indd 205 7/19/09 12:42:36 PM


EXAMPLES

1. Is this graph a function?

Solution

A vertical line only cuts the graph once. So the graph is a function.

2. Is this circle a function?

Solution

A vertical line can cut the curve in more than one place. So the circle is not a function.

You will learn how to sketch these graphs later in this chapter.

ch5.indd 206 7/19/09 12:42:46 PM


3. Does this set of ordered pairs represent a function? , , , , , , , , ,2 3 1 4 0 5 1 3 2 4- -^ ^ ^ ^ ^h h h h h

Solution

For each x value there is only one y value, so this set of ordered pairs is a function.

4. Is this a function?

y

x3

Solution

y

x3

Although it looks like this is not a function, the open circle at x 3= on the top line means that x 3= is not included, while the closed circle on the bottom line means that x 3= is included on this line.

So a vertical line only touches the graph once at x 3= . The graph is a function.

ch5.indd 207 7/19/09 12:42:57 PM


1.

2.

3.

4.

5.

6.

7.

8.

9. , , , ,, , ,1 3 2 1 3 3 4 0-^ ^ ^ ^h h h h

10. , , , , , ,,1 3 2 1 2 7 4 0-^ ^ ^ ^h h h h

11.

1

2

3

4

5

1

2

3

4

5

12. 1

2

3

4

5

1

2

3

4

5

13.

1

2

3

4

5

1

2

3

4

5

5.1 Exercises

Which of these curves are functions?

ch5.indd 208 7/19/09 12:43:06 PM


14. Name Ben Paul Pierre Hamish Jacob Lee Pierre Lien

Sport Tennis Football Tennis Football Football Badminton Football Badminton

15. A 3

B 4

C 7

D 3

E 5

F 7

G 4

Function notation

If y depends on what value we give x in a function, then we can say that y is a function of x . We can write this as y f x= ] g .

Notice that these two examples are asking for the same value and f (3) is the value of the function when x 3= .

EXAMPLES

1. Find the value of y when x 3= in the equation y x 1= + .

Solution

When :x

y x

3

1

3 1

4

=

= +

= +

=

2. If f x x 1= +] g , evaluate f (3).

Solution

f x x

f

1

3 3 14

= +

= +

=

]

]

g

g

If y f x= ] g then f ( a ) is the value of y at the point on the function where x a=

ch5.indd 209 7/19/09 1:25:30 PM


EXAMPLES

1. If ,f x x x3 12= + +] g fi nd .f 2-] g

Solution

( ) ( )f 2 2 3 2 1

4 6 1

1

2- = - + - +

= - +

= -

] g

2. If ,f x x x3 2= -] g fi nd the value of .f 1-] g

Solution

( )

( )

f x x x

f 1 1 11 1

2

3 2

3

= -

- = - - -

= - -

= -

2] ]g g

3. Find the values of x for which ,f x 0=] g given that .f x x x3 102= + -] g

Solution

( )

i.e.( ) ( )

,

f x

x xx x

x x

x x

0

3 10 05 2 0

5 0 2 0

5 2

2

=

+ - =

+ - =

+ = - =

= - =

4. Find , ,f f f3 2 0] ] ]g g g and iff f x4-] ]g g is defi ned as

when

when .f x

x x

x x

3 4 2

2 21

$=

+

-] g )

Solution

since 4 21-

( ) ( ) since

( ) ( ) since

( ) ( ) since

( ) ( )

f

f

f

f

3 3 3 4 3 2

13

2 3 2 4 2 2

10

0 2 0 0 2

0

4 2 4

8

1

$

$

= +

=

= +

=

= -

=

- = - -

=

5. Find the value of g g g1 2 3+ - -] ] ]g g g if

whenwhen

when

xx

x

21 2

1

2

1

# #-

-

g xx

x2 1

5

2

= -] g *

This is the same as fi nding y when 2.x -=

Putting (x) 0=f is different from fi nding (0) .f Follow this example carefully.

Use f (x) 3x 4= + when x is 2 or more, and use f (x) 2x= - when x is less than 2.

ch5.indd 210 7/19/09 12:43:27 PM


Solution

( ) ( )

( )

( )

g

g

g

1 2 1 1 1 1 2

1

2 5 2 1

3 3 3 29

since

since

since2

1

2

# #= - -

=

- = - -

=

=

( ) ( ) ( )g g g1 2 3 1 5 9

3

So + - - = + -

= -

DID YOU KNOW?

Leonhard Euler (1707–83), from Switzerland, studied functions and invented the term f (x) for function notation. He studied theology, astronomy, medicine, physics and oriental languages as well as mathematics, and wrote more than 500 books and articles on mathematics. He found time between books to marry and have 13 children, and even when he went blind he kept on having books published.

1. Given ,f x x 3= +] g fi nd f 1] g and .f 3-] g

2. If ,h x x 22= -] g fi nd ,h h0 2] ]g g and .h 4-] g

3. If ,f x x2= -] g fi nd , ,f f f5 1 3-] ] ]g g g and .f 2-] g

4. Find the value of f f0 2+ -] ]g g if .f x x x 14 2= - +] g

5. Find f 3-] g if .f x x x2 5 43= - +] g

6. If ,f x x2 5= -] g fi nd x when .f x 13=] g

7. Given ,f x x 32= +] g fi nd any values of x for which .f x 28=] g

8. If ,f x 3x=] g fi nd x when .f x

271

=] g

9. Find values of z for which f z 5=] g given .f z z2 3= +] g

10. If ,f x x2 9= -] g fi nd f p^ h and .f x h+] g

11. Find g x 1-] g when .g x x x2 32= + +] g

12. If ,f x x 13= -] g fi nd f k] g as a product of factors.

13. Given ,f t t t2 12= + +] g fi nd t when .f t 0=] g Also fi nd any values of t for which .f t 9=] g

14. Given ,f t t t 54 2= + -] g fi nd the value of .f b f b- -] ]g g

15. f xx xx x

11

forfor

3 2#

=] g )

Find ,f f5 1] ]g g and .1-] g

16. f x

x x

x x

x x

2 4 1

3 1 1

1

if

if

if2

1 1

$

#

=

-

+ -

-

] g

Z

[

\

]]

]]

Find the values of

.f f f2 2 1- - + -] ] ]g g g

5.2 Exercises

We can use pronumerals other than f for functions.

ch5.indd 211 7/19/09 12:43:36 PM


17. Find g g g3 0 2+ + -] ] ]g g g if

g xx x

x x

1 0

2 1 0

when

when 1

$=

+

- +] g )

18. Find the value of f f f3 2 2 3- + -] ] ]g g g when

f x

x x

x xx

2

2 24 2

for

forfor

2

2

1# #= -

-

] g *

19. Find the value of f f1 3- -] ]g g

if ( )1 2

2 3 1 2f x

x x

x x x

for

for

3

2 1

$=

-

+ -*

20. If f xx

x x3

2 32

=-

- -] g

evaluate (a) f (2) explain why the function (b)

does not exist for x 3= by taking several (c) x values

close to 3, fi nd the value of y that the function is moving towards as x moves towards 3.

21. If –f x x x5 42= +] g , fi nd f x h f x+ -] ]g g in its simplest form.

22. Simplify h

f x h f x+ -] ]g g where

f x x x2 2= +] g

23. If f x x5 4= -] g , fi nd f x f c-] ]g g in its simplest form.

24. Find the value of f k2^ h if

f xx x

x x

3 5 0

0

for

for2 1

$=

+] g *

25. If f x

x x

x x x

3

2 0

when

when

3

2

$

#

=

- +

x5 0 3when 1 1] g

Z

[

\

]]

]]

evaluate (a) f (0) (b) f f2 1-] ]g g

(c) f n2-^ h

Graphing Techniques

You may have previously learned how to draw graphs by completing a table of values and then plotting points. In this course, you will learn some other techniques that will allow you to sketch graphs by showing their important features.

Intercepts

One of the most useful techniques is to fi nd the x- and y -intercepts.

For x -intercept, y 0= For y -intercept, x 0=

Everywhere on the x-axis, 0=y and everywhere on the y-axis 0=x .

ch5.indd 212 7/31/09 4:30:32 PM


EXAMPLE

Find the x - and y -intercepts of the function .f x x x7 82= + -] g

Solution

For x -intercept: y 0=

,

,

x xx x

x x

x x

0 7 88 1

8 0 1 0

8 1

2= + -

= + -

+ = - =

= - =

] ]g g

For y -intercept: x 0=

y 0 7 0 8

8

2= + -

= -

] ]g g

This is the same as .y x x7 82= + -

You will use the intercepts to draw graphs in the next

section in this chapter .

Domain and range

You have already seen that the x -coordinate is called the independent variable and the y -coordinate is the dependent variable.

The set of all real numbers x for which a function is defi ned is called the domain . The set of real values for y or f ( x ) as x varies is called the range (or image) of f .

EXAMPLE

Find the domain and range of .f x x2=] g

Solution

You can see the domain and range from the graph, which is the parabola .y x2=

x

y

CONTINUED

ch5.indd 213 7/19/09 12:43:53 PM


Notice that the parabola curves outwards gradually, and will take on any real value for x . However, it is always on or above the x -axis.

Domain: {all real x } Range: { y : y 0$ }

You can also fi nd the domain and range from the equation y x2= . Notice that you can substitute any value for x and you will fi nd a value of y . However, all the y -values are positive or zero since squaring any number will give a positive answer (except zero).

Odd and even functions

When you draw a graph, it can help to know some of its properties, for example, whether it is increasing or decreasing on an interval or arc of the curve (part of the curve lying between two points) .

If a curve is increasing, as x increases, so does y , and the curve is moving upwards, looking from left to right.

If a curve is decreasing, then as x increases, y decreases and the curve moves downwards from left to right.

ch5.indd 214 7/19/09 12:44:02 PM


EXAMPLES

1. State the domain over which each curve is increasing and decreasing .

xx3x2x1

y

Solution

The left-hand side of the parabola is decreasing and the right side is increasing.

So the curve is increasing for x 2 x 2 and the curve is decreasing when x 1 x 2.

2.

xx3x2x1

y

Solution

The left-hand side of the curve is increasing until it reaches the y -axis (where x 0= ). It then turns around and decreases until x 3 and then increases again.

So the curve is increasing for ,x x x0 31 2 and the curve is decreasing for .x x0 31 1

The curve isn’t increasing or decreasing at x 2 . We say that it is stationary at that point. You will

study stationary points and further curve sketching in the HSC Course.

Notice that the curve is stationary at x 0= and .x x3=

ch5.indd 215 7/19/09 12:44:13 PM


Functions are odd if they have point symmetry about the origin. A graph rotated 180° about the origin gives the original graph.

This is an odd function:

x

y

For even functions, f x f x= -] ]g g for all values of x .

For odd functions, f x f x- = -] ]g g for all values of x in the domain.

As well as looking at where the curve is increasing and decreasing, we can see if the curve is symmetrical in some way. You have already seen that the parabola is symmetrical in earlier stages of mathematics and you have learned how to fi nd the axis of symmetry. Other types of graphs can also be symmetrical.

Functions are even if they are symmetrical about the y -axis. They have line symmetry (refl ection) about the y -axis.

This is an even function:

x

y

ch5.indd 216 7/19/09 12:44:23 PM


EXAMPLES

1. Show that f x x 32= +] g is an even function.

Solution

f x x

xf x

f x x

3

3

3 is an even function

2

2

2`

- = - +

= +

=

= +

] ]

]

]

g g

g

g

2. Show that f x x x3= -] g is an odd function.

Solution

f x x x

x x

x xf x

f x x x is an odd function

3

3

3

3`

- = - - -

= - +

= - -

= -

= -

] ] ]

^

]

]

g g g

h

g

g

Investigation

Explore the family of graphs of f x xn=] g .

For what values of n is the function even?

For what values of n is the function odd?

Which families of functions are still even or odd given k ? Let k take on different values, both positive and negative.

1. f x kxn=] g

2. f x x kn= +] g

3. f x x k n= +] ]g g

k is called a parameter. Some graphics calculators

and computer programs use parameters to show how

changing values of k change the shape of graphs .

1. Find the x - and y -intercept of each function.

(a) y x3 2= - (b) x y2 5 20 0- + = (c) x y3 12 0+ - =

(d) f x x x32= +] g (e) f x x 42= -] g (f) p x x x5 62= + +] g (g) y x x8 152= - + (h) p x x 53= +] g

5.3 Exercises

ch5.indd 217 7/19/09 12:44:32 PM


(i) y xx x3 0!=+

] g

(j) g x x9 2= -] g

2. Show that f x f x= -] ]g g where f x x 22= -] g . What type of function is it?

3. If f x x 13= +] g , fi nd (a) f x2^ h (b) ( )f x 26 @ (c) f x-] g Is it an even or odd function? (d)

4. Show that g x x x x3 28 4 2= + -] g is an even function .

5. Show that f ( x ) is odd, where .f x x=] g

6. Show that f x x 12= -] g is an even function.

7. Show that f x x x4 3= -] g is an odd function.

8. Prove that f x x x4 2= +] g is an even function and hence fi nd .f x f x- -] ]g g

9. Are these functions even, odd or neither?

(a) yx x

x4 2

3

=-

(b) yx 1

13

=-

(c) f xx 4

32

=-

] g

(d) yxx

33

=+

-

(e) f xx x

x5 2

3

=-

] g

10. If n is a positive integer, for what values of n is the function f x xn=] g

even? (a) odd? (b)

11. Can the function f x x xn= +] g ever be

even? (a) odd? (b)

12. For the functions below, state (i) the domain over which the graph is increasing (ii) the domain over which the graph is decreasing (iii) whether the graph is odd, even or neither.

x

y(a)

x4

y(b)

2-2x

y(c)

ch5.indd 218 7/19/09 12:44:41 PM


Investigation

Use a graphics calculator or a computer with graphing software to sketch graphs and explore what effect different constants have on each type of graph.

If your calculator or computer does not have the ability to use parameters (this may be called dynamic graphing), simply draw different graphs by choosing several values for k . Make sure you include positive and negative numbers and fractions for k .

Alternatively, you may sketch these by hand.

Sketch the families of graphs for these graphs with parameter 1. k.

y kx

y kx

y kx

y kx

y xk

(a)

(b)

(c)

(d)

(e)

2

3

4

=

=

=

=

=

What effect does the parameter k have on these graphs? Could you give a general comment about y k f x= ] g ?

Sketch the families of graphs for these graphs with parameter 2. k. y x k

y x k

y x k

y x k

y x k1

(a)

(b)

(c)

(d)

(e)

2

2

3

4

= +

= +

= +

= +

= +

] g

What effect does the parameter k have on these graphs? Could you give a general comment about y f x k= +] g ?

-2

1 2

-4

-1-2

2

4

y

x

(d) y

x

(e)

CONTINUED

ch5.indd 219 7/19/09 12:44:51 PM


Gradient form: y mx b= + has gradient m and y -intercept b General form: ax by c 0+ + =

Investigation

Are straight line graphs always functions? Can you fi nd an example of a straight line that is not a function?

Are there any odd or even straight lines? What are their equations?

For the family of functions y k f x= ] g , as k varies, the function changes its slope or steepness. For the family of functions ,y f x k= +] g as k varies, the graph moves up or down (vertical translation). For the family of functions y f x k= +] g , as k varies, the graph moves left or right (horizontal translation).

Sketch the families of graphs for these graphs with parameter 3. k.

y x k

y x k

y x k

y x k

yx k

1

(a)

(b)

(c)

(d)

(e)

2

3

4

= +

= +

= +

= +

=+

]

]

]

g

g

g

What effect does the parameter k have on these graphs? Could you give a general comment about y f x k= +] g ?

When 0 ,k2 the graph moves to the left and when

0 ,k1 the graph moves to the right.

Notice that the shape of most graphs is generally the same regardless of the parameter k . For example, the parabola still has the same shape even though it may be narrower or wider or upside down.

This means that if you know the shape of a graph by looking at its equation, you can sketch it easily by using some of the graphing techniques in this chapter rather than a time-consuming table of values. It also helps you to understand graphs more and makes it easier to fi nd the domain and range.

You have already sketched some of these graphs in previous years.

Linear Function

A linear function is a function whose graph is a straight line.

ch5.indd 220 7/19/09 12:45:00 PM


EXAMPLE

Sketch the function f x x3 5= -] g and state its domain and range.

Solution

This is a linear function. It could be written as .y x3 5= - Find the intercepts For x -intercept: y 0=

0 3 5

5 3

1

x

x

x32

=

=

=

-


3 5

5y 0=

= -

-] g

-1

-2

y

5

4

3

2

1 1 23

6

-3

-4

-5

1 4-1-2 32-3-4x

Notice that the line extends over the whole of the number plane, so that it covers all real numbers for both the domain and range. Domain: {all real x } Range: {all real y }

Notice too, that you can substitute any real number

into the equation of the function for x, and any real

number is possible for y.

The linear function ax by c 0+ + = has domain {all real x } and range {all real y } where a and b are non-zero

Special lines

Horizontal and vertical lines have special equations.

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a linear function, or choose different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k 1. y kx= 2. y x k= + 3. y mx b= + where m and b are both parameters

What effect do the parameters m and b have on these graphs?

ch5.indd 221 7/19/09 12:45:09 PM


EXAMPLES

1. Sketch y 2= on a number plane. What is its domain and range?

Solution

x can be any value and y is always 2. Some of the points on the line will be (0, 2), (1, 2) and (2, 2). This gives a horizontal line with y -intercept 2.

-1

-3

y

4

3

2

1

5

-2

-4

-5

1 4-1-2x

32-3-4

Domain: xall real" , Range: : 2y y =" ,

2. Sketch x 1= - on a number plane and state its domain and range.

Solution

y can be any value and x is always .1- Some of the points on the line will be , , ,1 0 1 1- -^ ^h h and , .1 2-^ h This gives a vertical line with x -intercept .1-

Domain: : 1x x = -" , Range: yall real" ,

-1

-3

4

3

2

1

5

-2

-4

-5

1 4--1-2 3-3-4

y

x

ch5.indd 222 7/19/09 12:45:18 PM


x a= is a vertical line with x -intercept a Domain: :x x a=! + Range: {all real y }

y b= is a horizontal line with y -intercept b Domain: {all real x } Range: :y y b=" ,

5.4 Exercises

1. Find the x - and y -intercepts of each function.

(a) y x 2= - (b) f x x2 3= +] g (c) x y2 1 0+ =- (d) x y 3 0+ =- (e) x y3 6 2 0=- -

2. Draw the graph of each straight line.

(a) x 4= (b) x 3 0=- (c) y 5= (d) y 1 0+ = (e) f x x2 1= -] g (f) y x 4= + (g) f x x3 2= +] g (h) x y 3+ = (i) x y 1 0=- - (j) x y2 3 0+ =-

3. Find the domain and range of (a) x y3 2 7 0+ =- (b) y 2= (c) x 4= - (d) x 2 0=- (e) y3 0=-

4. Which of these linear functions are even or odd?

(a) y x2= (b) y 3= (c) x 4= (d) y x= - (e) y x=

5. By sketching x y 4 0=- - and x y2 3 3 0+ =- on the same set of axes, fi nd the point where they meet.

ch5.indd 223 7/31/09 4:30:33 PM


Applications

The parabola shape is used in many different applications as it has special properties that are very useful. For example if a light is placed inside the parabola at a special place (called the focus), then all light rays coming from this light and bouncing off the parabola shape will radiate out parallel to each other, giving a strong light. This is how car headlights work. Satellite dishes also use this property of the parabola, as sound coming in to the dish will bounce back to the focus.

The pronumeral a is called the coeffi cient of .x2

Quadratic Function

The quadratic function gives the graph of a parabola.

f x ax bx c2= + +] g is the general equation of a parabola. If a 02 the parabola is concave upwards

If a 01 the parabola is concave downwards

ch5.indd 224 7/19/09 12:45:36 PM


The lens in a camera and glasses are also parabola shaped. Some bridges look like they are shaped like a parabola, but they are often based on the catenary. Research the parabola and catenary on the Internet for further information.

Investigation

Is the parabola always a function? Can you fi nd an example of a parabola that is not a function?

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a quadratic function, or choose different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k . 1. y kx2= 2. y x k2= + 3. y x k 2= +] g 4. y x kx2= +

What effect does the parameter k have on these graphs?

Which of these families are even functions? Are there any odd quadratic functions?

ch5.indd 225 7/31/09 4:30:33 PM


EXAMPLES

1. (a) Sketch the graph of ,y x 12= - showing intercepts. (b) State the domain and range.

Solution

This is the graph of a parabola. Since (a) ,a 02 it is concave upward

For x -intercept: y 0=

x

xx

0 1

11

2

2

!

= -

=

=


0 1

1y 2= -

= -

From the graph, the curve is moving outwards and will extend(b) to all real x values. The minimum y value is .1-

Domain: xall real" , Range: :y y 1$ -" ,

2. Sketch .f x x 1 2= +] ]g g

Solution

This is a quadratic function. We fi nd the intercepts to see where the parabola will lie. Alternatively, you may know from your work on parameters that f x x a 2= +] ]g g will move the function f x x2=] g horizontally a units to the left. So f x x 1 2= +] ]g g moves the parabola f x x2=] g 1 unit to the left. For x -intercept: y 0=

01 0

1

xx

x

1 2= +

+ =

= -

] g


1

y 0 1 2= +

=

] g

-1

-3

4

3

2

1

5

-2

-4

-5

-6

1 4-1-2 53-3-4

y

x

ch5.indd 226 7/19/09 1:37:58 PM


3. For the quadratic function f x x x 62= + -] g Find the (a) x - and y -intercepts Find the minimum value of the function (b) State the domain and range (c) For what values of (d) x is the curve decreasing?

Solution

For (a) x -intercept: y 0= This means f x 0=] g

,

,

x xx x

x x

x x

0 63 2

3 0 2 0

3 2

2= + -

= + -

+ = - =

= - =

] ]g g


f 0 0 0 66

2= + -

= -

] ] ]g g g

Since (b) ,a 02 the quadratic function has a minimum value. Since the parabola is symmetrical, this will lie halfway between the x -intercepts.

Halfway between 3x = - and 2:x =

2

3 221- +

= -

Minimum value is f21

-c m

f21

21

21 6

41

21 6

641

2

- = - + - -

= - -

= -

c c cm m m

So the minimum value is .641

-

CONTINUED

You will learn more about this in Chapter 9.

-1

-3

4

3

2

1

5

-2

-4

-5

1 4-1-2 32-3-4

y

x

ch5.indd 227 7/31/09 4:30:36 PM


Sketching the quadratic function gives a concave upward parabola. (c)

From the graph, notice that the parabola is gradually going outwards and will include all real x values. Since the minimum value is 6

41

- , all y values are greater than this. Domain: xall real" ,

Range: : 6yy41

$ -' 1

The curve decreases down to the minimum point and then (d)

increases. So the curve is decreasing for all .x21

1-

4. (a) Find the x - and y -intercepts and the maximum value of the quadratic function .f x x x4 52= - + +] g (b) Sketch the function and state the domain and range. (c) For what values of x is the curve increasing?

Solution

For (a) x -intercept: 0y = So f x 0=] g

0 4 5

4 5 00

x x

x xx x5 1

2

2

= - + +

=

+ =

- -

-] ]g g

,

,

x x

x x

5 0 1 0

5 1

- = + =

= = -

For y -intercept: 0x =

f 0 0 4 0 55

2= - + +

=

] ] ]g g g

-1

-3

4

3

2

1

5

-2

-4

-5

-6

y

1 4-1-2 53-3-4x

- 6-12

14

,

ch5.indd 228 7/31/09 4:30:37 PM


Since ,a 01 the quadratic function is concave downwards and has a maximum value halfway between the x -intercepts 1x = - and .x 5=

21 5 2- +

=

f 2 2 4 2 5

9= - + +

=

2] ] ]g g g

So the maximum value is 9. Sketching the quadratic function gives a concave downward parabola. (b)

From the graph, the function can take on all real numbers for x , but the maximum value for y is 9. Domain: xall real" , Range: : 9y y #" ,

From the graph, the function is increasing on the left of the(c) maximum point and decreasing on the right. So the function is increasing when .x 21


(a) 2y x x2= + (b) 3y x x2= - + (c) f x x 12= -] g (d) y x x 22= - - (e) y x x9 82= +-

2. Sketch (a) 2y x2= + (b) y x 12= - + (c) f x x 42= -] g (d) 2y x x2= + (e) y x x2= - - (f) f x x 3= - 2] ]g g

5.5 Exercises

-1

987

5432

6

1

-2-3-4-5

y

2 51 643-1-2-3-4x

ch5.indd 229 7/19/09 12:46:30 PM


EXAMPLES

1. Sketch f x x 1= -] g and state its domain and range.

Solution

Method 1: Table of values When sketching any new graph for the fi rst time, you can use a table of values. A good selection of values is x3 3# #- but if these don’t give enough information, you can fi nd other values.

Absolute Value Function

You may not have seen the graphs of absolute functions before. If you are not sure about what they look like, you can use a table of values or look at the defi nition of absolute value.

(g) f x x 1 2= +] ]g g (h) y x x3 42= + - (i) y x x2 5 32= - +

(j) f x x x3 22= - + -] g

3. For each parabola, fi nd the (i) x - and y -intercepts the domain and range (ii)

(a) –y x x7 122= + (b) f x x x42= +] g (c) y x x2 82= - - (d) y x x6 92= +- (e) f t t4 2= -] g

4. Find the domain and range of (a) y x 52= - (b) f x x x62= -] g (c) f x x x 22= - -] g (d) y x2= - (e) f x x 7 2= -] ]g g

5. Find the range of each function over the given domain.

(a) y x2= for x0 3# # (b) y x 42= - + for x1 2# #- (c) f x x 12= -] g for x2 5# #- (d) y x x2 32= + - for x2 4# #- (e) y x x 22= - +- for x0 4# #

6. Find the domain over which each function is

increasing (i) decreasing (ii)

(a) y x2= (b) y x2= - (c) f x x 92= -] g (d) y x x42= - + (e) f x x 5 2= +] ]g g

7. Show that f x x2= -] g is an even function.

8. State whether these functions are even or odd or neither.

(a) y x 12= + (b) f x x 32= -] g (c) y x2 2= - (d) f x x x32= -] g (e) f x x x2= +] g (f) y x 42= - (g) y x x2 32= - - (h) y x x5 42= +- (i) p x x 1 2= +] ]g g (j) y x 2= - 2] g

ch5.indd 230 7/19/09 12:46:40 PM


CONTINUED

e.g. When :x 3= -

| |y 3 13 12

= - -

= -=

x -3 -2 -1 0 1 2 3

y 2 1 0 -1 0 1 2

This gives a v-shaped graph.

y

-2

4

3

2

1

5

-1

-3

-4

-5

1 4-1-2 32-3-4x

Method 2: Use the defi nition of absolute value

| |y x x xx x1 1 0

1 0whenwhen 1

$= - =

-- -&

This gives 2 straight line graphs: y x x1 0$= - ] g

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4x

y = x - 1

ch5.indd 231 7/19/09 12:46:49 PM


y x 1= - - x 01] g

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4x

y = - x - 1

Draw these on the same number plane and then disregard the dotted lines to get the graph shown in method 1.

-3

4

3

2

1

5

-2

-1

-4

-5

yy

3-1-2 421-3-4x

y = -x - 1y = x - 1

Method 3: If you know the shape of the absolute value functions, fi nd the intercepts. For x -intercept: 0y = So f x 0=] g

| |

| |

x

x

x

0 1

1

1` !

= -

=

=


( ) | |f 0 0 1

1= -

= -

ch5.indd 232 7/19/09 12:46:58 PM


The graph is V -shaped, passing through these intercepts.

-3

4

3

2

1

5

-2

-1

-4

-5

y

4-2 5321-1-3-4x

From the graph, notice that x values can be any real number while the minimum value of y is .1- Domain: {all real x } Range: { y : y 1$ - }

2. Sketch .| |y x 2= +

Solution

Method 1: Use the defi nition of absolute value.

| | ( )y x x xx x2 2 2 0

2 2 0whenwhen 1

$= + =

+ +- + +'

This gives 2 straight lines: 2y x= + when x 2 0$+

x 2$ -

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4x

y = x + 2

If you already know how to sketch the graph of y | x |= , translate the

graph of y | x | 1= - down 1 unit, giving it a

y -intercept of .1-

CONTINUED

ch5.indd 233 7/19/09 12:47:09 PM


2y x= - +] g when x 2 01+ i.e. y x 2= - - when x 21 -

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4x

y = -x - 2

Draw these on the same number plane and then disregard the dotted lines.

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4x

y = -x - 2

y = x + 2

Method 2: Find intercepts For x -intercept: 0y = So 0f x =] g

0 | 2 |

0 2

2

x

x

x

= +

= +

- =


(0) | 0 2 |

2

f = +

=

There is only one solution for the equation | x 2 | 0.+ = Can you see why?

ch5.indd 234 7/19/09 12:47:23 PM


The graph is V -shaped, passing through these intercepts.

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4x

If you know how to sketch the graph of

y | x |= , translate it 2 places to the left for the graph of .y | x 2 |= +

Investigation

Are graphs that involve absolute value always functions? Can you fi nd an example of one that is not a function?

Can you fi nd any odd or even functions involving absolute values? What are their equations?

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on an absolute value function, or choose different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k

1. | |f x k x=] g 2. | |f x x k= +] g 3. | |f x x k= +] g


The equations and inequations involving absolute values that you studied in Chapter 3 can be solved graphically.

ch5.indd 235 7/19/09 12:47:33 PM


EXAMPLES

Solve 1. |2 1 | 3x - =

Solution

Sketch | 2 1 |y x= - and 3y = on the same number plane.

The solution of |2 1 | 3x - = occurs at the intersection of the graphs, that is, , .x 1 2= -

2. |2 1 | 3 2x x= -+

Solution

Sketch | 2 1 |y x= + and 3 2y x= - on the same number plane.

The solution is 3.x =

3. | 1 | 2x 1+

Solution

Sketch | 1 |y x= + and 2y = on the same number plane.

The graph shows that there is only one solution. Algebraically, you need to fi nd the 2 possible solutions and then check them.

ch5.indd 236 7/19/09 12:47:44 PM


The solution of | 1 | 2x 1+ is where the graph | 1 |y x= + is below the graph 2,y = that is, .x3 11 1-


(a) | |y x= (b) | |f x x 7= +] g (c) | |f x x 2= -] g (d) 5 | |y x= (e) | |f x x 3= - +] g (f) | 6 |y x= + (g) | |f x x3 2= -] g (h) | 5 4 |y x= + (i) | 7 1 |y x= - (j) | |f x x2 9= +] g

2. Sketch each graph on a number plane.

(a) | |y x= (b) | |f x x 1= +] g (c) | |f x x 3= -] g (d) 2 | |y x= (e) | |f x x= -] g (f) | 1 |y x= + (g) | |f x x 1= - -] g (h) | 2 3 |y x= - (i) | 4 2 |y x= + (j) | |f x x3 1= +] g

3. Find the domain and range of each function.

(a) | 1 |y x= - (b) | |f x x 8= -] g

(c) | |f x x2 5= +] g (d) 2 | | 3y x= - (e) | |f x x 3= - -] g

4. Find the domain over which each function is

increasing (i) decreasing (ii)

(a) | 2 |y x= - (b) | |f x x 2= +] g (c) | |f x x2 3= -] g (d) 4 | | 1y x= - (e) | |f x x= -] g

5. For each domain, fi nd the range of each function.

(a) | |y x= for x2 2# #- (b) | |f x x 4= - -] g for

x4 3# #- (c) | |f x x 4= +] g for x7 2# #- (d) | 2 5 |y x= - for x3 3# #- (e) | |f x x= -] g for x1 1# #-

6. For what values of x is each function increasing?

(a) | 3 |y x= + (b) | |f x x 4= - +] g (c) | |f x x 9= -] g (d) | |y x 2 1= - - (e) | |f x x 2= - +] g

5.6 Exercises

ch5.indd 237 7/19/09 12:47:53 PM


7. Solve graphically (a) | | 3x = (b) | |x 12 (c) | |x 2# (d) | 2 | 1x + = (e) | 3 | 0x - = (f) |2 3 | 1x - = (g) | |x 1 41- (h) | |x 1 3#+ (i) | |x 2 22- (j) | |x 3 1$-

(k) | |x2 3 5#+ (l) | |x2 1 1$- (m) |3 1 | 3x x- = + (n) |3 2 | 4x x- = - (o) | 1 | 1x x- = + (p) | 3 | 2 2x x+ = + (q) |2 1 | 1x x+ = - (r) |2 5 | 3x x- = - (s) | 1 | 2x x- = (t) |2 3 | 3x x- = +

The Hyperbola

A hyperbola is a function with its equation in the form .xy a y xaor= =

EXAMPLE

Sketch 1 .y x=

Solution

1y x= is a discontinuous curve since the function is undefi ned at .x 0=

Drawing up a table of values gives:

x -3 -2 -121

-41

- 041

21 1 2 3

y31

-21

- -1 -2 -4 — 4 2 121

31

Class Discussion

What happens to the graph as x becomes closer to 0? What happens as x becomes very large in both positive and negative directions? The value of y is never 0. Why?

ch5.indd 238 7/19/09 12:48:02 PM


To sketch the graph of a more general hyperbola, we can use the domain and range to help fi nd the asymptotes (lines towards which the curve approaches but never touches).

The hyperbola is an example of a discontinuous graph, since it has a gap in it and is in two separate parts.

Investigation

Is the hyperbola always a function? Can you fi nd an example of a hyperbola that is not a function?

Are there any families of odd or even hyperbolas? What are their equations?

Use a graphics calculator or a computer with dynamic graphing capability to explore the effect of a parameter on a hyperbola, or choose different values of k (both positive and negative).

Sketch the families of graphs for these graphs with parameter k

1. y xk

=

2. 1y x k= +

3. 1yx k

=+


EXAMPLES

1. (a) Find the domain and range of .f xx 3

3=

-] g

Hence sketch the graph of the function. (b)

Solution

This is the equation of a hyperbola. To fi nd the domain, we notice that .x 3 0!- So x 3! Also y cannot be zero (see example on page 238). Domain: {all real x : x 3! } Range: {all real y : y 0! } The lines 3x = and 0y = (the x -axis) are called asymptotes.

The denominator cannot be zero.

CONTINUED

ch5.indd 239 7/19/09 8:17:45 PM


To make the graph more accurate we can fi nd another point or two. The easiest one to fi nd is the y -intercept. For y -intercept, 0x =

1

y0 3

3=

-

= -

-3

4

3

2

1

5

-2

-1

-4

-5

y

-1-2 4 521-3-4x

x = 3

y = 0

Asymptotes

3

2. Sketch .yx2 4

1= -

+

Solution

This is the equation of a hyperbola. The negative sign turns the hyperbola around so that it will be in the opposite quadrants. If you are not sure where it will be, you can fi nd two or three points on the curve. To fi nd the domain, we notice that .x2 4 0!+

x

x

2 4

2

!

!

-

-

For the range, y can never be zero. Domain: {all real x : x 2! - } Range: {all real y : y 0! } So there are asymptotes at x 2= - and y 0= (the x -axis). To make the graph more accurate we can fi nd the y -intercept. For y -intercept, x 0=

( )y

2 0 41

41

= -+

= -

Notice that this graph is

a translation of 3

yx

=

three units to the right.

ch5.indd 240 7/19/09 12:48:21 PM


y

-2x

- 14

The function f xbx c

a=

+] g is a hyperbola with

domain :x xbcall real ! -& 0 and

range {all real :y y 0! }

1. For each graph State the domain and range. (i) Find the (ii) y -intercept if it

exists. Sketch the graph. (iii)

(a) 2y x=

(b) 1y x= -

(c) f xx 1

1=

+] g

(d) f xx 2

3=

-] g

(e) 3 6

1yx

=+

(f) f xx 3

2= -

-] g

(g) f xx 1

4=

-] g

(h) 1

2yx

= -+

(i) f xx6 3

2=

-] g

(j) 2

6yx

= -+

2. Show that f x x2

=] g is an odd function.

3. Find the range of each function over the given domain.

(a) f xx2 5

1=

+] g for x2 2# #-

(b) 3

1yx

=+

for x2 0# #-

(c) f xx2 4

5=

-] g for x3 1# #-

5.7 Exercises

ch5.indd 241 7/19/09 1:26:13 PM


(d) f xx 4

3= -

-] g for x3 3# #-

(e) 3 1

2yx

= -+

for x0 5# #

4. Find the domain of each function over the given range.

(a) 3y x= for y1 3# #

(b) 2y x= - for y221

# #- -

(c) f xx 1

1=

-] g for y1

71

# #- -

(d) f xx2 1

3= -

+] g for

y131

# #- -

(e) 3 2

6yx

=-

for y121 6# #

Circles and Semi-circles

The circle is used in many applications, including building and design.

Circle gate

A graph whose equation is in the form 0x ax y by c2 2+ + + + = has the shape of a circle.

There is a special case of this formula:

The graph of x y r2 2 2+ = is a circle, centre 0, 0^ h and radius r

Proof

r y

x

(x, y)

y

x

ch5.indd 242 7/19/09 1:26:26 PM


Given the circle with centre (0, 0) and radius r : Let ( x , y ) be a general point on the circle, with distances from the origin x

on the x -axis and y on the y -axis as shown. By Pythagoras’ theorem:

c a b

r x y

2 2 2

2 2 2`

= +

= +

EXAMPLE

Sketch the graph of (a) 4.x y2 2+ = Is it a function? State its domain and range. (b)

Solution

This is a circle with radius 2 and centre (0, 0). (a)

y

x

-2

-2 2

2

The circle is not a function since a vertical line will cut it in more than one place.

y

x2

2

-2

-2

The radius is .4

CONTINUED

ch5.indd 243 7/19/09 12:48:53 PM


Notice that the (b) x -values for this graph lie between 2- and 2 and the y -values also lie between 2- and 2. Domain: : 2 2{ }x x# #- Range: : 2 2{ }y y# #-

The circle x y r2 2 2+ = has domain: :x r x r# #-! + and range: :y r y r# #-" ,

The equation of a circle, centre ( a , b ) and radius r is – –x a y b r2 2 2+ =] ^g h

We can use Pythagoras’ theorem to fi nd the equation of a more general circle.

Proof

Take a general point on the circle, ( x , y ) and draw a right-angled triangle as shown.

y

x(a, b)

x

y

r

(x, y)

a

b x - a

y - b

Notice that the small sides of the triangle are –x a and –y b and the hypotenuse is r , the radius.


– –

c a b

r x a y b

2 2 2

2 2 2

= +

= +] ^g h

ch5.indd 244 7/19/09 12:49:01 PM


EXAMPLES

1. (a) Sketch the graph of .x y 812 2+ = (b) State its domain and range.

Solution

The equation is in the form (a) .x y r2 2 2+ = This is a circle, centre (0, 0) and radius 9.

y

x9

9

-9

-9

From the graph, we can see all the values that are possible for (b) x and y for the circle. Domain: : 9 9{ }x x# #- Range: : 9 9{ }y y# #-

2. (a) Sketch the circle – .x y1 2 42 2+ + =] ^g h (b) State its domain and range.

Solution

The equation is in the form (a) – – .x a y b r2 2 2+ =] ^g h

–

– –

x y

x y

1 2 4

1 2 2

2

2 2

+ + =

+ - =

2

2

] ^

] ]_

g h

g gi

So 1, 2a b= = - and 2r =

CONTINUED

ch5.indd 245 7/19/09 12:49:12 PM


This is a circle with centre ,1 2-^ h and radius 2. To draw the circle, plot the centre point ,1 2-^ h and count 2 units up, down, left and right to fi nd points on the circle.

y

x

1

1

-2 2 3 4-1

-3

-4

-5

2

3

4

5

-1

-2

-3-4

(1, -2)

From the graph, we can see all the values that are possible for (b) x and y for the circle.

Domain: : 1 3{ }x x# #- Range: : 4 0{ }y y# #-

3. Find the equation of a circle with radius 3 and centre ,2 1-^ h in expanded form.

Solution

This is a general circle with equation – –x a y b r2 2 2+ =] ^g h where ,a b2 1= - = and .r 3= Substituting:

– –

––

x a y b r

x yx y

2 1 32 1 9

2 2 2

2 2 2

2 2

+ =

- - + =

+ + =

] ^

]] ^

] ^

g h

gg h

g h

Remove the grouping symbols.

–

–

a b a ab b

x x x

x x

a b a ab b

y y y

y y

2

2 2 2 2

4 4

2

1 2 1 1

2 1

So

So

2 2 2

2 2 2

2

2 2 2

2 2 2

2

+ = + +

+ = + +

= + +

= - +

= - +

= - +

]

] ] ]

]

^ ^ ]

g

g g g

g

h h g

The equation of the circle is:

–

x x y y

x x y y

x x y y

x x y y

4 4 2 1 9

4 2 5 9

4 2 5 9

4 2 4 0

9 9

2

2

2

2

+ + + - + =

+ + - + =

+ + + =

+ + - - =

- -

You may need to revise this in Chapter 2.

ch5.indd 246 7/19/09 12:49:22 PM


Investigation

The circle is not a function. Could you break the circle up into two functions?

Change the subject of this equation to y .

What do you notice when you change the subject to y ? Do you get two functions? What are their domains and ranges?

If you have a graphics calculator, how could you draw the graph of a circle?

The equation of the semi-circle above the x -axis with centre (0, 0) and radius r is y r x2 2= -

The equation of the semi-circle below the x -axis with centre (0, 0) and radius r is y r x2 2= - -

y r x2 2= - is the semi-circle above the x -axis since its range is y $ 0 for all values.

y

xr

r

-r

The domain is { :x r x r# #- } and the range is { :y y r0# # }

Proof

–

x y r

y r x

y r x

2 2 2

2 2 2

2 2!

+ =

=

= -

This gives two functions:

By rearranging the equation of a circle, we can also fi nd the equations of semi-circles.

ch5.indd 247 7/19/09 12:49:32 PM


y r x2 2= - - is the semi-circle above the x -axis since its range is y 0# for all values.

y

xr

-r

-r

The domain is { :x r x r# #- } and the range is { : }y r y 0# #-

EXAMPLES

Sketch each function and state the domain and range. 1. f x x9 2= -] g

Solution

This is in the form f x r x2 2= -] g where .r 3= It is a semi-circle above the x -axis with centre (0, 0) and radius 3.

y

x3

3

-3

Domain: : 3 3{ }x x# #- Range: : 0 3{ }y y# #

ch5.indd 248 7/19/09 12:49:41 PM


2. y x4 2= - -

Solution

This is in the form y r x2 2= - - where .r 2= It is a semi-circle below the x -axis with centre (0, 0) and radius 2.

y

x2

-2

-2

Domain: : 2 2{ }x x# #- Range: : 2 0{ }y y# #-

1. For each of the following sketch each graph (i) state the domain and (ii)

range. (a) 9x y2 2+ = (b) x y 16 02 2+ =- (c) – –x y2 1 42 2+ =] ^g h (d) 1 9x y2 2+ + =] g (e) –x y2 1 12 2+ + =] ^g h

2. For each semi-circle state whether it is above or (i)

below the x -axis sketch the function (ii) state the domain and (iii)

range.

(a) 25y x2= - - (b) 1y x2= - (c) 36y x2= - (d) 64y x2= - - (e) 7y x2= - -

3. Find the length of the radius and the coordinates of the centre of each circle.

(a) 100x y2 2+ =

(b) 5x y2 2+ =

(c) – –x y4 5 162 2+ =] ^g h

(d) –x y5 6 492 2+ + =] ^g h

(e) –x y 3 812 2+ =^ h

5.8 Exercises

ch5.indd 249 7/19/09 1:26:38 PM


4. Find the equation of each circle in expanded form (without grouping symbols).

Centre (0, 0) and radius 4 (a) Centre (3, 2) and radius 5 (b) Centre (c) ,1 5-^ h and radius 3 Centre (2, 3) and radius 6 (d)

Centre (e) ,4 2-^ h and radius 5 Centre (f) ,0 2-^ h and radius 1 Centre (4, 2) and radius 7 (g) Centre (h) ,3 4- -^ h and radius 9 Centre (i) ,2 0-^ h and radius 5 Centre (j) ,4 7- -^ h and radius 3

Other Graphs

There are many other different types of graphs. We will look at some of these graphs and explore their domain and range.

Exponential and logarithmic functions

EXAMPLES

1. Sketch the graph of f x 3x=] g and state its domain and range.

Solution

If you do not know what this graph looks like, draw up a table of values. You may need to revise the indices that you studied in Chapter 1. e.g. When 0:x =

y 3

1

c=

=

:x

y

1

3

31

31

When1

1

= -

=

=

=

-

x 3- 2- 1- 0 1 2 3

y271

91

31

1 3 9 27

If you already know what the shape of the graph is, you can draw it just using 2 or 3 points to make it more accurate.

You will meet these graphs again in the HSC Course.

ch5.indd 250 7/19/09 1:26:50 PM


This is an exponential function with y -intercept 1. We can fi nd one other point.

When

x

y

1

33

1

=

=

=

y

x

1

2

1

3

From the graph, x can be any real value (the equation shows this as well since any x value substituted into the equation will give a value for y ). From the graph, y is always positive, which can be confi rmed by substituting different values of x into the equation. Domain: xall real" , Range: :y y 02" ,

2. Sketch logf x x=] g and state the domain and range.

Solution

Use the LOG key on your calculator to complete the table of values. Notice that you can’t fi nd the log of 0 or a negative number.

x −2 −1 0 0.5 1 2 3 4

y # # # −0.3 0 0.3 0.5 0.6

y

x

1

2

1 2 3 4-1

From the graph and by trying different values on the calculator, y can be any real number while x is always positive. Domain: :x x 02! + Range: yall real" ,

You learned about exponential graphs in earlier

stages of maths.

ch5.indd 251 7/19/09 12:50:13 PM


The exponential function y ax= has domain {all real x } and range { :y y 02 }

The logarithmic function logy xa= has domain :x x 02! + and range {all real y }

Cubic function

A cubic function has an equation where the highest power of x is .x3

EXAMPLE

1. Sketch the function f x x 23= +] g and state its domain and range.

Solution

Draw up a table of values.

x −3 −2 −1 0 1 2 3

y −25 −6 1 2 3 10 29

y

x1

1

-2 2 3 4-1

-3

-4

-5

2

3

4

5

-1

-2

-3-4

The function can have any real x or y value: Domain: xall real" , Range: yall real" ,

If you already know the shape of

, ( )y x f x x 23= = +3 has the same shape as ( )f x x= 3 but it is translated 2 units up (this gives a y-intercept of 2).

ch5.indd 252 7/19/09 12:50:21 PM


Domain and range

Sometimes there is a restricted domain that affects the range of a function.

EXAMPLE

1. Find the range of f x x 23= +] g over the given domain of .x1 4# #-

Solution

The graph of f x x 23= +] g is the cubic function in the previous example. From the graph, the range is {all real y }. However, with a restricted domain of x1 4# #- we need to see where the endpoints of this function are.

f

f

1 1 21 2

1

4 4 264 2

66

3

3

- = - +

= - +

=

= +

= +

=

] ]

] ]

g g

g g

Sketching the graph, we can see that the values of y all lie between these points.

y

x(-1, 1)

(4, 66)

Range: 1 66: yy # #" ,

ch5.indd 253 7/19/09 12:50:31 PM


You may not know what a function looks like on a graph, but you can still fi nd its domain and range by looking at its equation.

When fi nding the domain, we look for values of x that are impossible. For example, with the hyperbola you have already seen that the denominator of a fraction cannot be zero.

For the range, we look for the results when different values of x are substituted into the equation. For example, x 2 will always give zero or a positive number.

EXAMPLE

Find the domain and range of .f x x 4= -] g

Solution

We can only fi nd the square root of a positive number or zero.

– 4 0xx 4

So $$

When you take the square root of a number, the answer is always positive (or zero). So y 0$ Domain: :x x 4$! + Range: :y y 0$" ,

5.9 Exercises

1. Find the domain and range of (a) 4 3y x= + (b) f x 4= -] g

(c) 3x =

(d) –f x x4 12=] g

(e) –p x x 23=] g

(f) f x xx 12 2= - -] g

(g) 64x y2 2+ =

(h) f tt 4

3=

-] g

(i) ( )g 2 5zz

= +

(j) | |f x x=] g

2. Find the domain and range of (a) y x=

(b) 2y x= -

(c) | |f x x2 3= -] g

(d) | | 2y x= -

(e) f x x2 5= - +] g

(f) | |y x5= -

(g) 2y x=

(h) y 5x= -

(i) f x xx 1

=+

] g

(j) 2

4 3yx

x=

-

3. Find the x -intercepts of (a) y x x 5 2= -] g

(b) – –f x x x x1 2 3= +] ] ] ]g g g g

(c) y x x x6 83 2= +-

(d) g x x x164 2= -] g

(e) 49x y2 2+ =

You may like to simplify the function by dividing by x.

ch5.indd 254 7/19/09 12:50:39 PM


4. (a) Solve .x1 02$- (b) Find the domain of

.f x x1 2= -] g

5. Find the domain of (a) 2y x x2= - -

(b) g t t t62= +] g

6. Each of the graphs has a restricted domain. Find the range in each case.

(a) y x2 3= - in the domain x3 3# #-

(b) y x2= in the domain x2 3# #-

(c) f x x3=] g in the domain x2 1# #-

(d) 1y x= in the domain

x1 5# #

(e) | |y x= in the domain 0 4x# #

(f) y x x22= - in the domain x3 3# #-

(g) y x2= - in the domain x1 1# #-

(h) y x 12= - in the domain x2 3# #-

(i) y x x2 32= - - in the domain x4 4# #-

(j) y x x7 62= - + - in the domain 0 7x# #

7. (a) Find the domain for the

function .yx 1

3=

+

Explain why there is no (b) x - intercept for the function.

State the range of the (c) function.

8. Given the function f x xx

=] g

fi nd the domain of the (a) function

fi nd its range. (b)

9. Draw each graph on a number plane

(a) f x x4=] g

(b) y x3= -

(c) y x 34= -

(d) 2p x x3=] g

(e) 1g x x3= +] g

(f) 100x y2 2+ =

(g) 2 1y x= +

10. (a) Find the domain and range of .y x 1= -

(b) Sketch the graph of .y x 1= -

11. Sketch the graph of .y 5x=

12. For each function, state its domain and range (i) the domain over which the (ii)

function is increasing the domain over which the (iii)

function is decreasing. (a) y x2 9= -

(b) f x x 22= -] g

(c) 1y x=

(d) f x x3=] g

(e) f x 3x=] g

13. (a) Solve .x4 02$-

(b) Find the domain and range of (i) 4y x2= - (ii) .y x4 2= - -

ch5.indd 255 7/19/09 12:50:47 PM


DID YOU KNOW?

A lampshade can produce a hyperbola where the light meets the fl at wall. • Can you fi nd any other shapes made by

a light?

Lamp casting its light

Limits and Continuity

Limits

The exponential function and the hyperbola are examples of functions that approach a limit. The curve y ax= approaches the x -axis when x approaches very large negative numbers, but never touches it.

That is, when , .x a 0x" "3-

Putting a 3- into index form gives

aa1

1

03

Z

=

=

33

-

We say that the limit of ax as x approaches 3- is 0. In symbols, we write .lim a 0x =x " 3-

A line that a graph approaches but never touches is called an asymptote.

EXAMPLES

1. Find .lim xx x5

x 0

2 +"

Solution

Substituting 0x = into the function gives 00, which is undefi ned.

Factorising and cancelling help us fi nd the limit.

( )

lim lim

lim

xx x

x

x x

x

5 5

5

5

x x

x

0

2

0 1

1

0

+=

+

= +

=

" "

"

] g

ch5.indd 256 8/1/09 5:23:37 PM


2. Find .limxx

42

2 -

-x 2"

Solution

Substituting 2x = into the function gives 00, which is undefi ned.

lim lim

lim

xx

x x

x

x

42

2 2

2

21

41

2 1

1

-

-=

+ -

-

=+

=

x x2 2" "

x 2"

^ _h i

3. Find .lim

hh x hx h2 72 2+ -

h 0"

Solution

lim lim

limh

h x hx hh

h hx x

hx x

x

2 7 2 7

2 7

7

2 2 2

2

2

+ -=

+ -

= + -

= -h 0"

h h0 0" "

^ h

Continuity

Many functions are continuous. That is, they have a smooth, unbroken curve (or line). However, there are some discontinuous functions that have gaps in their graphs. The hyperbola is an example.

If a curve is discontinuous at a certain point, we can use limits to fi nd the value that the curve approaches at that point.

EXAMPLES

1. Find limxx

112

-

-

x 1" and hence describe the domain and range of the curve

11 .y

xx2

=-

- Sketch the curve.

Solution

Substituting 1x = into 11

xx2

-

- gives 00

CONTINUED

ch5.indd 257 7/19/09 12:51:05 PM


( )

lim lim

limxx

xx x

x11

11 1

1

2

x x

x

1

2

1

1

-

-=

-

+ -

= +

=

"

"

-

] ]g g

11y

xx2

=-

- is discontinuous at 1x = since y is undefi ned at that point.

This leaves a gap in the curve. The limit tells us that y 2" as 1,x" so the gap is at , .1 2^ h Domain: : , 1x x xall real !" ,

Range: : , 2y y yall real !" ,

yxx

x

x x

x

11

1

1 1

1

2

=-

-

=+

= +

-

-^ ^h h

` the graph is y x 1= + where x 1!

2. Find limx

x x2

2x 2

2

+

+ -"-

and hence sketch the curve .yx

x x2

22

=+ -

+

Solution

Substituting x 2= - into x

x x2

22

+

+ - gives 00

lim lim

limx

x xx

x x

x

22

2

1 2

1

3

x x

x

2

2

2

2

+

+ -=

+

- +

=

= -

-

" "

"

- -

-

^

^ ^

^

h

h h

h

2yx

x x x

yx

x

x

x2

2

2

2

1

1

is discontinuous at2

=+

+ -= -

=+

= -

+ -^ ^h h

So the function is y x 1= - where .x 2! - It is discontinuous at , .2 3- -^ h

Remember that .x 1!

ch5.indd 258 7/19/09 12:51:14 PM


1. Find (a) lim x 52 +x 4"

(b) lim t 7-t 3"-

(c) lim x x2 43 + -x 2"

(d) lim xx x32 +

x 0"

(e) limh

h h2

22

-

- -h 2"

(f) limy

y

51253

-

-

y 5"

(g) limx

x x1

2 12

+

+ +x 1"-

(h) limx

x x4

2 82

+

+ -x 4"-

(i) limcc

42

2 -

-c 2"

(j) limx xx 1

2 -

-x 1"

(k) limh

h h h2 73 2+ -h 0"

(l) limh

hx hx h32 2- +

h 0"

(m) limh

hx h x hx h2 3 53 2 2- + -

h 0"

(n) lim x cx c3 3

--

x c"

2. Determine which of these functions are discontinuous and fi nd x values for which they are discontinuous.

(a) 3y x2= -

(b) 1

1yx

=+

(c) f x x 1= -] g

(d) 4

1yx2

=+

(e) 4

1yx2

=-

3. Sketch these functions, showing any points of discontinuity.

(a) 3y xx x2

=+

(b) 33y

xx x2

=+

+

(c) 1

5 4yx

x x2

=+

+ +

5.10 Exercises

ch5.indd 259 7/19/09 12:51:22 PM


Regions

Class Investigation

How many solutions are there for ?y x 2$ + How would you record them all?

Inequalities can be shown as regions in the Cartesian plane. You can shade regions on a number plane that involve either linear or

non-linear graphs. This means that we can have regions bounded by a circle or a parabola, or any of the other graphs you have drawn in this chapter.

Regions can be bounded or unbounded. A bounded region means that the line or curve is included in the region.

EXAMPLE

Sketch the region x 3# .

Solution

x 3# includes both 3x = and x 31 in the region. Sketch 3x = as an unbroken or fi lled in line, as it will be included in the region. Shade in all points where x 31 as shown.

y

x=3

x1

1

-2 2 3 4-1

-3

-4

-5

2

3

4

5

-1

-2

-3-4

Remember that x 3= is a vertical line with x -intercept 3.

ch5.indd 260 7/19/09 12:51:32 PM


An unbounded region means that the line or curve is not included in the region.

EXAMPLE

Sketch the region y 12 - .

Solution

y 12 - doesn’t include .y 1= - When this happens, it is an unbounded region and we draw the line y 1= - as a broken line to show it is not included. Sketch y 1= - as a broken line and shade in all points where y 12 - as shown.

y

x1

1

-2 2 3 4

-3

-4

-5

2

3

4

5

-1

-2

-3-4-1y = -1

Remember that y 1= - is a horizontal line with

y -intercept .1-

For lines that are not horizontal or vertical, or for curves, we need to check a point to see if it lies in the region.

ch5.indd 261 7/19/09 12:51:40 PM


EXAMPLES

Find the region defi ned by 1. y x 2$ +

Solution

First sketch 2y x= + as an unbroken line. On one side of the line, y x 22 + and on the other side, .y x 21 + To fi nd which side gives ,y x 22 + test a point on one side of the

line (not on the line). For example, choose 0, 0^ h and substitute into

y x 2

0 0 2

0 2 (false)

$

$

$

+

+

This means that 0, 0^ h does not lie in the region .y x 2$ + The region is on the other side of the line.

2. x y2 3 61-

Solution

First sketch 2 3 6x y- = as a broken line, as it is not included in the region.

To fi nd which side of the line gives ,x y2 3 61- test a point on one side of the line.

For example, choose 0, 1^ h and substitute into

( )

x y2 3 6

2 0 3 1 63 6 (true)

1

1

1

-

-

-

] g

Any point in the region will make the inequality true. Test one to see this.

ch5.indd 262 7/19/09 12:51:56 PM


This means that 0, 1^ h lies in the region .x y2 3 61-

3. x y 12 22+

Solution

The equation 1x y2 2+ = is a circle, radius 1 and centre , .0 0^ h Draw 1x y2 2+ = as a broken line, since the region does not include

the curve. Choose a point inside the circle, say 0, 0^ h

x y 1

0 0 10 1 (false)

2 2

2 2

2

22

+

+

So the region lies outside the circle.

4. y x2$

Solution

The equation y x2= is a parabola. Sketch this as an unbroken line, as it is included in the region.

2x22 -3y =6

CONTINUED

ch5.indd 263 8/1/09 8:07:57 PM


Choose a point inside the parabola, say 1, 3^ h .

3 1

y x

3 1(true)

2

222

$

So 1, 3^ h lies in the region.

Sometimes a region includes two or more inequalities . When this happens, sketch each region on the number plane, and the fi nal region is where they overlap (intersect).

EXAMPLE

Sketch the region ,x y4 22# - and .y x2#

Solution

Draw the three regions, either separately or on the same set of axes, and see where they overlap.

y= x2

ch5.indd 264 7/19/09 8:19:30 PM


If you are given a region, you should also be able to describe it algebraically.

Put the three regions together.

EXAMPLES

Describe each region . 1. y

x1

1

-2 2 3 4-1

-3

-4

2

3

4

5

-1

-2

-3-4

6

Solution

The shaded area is below and including 6y = so can be described as y 6# .

It is also to the left of, but not including the line 4,x = which can be described as .x 41

The region is the intersection of these two regions: y 6# and x 41

CONTINUED

ch5.indd 265 7/19/09 12:52:27 PM


2.

y

x2

2

-2

-2

Solution

The shaded area is the interior of the circle, centre (0, 0) and radius 2 but it does not include the circle. The equation of the circle is 2 4.x y x yor2 2 2 2 2+ = + =

You may know (or guess) the inequality for the inside of the circle. If you are unsure, choose a point inside the circle and substitute into the equation e.g. (0, 0) .

x y

0 00

4

LHS

RHS

2 2

2 2

1

= +

= +

=

] g

So the region is x y 42 21+ .

5.11 Exercises

1. Shade the region defi ned by (a) x 2# (b) x 12 (c) y 0$ (d) y 51 (e) y x 1# +

(f) y x2 3$ - (g) x y 12+ (h) 3 6 0x y 1- - (i) x y2 2 0$+ - (j) x2 1 01-

ch5.indd 266 7/19/09 12:52:37 PM


2. Write an inequation to describe each region .

(a) y

x1

1

-2 2 3 4-1

-3

-4

2

3

4

5

-1

-2

-3-4

6

(b) y

x1

1

-2 2 3 4-1

-3

-4

2

3

4

5

-1

-2

-3-4

6

(c) y

x1

1

-2 2 3 4-1

-3

-4

2

3

4

5

-1

-2

-3-4

6

y = x + 1

(d) y

1

1

-2 2 3 4 5-1

-3

-4

-5

2

3

4

5

-1

-2

-3-4

y = x2 - 4

(e) y

x1

1

2

3

y = 2x

3. Shade each region described . (a) –y x 122 (b) x y 92 2#+ (c) x y 12 2$+ (d) y x2# (e) y x31

4. Describe as an inequality the set of points that lie (a)

below the line y x3 2= - the set of points that lie (b)

inside the parabola 2y x2= + the interior of a circle with (c)

radius 7 and centre (0, 0) the exterior of a circle with (d)

radius 9 and centre (0, 0) the set of points that lie to (e)

the left of the line 5x = and above the line 2y =

ch5.indd 267 7/19/09 12:52:46 PM


5. Shade the region (a) x y2 42 2#- +] g (b) x y1 2 12 2#- + -] ^g h (c) 1x y2 92 22+ + -] ^g h

6. Shade the intersection of these regions .

(a) ,x 3# y 1$ - (b) ,x y x3 32$ - - (c) ,y 1# y x3 5$ - (d) ,y x y x1 32 #+ - (e) ,y x y1 92 2# #+ (f) ,x x y1 42 22 1- + (g) ,y y x4 2# $ (h) , ,x y y x2 3 31 2#- (i) ,y x y0 12 2# $+ (j) x y1 21 #- -

7. Shade the region bounded by the curve (a) ,y x2= the x -axis

and the lines 1x = and 3x = the curve (b) ,y x3= the y -axis

and the lines 0y = and 1y =

the curve (c) 4,x y2 2+ = the x -axis and the lines 0x = and 1x = in the fi rst quadrant

the curve (d) 2,y x= the x -axis

and the lines 1x = and 4x =

the curve (e) 2

1 ,yx

=+

the

x -axis and the lines 0x = and

2x =

8. Shade the regions bounded by the intersection of

(a) ,x y2 511 and y x2# (b) , ,x y y x3 1 21 $ #- - (c) , ,y x y x x y1 2 1 2 3 6# # #- + - (d) , ,x y x y3 2 92 2$ # $- + (e) , , | |x y y x2 31 # $

The fi rst quadrant is where x and y values are both positive.

Application

Regions are used in business applications to fi nd optimum profi t. Two (or more) equations are graphed together, and the region where a profi t is made is shaded.

The optimum profi t occurs at the endpoints (or vertices) of the region.

EXAMPLE

A company makes both roller skates ( X ) and ice skates ( Y ). Roller skates make a $25 profi t, while ice skates make a profi t of $21. Each pair of roller skates spends 2 hours on machine A (available 12 hours per day) and 2 hours on machine B (available 8 hours per day). Each pair of ice skates spends 3 hours on machine A and 1 hour on machine B.

How many skates of each type should be made each day to give the greatest

profi t while making the most effi cient use of the machines?

ch5.indd 268 7/19/09 8:22:29 PM


SOLUTION

Profit P $25 $21

Machine A: 2 3 12

Machine B: 2 8

X Y

X Y

X Y

+

+

+

=

#

#

Sketch the regions and fi nd the

point of intersection of the lines.

The shaded area shows all possible ways of making a profi t. Optimum profi t occurs at one of the endpoints of the regions.

(0, 4): $25 0 $21 4 $84(4, 0): $25 4 $21 0 $100(3, 2): $25 3 $21 2 $117

PPP

= + == + == + =

] ]

] ]

] ]

g g

g g

g g

3, 2^ h gives the greatest profi t, so 3 pairs of roller skates and 2 pairs of ice

skates each day gives optimum profi t.

ch5.indd 269 7/19/09 12:53:05 PM


Test Yourself 5

1. If ,f x x x3 42= - -] g fi nd (a) f 2-] g (b) f a] g (c) x when f x 0=] g

2. Sketch each graph (a) 3 4y x x2= - - (b) f x x3=] g (c) 1x y2 2+ = (d) 1y x2= - (e) 1y x2= - -

(f) 2y x=

(g) 2 5 10 0x y- + = (h) | 2 |y x= +

3. Find the domain and range of each graph in question 2.

4. If f xx x

x x

2 1

3 1

if

if2 1

$=

-] g *

fi nd f f f5 0 1- +] ] ]g g g

5. Given f x

x

x xx x

3 3

1 32 1

if

ifif

2

2

1# #=

-

] g *

fi nd (a) f 2] g (b) f 3-] g (c) f 3] g (d) f 5] g (e) f 0] g

6. Shade the region y x2 1$ + .

7. Shade the region where x y3 1and1 $ - .

8. Shade the region given by x y 12 2$+ .

9. Shade the region given by 2 3 6 0 2x y xand# $+ - - .

10. Shade the region y x 12 + and x y 2#+ .

11. Describe each region (a)

(b)

(c)

12. (a) Write down the domain and range of

the curve 3

2yx

=-

.

(b) Sketch the graph of 3

2yx

=-

.

ch5.indd 270 7/19/09 12:53:16 PM


13. (a) Sketch the graph | 1 |y x= + . From the graph, solve (b)

(i) | 1 | 3x + = (ii) | |x 1 31+ (iii) | |x 1 32+

14. If ,f x x3 4= -] g fi nd (a) f 2] g (b) x when f x 7=] g (c) x when f x 0=] g

15. Find the x - and y -intercepts of (a) 2 5 20 0x y- + = (b) 5 14y x x2= - -

16. State which functions are (i) even (ii) odd (iii) neither even nor odd .

(a) 1y x2= - (b) 1y x= + (c) y x3= (d) y x4= (e) 2y x=

17. Find

(a) limx

x x3

2 3x 3

2

-

- -"

(b) limx x

x5

2x 0 2 +"

(c) limxx

11

x 1 2

3

-

+"-

(d) limh

xh h2 3h 0

2 +"

18. Sketch 10 , logy y x y xandx= = = on the same number plane.

19. (a) State the domain and range of 2 4y x= - .

Sketch the graph of (b) 2 4y x= - .

20. Show that (a) f x x x3 14 2= + -] g is even

(b) f x x x3= -] g is odd.


1. Find the values of b if f x x x3 7 12= - +] g and .f b 7=] g

2. Sketch 2 1y x 2= + -] g in the domain .x3 0# #-

3. Sketch the region y x4 22 - in the fi rst quadrant.

4. Draw the graph of | | 3 4.y x x= + -

5. f x

x x

x

x x

2 3 2

1 2 2

2

when

when

when2

2

1

# #=

+

-

-

] g

Z

[

\

]]

]]

Find , ,f f f3 4 0-] ] ]g g g and sketch the curve.

6. Find the domain and range of

1

1 .yx2

=-

7. Sketch the region , ,x y x y2 61 1+ 2 4 0.x y $+ -

8. Find the domain and range of x y2 = in the fi rst quadrant.

9. If ,f x x x x2 2 123 2= - -] g fi nd x when .f x 0=] g

10. Sketch the region defi ned by yx 2

12

+

in the fi rst quadrant.

ch5.indd 271 7/19/09 12:53:24 PM


11. If h tt t

t t

1 1

1 1

if

if

2

2

2

#=

-

-] g )

fi nd the value of h h h2 1 0-+ -] ] ]g g g and sketch the curve.

12. Sketch 1y x2= - in the fi rst quadrant.

13. Sketch the region , .y x y x x5 21$ - +

14. If ,f x x2 1= -] g show that 2( )f a f a2 = -^ _h i for all real a .

15. Find the values of x for which f x 0=] g when f x x x2 52= - -] g (give exact answers).

16. (a) Show that 3

2 7 23

1 .xx

x+

+= +

+

(b) Find the domain and range of

3

2 7 .yxx

=+

+

(c) Hence sketch the graph of

3

2 7 .yxx

=+

+

17. Sketch 2 .y 1x= -

18. Sketch | |

.yx

x2

=

19. Find the domain and range of .xf x2 6= -] g

20. What is the domain of 4

1 ?yx2

=-

21. Sketch .f xx

1 12

= -] g

ch5.indd 272 7/19/09 12:53:33 PM

ch5.indd 273 7/19/09 12:53:42 PM

TERMINOLOGY

6 Trigonometry

Angle of depression: The angle between the horizontal and the line of sight when looking down to an object below

Angle of elevation: The angle between the horizontal and the line of sight when looking up to an object above

Angles of any magnitude: Angles can be measured around a circle at the centre to fi nd the trigonometric ratios of angles of any size from 0c to 360c and beyond

Bearing: The direction relative to north. Bearings may be written as true bearings (clockwise from North) or as compass bearings (using N, S, E and W)

Complementary angles: Two or more angles that add up to 90c

Cosecant: The reciprocal ratio of sine (sin). It is the hypotenuse over the opposite side in a right triangle

Cotangent: The reciprocal ratio of tangent (tan). It is the adjacent over the opposite side in a right triangle

Secant: The reciprocal ratio of cosine (cos). It is the hypotenuse over the adjacent side in a right triangle

Trigonometric identities: A statement that is true for all trigonometric values in the domain. Relationships between trigonometric ratios

ch6.indd 274 7/19/09 9:22:21 AM

275Chapter 6 Trigonometry

INTRODUCTION

TRIGONOMETRY IS USED IN many fi elds, such as building, surveying and navigating. Wave theory also uses trigonometry.

This chapter revises basic right-angled triangle problems and applies them to real-life situations. Some properties of trigonometric ratios, angles greater than 90c and trigonometric equations are introduced. You will also study trigonometry in non-right-angled triangles.

Ptolemy (Claudius Ptolemaeus), in the second century, wrote He mathe matike syntaxis (or Almagest as it is now known) on astronomy. This is considered to be the fi rst treatise on trigonometry, but was based on circles and spheres rather than on triangles. The notation ‘chord of an angle’ was used rather than sin, cos or tan.

Ptolemy constructed a table of sines from 0c to 90c in steps of a quarter of a degree. He also calculated a value of r to 5 decimal places, and established the relationship for sin ( )X Y! and cos ( ) .X Y!

DID YOU KNOW?

Trigonometric Ratios

In similar triangles, pairs of corresponding angles are equal and sides are in proportion. For example:

the • hypotenuse is the longest side, and is always opposite the right angle the • opposite side is opposite the angle marked in the triangle the • adjacent side is next to the angle marked

In any triangle containing an angle of ,30c the ratio of : : .AB AC 1 2= Similarly, the ratios of other corresponding sides will be equal. These ratios of sides form the basis of the trigonometric ratios.

In order to refer to these ratios, we name the sides in relation to the angle being studied:

You studied similar triangles in Geometry 1 in Chapter 4.

ch6.indd 275 8/7/09 12:07:25 PM


The opposite and adjacent sides vary according to where the angle is marked. For example:

The trigonometric ratios are

You can learn these by their initials SOH , CAH , TOA .

What about S ome O ld H ags C an’t A lways H ide T heir O ld A ge?

DID YOU KNOW? Trigonometry, or triangle measurement , progressed from the study of geometry in ancient Greece. Trigonometry was seen as applied mathematics. It gave a tool for the measurement of planets and their motion. It was also used extensively in navigation, surveying and mapping, and it is still used in these fi elds today.

Trigonometry was crucial in the setting up of an accurate calendar, since this involved measuring the distances between the Earth, sun and moon.

sin

cos

tan

hypotenuse

opposite

hypotenuse

adjacent

adjacent

opposite

Sine

Cosine

Tangent

i

i

i

=

=

=

As well as these ratios, there are three inverse ratios,

cosecsin

seccos

cottan

1

1

1

oppositehypotenuse

adjacent

hypotenuse

oppositeadjacent

Cosecant

Secant

Cotangent

ii

ii

ii

= =

= =

= =

f

f

f

p

p

p

ch6.indd 276 7/19/09 9:22:41 AM


EXAMPLES

1. Find ,sina tana and .seca

Solution

sin

tan

sec cos

AB

BC

AC

5

3

4

53

43

1

45

hypotenuse

opposite side

adjacent side

hypotenuse

opposite

adjacent

opposite

adjacent

hypotenuse

a

a

a a

= =

= =

= =

=

=

=

=

=

=

=

2. If 72,sini = fi nd the exact ratios of ,cosi tani and .coti

Solution


7 2

49 4

45

c a b

a

a

a

a 45

2 2 2

2 2 2

2

2

`

= +

= +

= +

=

=

CONTINUED

To fi nd the other ratios you need to fi nd the

adjacent side.

ch6.indd 277 7/19/09 9:22:43 AM


cos

tan

cottan

745

452

1

245

hypotenuse

adjacent

adjacent

opposite

i

i

ii

=

=

=

=

=

=

Complementary angles

, ,ABC B A 90In if then c+ +i iD = = -

sin

cos

tan

sec

cosec

cot

cb

ca

ab

ac

bc

ba

i

i

i

i

i

i

=

=

=

=

=

=

(angle sum of a Δ)

( )

( )

( )

( )

( )

( )

sin

cos

tan

sec

cosec

cot

ca

cb

ba

bc

ac

ab

90

90

90

90

90

90

c

c

c

c

c

c

i

i

i

i

i

i

- =

- =

- =

- =

- =

- =

From these ratios come the results.

( )

( )

( )

( )

( )

( )

°

°

°

°

°

°

sin cos

cos sin

sec cosec

cosec sec

tan cot

cot tan

90

90

90

90

90

90

i i

i i

i i

i i

i i

i i

= -

= -

= -

= -

= -

= -

ch6.indd 278 7/19/09 9:22:44 AM


1. Write down the ratios of ,cos sini i and .tani

2. Find ,sin cotb b and .secb

3. Find the exact ratios of ,sin tanb b and .cosb

4. Find exact values for ,cos tanx x and .cosecx

EXAMPLES

1. Simplify 50 40 .tan cotc c-

Solution

tan cotcot

tan cot tan tan

50 90 5040

50 40 50 50

0

`

c c c

c

c c c c

= -

=

- = -

=

] g

2. Find the value of m if .sec cosec m55 2 15c c= -] g

Solution

90 5535

sec coseccosec

m

m

m

55

2 15 35

2 50

25

`

c c c

c

= -

=

- =

=

=

] g

Check this by substituting m into the equation.

6.1 Exercises

Check this answer on your calculator.

ch6.indd 279 7/19/09 9:22:44 AM


5. If ,tan34

i = fi nd cos i and .sin i

6. If 32,cosi = fi nd exact values for

,tani seci and .sini

7. If 61,sini = fi nd the exact ratios

of cosi and .tani

8. If 0.7,cosi = fi nd exact values for tani and .sini

9. ABCD is a right-angled isosceles triangle with ABC 90c+ = and 1.AB BC= =

Find the exact length of (a) AC . Find (b) .BAC+ From the triangle, write down (c)

the exact ratios of 45 , 45sin cosc c and .ctan45

10.

Using Pythagoras’ theorem, (a) fi nd the exact length of AC .

Write down the exact ratios of (b) 30 , 30sin cosc c and 30 .tan c

Write down the exact ratios of (c) 60 , 60sin cosc c and 60 .tan c

11. Show .sin cos67 23c c=

12. Show .sec cosec82 8c c=

13. Show .tan cot48 42c c=

14. Simplify (a) cos sin61 29c c+

(b) 90sec cosec ci i- -] g

(c) 70 20 2 70tan cot tanc c c+ -

(d) 3555

cossin

c

c

(e) 25

25 65cot

cot tanc

c c+

15. Find the value of x if .sin cos x80 90c c= -] g

16. Find the value of y if .tan cot y22 90c c= -^ h

17. Find the value of p if .cos sin p49 10c c= +^ h

18. Find the value of b if .sin cos b35 30c c= +] g

19. Find the value of t if .cot tant t2 5 3 15c c+ = -] ]g g

20. Find the value of k if .tan cotk k15 2 60c c- = +] ]g g

Hint: Change 0.7 to a fraction.

Trigonometric ratios and the calculator

Angles are usually given in degrees and minutes. In this section you will practise rounding off angles and fi nding trigonometric ratios on the calculator.

Angles are usually given in degrees and minutes in this course. The calculator uses degrees, minutes and seconds, so you need to round off.

utes ree

onds ute( )

60 1 (60 1)min degsec min

60 1 60 1c= =

= =

lm l

In normal rounding off, you round up to the next number if the number to the right is 5 or more. Angles are rounded off to the nearest degree by rounding up if there are 30 minutes or more. Similarly, angles are rounded off to the nearest minute by rounding up if there are 30 seconds or more.

ch6.indd 280 7/19/09 9:22:45 AM


EXAMPLES

Round off to the nearest minute. 1. 23 12 22c l m

Solution

23 12 22 23 12c c=l m l

2. 59 34 41c l m

Solution

59 34 41 59 35c c=l m l

3. 16 54 30c l m

Solution

16 54 30 16 55c c=l m l

, ,,% KEY

This key changes decimal angles into degrees, minutes and seconds

and vice versa.

Some calculators have

deg or dms keys.

EXAMPLES

1. Change 58 19c l into a decimal.

Solution

, ,, , ,, , ,,% % %58 19Press = So .58 19 58 31666667c =l

2. Change 45.236c into degrees and minutes.

Solution

, ,,%.45 236Press SHIFT= So .45 236 45 14c c= l

If your calculator does not give these answers, check

the instructions for its use.

Because 30 seconds is half a minute, we round

up to the next minute.

ch6.indd 281 7/19/09 9:22:45 AM


In order to use trigonometry in right-angled triangle problems, you need to fi nd the ratios of angles on your calculator.

EXAMPLES

1. Find ,cos 58 19c l correct to 3 decimal places.

Solution

, ,, , ,,% %58 19Press COS = So .cos 58 19 0 525c =l

2. Find ,sin 38 14c l correct to 3 decimal places.

Solution

, ,, , ,,% %38 14Press SIN = So .sin 38 14 0 619c =l

3. If 0.348,tani = fi nd i in degrees and minutes.

Solution

This is the reverse of fi nding trigonometric ratios. To fi nd the angle, given the ratio, use the inverse key .tan 1-^ h , ,,%TAN .0 348Press SHIFT SHIFT1 =-

.

( . )

tan

tan

0 348

0 34819 11

1

c

i

i

=

=

=

-

l

4. Find i in degrees and minutes if . .cos 0 675i =

Solution

, ,,%.0 675Press SHIFT COS SHIFT1 =-

.

( . )cos

cos0 675

0 67547 33

1

c

ii

===

-

l

6.2 Exercises

1. Round off to the nearest degree. (a) °47 13 12l m (b) °81 45 43l m (c) °19 25 34l m (d) °76 37 19l m (e) °52 29 54l m

2. Round off to the nearest minute. (a) °47 13 12l m (b) °81 45 43l m (c) °19 25 34l m (d) °76 37 19l m (e) °52 29 54l m

If your calculator doesn't give this answer, check that it is in degree mode.

ch6.indd 282 7/19/09 9:22:46 AM


3. Change to a decimal. (a) 77 45c l (b) 65 30c l (c) 24 51c l (d) 68 21c l (e) 82 31c l

4. Change into degrees and minutes. (a) 59.53c (b) 72.231c (c) 85.887c (d) 46.9c (e) 73.213c

5. Find correct to 3 decimal places. (a) 39 25sin c l (b) cos 45 51c l (c) 18 43tan c l (d) 68 06sin c l (e) 54 20tan c l

6. Find i in degrees and minutes if (a) .sin 0 298i = (b) .tan 0 683i = (c) .cos 0 827i = (d) .tan 1 056i = (e) .cos 0 188i =

Right-angled Triangle Problems

Trigonometry is used to fi nd an unknown side or angle of a triangle.

Finding a side

We can use trigonometry to fi nd a side of a right-angled triangle.

EXAMPLES

1. Find the value of x , correct to 1 decimal place.

Solution

°.

°.

. °

.

. .

cos

cos

cos

cos

x

x

x

x

23 4911 8

23 4911 8

11 8 23 49

10 8 1

11 8 11 8

hypotenuse

adjacent

cm to decimal point`

# #

i =

=

=

=

=

l

l

l

^ h

CONTINUED

ch6.indd 283 7/19/09 9:22:46 AM


2. Find the value of y , correct to 3 signifi cant fi gures.

Solution

c15c

15

15

15

c

c

c

c

15

15

c

.

.

.

.

.

.

sin

sin

sin

sin

sin

sin

sin sin

y

yy

y

y

y y

41 15 9 7

41 9 7

41 9 7

41 9 7

419 7

14 7 3

41 41

hypotenuse

opposite

m to significant figures

# #

i =

=

=

=

=

=

=

l

l

l

l

l

l l

^ h

6.3 Exercises

1. Find the values of all pronumerals, correct to 1 decimal place.

(a)

(b)

(c)

(d)

ch6.indd 284 7/19/09 9:22:47 AM


(e)

(f)

(g)

(h)

(i)

(j)

(k)

x

5.4 cm

31c12l

(l)

x4.7 cm

37c22l

(m) x

6.3 cm

72c18l

(n)

23 mm

63c14l

x

(o)

3.7 m

39c47l

y

(p)

14.3 cm46c5l

k

(q)

4.8 m74c29l

h

ch6.indd 285 7/19/09 2:14:10 PM


(r) 0.45 m

68c41ld

(s) 5.75 cm

19c17l

x

(t) 17.3 m

6c3l

b

2. A roof is pitched at 60c. A room built inside the roof space is to have a 2.7 m high ceiling. How far in from the side of the roof will the wall for the room go?

60c

2.7 m

x

3. A diagonal in a rectangle with breadth 6.2 cm makes an angle of 73c with the vertex as shown. Find the length of the rectangle correct to 1 decimal place.

73c

6.2 cm

4. Hamish is standing at an angle of 67c from a goalpost and 12.8 m away as shown. How far does he need to kick a football for it to reach the goal?

x

67c

12.8 m

5. Square ABCD with side 6 cm has line CD produced to E as shown so that EAD 64 12c+ = l . Evaluate the length, correct to 1 decimal place, of

(a) CE (b) AE

E

6 cm

64c12l

B

A

C

D

6. A right-angled triangle with hypotenuse 14.5 cm long has one interior angle of 43 36c l. Find the lengths of the other two sides of the triangle.

ch6.indd 286 7/19/09 9:22:49 AM


7. A right-angled triangle ABC with the right angle at A has B 56 44c+ = l and 26AB = mm. Find the length of the hypotenuse.

8. A triangular fence is made for a garden inside a park. Three holes A , B and C for fence posts are made at the corners so that A and B are 10.2 m apart, AB and CB are perpendicular, and angle CAB is 59 54c l. How far apart are A and C ?

9. Triangle ABC has 46BAC c+ =

and .ABC 54c+ = An altitude is drawn from C to meet AB at point D . If the altitude is 5.3 cm long, fi nd, correct to 1 decimal place, the length of sides

(a) AC (b) BC (c) AB

10. A rhombus has one diagonal 12 cm long and the diagonal makes an angle of 28 23c l with the side of the rhombus.

Find the length of the side of (a) the rhombus.

Find the length of the other (b) diagonal.

11. Kite ABCD has diagonal 15.8BD = cm as shown. If ABD+ = 57 29 andc l

72 51DBC c+ = l, fi nd the length of the other diagonal AC.

B

A

C

D72c51l

57c29l

15.8 cm

Finding an angle

Trigonometry can also be used to fi nd one of the angles in a right-angled triangle.

EXAMPLES

1. Find the value of ,i in degrees and minutes.

CONTINUED

ch6.indd 287 7/19/09 9:22:50 AM


Solution

..

7.35.8

cos

7 35 8hypotenuse

adjacent

1

i =

-cos

37 23

`

c

i =

=

=

l

c m

2. Find the value of ,a in degrees and minutes.

Solution

..

.

.

tan

tan

2 14 9

2 14 9

66 48

adjacent

opposite

1`

c

a

a

=

=

=

=

-

l

c m

6.4 Exercises

1. Find the value of each pronumeral, in degrees and minutes.

(a)

(b)

ch6.indd 288 7/19/09 10:55:32 AM


(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

2.4 cm

3.8 cm

a

(l)

8.3 cm

5.7 cm

i

(m)

6.9 mm

11.3 mm

i

(n)

3 m

7 m

i

ch6.indd 289 7/19/09 9:22:51 AM


(o)

5.1 cm

11.6 cm

b

(p)

15 m

13 ma

(q)

7.6 cm

4.4 cmi

(r)

14.3 cm8.4 cm

a

(s)

3 m

5 m

i

(t)

10.3 cm

18.9 cmc

2. A kite is fl ying at an angle of i above the ground as shown. If the kite is 12.3 m above the ground and has 20 m of string, fi nd angle i .

12.3 m20 m

i

3. A fi eld is 13.7 m wide and Andre is on one side. There is a gate on the opposite side and 5.6 m along from where Andre is. At what angle will he walk to get to the gate?

Gate

Andre

5.6 m

13.7 m

i

4. A 60 m long bridge has an opening in the middle and both sides open up to let boats pass underneath. The two parts of the bridge fl oor rise up to a height of 18 m. Through what angle do they move?

18 m

i60 m

5. An equilateral triangle ABC with side 7 cm has an altitude AD that is 4.5 cm long. Evaluate the angle the altitude makes with vertex A DAB+] g .

ch6.indd 290 7/31/09 4:45:58 PM


6. Rectangle ABCD has dimensions 18 cm # 7 cm. A line AE is drawn so that E bisects DC .

How long is line (a) AE ? (Answer to 1 decimal place).

Evaluate (b) DEA+ .

7. A 52 m tall tower has wire stays on either side to minimise wind movement. One stay is 61.3 m long and the other is 74.5 m long as shown. Find the angles that the tower makes with each stay.

52 m

61.3 m 74.5 m

ba

8. (a) The angle from the ground up to the top of a pole is 41c when standing 15 m on one side of it. Find the height h of the pole, to the nearest metre.

If Seb stands 6 m away on the (b) other side, fi nd angle i .

41c

h

6 m 15 mi

9. Rectangle ABCD has a line BE drawn so that AEB 90c+ = and 1DE = cm. The width of the rectangle is 5 cm.

5 cm

BA

CE

D1 cm

Find (a) BEC+ . Find the length of the (b)

rectangle.

10. A diagonal of a rhombus with side 9 cm makes an angle of 16cwith the side as shown. Find the lengths of the diagonals.

16c

9 cm

11. (a) Kate is standing at the side of a road at point A , 15.9 m away from an intersection. She is at an angle of 39c from point B on the other side of the road. What is the width w of the road?

(b) Kate walks 7.4 m to point C . At what angle is she from point B ?

w

B

CA7.4 m

15.9 m

39c i

ch6.indd 291 7/19/09 9:22:53 AM


Applications

DID YOU KNOW?

The Leaning Tower of Pisa was built as a belfry for the cathedral nearby. Work started on the tower in 1174, but when it was only half completed the soil underneath one side of it subsided. This made the tower lean to one side. Work stopped, and it wasn’t until 100 years later that architects found a way of completing the tower. The third and fi fth storeys were built close to the vertical to compensate for the lean. Later a vertical top storey was added.

The tower is about 55 m tall and 16 m in diameter. It is tilted about 5 m from the vertical, and tilts by an extra 0.6 cm each year.

Class Investigation

Discuss some of the problems with the Leaning Tower of Pisa.

Find the angle at which it is tilted from the vertical. • Work out how far it will be tilted in 10 years. • Use research to fi nd out if the tower will fall over, and if so, when. •

Angle of elevation

The angle of elevation is used to measure the height of tall objects that cannot be measured directly, for example a tree, cliff, tower or building.

ch6.indd 292 7/19/09 9:22:54 AM


Class Exercise

Stand outside the school building and look up to the top of the building. Think about which angle your eyes pass through to look up to the top of the building.

The angle of elevation, ,i is the angle measured when looking from the ground up to the top of the object. We assume that the ground is horizontal.

EXAMPLE

The angle of elevation of a tree from a point 50 m out from its base is .38 14c l Find the height of the tree, to the nearest metre.

Solution

We assume that the tree is vertical!

tan

tan

tan

h

h

h

h

38 1450

38 1450

50 38 1439

50 50# #

c

c

c

Z

=

=

=

l

l

l

So the tree is 39 m tall, to the nearest metre.

A clinometer is used to measure the angle of

elevation or depression.

ch6.indd 293 7/19/09 9:22:57 AM


Angle of depression

The angle of depression is the angle formed when looking down from a high place to an object below.

Class Exercise

If your classroom is high enough, stand at the window and look down to something below the window. If the classroom is not high enough, fi nd a hill or other high place. Through which angle do your eyes pass as you look down?

The angle of depression, ,i is the angle measured when looking down from the horizontal to an object below.

EXAMPLES

1. The angle of depression from the top of a 20 m building to a boy below is .c 961 3 l How far is the boy from the building, to 1 decimal place?

Solution

ch6.indd 294 7/19/09 9:22:57 AM


39

39c

39

39

39

3939

c

c

c

c

c

c c

( , )

61 39.

61

tan

tan

tan

tan

tan

tan tan

DAC ACB

AD BC

x

xx

x

x

x x

61

61 20

61 20

61 20

61 20

20

10 8

61

alternate angles

# #

+ +

<

Z

=

=

=

=

=

=

=

l

l

l

l

l

l

l l

So the boy is 10.8 m from the building.

2. A bird sitting on top of an 8 m tall tree looks down at a possum 3.5 m out from the base of the tree. Find the angle of elevation to the nearest minute.

Solution

3.5 m

8 m

AB

C D

i

The angle of depression is i

AB DCBDC

Since horizontal linesalternate angles+

<i=

]

^

g

h

.

.

tan

tan

3 58

3 58

66

1`

c

i

i

=

=

=

-

22l

c m

ch6.indd 295 7/19/09 9:23:00 AM


Bearings

Bearings can be described in different ways: For example, N70 Wc :

Start at north and measure 70o around towards the west.

True bearings measure angles clockwise from north

EXAMPLES

1. Sketch the diagram when M is on a bearing of 315c from P .

Solution

2. X is on a bearing of 030c from Y . Sketch this diagram.

Solution

3. A house is on a bearing of 305c from a school. What is the bearing of the school from the house?

Measure clockwise, starting at north.

All bearings have 3 digits so 30° becomes 030° for a bearing.

We could write 315o T for true bearings.

ch6.indd 296 7/19/09 9:23:01 AM


Solution

The diagram below shows the bearing of the house from the school.

North

School

House

305c

To fi nd the bearing of the school from the house, draw in North from the house and use geometry to fi nd the bearing as follows:

S

H

N1

N2NN

305c

The bearing of the school from the house is N HS2+ .

360 305

180 55 ( )

N SH

N HS N H N S

55

125

angle of revolution

cointerior angles,

1

2 2 1

c c

c

c c

c

+

+ <

= -

=

= -

=

^ h

So the bearing of the school from the house is 125c .

4. A plane leaves Sydney and fl ies 100 km due east, then 125 km due north. Find the bearing of the plane from Sydney, to the nearest degree.

CONTINUED

ch6.indd 297 7/19/09 9:23:05 AM


Solution

c

.

( . )51 ( )

tan

tan

x

x

x

100125

1 25

1 25

90

90 51

39

to the nearest degree

1

c c

c c

c

i

=

=

=

=

= -

= -

=

-

So the bearing of the plane from Sydney is .°039

5. A ship sails on a bearing of °140 from Sydney for 250 km. How far east of Sydney is the ship now, to the nearest km?

Solution

cos

cos

cos

x

x

x

x

140 90

50

50250

50250

250 50

161

250 250# #

c c

c

c

c

c

Z

i = -

=

=

=

=

So the ship is 161 km east of Sydney, to the nearest kilometre.

A navigator on a ship uses a sextant to measure angles.

Could you use a different triangle for this question?

ch6.indd 298 7/19/09 9:23:12 AM


6.5 Exercises

1. Draw a diagram to show the bearing in each question .

A boat is on a bearing of 100(a) c from a beach house.

Jamie is on a bearing of 320(b) c from a campsite.

A seagull is on a bearing of (c) 200c from a jetty.

Alistair is on a bearing of (d) 050c from the bus stop.

A plane is on a bearing of (e) 285c from Broken Hill .

A farmhouse is on a bearing (f) of 012c from a dam.

Mohammed is on a bearing of (g) 160c from his house.

A mine shaft is on a bearing (h) of 080c from a town.

Yvonne is on a bearing of (i) 349c from her school.

A boat ramp is on a bearing of (j) 280c from an island.

2. Find the bearing of X from Y in each question in 3 fi gure (true) bearings .

X

Y

North

112c

(a)

X

35c

Y

North

South

EastWest

(b)

X

10cY

North

South

EastWest

(c)

23c

X

Y

North

South

EastWest

(d)

X

Y

North

South

EastWest

(e)

ch6.indd 299 7/19/09 9:23:16 AM


3. Jack is on a bearing of 260c from Jill. What is Jill’s bearing from Jack?

4. A tower is on a bearing of 030c from a house. What is the bearing of the house from the tower?

5. Tamworth is on a bearing of 340c from Newcastle. What is the bearing of Newcastle from Tamworth?

6. The angle of elevation from a point 11.5 m away from the base of a tree up to the top of the tree is 42 12c l. Find the height of the tree to one decimal point.

7. Geoff stands 25.8 m away from the base of a tower and measures the angle of elevation as .39 20c l Find the height of the tower to the nearest metre.

8. A wire is suspended from the top of a 100 m tall bridge tower down to the bridge at an angle of elevation of 52c. How long is the wire, to 1 decimal place?

9. A cat crouches at the top of a 4.2 m high cliff and looks down at a mouse 1.3 m out from the foot (base) of the cliff. What is the angle of depression, to the nearest minute?

10. A plane leaves Melbourne and fl ies on a bearing of 065c for 2500 km.

How far north of Melbourne (a) is the plane?

How far east of Melbourne (b) is it?

What is the bearing of (c) Melbourne from the plane?

11. The angle of elevation of a tower is 39 44c l when measured at a point 100 m from its base. Find the height of the tower, to 1 decimal place.

12. Kim leaves his house and walks for 2 km on a bearing of .155c How far south is Kim from his house now, to 1 decimal place?

13. The angle of depression from the top of an 8 m tree down to a rabbit is .43 52c l If an eagle is perched in the top of the tree, how far does it need to fl y to reach the rabbit, to the nearest metre?

14. A girl rides a motorbike through her property, starting at her house. If she rides south for 1.3 km, then rides west for 2.4 km, what is her bearing from the house, to the nearest degree?

15. A plane fl ies north from Sydney for 560 km, then turns and fl ies east for 390 km. What is its bearing from Sydney, to the nearest degree?

16. Find the height of a pole, correct to 1 decimal place, if a 10 m rope tied to it at the top and stretched out straight to reach the ground makes an angle of elevation of .67 13c l

ch6.indd 300 7/19/09 9:23:17 AM


17. The angle of depression from the top of a cliff down to a boat 100 m out from the foot of the cliff is .59 42c l How high is the cliff, to the nearest metre?

18. A group of students are bushwalking. They walk north from their camp for 7.5 km, then walk west until their bearing from camp is .320c How far are they from camp, to 1 decimal place?

19. A 20 m tall tower casts a shadow 15.8 m long at a certain time of day. What is the angle of elevation from the edge of the shadow up to the top of the tower at this time?

15.8 m

20 m

20. A fl at verandah roof 1.8 m deep is 2.6 m up from the ground. At a certain time of day, the sun makes an angle of elevation of .72 25c l How much shade is provided on the ground by the verandah roof at that time, to 1 decimal place?

21. Find the angle of elevation of a .15 9 m cliff from a point 100 m

out from its base.

22. A plane leaves Sydney and fl ies for 2000 km on a bearing of 195 .c How far due south of Sydney is it?

23. The angle of depression from the top of a 15 m tree down to a pond is .25 41c l If a bird is perched in the top of the tree, how far does it need to fl y to reach the pond, to the nearest metre?

24. A girl starting at her house, walks south for 2.7 km then walks east for 1.6 km. What is her bearing from the house, to the nearest degree?

25. The angle of depression from the top of a tower down to a car 250 m out from the foot of the tower is .38 19c l How high is the tower, to the nearest metre?

26. A hot air balloon fl ies south for 3.6 km then turns and fl ies east until it is on a bearing of 127c from where it started. How far east does it fl y?

27. A 24 m wire is attached to the top of a pole and runs down to the ground where the angle of elevation is .22 32c l Find the height of the pole.

ch6.indd 301 7/31/09 4:46:00 PM


28. A train depot has train tracks running north for 7.8 km where they meet another set of tracks going east for 5.8 km into a station. What is the bearing of the depot from the station, to the nearest degree?

29. Jessica leaves home and walks for 4.7 km on a bearing of .075c She then turns and walks for 2.9 km on a bearing of 115c and she is then due east of her home.

How far north does Jessica (a) walk?

How far is she from home? (b)

30. Builder Jo stands 4.5 m out from the foot of a building and looks up at to the top of the building where the angle of elevation is 71c. Builder Ben stands at the top of the building looking down at his wheelbarrow that is 10.8 m out from the foot of the building on the opposite side from where Jo is standing.

Find the height of the (a) building.

Find the angle of depression (b) from Ben down to his wheelbarrow.

Exact Ratios

A right-angled triangle with one angle of °45 is isosceles. The exact length of its hypotenuse can be found.

c a b

AC

AC

1 12

2

2 2 2

2 2 2

= +

= +

=

=

This means that the trigonometric ratios of 45c can be written as exact ratios.

Pythagoras’ theorem is used to fi nd the length of the hypotenuse.

sin

cos

tan

452

1

452

1

45 1

c

c

c

=

=

=

ch6.indd 302 7/19/09 9:23:19 AM


This angle is commonly used; for example, °45 is often used for the pitch of a roof. The triangle with angles of °60 and °30 can also be written with exact sides.

2 13

AD

AD 3

2 2 2= -

=

=

Halve the equilateral triangle to get .ABDT

60

60

60

°

°

°

sin

cos

tan

23

21

3

=

=

=

30sin

cos

tan

21

3023

303

1

c

c

c

=

=

=

It may be easier to remember the triangle

rather than all these ratios.

DID YOU KNOW?

The ratios of all multiples of these angles follow a pattern:

A 0c 30c 45c 60c 90c 120c 135c 150c

sin A 20

21

22

23

24

23

22

21

cos A 24

23

22

21

20

2

1-

22-

2

3-

The rules of the pattern are:

for sin • A , when you reach 4, reverse the numbers

for cos • A , when you reach 0, change signs and reverse

ch6.indd 303 7/19/09 9:23:19 AM


EXAMPLES

1. Find the exact value of °.sec 45

Solution

°°

seccos

4545

1

211

2

=

=

=

2. A boat ramp is to be made with an angle of 30c and base length 5 m. What is the exact length of the surface of the ramp?

Solution

cos

cos

cos

xx

x

30 5

30 5

305

23

5

53

2

310

310 3

#

c

c

c

=

=

=

=

=

=

=

So the exact length of the ramp is .3

10 3m

ch6.indd 304 7/19/09 9:23:19 AM


6.6 Exercises

Find the exact value in all questions, with rational denominator where relevant.

1. Evaluate (a) sin cos60 60c c+

(b) cos sin45 452 2c c+ (c) cosec 45c (d) sec2 60c (e) cot cot30 60c c+

(f) tan tan60 30c c-

(g) sin sin60 452 2c c+

(h) sin cos cos sin45 30 45 30c c c c+

(i) tan3 30c

(j) tan tan

tan tan1 45 60

45 60c c

c c

-

+

(k) cos cos sin sin30 60 30 60c c c c-

(l) cos sin30 302 2c c+ sec cosec2 45 30c c-(m)

(n) sinsin

452 60

c

c

(o) tan1 302 c+

(p) coscos

1 451 45

c

c

+

-

(q) seccot

6030c

c

(r) sin 45 12 c -

(s) cosec5 602 c

(t) sec

tan45

2 602 c

c-

2. Find the exact value of all pronumerals

(a)

(b)

(c)

3. A 2.4 m ladder reaches 1.2 m up a wall. At what angle is it resting against the wall?

4. A 2-person tent is pitched at an angle of .45c Each side of the tent is 2 m long. A pole of what height is needed for the centre of the tent?

5. If the tent in the previous question was pitched at an angle of ,60c how high would the pole need to be?

6. The angle of elevation from a point 10 m out from the base of a tower to the top of the tower is .30c Find the exact height of the tower, with rational denominator.

cos 45 ( )cos 452 2c c=

ch6.indd 305 7/19/09 9:23:20 AM


7. The pitch of a roof is 45c and spans a length of 12 .m

What is the length (a) l of the

roof? If a wall is placed inside the (b)

roof one third of the way along from the corner, what height will the wall be?

8. A 1.8 m ladder is placed so that it makes a 60c angle where it meets

the fl oor. How far out from the wall is it?

9. Find the exact length of AC .

10. The angle of depression from the top of a 100 m cliff down to a boat at the foot of the cliff is 30 .c How far out from the cliff is theboat?

Angles of Any Magnitude

The angles in a right-angled triangle are always acute. However, angles greater than 90c are used in many situations, such as in bearings. Negative angles are also used in areas such as engineering and science.

We can use a circle to fi nd trigonometric ratios of angles of any magnitude (size) up to and beyond 360 .c

Investigation

(a) Copy and complete the table for these acute angles 1. (between 0c and 90c).

x 0c 10c 20c 30c 40c 50c 60c 70c 80c 90c

sin x

cos x

tan x

(b) Copy and complete the table for these obtuse angles (between 90c and 180c).

x 100c 110c 120c 130c 140c 150c 160c 170c 180c

sin x

cos x

tan x

ch6.indd 306 7/19/09 9:23:20 AM


(c) Copy and complete the table for these refl ex angles (between 180c and 270c).

x 190c 200c 210c 220c 230c 240c 250c 260c 270c

sin x

cos x

tan x

(d) Copy and complete the table for these refl ex angles (between 270c and 360c).

x 280c 290c 300c 310c 320c 330c 340c 350c 360c

sin x

cos x

tan x

What do you notice about their signs? Can you see any patterns? 2. Could you write down any rules for the sign of sin, cos and tan for different angle sizes? Draw the graphs of 3. siny x= , cosy x= and tany x= for .x0 360c c# # For tany x= , you may need to fi nd the ratios of angle close to and either side of 90c and 270c.

Drawing the graphs of the trigonometric ratios can help us to see the change in signs as angles increase.

We divide the domain 0c to 360c into 4 quadrants:

1 st quadrant: 0c to 90c 2 nd quadrant: 90c to 180c 3 rd quadrant: 180c to 270c 4 th quadrant: 270c to 360c

EXAMPLES

1. Describe the sign of sin x in each section (quadrant) of the graph .siny x=

Solution

We can sketch the graph using the table below or using the values from the tables in the investigation above for more accuracy.

x 0c 90c 180c 270c 360c

y 0 1 0 -1 0

CONTINUED

ch6.indd 307 7/19/09 9:23:21 AM


y

90c 180c 270c 360c

1

-1

y = sin x

x

The graph is above the x -axis for the fi rst 2 quadrants, then below for the 3 rd and 4 th quadrants. This means that sin x is positive in the 1 st and 2 nd quadrants and negative in the 3 rd and 4 th quadrants.

2. Describe the sign of cos x in each section (quadrant) of the graph of .cosy x=

Solution


x 0c 90c 180c 270c 360c

y 1 0 -1 0 1

y

90c 180c 270c 360c

1

-1

y = cos x

x

The graph is above the x -axis in the 1 st quadrant, then below for the 2 nd and 3 rd quadrants and above again for the 4 th quadrant.

ch6.indd 308 7/19/09 9:23:21 AM


This means that cos x is positive in the 1 st and 4 th quadrants and negative in the 2 nd and 3 rd quadrants.

3. Describe the sign of tan x in each section (quadrant) of the graph tany x= .

Solution


x 0c 90c 180c 270c 360c

y 0 No result 0 No result 0

Neither tan 90c nor tan 270c exists (we say that they are undefi ned). Find the tan of angles close to these angles, for example tan 89c 59l and tan 90c 01l, tan 279c 59l and tan 270 .01c l

There are asymptotes at 90c and 270 .c On the left of 90c and 270c, tan x is positive and on the right, the ratio is negative.

y

x90c 180c 270c 360c

y= tanx

The graph is above the x -axis in the 1 st quadrant, below for the 2 nd , above for the 3 rd and below for the 4 th quadrant.

This means that tan x is positive in the 1 st and 3 rd quadrants and negative in the 2 nd and 4 th quadrants.

You will see why these ratios are undefi ned later

on in this chapter.

To show why these ratios have different signs in different quadrants, we look at angles around a unit circle (a circle with radius 1 unit).

We use congruent triangles when fi nding angles of any magnitude. Page 310 shows an example of congruent triangles all with angles of 20c inside a circle with radius 1 unit.

ch6.indd 309 7/31/09 4:46:01 PM


y

x

1 unit 1 unit

20c20c 20c

20c

1 unit 1 unit

If we divide the circle into 4 quadrants, we notice that the x - and y -values have different signs in different quadrants. This is crucial to notice when looking at angles of any magnitude and explains the different signs you get when fi nding sin, cos and tan for angles greater than 90c.

Quadrant 1

Looking at the fi rst quadrant (see diagram below), notice that x and y are both positive and that angle i is turning anticlockwise from the x -axis.

(x, y)

1 unit

First quadrant

y

x

y

xi

Point ( x , y ) forms a triangle with sides 1, x and y , so we can fi nd the trigonometric ratios for angle i .

The angle at the x -axis is 0 and the angle at the y -axis is 90c, with all other angles in this quadrant between these two angles .

ch6.indd 310 7/19/09 9:23:24 AM


Investigation

Since cos xi = and sin yi = , we can write the point ( x , y ) as (cos i , sin i ) .

The polar coordinates (cos i , sin i ) give a circle.

The polar coordinates ,sin sinA a c B bi i+] ]g g6 @ form a shape called a Lissajous fi gure. These are sometimes called a Bowditch curve and they are often used as logos, for example the ABC logo.

Use the Internet to research these and other similar shapes.

Use a graphics calculator or a computer program such as Autograph to draw other graphs with polar coordinates using variations of sin i and cos i .

These are called polar coordinates.

Quadrant 2

In the second quadrant, angles are between 90c and 180 .c If we take the 1 st quadrant coordinates ( x , y ), where x 02 and 0y2 and

put them in the 2 nd quadrant, we notice that all x values are negative in the second quadrant and y values are positive.

So the point in the 2 nd quadrant will be (- x , y )

x

y

0c

90c

180c

(-x, y)

1 unit

Second quadrant

y

x

180c-i

i

siny

y1

i =

=

cos x

x1

i =

=

tan xy

i =

ch6.indd 311 8/1/09 5:38:54 PM


Since cos xi = , cos i will negative in the 2 nd quadrant. Since sin yi = , sin i will be positive in the 2 nd quadrant.

tan xy

i = so it will be negative (a positive number divided by a negative number).

To have an angle of i in the triangle, the angle around the circle is 180c - i .

Quadrant 3

In the third quadrant, angles are between 180c and 270 .c

90c

270c

x

y

0c180c

(-x, -y)

1 unit

Third quadrant

i

y

x180c + i

Notice that x and y are both negative in the third quadrant, so cos i and sin i will be both negative.

tan xy

i = so will be positive (a negative divided by a negative number).

To have an angle of i in the triangle, the angle around the circle is 180c + i .

Quadrant 4

In the fourth quadrant, angles are between 270c and 360 .c

90c

270c

x

y

0c180c

(x, -y)

1 unit

Fourth quadrant

y

x

360c - i360ci

ch6.indd 312 7/19/09 9:25:26 AM


While y remains negative in the fourth quadrant, x is positive again, so sin i is negative and cos i is positive.

tan xy

i = so will be negative (a negative divided by a positive number)

For an angle i in the triangle, the angle around the circle is 360c - i .

ASTC rule

Putting all of these results together gives a rule for all four quadrants that we usually call the ASTC rule.

4th quadrant

1st quadrant

3rd quadrant

2nd quadrant

S A

T C180c + i 360c - i

360c

180c - i i

90c

270c

0c180c

y

x

You could remember this rule as A ll S tations

T o C entral or A S illy T rigonometry C oncept, or

you could make up your own!

This rule also works for the reciprocal trigonometric ratios. For example, where cos is positive, sec is also positive, where sin is positive, so is cosec and where tan is positive, so is cot.

We can summarise the ASTC rules for all 4 quadrants:

A: ALL ratios are positive in the 1 st quadrant S: Sin is positive in the 2 nd quadrant (cos and tan are negative) T: Tan is positive in the 3 rd quadrant (sin and cos are negative) C: Cos is positive in the 4 th quadrant (sin and tan are negative)

First quadrant: Angle i : sin i is positive cos i is positive tan i is positive

ch6.indd 313 7/19/09 9:25:27 AM


Second quadrant: Angle 180c i- : sin sin180c i i- =] g cos cos180c i i- = -] g tan tan180c i i- = -] g

Third quadrant: Angle 180c i+ : sin sin180c i i+ = -] g cos cos180c i i+ = -] g tan tan180c i i+ =] g

Fourth quadrant: Angle 360c i- : sin sin360c i i- = -] g cos cos360c i i- =] g tan tan360c i i- = -] g

EXAMPLES

1. Find all quadrants where (a) sin 02i (b) cos 01i (c) tan cos0 0and1 2i i

Solution

(a) sin 02i means sin i is positive. Using the ASTC rule, sin i is positive in the 1 st and 2 nd quadrants.

cos (b) i is positive in the 1 st and 4 th quadrants, so cos i is negative in the 2 nd and 3 rd quadrants.

tan (c) i is positive in the 1 st and 3 rd quadrants so tan i is negative in the 2 nd and 4 th quadrants. Also cos i is positive in the 1 st and 4 th quadrants. So tan i 1 0 and cos i 2 0 in the 4 th quadrant.

ch6.indd 314 7/19/09 9:26:07 AM


2. Find the exact ratio of tan 330c .

Solution

First we fi nd the quadrant that 330c is in. It is in the 4 th quadrant.

The angle inside the triangle in the 4 th quadrant is 30c and tan is negative in the 4 th quadrant.

tan tan330 30

31

c c= -

= -

3. Find the exact value of sin 225c .

Solution

The angle in the triangle in the 3 rd quadrant is 45c and sin is negative in the 3 rd quadrant.

CONTINUED

Notice that 30 3 0 .360 3c c c=-

Notice that 1 0 .45 2258 c c c=+

330c 30c

y

x

60c

30c

2

1

:3

ch6.indd 315 7/19/09 9:45:11 AM


225c45c

y

x

sin sin225 45

21

c c= -

= -

4. Find the exact value of cos 510c .

Solution

To fi nd cos 510c, we move around the circle more than once.

510c

150c30c

y

x

510 360 150

510 360 150

So

c c c

c c c

- =

= +

45c

45c

1

1:2

ch6.indd 316 7/19/09 9:45:45 AM


The angle is in the 2 nd quadrant where cos is negative. The triangle has 30c in it.

cos cos510 30

23

c c= -

= -

5. Simplify cos (180c + x ) .

Solution

180c + x is an angle in the 3 rd quadrant where cos is negative. So cos cosx x180c + = -] g

6. If sin 53x = - and cos x 2 0, fi nd the value of tan x and sec x .

Solution

sin x 1 0 in the 3 rd and 4 th quadrants and cos x 2 0 in the 1 st and 4 th quadrants. So sin x 1 0 and cos x 2 0 in the 4 th quadrant. This means that tan x 1 0 and sec x 2 0.

sin xhypotenuse

opposite=

So the opposite side is 3 and the hypotenuse is 5.

35

y

xx

By Pythagoras’ theorem, the adjacent side is 4.

sec x is the reciprocal of cos x so is positive in the

4 th quadrant .

This is a 3-4-5 triangle .

Notice that 180 .30 150c c c=-

CONTINUED

60c

30c

2

1

:3

ch6.indd 317 7/19/09 10:08:53 AM


So tan x43

= -

sec cosx x

1

45

=

=

The ASTC rule also works for negative angles. These are measured in the opposite way (clockwise) from positive angles as shown.

4th quadrant

1st quadrant

3rd quadrant

2nd quadrant

S A

T C-i

0

-(180c+ i )

-(180c- i )

-(360c- i )

-180c

y

x

-90c

-270c

-360c

The only difference with this rule is that the angles are labelled differently.

EXAMPLE

Find the exact value of tan (-120c) .

Solution

Moving around the circle the opposite way, the angle is in the 3 rd quadrant, with 60c in the triangle.

y

x120c

60c

Notice that 180 0 1 0 .( 6 ) 2c c c=- - -

ch6.indd 318 7/19/09 10:12:04 AM


Tan is positive in the 3 rd quadrant.

tan tan120 60

3

c c- =

=

] g

6.7 Exercises

1. Find all quadrants where (a) cos 02i

(b) tan 02i

(c) sin 02i

(d) tan 01i

(e) sin 01i

(f) cos 01i

(g) sin 01i and tan 02i

(h) cos 01i and tan 02i

(i) sin 02i and tan 01i

(j) sin 01i and tan 01i

2. (a) Which quadrant is the angle 240c in?

Find the exact value of cos (b) 240c .


Find the exact value of sin (b) 315c .


Find the exact value of (b) tan 120c .

5. (a) Which quadrant is the angle -225c in?

Find the exact value of (b) sin (-225c) .

6. (a) Which quadrant is the angle -330c in?

Find the exact value of (b) cos (-330c) .

7. Find the exact value of each ratio. tan 225(a) c cos 315(b) c tan 300(c) c sin 150(d) c cos 120(e) c sin 210(f) c cos 330(g) c tan 150(h) c sin 300(i) c cos 135(j) c

8. Find the exact value of each ratio. cos ((a) -225c) cos ((b) -210c) tan ((c) -300c) cos ((d) -150c) sin ((e) -60c) tan ((f) -240c) cos ((g) -300c) tan ((h) -30c) cos ((i) -45c) sin ((j) -135c)

60c

30c

2

1

:3

ch6.indd 319 7/19/09 10:56:10 AM


Trigonometric Equations

Whenever you fi nd an unknown angle in a triangle, you solve a trigonometric equation e.g. .cos x 0 34= . You can fi nd this on your calculator.

Now that we know how to fi nd the trigonometric ratios of angles of any magnitude, there can be more than one solution to a trigonometric equation if we look at a larger domain.

9. Find the exact value of cos 570(a) c tan 420(b) c sin 480(c) c cos 660(d) c sin 690(e) c tan 600(f) c sin 495(g) c cos 405(h) c tan 675(i) c sin 390(j) c

10. If tan43

i = and cos 01i , fi nd

sin i and cos i as fractions.

11. Given sin74

i = and tan 01i ,

fi nd the exact value of cos i and tan i .

12. If sin x 1 0 and tan x85

= - , fi nd

the exact value of cos x and cosec x.

13. Given cos 52x = and ,tan x 01

fi nd the exact value of cosec x , cot x and tan x .

14. If cos x 1 0 and sin x 1 0, fi nd cos x and sin x in surd form with

rational denominator if tan x75

= .

15. If sin94

i = - and

270 360c c1 1i , fi nd the exact

value of tan i and sec i .

16. If cos83

i = - and

° °180 2701 1i , fi nd the exact value of tan x , sec x and cosec x .

17. Given sin 0.3x = and tan x 1 0, express sin (a) x as a fraction fi nd the exact value of cos (b) x

and tan x .

18. If tan 1.2a = - and ° °270 3601 1i , fi nd the exact values of cot a , sec a and cosec a .

19. Given that 0.7cos i = - and 90 180c c1 1i , fi nd the exact value of sin i and cot i .

20. Simplify (a) sin 180c i-] g

(b) cos x360c -] g

(c) tan 180c b+^ h

(d) sin 180c a+] g

(e) tan 360c i-] g

(f) sin i-] g

(g) cos a-] g

(h) tan x-] g

Use Pythagoras’ theorem to fi nd the third side .

This is called the principle solution.

ch6.indd 320 7/19/09 10:30:18 AM


EXAMPLES

1. Solve cos x23

= in the domain ° °x0 360# # .

Solution

23

is a positive ratio and cos is positive in the 1 st and 4 th quadrants .

So there are two possible answers. In the 1 st quadrant, angles are in the form of i and in the 4 th quadrant angles are in the form of 360c - i .

cos 3023

c =

But there is also a solution in the 4 th quadrant where the angle is 360c - i .

cos x23

For =

,,

x 30 360 3030 330c c cc c

= -=

2. Solve sin x2 1 02 =- for .x0 360c c# #

Solution

sin

sin

sin

sin

x

x

x

x

2 1 0

2 1

21

2

1

21

2

2

2

!

!

- =

=

=

=

=

Since the ratio could be positive or negative, there are solutions in all 4 quadrants. 1 st quadrant: angle i 2 nd quadrant: angle 180c - i 3 rd quadrant: angle 180c + i 4 th quadrant: angle 360c - i

60c

30c

2

1

:3

This is called the principle solution.

CONTINUED

ch6.indd 321 7/19/09 10:30:27 AM


, , ,

, , ,

sin

x

452

1

45 180 45 180 45 360 45

45 135 225 315

c

c c c c c c c

c c c c

=

= - + -

=

3. Solve tan x 3= for x180 180c c# #- .

Solution

3 is a positive ratio and tan is positive in the 1 st and 3 rd quadrants . So there are two possible answers. In the domain x180 180c c# #- , we use positive angles for x0 180c c# # and negative angles for .x180 0c c# #-

In the 1 st quadrant, angles are in the form of i and in the 3 rd quadrant angles are in the form of 180c i- -^ h . tan 60 3c = But there is also a solution in the 3 rd quadrant where the angle is

180c i- -^ h .

120c

,,

tan xx

360 180 6030

Forc c c

c

=

= - -

= -

] g

45c

45c

1

1:2

60c

30c

2

1

:3

4th quadrant

1st quadrant

3rd quadrant

2nd quadrant

S A

T C-(180c - i) -i

180c - i

90c

-90c

0c0c

180c-180c

y

x

i

ch6.indd 322 7/19/09 10:30:52 AM


4. Solve sin x2 2 1 0- = for 0 360xc c# # .

Solution

Notice that the angle is 2 x but the domain is for x . If 0 360xc c# # then we multiply each part by 2 to get the domain for 2 x .

0 2 720xc c# #

This means that we can fi nd the solutions by going around the circle twice!

sin

sin

sin

sin

x

x

x

2 2 1 0

2 2 1

221

3021

c

- =

=

=

=

Sin is positive in the 1 st and 2 nd quadrants. First time around the circle, 1 st quadrant is i and the 2 nd quadrant is 180c i- . Second time around the circle, we add 360c to the angles. So 1 st quadrant answer is 360c i+ and the 2 nd quadrant answer is 360 180c c i+ -] g or 540c i- .

,, ,

, , ,

, , ,

x

x

2 30 180 30 360 30 540 30

30 150 390 510

15 75 195 255

So c c c c c c c

c c c c

c c c c

= - + -

=

=`

The trigonometric graphs can also help solve some trigonometric equations.

EXAMPLE

Solve cos x 0= for 0 360xc c# # . cos 90 0c = However, looking at the graph of cosy x= shows that there is another solution in the domain 0 360xc c# # .

0

90 , 270

cos x

x

For

c c

=

=

y

90c 180c 270c 360c

1

-1

x

Notice that these solutions lie inside the original domain of

.0 x 360c c# #

60c

30c

2

1

:3

ch6.indd 323 7/19/09 10:56:26 AM


Investigation

Here are the 3 trigonometric graphs that you explored earlier in the chapter.

siny x=

cosy x=

tany x=

Use the values in the sin, cos and tan graphs to fi nd values for the inverse trigonometric functions in the tables below and then sketch the inverse trigonometric functions.

For example sin °270 1= -

cosec 270

11

1

So c =-

= -

Some values will be undefi ned, so you will need to fi nd values near them in order to see where the graph goes.

cosecy x=

x 0c 90c 180c 270c 360c

sin x

cosec x

ch6.indd 324 7/19/09 10:31:18 AM


Here are the graphs of the inverse trigonometric functions.

cosecy x=

secy x=

coty x=

secy x=

x 0c 90c 180c 270c 360c

cos x

sec x

coty x=

x 0c 90c 180c 270c 360c

tan x

cot x

y

y = cotx

x90c 180c 270c 360c

360c-1

1

0

ch6.indd 325 7/19/09 10:31:30 AM


1. Solve for .0 360c c# #i (a) .sin 0 35i =

(b) cos21

i = -

(c) tan 1i = -

(d) sin23

i =

(e) tan3

1i = -

(f) cos2 3i =

(g) tan 2 3i =

(h) sin2 3 1i = -

(i) cos2 2 1 0i - =

(j) tan 3 12 i =

2. Solve for .180 180c c# #i- (a) .cos 0 187i =

(b) sin21

i =

(c) tan 1i =

(d) sin23

i = -

(e) tan3

1i = -

(f) tan3 12 i =

(g) tan 2 1i =

(h) sin2 3 12 i =

(i) 1 0tani + =

(j) tan 2 32 i =

3. Sketch cosy x= for 0 360 .xc c##

4. Evaluate .sin 270c

5. Sketch tany x= for 0 360 .xc c# #

6. Solve 0tanx = for 0 360 .xc c# #

7. Evaluate .cos180c

8. Find the value of .sin90c

9. Solve cosx 1= for .x0 360c c# #

10. Sketch siny x= for .x180 180c c# #-

11. Evaluate .cos 270c

12. Solve sin x 1 0+ = for .x0 360c c# #

13. Solve cos x 12 = for .x0 360c c# #

14. Solve sin x 0= for .x0 360c c# #

15. Solve sin x 1= for .x360 360c c# #-

16. Sketch secy x= for .x0 360c c# #

17. Sketch coty x= for

.x0 360c c# #

6.8 Exercises

Trigonometric Identities

Trigonometric identities are statements about the relationships of trigonometric ratios. You have already met some of these—the reciprocal ratios, complementary angles and the rules for the angle of any magnitude.

ch6.indd 326 7/19/09 10:31:54 AM


cosecsin

seccos

cottan

1

1

1

ii

ii

ii

=

=

=

Reciprocal ratios

Complementary angles

sin cos

cosec sec

tan cot

90

90

90

c

c

c

i i

i i

i i

= -

= -

= -

]

]

]

g

g

g

Angles of any magnitude

sin sin

cos cos

tan tan

180

180

180

c

c

c

i i

i i

i i

- =

- = -

- = -

]

]

]

g

g

g

( )

( )

( )

sin sin

cos cos

tan tan

180

180

180

c

c

c

i i

i i

i i

+ = -

+ = -

+ =

( )

( )

( )

sin sin

cos cos

tan tan

360

360

360

c

c

c

i i

i i

i i

- = -

- =

- = -

)

)

cos

tan

i i

i i

=

= -

) sini i= -(

(

(

sin

cos

tan

-

-

-

In this section you will learn some other identities, based on the unit circle. In the work on angles of any magnitude, we defi ned sin i as the y -coordinate of P and cos i as the x -coordinate of P .

ch6.indd 327 7/19/09 10:32:04 AM


tan

cossinxy

i

ii

=

=

tancossin

iii

=

cot

tan

sincos

1i

i

ii

=

=

cotsincos

iii

=

Pythagorean identities

The circle has equation .x y 12 2+ = Substituting cosx i= and siny i= into 1x y2 2+ = gives

cos sin 12 2i i+ =

This is an equation so can be rearranged to give

sin coscos sin

11

2 2

2 2i ii i==

--

There are two other identities that can be derived from this identity.

tan sec1 2 2i i+ =

Remeber that cos2 i means (cos ) 2i .

ch6.indd 328 7/19/09 10:32:13 AM


Proof

cos sin

coscos

cossin

costan sec

1

1

1

2 2

2

2

2

2

2

2 2

i i

i

i

i

i

i

i i

+ =

+ =

+ =

This identity can be rearranged to give

tan sec

sec tan

1

1

2 2

2 2

i i

i i

=

=

-

-

cot cosec12 2i i+ =

Proof

cos sin

sincos

sinsin

sincot cosec

1

1

1

2 2

2

2

2

2

2

2 2

i i

i

i

i

i

i

i i

+ =

+ =

+ =

This identity can be rearranged to give

cot cosec

cosec cot

1

1

2 2

2 2

i i

i i

= -

= -

These are called Pythagorean identities since the equation

of the circle comes from Pythagoras’ rule (see Chapter 5).

EXAMPLES

1. Simplify .sin coti i

Solution

sin cot sin

sincos

cos

#i i iii

i

=

=

2. Simplify sin sec90c b b-^ h where b is an acute angle .

Solution

1

sin sec coscos

90 1#c b b b

b- =

=

^ h

CONTINUED

ch6.indd 329 7/19/09 10:32:22 AM


3. Simplify .sin sin cos4 2 2i i i+

Solution

sin sin cos sin sin cos

sin

sinsin

1

4 2 2 2 2 2

2

2

i i i i i i

i

ii

+ = +

=

=

=

^

]

h

g

4. Prove .cot tan cosec secx x x x+ =

Solution

cot tan

sincos

cossin

sin coscos sin

sin cos

sin coscosec sec

x x

xx

xx

x xx x

x x

x xx x

1

1 1

LHS

RHS

2 2

#

= +

= +

=+

=

=

=

=

cot tan cosec secx x x x+ =`

5. Prove that .sin

coscosx

xx

11

12

-=

+

Solution

sincos

coscos

cos coscos

cos

xx

xx

x xx

x

1

11

1 11

11

LHS

RHS

2

2

=-

=-

-

=+ -

-

=+

=

] ]g g

11sin

coscosx

xx

12

`-

+=

ch6.indd 330 7/31/09 6:47:01 PM


6.9 Exercises

1. Simplify (a) sin 90c i-] g (b) tan 360c i-] g (c) cos i-] g (d) cot 90c i-] g (e) sec 180c a+] g

2. Simplify (a) tan cosi i (b) tan coseci i (c) sec cotx x (d) sin x1 2-

(e) cos1 2 a-

(f) cot x 12 +

(g) tan x1 2+

(h) 1sec2 i -

(i) cot5 52 i +

(j) cosec x

12

(k) sin cosec2 2a a

(l) cot cot cos2i i i-

3. Prove that (a) cos sinx x12 2- = -

(b) sec tancos

sin1i i

ii

+ =+

(c) tansin

3 31

322

aa

+ =-

(d) sec tanx x2 2-

cosec cotx x2 2= -

(e) sin cosx x 3-] g

+2sin cosx xsin cos sin cosx x x x2 2

2= - -

(f) cot sec2i i+

sin cossin sin1 22

i ii i

=- +

(g) cos cot902 c i i-] g

= sin i cos i

(h) ( )( )cosec cot cosec cotx x x x 1+ - =

( )cos

sin cos

tan cos

1i2

2 2

2 2

i

i i

i i

-

= +

( )cosec

cotcos

tan cot

sec

1j

b

bb

b b

b

+-

=+

4. If cosx 2 i= and siny 2 i= , show that 4x y2 2+ = .

5. Show that 81x y2 2+ = if

cosx 9 i= and y = 9 sin i.

Non-right-angled Triangle Results

A non-right-angled triangle is named so that its angles and opposite sides have the same pronumeral. There are two rules in trigonometry that refer to non-right-angled triangles. These are the sine rule and the cosine rule .

ch6.indd 331 7/19/09 10:32:41 AM


Proof

In ,ABCT draw perpendicular AD and call it h .

From ,ABDT

sinsin

B ch

h c B`

=

=

(1)

From ,ACDT

sin

sin

Cbh

h b C`

=

=

(2)

From (1) and (2),

sin sin

sin sin

c B b C

bB

cC

=

=

Similarly, drawing a perpendicular from C it can be proven that

.sin sina

Ab

B=

or

sin sin sina

Ab

Bc

C= = Use this rule for fi nding an angle.

Use this rule for fi nding a side.

sin sin sinAa

Bb

Cc

= =

Sine rule

ch6.indd 332 7/19/09 10:32:53 AM


EXAMPLES

1. Find the value of x , correct to 1 decimal place.

Solution

Name the sides a and b, and angles A and B.

.

.

.

.

sin sin

sin sin

sin sin

sinsin

sin sin

Aa

Bb

x

x

x

43 21 79 1210 7

43 21 79 1210 7

79 1210 7 43 21

7 5

43 21 43 21

cm

# #

c c

c c

c

c

c c

Z

=

=

=

=

l l

l l

l

l

l l

2. Find the value of y , to the nearest whole number.

Solution

( )

sin sin

sin sin

sin sin

sinsin

sin sin

Y

Aa

Bb

y

y

y

180 53 24

103

103 538

103 538

538 103

10

103 103# #

c c c

c

c c

c c

c

c

c c

+

Z

= - +

=

=

=

=

=

You can rename the triangle ABC or just make

sure you put sides with their opposite angles

together.

The sine rule uses 2 sides and 2 angles, with 1

unknown.

You need to fi nd Y+ fi rst, as it is opposite y .

CONTINUED

ch6.indd 333 7/19/09 10:33:05 AM


3. Find the value of ,i in degrees and minutes.

Solution

1-

. .

. .86 11

..

..

6.7 6.7

sin sin

sin sin

sin sin

sin sin

sin sin

aA

bB

6 7 8 386 11

6 7 8 3

8 36 7 86 11

8 36 7 86 11

53 39

# #

c

c

c

c

cZ

i

i

i

i

=

=

=

=

=

l

l

l

l

l

c m

Since sin x is positive in the fi rst 2 quadrants, both acute angles (between 0c and 90c) and obtuse angles (between 90c and 180c) give positive sin ratios. e.g. .sin50 0 766c= and .sin130 0 766c = This affects the sine rule, since there is no way of distinguishing between an acute angle and an obtuse angle. When doing a question involving an obtuse angle, we need to use the 2 nd quadrant angle of 180c - i rather than relying on the calculator to give the correct answer.

EXAMPLE

Angle i is obtuse. Find the value of ,i in degrees and minutes.

ch6.indd 334 7/19/09 10:33:15 AM


6.10 Exercises

1. Evaluate all pronumerals, correct to 1 decimal place.

(a)

(b)

(c)

(d)

(e)

Solution

. .

. .

..

..

. .

sin sin

sin sin

sin sin

sin sin

sin sin

aA

bB

11 9 5 415 49

11 9 5 415 49

5 411 9 15 49

5 411 9 15 49

36 55

180 36 55

143 05

11 9 11 9

acute angleBut is obtuse

1

`

# #

c

c

c

c

c

c c

c

i

i

i

i

i

i

=

=

=

=

=

=

= -

=

-

l

l

l

l

l

l

l

c

^

m

h

ch6.indd 335 7/19/09 10:33:24 AM


2. Find the value of all pronumerals, in degrees and minutes.

(a)

(b)

(c)

(d)

((e) i is obtuse)

3. Triangle ABC has an obtuse angle at A . Evaluate this angle to the nearest minute if AB = 3.2 cm,

BC = 4.6 cm and .ACB 33 47c+ = l

4. Triangle EFG has FEG 48c+ = , EGF 32c+ = and FG = 18.9 mm. Find the length of

the shortest side (a) the longest side. (b) .

5. Triangle XYZ has ,XYZ 51c+ = YXZ 86c+ = and XZ = 2.1 m. Find the length of

the shortest side (a) the longest side .(b)

6. Triangle XYZ has XY = 5.4 cm, ZXY 48c+ = and .XZY 63c+ = Find the length of XZ .

7. Triangle ABC has BC = 12.7 m, ABC 47c+ = and ACB 53c+ = as shown. Find the lengths of

(a) AB (b) AC.

53c47c

12.7 mCB

A

8. Triangle PQR has sides PQ = 15 mm, QR = 14.7 mm and PRQ 62 29c+ = l . Find to the nearest minute

(a) QPR+ (b) .PQR+

9. Triangle ABC is isosceles with AB = AC . BC is produced to D as shown. If AB = 8.3 cm, BAC+ = 52c and ADC 32c+ = fi nd the length of

4.9

3.7

21c31l i

The shortest side is opposite the smallest angle and the longest side is opposite the largest angle .

ch6.indd 336 7/19/09 10:33:33 AM


(a) AD (b) BD.

32c

52c8.3 cm

DBC

A

10. Triangle ABC is equilateral with side 63 mm. A line is drawn from A to BC where it meets BC at D and .DAB 26 15c+ = l Find the length of

(a) AD (b) DC.

Cosine rule

cosc a b ab C22 2 2= + -

Similarly

cosa b c bc A22 2 2= + -

cosb a c ac B22 2 2= + -

Proof

BCD

cbp

x a - x

A

In triangle ABC , draw perpendicular CD with length p and let CD = x . Since BC = a , BD = a - x From triangle ACD

b x p2 2 2= + (1)

cos

cos

Cbx

b C x`

=

=

(2)

From triangle DAB

c p a xp a ax xp x a ax

22

2 2 2

2 2 2

2 2 2

= + -

= + - +

= + + -

] g

(3)

ch6.indd 337 7/19/09 10:33:42 AM


Substitute (1) into (3):

c b a ax22 2 2= + - (4)

Substituting (2) into (4):

cos

cos

c b a a b C

b a ab C

2

2

2 2 2

2 2

= + -

= + -

] g

DID YOU KNOW?

Pythagoras’ theorem is a special case of the cosine rule when the triangle is right angled.

cosc a b 2ab C2 2 2= -+

When C = 90c

2 cos 90

2 0

c a b ab

a b ab

a b

2 2 2

2 2

2 2

=

=

- c+

+ -

= +

] g

EXAMPLE

Find the value of x , correct to the nearest whole number.

Solution

.99 79Z

10Z

5.6 6.4 2(5.6)(6.4) 112 32

.

cos

cos

c a b ab C

x

x

2

99 79

2 2 2

2 2 2 c

= + -

= + -

=

l

The cosine rule uses 3 sides and 1 angle, with 1 unknown.

, ,, , ,,% %

Press 5.6 6.4 2 5.6 6.4

cos 112 32

x x2 2# #

#

+ -

= =

ch6.indd 338 7/19/09 10:33:51 AM


When fi nding an unknown angle, it is easier to change the subject of this formula to cos C.

cos

cos

cos

cos

cos

cos

cos cos

c a b ab C

c a b ab C

c ab C a b

c ab C a b

ab C a b c

ab C a b c

ab C ab C

c c

ab ab

2

2

2

2

2

2

2 2

2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

2 2

= + -

= + -

+ = +

+ = +

= + -

=+ -

+ +

- -

Subtract the square of the side opposite the

unknown angle.

EXAMPLES

1. Find ,i in degrees and minutes.

Solution

cos

cos

cos

Cab

a b c2

2 5 65 6 3

6052

6052

29 56

2 2 2

2 2 2

1

cZ

i

i

=+ -

=+ -

=

= -

l

] ]

c

g g

m

2. Evaluate BAC+ in degrees and minutes .

C

4.5 cm

A

B8.4 cm

6.1 cm

CONTINUED

cosCab

a b c2

2 2 2

=+ -

Similarly

cos Abc

b c a2

2 2 2

=+ -

cos Bac

a c b2

2 2 2

=+ -

ch6.indd 339 7/19/09 10:33:59 AM


Solution

1-

. .. . .

.

.

cos

cos

cos

Cab

a b c

BAC

BAC

2

2 4 5 6 14 5 6 1 8 4

0 2386

0 2386103 48

2 2 2

2 2 2

c

+

+

=+ -

=+ -

= -

= -

= l

] ]

]

g g

g

Notice that the negative sign tells us that the angle will be obtuse.

6.11 Exercises

1. Find the value of all pronumerals, correct to 1 decimal place.

(a)

(b)

(c)

(d)

(e)

2. Evaluate all pronumerals correct to the nearest minute

(a)

(b)

(c)

ch6.indd 340 7/19/09 10:34:10 AM


(d)

(e)

3. Kite ABCD has AB = 12.9 mm, CD = 23.8 mm and ABC 125c+ = as shown. Find the length of diagonal AC.

12.9 mm 125c

23.8 mm

A

B

C

D

4. Parallelogram ABCD has sides 11 cm and 5 cm, and one interior angle .79 25c l Find the length of the diagonals .

5. Quadrilateral ABCD has sides AB = 12 cm, BC = 10.4 cm, CD = 8.4 cm and AD = 9.7 cm with .ABC 63 57c+ = l

Find the length of diagonal (a) AC Find (b) DAC+ Find (c) .ADC+

6. Triangle XYZ is isosceles with XY = XZ = 7.3 cm and

YZ = 5.9 cm. Find the value of all angles, to the nearest minute .

7. Isosceles trapezium MNOP has MP = NO = 12 mm, MN = 8.9 mm, OP = 15.6 mm and 11 15 .NMP 9c l+ =

Find the length of diagonal (a) NP. Find (b) .NOP+

8. Given the fi gure below, fi nd the length of

(a) AC (b) AD.

8.4 cm

42c8l

A

B

C

D

101c38l

3.7 cm

9.9 cm

9. In a regular pentagon ABCDE with sides 8 cm, fi nd the length of diagonal AD .

10. A regular hexagon ABCDEF has sides 5.5 cm.

Find the length of (a) AD. Find (b) .ADF+

ch6.indd 341 7/19/09 10:34:21 AM


Application s

The sine and cosine rules can be used in solving problems.

Use the sine rule to fi nd:

a 1. side, given one side and two angles an 2. angle, given two sides and one angle

Use the cosine rule to fi nd: a 1. side, given two sides and one angle an 2. angle, given three sides

EXAMPLES

1. The angle of elevation of a tower from point A is .72c From point B , 50 m further away from the tower than A , the angle of elevation is .47c

Find the exact length of (a) AT . Hence, or otherwise, fi nd the height (b) h of the tower to 1 decimal place.

Solution

)

(a)

(

sin sin

sin sin

sinsin

BAT

BTA

Aa

Bb

AT

AT

180 72108

180 47 10825

47 2550

2550 47

straight angle

angle sum of

`

c c

c

c c c

c

c c

c

c

+

+ T

= -

=

= - +

=

=

=

=

^

]

h

g

Use BTAT to fi nd AT.

ch6.indd 342 7/19/09 10:34:31 AM


( )

50 47

.

sin

sin

sinsin sin

ATh

h AT

72

72

2572

82 3

b

m

`

#

c

c

c

cc

Z

=

=

=

2. A ship sails from Sydney for 200 km on a bearing of ,040c then sails on a bearing of 157c for 345 km.

How far from Sydney is the ship, to the nearest km? (a) What is the bearing of the ship from Sydney, to the nearest degree? (b)

Solution

)

( )

(

SAN

SAB

180 40140

360 140 15763

a cointerior angles

angle of revolution`

c c

c

c c c

c

+

+

= -

=

= - +

=

^

^

h

h

( ) ( )

.

.

cos

cos

c a b ab C

x

x

2

200 345 2 200 345 6396374 3

96374 3310

2 2 2

2 2 2 c

Z

Z

= + -

= + -

=

So the ship is 310 km from Sydney.

( )

.

sin sin

sin sin

sin sin

b aA

bB

345 31063

310345 63

0 99

82

`

c

c

c

Z

Z

i

i

i

=

=

=

40 82

122

The bearing from Sydney c c

c

= +

=

Use right-angled ATOT to fi nd h . Do not use the

sine rule.

To fi nd the bearing, measure TSB.+

ch6.indd 343 7/19/09 10:34:40 AM


6.12 Exercises

1. Find the lengths of the diagonals of a parallelogram with adjacent sides 5 cm and 8 cm and one of its angles .32 42c l

2. A car is broken down to the north of 2 towns. The car is 39 km from town A and 52 km from town B .If A is due west of B and the 2 towns are 68 km apart, what is the bearing of the car from (a) town A (b) town B , to the nearest degree?

3. The angle of elevation to the top of a tower is 54 37c l from a point 12.8 m out from its base. The tower is leaning at an angle of 85 58c l as shown. Find the height of the tower.

54c37l 85c58l

12.8 m

4. A triangular park has sides 145.6 m, 210.3 m and 122.5 m. Find the size of the largest interior angle of the park.

5. A 1.5 m high fence leans outwards from a house at an angle of 102c. A boy sits on top of the fence and the angle of depression from him down to the house is .32 44c l How far from the fence is the house?

6. Football posts are 3.5 m apart. If a footballer is standing 8 m

from one post and 11 m from the other, fi nd the angle within which the ball must be kicked to score a goal, to the nearest degree.

7. A boat is sinking 1.3 km out to sea from a marina. Its bearing is 041c from the marina and 324c from a rescue boat. The rescue boat is due east of the marina.

How far, correct to 2 decimal (a) places, is the rescue boat from the sinking boat?

How long will it take the (b) rescue boat, to the nearest minute, to reach the other boat if it travels at 80 km/h?

8. The angle of elevation of the top of a fl agpole from a point a certain distance away from its base is .20c After walking 80 m towards the fl agpole, the angle of elevation is .75c Find the height of the fl agpole, to the nearest metre.

9. A triangular fi eld ABC has sides 85AB m= and 50 .AC m= If B is on a bearing of 065c from A and C is on a bearing of 166c from A , fi nd the length of BC , correct to the nearest metre.

10. (a) Find the exact value of AC in the diagram.

Hence, or otherwise, fi nd the (b) angle ,i correct to the nearest minute.

ch6.indd 344 7/19/09 10:34:48 AM


11. Find the value of h , correct to 1 decimal place.

12. A motorbike and a car leave a service station at the same time. The motorbike travels on a bearing of 080c and the car travels for 15.7 km on a bearing of 108c until the bearing of the motorbike from the car is .310c How far, correct to 1 decimal place, has the motorbike travelled?

13. A submarine is being followed by two ships, A and B , 3.8 km apart, with A due east of B . If A is on a bearing of 165c from the submarine and B is on a bearing of 205c from the submarine, fi nd the distance from the submarine to both ships.

14. A plane fl ies from Dubbo on a bearing of 1 93 c for 852 km, then turns and fl ies on a bearing of 2 58 cuntil it is due west of Dubbo. How far from Dubbo is the plane, to the nearest km?

15. A triangular roof is 16.8 m up to its peak, then 23.4 m on the other side with a 125c angle at the peak as shown. Find the length of the roof.

125c23.4 m16.8 m

16. Rhombus ABCD with side 8 cm has diagonal BD 11.3 cm long. Find .DAB+

17. Zeke leaves school and runs for 8.7 km on a bearing of 338c, then turns and runs on a bearing of 061c until he is due north of school. How far north of school is he?

18. A car drives due east for 83.7 km then turns and travels for 105.6 km on a bearing of 029c. How far is the car from its starting point?

19. The fi gure below shows the diagram that a surveyor makes to measure a triangular piece of land. Find its perimeter.

58c1l132c31l

14.3 m

11.4 m

13.9 m

20. A light plane leaves Sydney and fl ies for 1280 km on a bearing of

.050c It then turns and fl ies for 3215 km on a bearing of .149c How far is the plane from Sydney, to the nearest km?

21. Trapezium ABCD has AD BC; , with AB = 4.6 cm, BC = 11.3 cm, CD = 6.4 cm, DAC 2 303c+ = l and ABC 78c+ = .

Find the length of (a) AC. Find (b) ADC+ to the nearest

minute .

22. A plane leaves Adelaide and fl ies for 875 km on a bearing of

.056c It then turns and fl ies on a bearing of i for 630 km until it is due east of Adelaide. Evaluate i to the nearest degree.

ch6.indd 345 7/19/09 10:34:58 AM


Similarly,

sin

sin

A ac B

A bc A

21

21

=

=

Proof

From ,BCDD

sin

sin

sin

C ah

h a C

A bh

ba C

21

21

`

=

=

=

=

23. Quadrilateral ABCD has AB = AD = 7.2 cm, BC = 8.9 cmand CD = 10.4 cm, with DAB 107c+ =

Find the length of diagonal (a) BD. Find (b) BCD+ .

24. Stig leaves home and travels on a bearing of 248c for 109.8 km. He then turns and travels for 271.8 km on a bearing of 143c. Stig then turns and travels home on a bearing of a .

How far does he travel on the (a) fi nal part of his journey?

Evaluate (b) a .

25. A wall leans inwards and makes an angle of 88c with the fl oor.

A 4 m long ladder leans against (a) the wall with its base 2.3 m out from the wall. Find the angle that the top of the ladder makes with the wall.

A longer ladder is placed the (b) same distance out from the wall and its top makes an angle of 31c with the wall.

How long is this (i) ladder?

How much further (ii) does it reach up the wall than the fi rst ladder?

Area

To fi nd the area of a triangle, you need to know its perpendicular height. Trigonometry allows us to fi nd this height in terms of one of the angles in the triangle.

sinA ab C21

=

ch6.indd 346 7/19/09 10:35:07 AM


EXAMPLE

Find the area of ABCD correct to 2 decimal places.

Solution

( . ) ( . )

.

sin

sin

A ab C21

21 4 3 5 8 112 34

11 52 units2

c

Z

=

= l

To fi nd the area, use 2 sides and their included angle.

6.13 Exercises

1. Find the area of each triangle correct to 1 decimal place. (a)

(b)

(c)

(d)

(e)

ch6.indd 347 7/19/09 10:35:15 AM


2. Calculate the exact area of .ABCD

3. Find the area of OABD correct to 1 decimal place ( O is the centre of the circle).

4. Find the area of a parallelogram with sides 3.5 cm and 4.8 cm, and one of its internal angles ,67 13c l correct to 1 decimal place.

5. Find the area of kite ABCD , correct to 3 signifi cant fi gures.

6. Find the area of the sail, correct to 1 decimal place.

7. Find the area of a regular hexagon with sides 4 cm, to the nearest .cm2

8. Calculate the area of a regular pentagon with sides 12 mm.

9. The fi gure below is made from a rectangle and isosceles triangle with AE = AB as shown.

10.5 cm

84c

A

B

CD

14.3 cm

E

Find the length of (a) AE. Find the area of the fi gure .(b)

10. Given the following fi gure,

9.4 cm

44c

A

B C D6.7 cm

58c

36c

Find the length of (a) AC Find the area of triangle (b) ACD Find the area of triangle (c) ABC .

ch6.indd 348 7/19/09 10:35:24 AM


1. Find the exact value of cosi and sini if

.tan53

i =

2. Simplify

(a) sin cotx x

(b) cos

cos sin40

40 50c

c c+

(c) cot A1 2+

3. Evaluate to 2 decimal places. (a) sin 39 54c l (b) tan 61 30c l (c) cos 19 2c l

4. Find i to the nearest minute if (a) .sin 0 72i = (b) .cos 0 286i =

(c) tan75

i =

5. Prove that .sin

cos sin12 2 2

2

ii

i-

= +

6. Find the value of b if .sin cosb b2 30 c= -] g

7. Find the exact value of (a) cos 315c (b) sin 60c-] g (c) tan 120c

8. Solve cos x2 1= - for .x0 360c c# #

9. Sketch the graph of ,cosy x= and hence solve cos x 0= for 0 360 .xc c##

10. A ship sails on a bearing of 215c from port until it is 100 km due south of port. How far does it sail, to the nearest km?

11. Find the length of AB as a surd.

12. Evaluate x , correct to 2 signifi cant fi gures. (a)

(b)

13. Evaluate i to the nearest minute.

(a)

(b)

(c)

14. Find the area of triangle MNO .

Test Yourself 6

ch6.indd 349 7/19/09 10:56:50 AM


15. Solve for x180 180c c# # .-

(a) sin x432 =

(b) tan x23

1=

(c) tan tanx x3 2 =

16. If sec45

i = - and ,tan 02i fi nd sin i and .cot i

17. Jacquie walks south from home for 3.2 km, then turns and walks west for 1.8 km. What is the bearing, to the nearest degree, of

Jacquie from her home? (a) her home from where Jacquie is now? (b)

18. The angle of elevation from point B to the top of a pole is 39 ,c and the angle of elevation from D , on the other side of the pole, is . B42c and D are 20 m apart.

(a) Find an expression for the length of AD . Find the height of the pole, to (b)

1 decimal place.

19. A plane fl ies from Orange for 1800 km on a bearing of .300c It then turns and fl ies for 2500 km on a bearing of .205c How far is the plane from Orange, to the nearest km?


1. Two cars leave an intersection at the same time, one travelling at 70 km/h along one road and the other car travelling at 80 km/h along the other road. After 2 hours they are 218 km apart. At what angle, to the nearest minute, do the roads meet at the intersection?

2. A ship sails from port on a bearing of ,055c then turns and sails on a bearing of 153c for 29.1 km, when it is due east of port. How far, to 1 decimal place, is the ship from its starting point?

3. Evaluate x correct to 3 signifi cant fi gures.

4. (a) Find an exact expression for the length of AC .

(b) Hence, or otherwise, fi nd the value of h correct to 1 decimal place.

5. A man walks 3.8 km on a bearing of 134c from a house. He then walks 2.9 km on a bearing of .029c How far is he from the house, to 1 decimal place?

ch6.indd 350 7/19/09 2:14:25 PM


6. Simplify .sin tanx x360 90$c c- -] ]g g

7. Find the exact area of .ABCD

8. Find the exact value of ( ) .cos 315c-

9. Solve 2 1 0tan x - = for .x0 360c c# #

10. Find i to the nearest minute.

11. The angle of depression from the top of a 4.5 m mast of a boat down to a fi sh is 56 28 .c l How far down, to 1 decimal place, does a pelican sitting at the top of the mast need to fl y to catch the fi sh?

12. Solve 2 ( 10 ) 1cos ci + = - for .0 360c c# #i

13. Two roads meet at an angle of 74 .c Find the distance, correct to 3 signifi cant fi gures, between two cars, one 6.3 km from the intersection along one road and the other 3.9 km along the other road.

14. Find the exact value of ,cosi given

sin95

i = and .cos 01i

15. From the top of a vertical pole the angle of depression to a man standing at the foot of the pole is 43 .c On the other side of the pole is another man, and the angle of depression from the top of the pole to this man is 52 .c The men are standing 58 m apart. Find the height of the pole, to the nearest metre.

16. Show that

.sin sin

cos sin costan

1 11

i i

i i ii

+ -

+= +

] ]

]

g g

g

17. If 3x = sin i and ,cosy 3 2i= - eliminate i to fi nd an equation relating x and y .

ch6.indd 351 7/19/09 2:18:12 PM

TERMINOLOGY

7 Linear Functions

Collinear points: Two or more points that lie on the same straight line

Concurrent lines: Two or more lines that intersect at a single point

Gradient: The slope of a line measured by comparing the vertical rise over the horizontal run. The symbol for gradient is m

Interval: A section of a straight line including the end points

Midpoint: A point lying exactly halfway between two points

Perpendicular distance: The shortest distance between a point and a line. The distance will be at right angles to the line

ch7.indd 352 7/18/09 3:14:41 PM

353Chapter 7 Linear Functions

INTRODUCTION

IN CHAPTER 5, YOU STUDIED functions and their graphs. This chapter looks at the linear function, or straight-line graph, in more detail. Here you will study the gradient and equation of a straight line, the intersection of two or more lines, parallel and perpendicular lines, the midpoint, distance and the perpendicular distance from a point to a line.

DID YOU KNOW?

Pierre de Fermat (1601–65) was a lawyer who dabbled in mathematics. He was a contemporary of Descartes, and showed the relationship between an equation in the form Dx By,= where D and B are constants, and a straight-line graph. Both de Fermat and Descartes only used positive values of x , but de Fermat used the x -axis and y -axis as perpendicular lines as we do today.

De Fermat’s notes Introduction to Loci, Method of Finding Maxima and Minima and Varia opera mathematica were only published after his death. This means that in his lifetime de Fermat was not considered a great mathematician. However, now he is said to have contributed as much as Descartes towards the discovery of coordinate geometry. De Fermat also made a great contribution in his discovery of differential calculus.

Class Assignment

Find as many examples as you can of straight-line graphs in newspapers and magazines.

Distance

The distance between two points (or the length of the interval between two points) is easy to fi nd when the points form a vertical or horizontal line.

ch7.indd 353 7/18/09 11:22:03 AM


EXAMPLES

Find the distance between 1. ,1 4-^ h and ,1 2- -^ h

Solution

Counting along the y -axis, the distance is 6 units.

2. ,3 2^ h and ,4 2-^ h

Solution

Counting along the x -axis, the distance is 7 units.

When the two points are not lined up horizontally or vertically, we use Pythagoras’ theorem to fi nd the distance.

ch7.indd 354 7/18/09 11:22:10 AM


EXAMPLE

Find the distance between points ,3 1-^ h and ,2 5-^ h.

Solution

5BC = and 6AC = By Pythagoras’ theorem,

5 625 36

61

7.81

c a b

AB

AB 61

2 2 2

2 2 2

`

Z

= +

= +

= +

=

=

You studied Pythagoras’ theorem in Chapter 4.

DID YOU KNOW?

Pythagoras made many discoveries about music as well as about mathematics. He found that changing the length of a vibrating string causes the tone of the music to change. For example, when a string is halved, the tone is one octave higher.

The distance between two points ,x y1 1_ i and ,x y2 2_ i is given by

d x x y y2 12

2 12= - + -_ _i i

ch7.indd 355 7/18/09 11:22:15 AM


Proof

Let ,x yA 1 1= _ i and ,x yB 2 2= _ i Length AC x x2 1= - and length BC y y2 1= - By Pythagoras’ theorem

AB AC BC

d x x y y

d x x y y

2 2 2

22 1

22 1

2

2 12

2 12

`

= +

= - + -

= - + -

_ _

_ _

i i

i i

EXAMPLES

1. Find the distance between the points ,1 3^ h and ,3 0-^ h .

Solution

Let ,1 3^ h be ,x y1 1_ i and ,3 0-^ h be ,x y2 2_ i

d x x y y

3 1 0 3

4 316 9

255

2 12

2 12

2 2

2 2

= - + -

= - - + -

= - + -

= +

=

=

_ _

] ]

] ]

i i

g g

g g

So the distance is 5 units.

2. Find the exact length of AB given that ,A 2 4- -= ^ h and ,B 1 5-= ^ h .

Solution

Let ,2 4- -^ h be ,x y1 1_ i and ,1 5-^ h be ,x y2 2_ i

4

d x x y y

1 2 5

1 91 81

82

2

2 12

2 12

2

2 2

= - + -

= - - - + - -

= +

= +

=

_ _

^ ^

i i

h h6 6@ @

If points A and B were changed around, the formula would be

( ) ( ) ,d x x y y1 22

1 22= - + -

which would give the same answer.

You would still get 82 if you used )( 2, 4- - as ( , )x y2 2 and ( ),1 5- as ( , )x y1 1 .

ch7.indd 356 7/18/09 11:22:17 AM


7.1 Exercises

1. Find the distance between points (a) ,0 2^ h and 3,6^ h (b) ,2 3-^ h and ,4 5-^ h (c) ,2 5-^ h and ,3 7-^ h

2. Find the exact length of the interval between points

(a) 2, 3^ h and ,1 1-^ h (b) ,5 1-^ h and 3, 0^ h (c) ,2 3- -^ h and 4,6-^ h (d) ,1 3-^ h and ,7 7-^ h

3. Find the distance, correct to 2 decimal places, between points

(a) ,1 4-^ h and 5,5^ h (b) 0, 4^ h and ,3 2-^ h (c) ,8 1-^ h and ,7 6-^ h

4. Find the perimeter of ABCD with vertices , ,,A B3 1 1 1-^ ^h h and , .C 1 2- -^ h

5. Prove that the triangle with vertices 3, 4^ h , ,2 7-^ h and ,6 1-^ h is isosceles.

6. Show that ,AB BC= where , ,,A B2 5 4 2= - -=^ ^h h and , .C 3 8= - -^ h

7. Show that points ,3 4-^ h and 8,1^ h are equidistant from point , .7 3-^ h

8. A circle with centre at the origin O passes through the point , .2 7_ i Find the radius of the circle, and hence its equation.

9. Prove that the points , ,,X Y2 3 1 10- -_ _i i and ,Z 6 5-_ i all lie on a circle with centre at the origin. Find its equation.

10. If the distance between ,a 1-^ h and 3, 4^ h is 5, fi nd the value of a .

11. If the distance between ,3 2-^ h and 4, a^ h is 7, fi nd the exact value of a .

12. Prove that , , ,A B1 4 1 2^ ^h h and ,C 1 3 3+_ i are the vertices of an equilateral triangle.

13. If the distance between , 3a^ h and 4, 2^ h is 37, fi nd the values of a .

14. The points , , ( , ),M N1 2 3 0- -^ h ,P 4 6^ h and ,Q 0 4^ h form a quadrilateral. Prove that MQ NP= and .QP MN= What type of quadrilateral is MNPQ?

15. Show that the diagonals of a square with vertices , , , , ,A B C2 4 5 4 5 3- -^ ^ ^h h h and ,D 2 3- -^ h are equal.

16. (a) Show that the triangle with vertices , , ,A B0 6 2 0^ ^h h and ,C 2 0-^ h is isosceles.

(b) Show that perpendicular ,OA where O is the origin, bisects BC .

17. Find the exact length of the diameter of a circle with centre ,3 4-^ h if the circle passes through the point ,7 5^ h .

18. Find the exact length of the radius of the circle with centre (1, 3) if the circle passes through the point ,5 2- -^ h .

19. Show that the triangle with vertices , , ,A B2 1 3 3-^ ^h h and ,C 7 7-^ h is right angled .

20. Show that the points , , ,X Y3 3 7 4-^ ^h h and ,Z 4 1-^ h form the vertices of an isosceles right-angled triangle .

ch7.indd 357 7/31/09 4:47:03 PM


Midpoint

The midpoint is the point halfway between two other points.

The midpoint of two points ,x y1 1_ i and ,x y2 2_ i is given by

,x x y y

M2 2

1 2 1 2=

+ +e o

Proof

Find the midpoint of points ,x yA 1 1_ i and ,x yB 2 2_ i . Let ,x yM = ^ h Then ABR;DAPQ <D

ARAQ

ABAP

` =

`

`

x xx x

x x x xx x x x

x x x

xx x

yy y

21

22 2

2

2

2Similarly,

2 1

1

1 2 1

1 2 1

1 2

1 2

1 2

-

-=

- = -

- = -

= +

=+

=+

_ i

Can you see why these triangles are similar?

EXAMPLES

1. Find the midpoint of ,1 4-^ h and 5, 2^ h .

Solution

xx x

21 2

=+

ch7.indd 358 7/18/09 11:22:20 AM


So ( , ) .

yy y

M

21 5

24

2

2

24 2

26

3

2 3

1 2

=- +

=

=

=+

=+

=

=

=

2. Find the values of a and b if ,2 3-^ h is the midpoint between ,7 8- -^ h and ,a b^ h .

Solution

So and .

xx x

a

a

a

yy y

b

b

b

a b

2

22

7

4 7

11

2

32

8

6 8

2

11 2

1 2

1 2

=+

=- +

= - +

=

=+

- =- +

- = - +

=

= =

Note that the x -coordinate of the midpoint is the average of x1 and .x2 The same applies to the y -coordinate.

PROBLEM

A timekeeper worked out the average time for 8 fi nalists in a race. The average was 30.55, but the timekeeper lost one of the fi nalist’s times. The other 7 times were 30.3, 31.1, 30.9, 30.7, 29.9, 31.0 and 30.3. Can you fi nd out the missing time?

ch7.indd 359 7/18/09 11:22:21 AM


7.2 Exercises

1. Find the midpoint of (a) ,0 2^ h and ,4 6^ h (b) ,2 3-^ h and ,4 5-^ h (c) ,2 5-^ h and ,6 7-^ h (d) ,2 3^ h and ,8 1-^ h (e) ,5 2-^ h and ,3 0^ h (f) ,2 2- -^ h and ,4 6-^ h (g) ,1 4-^ h and ,5 5^ h (h) 0, 4^ h and ,3 2-^ h (i) ,8 1-^ h and ,7 6-^ h (j) ,3 7^ h and ,3 4-^ h

2. Find the values of a and b if (a) ,4 1^ h is the midpoint of ,a b^ h

and ,1 5-^ h (b) ,1 0-^ h is the midpoint of

,a b^ h and ,3 6-^ h (c) ,a 2^ h is the midpoint of ( , b3 h

and ,5 6-^ h (d) ,2 1-^ h is the midpoint of

,a 4^ h and , b3-^ h (e) , b3^ h is the midpoint of ,a 2^ h

and ,0 0^ h

3. Prove that the origin is the midpoint of ,3 4-^ h and ,3 4-^ h .

4. Show that P Q= where P is the midpoint of ,2 3-^ h and ,6 5-^ h and Q is the midpoint of ,7 5- -^ h and ,11 3^ h .

5. Find the point that divides the interval between ,3 2-^ h and ,5 8^ h in the ratio of 1:1.

6. Show that the line 3x = is the perpendicular bisector of the interval between the points ,1 2-^ h and ,7 2^ h .

7. The points , , , ,A B1 2 1 5-^ ^h h ,C 6 5^ h and ,D 4 2^ h form a parallelogram. Find the midpoints of the diagonals AC and BD . What property of a parallelogram does this show?

8. The points , , , ,A B3 5 9 3-^ ^h h

,C 5 6-^ h and ,D 1 2-^ h form a quadrilateral. Prove that the diagonals are equal and bisect one another. What type of quadrilateral is ABCD ?

9. A circle with centre ,2 5-^ h has one end of a diameter at , .4 3-^ h Find the coordinates of the other end of the diameter.

10. A triangle has vertices at , ,,A B1 3 0 4-^ ^h h and ,C 2 2-^ h .

Find the midpoints (a) X , Y and Z of sides AB , AC and BC respectively.

Show that (b) ,XY BC21

=

XZ AC21

= and 21 .YZ AB=

11. Point ,x yP ^ h moves so that the midpoint between P and the origin is always a point on the circle 1.x y2 2+ = Find the equation of the locus of P .

12. Find the equation of the locus of the point ,x yP ^ h that is the midpoint between all points on the circle 4x y2 2+ = and the origin.

Gradient

The gradient of a straight line measures its slope. The gradient compares the vertical rise with the horizontal run.

The locus is the path that ( , )P x y follows.

ch7.indd 360 7/31/09 4:47:04 PM


Gradient runrise

=

On the number plane, this is a measure of the rate of change of y with respect to x .

The rate of change of y with respect to x is a very important measure of their relationship. In later chapters you will use the gradient for many purposes, including sketching curves, fi nding the velocity and acceleration of objects, and fi nding maximum and minimum values of formulae.

EXAMPLES

Find the gradient of each interval. 1.

Solution

Gradient run

rise

32

=

=

You will study the gradient at different points on a curve in

the next chapter.

CONTINUED

ch7.indd 361 7/18/09 11:22:22 AM


2.

Solution

In this case, x is 3- (the run is measured towards the left) .

Gradient runrise

32

32

=

=-

= -

Positive gradient leans to the right. Negative gradient leans to the left.

Gradient given 2 points

The gradient of the line between ,x y1 1_ i and ,x y2 2_ i is given by

m x xy y

2 1

2 1=

-

-

Proof

ch7.indd 362 7/18/09 11:22:23 AM


BC y y2 1= - and AC x x2 1= -

Gradient run

rise

x xy y

2 1

2 1

=

=-

-

This formula could also be

written mx x

y y

1 2

1 2=

-

-

EXAMPLES

1. Find the gradient of the line between points 2, 3^ h and , .3 4-^ h

Solution

Gradient: m x xy y

3 24 3

51

51

2 1

2 1=

-

-

=- -

-

=-

= -

2. Prove that points , ,,2 3 2 5- -^ ^h h and ,0 1-^ h are collinear.

Solution

To prove points are collinear, we show that they have the same gradient (slope).

Collinear points lie on the same line, so they have

the same gradients.

CONTINUED

ch7.indd 363 7/18/09 11:22:24 AM


Gradient of the interval between ,2 5- -^ h and ,0 1-^ h :

m x xy y

25

21 5

24

2

012 1

2 1=

-

-

=-

-

=- +

=

=

-

- -

]

]

g

g

Gradient of the interval between ,0 1-^ h and ,2 3^ h :

m x xy y

2 01

23 1

24

2

32 1

2 1=

-

-

=-

-

=+

=

=

- ] g

Since the gradient of both intervals is the same, the points are collinear.

Gradient given the angle at the x -axis

The gradient of a straight line is given by

tanm i=

where i is the angle the line makes with the x -axis in the positive direction

Proof

runrise

adjacent

opposite

tan

m

i

=

=

=

ch7.indd 364 7/18/09 11:22:26 AM


For an acute angle tan 02i . For an obtuse angle tan 01i .

Class Discussion

Which angles give a positive gradient? 1. Which angles give a negative gradient? Why? 2. What is the gradient of a horizontal line? What angle does it make 3. with the x -axis? What angle does a vertical line make with the 4. x -axis? Can you fi nd its gradient?

EXAMPLES

1. Find the gradient of the line that makes an angle of 135c with the x -axis in the positive direction.

Solution

tan

tan

m

135

1

c

i=

=

= -

2. Find the angle, in degrees and minutes, that a straight line makes with the x -axis in the positive direction if its gradient is 0.5.

Solution

.

tan

tan

m

0 5

26 34

`

c

i

i

i

=

=

= l

Can you see why the gradient is negative?

ch7.indd 365 7/18/09 11:22:26 AM


7.3 Exercises

1. Find the gradient of the line between

(a) ,3 2^ h and ,1 2-^ h (b) ,0 2^ h and ,3 6^ h (c) ,2 3-^ h and ,4 5-^ h (d) ,2 5-^ h and ,3 7-^ h (e) ,2 3^ h and ,1 1-^ h (f) ,5 1-^ h and ,3 0^ h (g) ,2 3- -^ h and ,4 6-^ h (h) ,1 3-^ h and ,7 7-^ h (i) ,1 4-^ h and ,5 5^ h (j) ,0 4^ h and ,3 2-^ h

2. If the gradient of , y8 1_ i and ,1 3-^ h is 2, fi nd the value of .y1

3. The gradient of ,2 1-^ h and ,x 0^ h is –5 . Find the value of x .

4. The gradient of a line is –1 and the line passes through the points ,4 2^ h and ,x 3-^ h . Find the value of x .

5. (a) Show that the gradient of the line through ,2 1-^ h and 3, 4^ h is equal to the gradient of the line between the points , ,and2 1 7 2-^ ^h h .(b) Draw the two lines on the number plane. What can you say about the lines?

6. Show that the points , , , , ,A B C1 2 1 5 6 5-^ ^ ^h h h and ,D 4 2^ h form a parallelogram. Find the gradients of all sides.

7. The points , , , , ,A B C3 5 9 3 5 6- -^ ^ ^h h h and

,D 1 2-^ hform a rectangle. Find the gradients of all the sides and the diagonals.

8. Find the gradients of the diagonals of the square with vertices , , , ,A B2 1 3 1-^ ^h h

, and , .C D3 6 2 6-^ ^h h

9. A triangle has vertices , ,,A B3 1 1 4- -^ ^h h and , .C 11 4-^ h

By fi nding the lengths of all (a) sides, prove that it is a right-angled triangle.

Find the gradients of sides (b) AB and BC .

10. (a) Find the midpoints F and G of sides AB and AC where ABC is a triangle with vertices , ,,A B0 3 2 7-^ ^h h and ,C 8 2-^ h .(b) Find the gradients of FG and BC .

11. The gradient of the line between a moving point ,P x y^ h and the point ,A 5 3^ h is equal to the gradient of line PB where B has coordinates ,2 1-^ h . Find the equation of the locus of P .

12. Prove that the points , , ,3 1 5 5-^ ^h h and ,2 4-^ h are collinear.

13. Find the gradient of the straight line that makes an angle of 45c with the x -axis in the positive direction.

14. Find the gradient, to 2 signifi cant fi gures, of the straight line that makes an angle of 42 51c l with the x -axis.

15. Find the gradient of the line that makes an angle of 87 14c l with the x -axis, to 2 signifi cant fi gures.

16. Find the angle, in degrees and minutes, that a line with gradient 1.2 makes with the x -axis.

17. What angle, in degrees and minutes does the line with gradient –3 make with the x -axis in the positive direction?

ch7.indd 366 7/31/09 4:47:04 PM


Gradient given an equation

In Chapter 5 you explored and graphed linear functions. You may have noticed a relationship between the graph and the gradient and y -intercept of a straight line.

18. Find the exact gradient of the line that makes an angle with the x -axis in the positive direction of

(a) 60c (b) 30c (c) .120c

19. Show that the line passing through ,4 2-^ h and ,7 5-^ h

makes an angle of 135c with the x -axis in the positive direction.

20. Find the exact value of x with rational denominator if the line passing through ,x 3^ h and ,2 1^ h makes an angle of 60c with the x -axis.

Investigation

1. (i) Draw the graph of each linear function. (ii) By selecting two points on the line, fi nd its gradient.

(a) y x= (b) 2y x= (c) 3y x= (d) y x= - (e) 2y x= -

Can you fi nd a pattern for the gradient of each line? Can you predict what the gradient of 5y x= and 9y x= - would be?

2. (i) Draw the graph of each linear function. (ii) Find the y -intercept.

(a) y x= (b) 1y x= + (c) 2y x= + (d) 2y x= - (e) 3y x= -

Can you fi nd a pattern for the y -intercept of each line? Can you predict what the y -intercept of 11y x= + and 6y x= - would be?

hasy mx b= + gradientm =

b y= -intercept

ch7.indd 367 7/18/09 11:22:31 AM


EXAMPLES

1. Find the gradient and y -intercept of the linear function 7 5y x= - .

Solution

The equation is in the form y mx b= + where 7m = and 5b = - .

Gradient 7=

y-intercept 5= -

2. Find the gradient of the straight line with equation .x y2 3 6 0+ - =

Solution

First, we change the equation into the form y mx b= + .

So the gradient is .

x y

x y

x y

x y

y x

x

y x

y x

x

m

x x

2 3 6 0

2 3 6 0

2 3 6

2 3 6

3 6 2

2 6

3 2 6

32

36

32 2

32

32

6 6

2 2

3 3

+ - =

+ - =

+ =

+ =

= -

= - +

=- +

=-

+

= - +

= -

-

+ +

- -

There is a general formula for fi nding the gradient of a straight line.

The gradient of the line 0ax by c+ + = is given by

mba

= -

Proof

0ax by c

by ax c

ybax

bc

+ + =

= - -

= - -

mba

` = -

ch7.indd 368 7/18/09 11:22:31 AM


EXAMPLE

Find the gradient of 3 2x y- = .

Solution

,

x y

x y

a b

mba

3 2

3 2 0

3 1

13

3

3gradient is`

- =

- - =

= = -

= -

= --

=

7.4 Exercises

1. Find (i) the gradient and (ii) the y -intercept of each linear function.

(a) 3 5y x= + (b) 2 1f x x= +] g (c) 6 7y x= - (d) y x= - (e) 4 3y x= - + (f) 2y x= - (g) 6 2f x x= -] g (h) 1y x= - (i) 9y x= (j) 5 2y x= -

2. Find (i) the gradient and (ii) the y -intercept of each linear function .

(a) 2 3 0x y+ - = (b) 5 6 0x y+ + = (c) 6 1 0x y- - = (d) 4 0x y- + = (e) 4 2 1 0x y+ - = (f) 6 2 3 0x y- + = (g) 3 6 0x y+ + = (h) 4 5 10 0x y+ - = (i) 7 2 1 0x y- - = (j) 5 3 2 0x y- + =

3. Find the gradient of the straight line .

(a) 4y x= (b) 2 1y x= - - (c) 2y = (d) 2 5 0x y+ - = (e) 1 0x y+ + = (f) 3 8x y+ = (g) 2 5 0x y- + = (h) 4 12 0x y+ - = (i) 3 2 4 0x y- + = (j) 5 4 15x y- =

(k) 32 3y x= +

(l) 2

y x=

(m) 5

1y x= -

(n) 72 5y x

= +

(o) 53 2y x

= - -

(p) 27 3

1y x= - +

(q) 35

8xy

- =

(r) 2 3

1x y+ =

(s) 32 4 3 0x y- - =

(t) 4 3

27 0x y

+ + =

ch7.indd 369 7/18/09 11:22:32 AM


Equation of a Straight Line

There are several different ways to write the equation of a straight line.

General form

0ax by c+ + =

Gradient form

y mx b= +

where gradientm = and b y= -intercept

Intercept form

1ax

b

y+ =

where a and b are the x -intercept and y -intercept respectively

Proof

,m ab b b

y ab x b

b

yax

ax

b

y

1

1`

`

= - =

= - +

= - +

+ =

Point-gradient formula

There are two formulae for fi nding the equation of a straight line. One of these uses a point and the gradient of the line.

The equation of a straight line is given by

x xy y m 11 -- = _ i

where ,x y1 1_ i lies on the line with gradient m This is a very useful formula as it is used in many topics in this course.

ch7.indd 370 7/18/09 11:22:33 AM


Proof

Given point ,x y1 1_ i on the line with gradient m

Let ,P x y= ^ h Then line AP has gradient

m x xy y

m x xy y

m x x y y

2 1

2 1

1

1

1 1

`

=-

-

=-

-

- = -_ i

Two-point formula

The equation of a straight line is given by

x xy y

x xy y

1

1

2 1

2 1

-

-=

-

-

where ,x y1 1_ i and ,x y2 2_ i are points on the line

Proof

ABRD;

,

So

P x y

APQ

AQPQ

ARBR

x xy y

x xy y

Let

i.e.1

1

2 1

2 1

<D

=

=

-

-=

-

-

^ h

The two-point formula is not essential. The right-hand side of it is the gradient of the line. Replacing this by m gives the point–gradient formula.

This formula is optional as you can

use the point–gradient formula for any

question.

The gradient is the same anywhere along

a straight line.

ch7.indd 371 7/18/09 11:22:33 AM


EXAMPLES

1. Find the equation of the straight line with gradient 4- and passing through the point , .2 3-^ h

Solution

, andm x y4 2 31 1= - = - =

Equation: ( )

[ ( )]

( )

(gradient form)

or (general form)

y y m x x

y x

x

x

y x

x y

3 4 2

4 2

4 8

4 5

4 5 0

1 1

`

- = -

- = - - -

= - +

= - -

= - -

+ + =

2. Find the equation of the straight line that passes through the points ,2 3-^ h and , .4 7- -^ h

Solution

By two-point formula:

x xy y

x xy y

x

y

x

y

x

y

y xy x

x y

x y

4

7

2 43 7

47

2 43 7

47

32

3 7 2 43 21 2 8

2 3 13 0

2 3 13 0or

1

1

2 1

2 1

-

-=

-

-

- -

- -=

- -

- - -

+

+=

+

- +

+

+=

+ = +

+ = +

- + + =

- - =

]

]

]

]

^ ]

g

g

g

g

h g

By point-gradient method:

m x xy y

2 43 7

2 43 7

32

2 1

2 1=

-

-

=- -

- - -

=+

- +

=

]

]

g

g

Use one of the points, say ,4 7- -^ h .

, 4 7m x y32 and1 1= = - = -

Equation: ( )

( ) ( )

y y m x x

y x732 4

1 1- = -

- - = - -6 @

ch7.indd 372 7/18/09 11:22:34 AM


( )

or

y x

y xy x

x y

x y

732 4

3 7 2 43 21 2 8

2 3 13 0

2 3 13 0

`

+ = +

+ = +

+ = +

- + + =

- - =

^ ]h g

3. Find the equation of the line with x- intercept 3 and y- intercept 2.

Solution

Intercept form is 1,ax

b

y+ = where a and b are the x- intercept and

y- intercept respectively.

1

2 3 6

2 3 6 0

x y

x y

x y

3 2`

`

+ =

+ =

+ - =

Again, the point-gradient formula can be used. The x -intercept and y -intercept are the points 3, 0^ h and , .0 2^ h

7.5 Exercises

1. Find the equation of the straight line

with gradient 4 and (a) y -intercept 1-

with gradient (b) 3- and passing through ,0 4^ h

passing through the origin (c) with gradient 5

with gradient 4 and (d) x -intercept 5-

with (e) x -intercept 1 and y -intercept 3

with (f) x -intercept 3, y -intercept 4-

with (g) y -intercept 1- and making an angle of 45c with the x -axis in the positive direction

with (h) y -intercept 5 and making an angle of 45c with the x -axis in the positive direction.

2. Find the equation of the straight line that makes an angle of 135c with the x -axis and passes through the point , .2 6^ h

3. Find the equation of the straight line passing through

(a) ,2 5^ h and ,1 1-^ h (b) ,0 1^ h and ,4 2- -^ h (c) ,2 1-^ h and ,3 5^ h (d) ,3 4^ h and ,1 7-^ h (e) ,4 1- -^ h and ,2 0-^ h .

4. What is the equation of the line with x -intercept 2 and passing through , ?3 4-^ h

5. Find the equation of the line parallel to the (a) x -axis and

passing through ,2 3^ h parallel to the (b) y -axis and

passing through ,1 2-^ h .

ch7.indd 373 7/31/09 4:47:05 PM


Parallel and Perpendicular Lines

Parallel lines

6. A straight line passing through the origin has a gradient of 2.- Find its equation.

7. A straight line has x -intercept 4 and passes through , .0 3-^ h Find its equation.

8. Find the equation of the straight line with gradient 2- that passes through the midpoint of ,5 2-^ h and , .3 4-^ h

9. What is the equation of the straight line through the point ,4 5-^ h and the midpoint of ,1 2^ h and , ?9 4-^ h

10. What is the equation of the straight line through the midpoint of ,0 1^ h and ,6 5-^ h and the midpoint of ,2 3^ h and , ?8 3-^ h

Class Investigation

Sketch the following straight lines on the same number plane . 1. y x2=

y x2 1= +2. y x2 3= -3. y x2 5= +4.

What do you notice about these lines?

If two lines are parallel, then they have the same gradient. That is, m m1 2=

Two lines that are parallel have equations 0ax by c1+ + = and 0ax by c2+ + =

ch7.indd 374 7/18/09 11:22:36 AM


Proof

0ax by c1+ + = has gradient mba

1 = -

0ax by c2+ + = has gradient mba

2 = -

Since ,m m1 2= the two lines are parallel.

EXAMPLES

1. Prove that the straight lines 5 2 1 0x y- - = and 5 2 7 0x y- + = are parallel.

Solution

x y

x y

x y

m

x y

x y

x y

m

5 2 1 0

5 1 2

25

21

25

5 2 7 0

5 7 2

25

27

25

1

2

`

`

- - =

- =

- =

=

- + =

+ =

+ =

=

25m m1 2= =

` the lines are parallel .

2. Find the equation of a straight line parallel to the line 2 3 0x y- - = and passing through , .1 5-^ h

Solution

2 3 0

2 3

2

x y

x y

m1`

- - =

- =

=

For parallel lines m m1 2= 2m2` =

Equation: ( )( ) ( )

y y m x xy x

y xx y

5 2 15 2 20 2 7

1 1- = -

- - = -

+ = -

= - -

Notice that the equations are both in the form

5x 2y k 0.- + =

ch7.indd 375 7/18/09 11:22:37 AM


DID YOU KNOW?

Parallel lines are usually thought of as lines that never meet. However, there is a whole branch of geometry based on the theory that parallel lines meet at infi nity. This is called affi ne geometry . In this geometry there are no perpendicular lines.

Perpendicular lines

Class Investigation

Sketch the following pairs of straight lines on the same number plane.

(a) 1. 3 4 12 0x y- + = (b) 4 3 8 0x y+ - = (a) 2. 2 4 0x y+ + = (b) 2 2 0x y- + =

What do you notice about these pairs of lines?

If two lines with gradients m1 and m2 respectively are perpendicular, then

1m m

m m1i.e.

1 2

21

= -

= -

Proof

Let line AB have gradient tanm1 a= . Let line CD have gradient tanm2 b= .

straight angle

tan

tan

cot

ECEB

CBE

EBEC

ECEB

180

180

180`

c

c

c

+

b

a

a

a

=

= -

- =

- =

^

]

]

h

g

g

Gradients of perpendicular lines are negative reciprocals of each other.

ch7.indd 376 7/18/09 11:22:38 AM


So

or

tan cotcot

tan

m mm m

180

1

1

1

21

1 2

` cb a

a

a

= -

= -

= -

= -

= -

] g

Perpendicular lines have equations in the form 0ax by c1+ + = and 0bx ay c2- + =

Proof

has gradient

has gradient

ax by c mba

bx ay c m ab

ab

0

0

1 1

2 2

+ + = = -

- + = = --

=

m mba

ab

1

1 2 #= -

= -

Since ,m m 11 2 = - the two lines are perpendicular .

EXAMPLES

1. Show that the lines 3 11 0x y+ - = and 3 1 0x y- + = are perpendicular.

Solution

3

1

x y

y x

m

x y

x y

x y

m

m m

3 11 0

3 11

3

3 1 0

1 3

31

31

31

31

1

2

1 2

`

`

#

+ - =

= - +

= -

- + =

+ =

+ =

=

= -

= -

the lines are perpendicular .

Notice that the equations are in the form

x y c3 01+ + = and .x y c3 02- + =

CONTINUED

ch7.indd 377 7/18/09 11:22:39 AM


2. Find the equation of the straight line through 2, 3^ h perpendicular to the line that passes through ,1 7-^ h and , .3 3^ h

Solution

Line through ,1 7-^ h and , :3 3^ h

1

m x xy y

m1 3

7 3

44

2 1

2 1

1

=-

-

=- -

-

=-

= -

For perpendicular lines, 1= -m m1 2

1 1m

m 1

i.e. 2

2

- = -

=

Equation through , :2 3^ h

( )

( )

y y m x x

y x

x

x y

3 1 2

2

0 1

1 1- = -

- = -

= -

= - +

7.6 Exercises

1. Find the gradient of the straight line

parallel to the line (a) 3 4 0x y+ - =

perpendicular to the line (b) 3 4 0x y+ - =

parallel to the line joining (c) ,3 5^ h and ,1 2-^ h

perpendicular to the line with (d) x -intercept 3 and y -intercept 2

perpendicular to the line (e) making an angle of 135c with the x -axis in the positive direction

perpendicular to the line (f) 6 5 4 0x y- - =

parallel to the line making an (g) angle of c30 with the x -axis

parallel to the line (h) 3 7 0x y- - =

perpendicular to the line (i) making an angle of c120 with the x -axis in the positive direction

perpendicular to the line (j) passing through ,4 2-^ h and , .3 3^ h

2. Find the equation of each straight line

passing through (a) ,2 3^ h and parallel to the line 6y x= +

through (b) ,1 5-^ h and parallel to the line 3 7 0x y- - =

with (c) x -intercept 5 and parallel to the line 4y x= -

through (d) ,3 4-^ h and perpendicular to the line 2y x=

through (e) ,2 1-^ h and perpendicular to the line 2 3 0x y+ + =

ch7.indd 378 7/18/09 11:22:41 AM


through (f) ,7 2-^ h and perpendicular to the line 3 5 0x y- - =

through (g) ,3 1- -^ h and perpendicular to the line .x y4 3 2 0- + =

3. Show that the straight lines 3 2y x= - and 6 2 9 0x y- - = are parallel .

4. Show that lines 5 0x y+ = and 5 3y x= + are perpendicular .

5. Show that lines 6 5 1 0x y- + = and 6 5 3 0x y- - = are parallel.

6. Show that lines 7 3 2 0x y+ + = and 3 7 0x y- = are perpendicular.

7. If the lines 3 2 5 0x y- + = and 1y kx= - are perpendicular, fi nd the value of k .

8. Show that the line joining ,3 1-^ h and ,2 5-^ h is parallel to the line 8 2 3 0.x y- - =

9. Show that the points , ,A 3 2- -^ h , ,B 1 4-^ h , ,C 7 1-^ h and ,D 5 7-^ h are the vertices of a parallelogram.

10. The points , ,A 2 0-^ h , ,B 1 4^ h ,C 6 4^ h and ,D 3 0^ h form a rhombus. Show that the diagonals are perpendicular.

11. Find the equation of the straight line

passing through the (a) origin and parallel to the line 3 0x y+ + =

through (b) ,3 7^ h and parallel to the line 5 2 0x y- - =

through (c) ,0 2-^ h and perpendicular to the line 2 9x y- =

perpendicular to the line (d) 3 2 1 0x y+ - = and passing through the point ,2 4-^ h .

12. Find the equation of the straight line passing through ,6 3-^ h that is perpendicular to the line joining ,2 1-^ h and , .5 7- -^ h

13. Find the equation of the line through ,2 1^ h that is parallel to the line that makes an angle of c135 with the x -axis in the positive direction.

14. Find the equation of the perpendicular bisector of the line passing through ,6 3-^ h and , .2 1-^ h

15. Find the equation of the straight line parallel to the line 2 3 1 0x y- - = and through the midpoint of ,1 3^ h and , .1 9-^ h

Intersection of Lines

Two straight lines intersect at a single point , .x y^ h The point satisfi es the equations of both lines. We fi nd this point by solving simultaneous equations.

You may need to revise simultaneous equations

from Chapter 3.

ch7.indd 379 7/18/09 11:22:42 AM


Concurrent lines meet at a single point. To show that lines are concurrent, solve two simultaneous equations to fi nd the point of intersection. Then substitute this point of intersection into the third and subsequent lines to show that these lines also pass through the point.

EXAMPLES

1. Find the point of intersection between lines x y3 3 02 - - = and .x y5 2 13 0- - =

Solution

Solve simultaneous equations:

:

:

:

x

x y

x y

x y

x y

xx

3

2 3 3 0 1

5 2 13 0 2

1 2 4 6 6 0 3

2 3 15 6 39 0 4

3 4 11 33 033 11

#

#

=

- - =

- - =

- - =

- - =

- - + =

=

^

^

^ ^

^ ^

^ ^

h

h

h h

h h

h h

ubstitute into :S x 3 1= ^ h

yy

y

y

2 3 3 3 03 3 0

3 3

1

- - =

- + =

=

=

^ h

So the point of intersection is , .3 1^ h

2. Show that the lines ,x y x y3 1 0 2 12 0- + = + + = and x y4 3 7 0- - = are concurrent.

Solution

Solve any two simultaneous equations:

:

:

x y

x y

x y

x y

x

3 1 0 1

2 12 0 2

4 3 7 0 3

1 2 6 2 2 0 4

2 4 7 14 0

#

- + =

+ + =

- - =

- + =

+ + =

^

^

^

^ ^

^ ^

h

h

h

h h

h h

You could use a computer spreadsheet to solve these simultaneous equations.

ch7.indd 380 7/18/09 11:22:44 AM


7 14

x

x

2= -

= -

ubstitute into :xS 2 1= - ^ h

yy

y

3 2 1 05 0

5

- - + =

- - =

- =

^ h

So the point of intersection of (1) and (2) is ,2 5- -^ h . Substitute ,2 5- -^ h into (3): x y4 3 7 0- - =

LHS 4 2

RHS

3 5 7

8 15 7

0

= - - -

= - + -

=

=

-^ ^h h

So the point lies on line (3) all three lines are concurrent .

Equation of a line through the intersection of 2 other lines

To fi nd the equation of a line through the intersection of 2 other lines, fi nd the point of intersection, then use it with the other information to fi nd the equation.

Another method uses a formula to fi nd the equation.

If a x b y c 01 1 1+ + = and a x b y c 02 2 2+ + = are 2 given lines then the equation of a line through their intersection is given by the formula

( ) ( )a x b y c k a x b y c 01 1 1 2 2 2+ + + + + = where k is a constant

Proof

Let l1 have equation .a x b y c 01 1 1+ + = Let l2 have equation .a x b y c 02 2 2+ + = Let the point of intersection of l1 and l2 be ,x yP 1 1^ h . Then P satisfi es l1

i.e. a x b y c 01 11 1 1+ + =

P also satisfi es l2

i.e. a x b y c 01 12 2 2+ + =

Substitute P into ( ) ( )a x b y c k a x b y c 01 1 1 2 2 2+ + + + + =

( ) ( )a x b y c k a x b y c

k

0

0 0 00 0

1 1 1 11 1 1 2 2 2+ + + + + =

+ =

=

^ h

if point P satisfi es both equations l1 and l2 then it satisfi es l kl 01 2+ = .

ch7.indd 381 7/18/09 11:22:45 AM


7.7 Exercises

EXAMPLE

Find the equation of the line through ,1 2-^ h that passes through the intersection of lines x y2 5 0+ - = and .x y3 1 0- + =

Solution

Using the formula: , , , ,a b c a b c2 1 5 1 3 11 1 1 2 2 2= = = - = = - =

a x b y c k a x b y c

x y k x y

0

2 5 3 1 01 1 1 2 2 2+ + + + + =

+ - + - + =

^ ^

^ ^

h h

h h

Since this line passes through , ,1 2-^ h substitute the point into the equation:

kk

k

k

2 2 5 1 6 1 05 6 0

5 6

65

- + - + - - + =

- - =

- =

- =

^ ^h h

So the equation becomes:

x y x y

x y x yx y x y

x y

x y

2 565 3 1 0

6 2 5 5 3 1 012 6 30 5 15 5 0

7 21 35 0

3 5 0

+ - - - + =

+ - - - + =

+ - - + - =

+ - =

+ - =

^ ^

^ ^

h h

h h

Another way to do this example is to fi nd the point of intersection, then use both points to fi nd the equation.

Substitute the value of k back into the equation.

1. Find the point of intersection of straight lines

(a) x y3 4 10 0+ + = and x y2 3 16 0- - =

x y5 2 11 0+ + =(b) and x y3 6 0+ + =

x y7 3 16- =(c) and x y5 2 12- =

x y2 3 6- =(d) and x y4 5 10- =

x y3 8 0- - =(e) and x y4 7 13 0+ - =

y x5 6= +(f) and y x4 3= - -

y x2 1= +(g) and x y5 3 6 0+ =-

x y3 7 12+ =(h) and x y4 1 06- =-

x y3 5 7- = -(i) and x y2 3 4- =

x y8 7 3 0- - =(j) and x y5 2 1 0- - =

2. Show that the lines x y2 11 0- - = and

x y2 10 0- =- intersect at the point , .3 4-^ h

3. A triangle is formed by 3 straight lines with equations ,x y2 1 0- + = x y2 09+ - =

ch7.indd 382 7/18/09 11:22:46 AM


and .x y2 5 3 0- - = Find the coordinates of its vertices.

4. Show that the lines ,x y5 17 0- - =

x y3 2 12 0- - = and x y5 7 0+ - = are concurrent.

5. Show that the lines ,x y4 5 0+ + = ,x y3 7 15 0- + =

x y2 10 0- =+ and x y6 5 30 0+ + = are concurrent.

6. Find the equation of the straight line through the origin that passes through the intersection of the lines x y5 2 14 0- + = and

x y3 4 7 0+ - = .

7. Find the equation of the straight line through ,3 2^ h that passes through the intersection of the lines x y5 2 01+ + = and

x y3 16 0- + = .

8. Find the equation of the straight line through ,4 1- -^ h that passes through the intersection of the lines x y2 1 0+ - = and .x y3 5 16 0+ + =

9. Find the equation of the straight line through ,3 4-^ h that passes through the intersection of the lines x y2 3 0+ - = and x y3 2 8 0- - = .

10. Find the equation of the straight line through ,2 2-^ h that passes through the intersection of the lines x y2 3 6 0+ - = and x y3 5 10 0+ =- .

11. Find the equation of the straight line through ,3 0^ h that passes through the intersection of the lines x y 1 0- + = and x y4 2 0- - = .

12. Find the equation of the straight line through ,1 2- -^ h that passes through the intersection of the lines x y2 6 0+ - = and .x y3 7 9 0+ - =

13. Find the equation of the straight line through ,1 2^ h that passes through the intersection of the lines x y2 10 0+ + = and .x y 02 5- =+

14. Find the equation of the straight line through ,2 0-^ h that passes through the intersection of the lines x y3 4 7 0+ - = and .x y3 2 1 0- - =

15. Find the equation of the straight line through ,3 2-^ h that passes through the intersection of the lines x y5 2 13 0+ - = and x y3 11 0- + = .

16. Find the equation of the straight line through ,3 2- -^ h that passes through the intersection of the lines x y 1 0+ + = and x y3 2 0+ = .

17. Find the equation of the straight line through ,3 1^ h that passes through the intersection of the lines x y3 4 0- + = and x y2 12 0- + = .

18. Find the equation of the straight line with gradient 3 that passes through the intersection of the lines x y2 1 0+ - = and x y3 5 16 0+ + = .

19. Find the equation of the straight line with gradient 2 that passes through the intersection of the lines x y5 2 3 0- - = and x y7 3 4 0- - = .

ch7.indd 383 7/18/09 11:22:47 AM


20. Find the equation of the straight line parallel to the line x y3 7 0- - = that passes through the intersection of the lines x y3 2 10 0- - = and

.x y4 17 0+ - =

21. Find the equation of the straight line perpendicular to the line x y5 1 0+ - = that passes through the intersection of lines x y3 5 3 0- - = and x y2 3 17 0+ + = .

Perpendicular Distance

The distance formula d x x y y2 12

2 12= - + -_ _i i is used to fi nd the distance

between two points. Perpendicular distance is used to fi nd the distance between a point and

a line. If we look at the distance between a point and a line, there could be many distances.

So we choose the shortest distance, which is the perpendicular distance.

The perpendicular distance from ,x y1 1_ i to the line 0ax by c+ + = is

given by | |

da b

ax by c2 2

1 1=

+

+ +

A distance is always positive, so take the absolute value.

Proof

ch7.indd 384 7/18/09 11:22:48 AM


Let d be the perpendicular distance of ,x y1 1_ i from the line .ax by c 0+ + =

,A ac 0= -b l ,C

bc0= -c m ,R x

b

ax c1

1=

- -e o

,ACOac

bc

a bc b c a

abc a b

In AC2

2

2

2

2 2

2 2 2 2

2 2

D = +

=+

=+

PR yb

ax c

b

ax by c

1

1

1 1

= -- -

=+ +

e o

ACOD is similar to PRQD

.AOPQ

ACPR

PQAC

AO PR

d ac

b

ax by c

abc a b

ab

c ax by c

c a b

ab

a b

ax by c

1 12 2

1 1

2 2

2 2

1 1

`

` # '

#

=

=

=+ + +

=+ +

+

=+

+ +

_ i

All points on one side of the line 0ax by c+ + = make the numerator of this formula positive. Points on the other side make the numerator negative.

Usually we take the absolute value of d . However, if we want to know if points are on the same side of a line or not, we look at the sign of d .

To fi nd A and C , substitute y 0= and x 0= into

.ax by c 0+ + =

Why?

EXAMPLES

1. Find the perpendicular distance of ,4 3-^ h from the line .x y3 4 1 0- - =

Solution

, , , ,

| |

| |

x y a b c

da b

ax by c

4 3 3 4 1

3 4

3 4 4 3 1

1 1

2 2

1 1

2 2

= = - = = - = -

=+

+ +

=+ -

+ - - + -

]

] ] ] ]

g

g g g g

CONTINUED

ch7.indd 385 7/18/09 11:22:50 AM


| |

.

25

12 12 1

523

4 6

=+ -

=

=

So the perpendicular distance is 4.6 units.

2. Prove that the line x y6 8 20 0+ + = is a tangent to the circle 4.x y2 2+ =

Solution

There are three possibilities for the intersection of a circle and a straight line.

The centre of the circle x y 42 2+ = is ,0 0^ h and its radius is 2 units. A tangent is perpendicular to the centre of the circle. So we prove that the

perpendicular distance from the line to the point ,0 0^ h is 2 units (the radius).

| |

| ( ) ( ) |

| |

da b

ax by c

6 8

6 0 8 0 20

100

20

1020

2

2 2

1 1

2 2

=+

+ +

=+

+ +

=

=

=

the line is a tangent to the circle.

3. Show that the points ,1 3-^ h and ,2 7^ h lie on the same side of the line .x y2 3 4 0- + =

ch7.indd 386 7/18/09 11:22:52 AM


Solution

To show that points lie on the same side of a line, their perpendicular distance must have the same sign. We use the formula without the absolute value sign.

, :

, :

da b

ax by c

d

d

1 3

2 3

2 1 3 3 4

4 92 9 4

137

2 7

2 3

2 2 3 7 4

4 94 21 4

1313

2 2

1 1

2 2

2

=+

+ +

-

=+ -

- - +

=+

- - +

=-

=+ -

- +

=+

- +

=-

2

^

]

] ]

^

]

] ]

h

g

g g

h

g

g g

Since the perpendicular distance for both points has the same sign, the points lie on the same side of the line.

1. Find the perpendicular distance between

(a) ,1 2^ h and x y3 4 2 0+ + = (b) ,3 2-^ h and 5 12 7 0x y+ + = (c) ,0 4^ h and 8 6 1 0x y- - = (d) 3, 2- -^ h and x y4 3 6 0- - = the origin and (e)

.x y12 5 8 0- + =

2. Find, correct to 3 signifi cant fi gures, the perpendicular distance between

(a) ,1 3^ h and 3 1 0x y+ + = (b) ,1 1-^ h and 2 5 4 0x y+ + = (c) ,3 0^ h and 5 6 12 0x y- - = (d) ,5 3-^ h and 4 2 0x y- - = (e) 6, 3- -^ h and .x y2 3 9 0- + =

3. Find as a surd with rational denominator the perpendicular distance between

the origin and the line (a) 3 2 7 0x y- + =

(b) ,1 4-^ h and 2 3 0x y+ + = (c) ,3 1-^ h and 3 14 1 0x y+ + = (d) 2, 6-^ h and 5 6 0x y- - = (e) 4, 1- -^ h and

.x y3 2 4 0- - =

4. Show that the origin is equidistant from the lines 7 24 25 0,x y+ + = 4 3 5 0x y+ - = and 12 5 13 0.x y+ - =

7.8 Exercises

ch7.indd 387 7/18/09 11:22:55 AM


Equidistant means that two or more objects are the same distance away from another object.

5. Show that points ,A 3 5-^ h and ,B 1 4-^ h lie on opposite sides of 2 3 0.x y- + =

6. Show that the points 2, 3-^ h and ,9 2^ h lie on the same side of the line .x y3 2 0- + =

7. Show that 3, 2-^ h and ,4 1^ h lie on opposite sides of the line .x y4 3 2 0- - =

8. Show that 0, 2-^ h is equidistant from the lines 3 4 2 0x y+ - = and .x y12 5 16 0- + =

9. Show that the points 8, 3-^ h and ,1 1^ h lie on the same side of the line .x y6 4 0- + =

10. Show that 3, 2-^ h and ,4 1^ h lie on opposite sides of the line .x y2 2 0+ - =

11. Show that the point ,3 2-^ h is the same distance from the line 6 8 6 0x y- + = as the point 4, 1- -^ h is from the line .x y5 12 20 0+ - =

12. Find the exact perpendicular distance with rational denominator from the point ,4 5^ h to the line with x -intercept 2 and y -intercept .1-

13. Find the perpendicular distance from ,2 2-^ h to the line passing through ,3 7^ h and , .1 4-^ h

14. Find the perpendicular distance between ,0 5^ h and the line through ,3 8-^ h parallel to 4 3 1 0.x y- - =

15. The perpendicular distance between the point , 1x -^ h and the line 3 4 7 0x y- + = is 8 units. Find two possible values of x .

16. The perpendicular distance between the point , b3^ h and the line 5 12 2 0x y- - = is 2 units. Find the values of b .

17. Find m if the perpendicular distance between ,m 7^ h and the line 9 12 6 0x y+ + = is 5 units.

18. Prove that the line 3 4 25 0x y- + = is a tangent to the circle with centre the origin and radius 5 units.

19. Show that the line 3 4 12 0x y- + = does not cut the circle 1.x y2 2+ =

20. The sides of a triangle are formed by the lines with equations 2 7 0, 3 5 4 0x y x y- - = + - = and 3 4 0.x y+ - =

Find the vertices of the (a) triangle.

Find the exact length of all (b) the altitudes of the triangle.

ch7.indd 388 7/18/09 11:22:56 AM


1. Find the distance between points ,1 2-^ h and , .3 7^ h

2. What is the midpoint of the origin and the point , ?5 4-^ h

3. Find the gradient of the straight line passing through (a) ,3 1-^ h and ,2 5-^ h with equation (b) 2 1 0x y- + = making an angle of (c) 30c with the

x -axis in the positive direction perpendicular to the line (d)

.x y5 3 8 0+ - =

4. Find the equation of the linear function passing through (a) ,2 3^ h and with

gradient 7 parallel to the line (b) 5 3 0x y+ - =

and passing through ,1 1^ h through the origin, and (c)

perpendicular to the line 2 3 6 0x y- + = through (d) ,3 1^ h and ,2 4-^ h with (e) x -intercept 3 and y -intercept – .1

5. Find the perpendicular distance between ,2 5^ h and the line 2 7 0x y- + = in surd form with rational denominator.

6. Prove that the line between ,1 4-^ h and ,3 3^ h is perpendicular to the line 4 6 0.x y- - =

7. Find the x - and y -intercepts of 2 5 10 0.x y- - =

8. (a) Find the equation of the straight line l that is perpendicular to the line

21 3y x= - and passes through , .1 1-^ h

(b) Find the x -intercept of l . (c) Find the exact distance from ,1 1-^ h

to the x -intercept of l .

9. Prove that lines 5 7y x= - and 10 2 1 0x y- + = are parallel.

10. Find the equation of the straight line passing through the origin and parallel to the line with equation 3 4 5 0.x y- + =

11. Find the point of intersection between lines 2 3y x= + and 5 6 0.x y- + =

12. The midpoint of ,a 3^ h and , b4-^ h is , .1 2^ h Find the values of a and b .

13. Show that the lines 4 0,x y- - =

,x y x y2 1 0 5 3 14 0+ + = - - = and 3 2 9 0x y- - = are concurrent.

14. A straight line makes an angle of 153 29c l with the x -axis in the positive direction. What is its gradient, to 3 signifi cant fi gures?

15. The perpendicular distance from ,3 2-^ h to the line 5 12 0x y c- + = is 2. Find 2 possible values of c .

16. Find the equation of the straight line through ,1 3^ h that passes through the intersection of the lines 2 5 0x y- + = and 2 5 0.x y+ - =

17. The gradient of the line through ,3 4-^ h and ,x 2^ h is −5. Evaluate x .

18. Show that the points ,2 1-^ h and ,6 3^ h are on opposite sides of the line .x y2 3 1 0- - =

19. Find the equation of the line with x -intercept 4 that makes an angle of 45c with the x -axis.

20. Find the equation of the line with y -intercept 2- and perpendicular to the line passing through ,3 2-^ h and , .0 5^ h

Test Yourself 7

ch7.indd 389 7/18/09 11:22:59 AM



1. If points , ,,k k k3 1 1 3- - -^ ^h h and ,k k4 5- -^ h are collinear, fi nd the value of k .

2. Find the equation, in exact form, of the line passing through ,3 2-_ i that makes an angle of 30c with the positive x -axis.

3. Find the equation of the circle whose centre is at the origin and with tangent 3 9 0.x y- + =

4. ABCD is a rhombus where , , , ,,A B C3 0 0 4 5 4= - = =^ ^ ^h h h and , .D 2 0= ^ h Prove that the diagonals are perpendicular bisectors of one another.

5. Prove that the points , ,1 2 2-_ i ,3 6-_ i and ,5 2-_ i all lie on a circle with centre the origin. What are the radius and equation of the circle?

6. Find the exact distance between the parallel lines 3 2 5 0x y+ - = and 3 2 1.x y+ =

7. A straight line has x -intercept ,aA 0^ h and y -intercept , ,B b0^ h where a and b are positive integers. The gradient of line AB is .1- Find OBA+ where O is the origin and hence prove that .a b=

8. Find the exact perpendicular distance between the line 2 3 1 0x y+ + = and the point of intersection of lines 3 7 15x y- = and 4 5.x y- = -

9. Find the magnitude of the angle, in degrees and minutes, that the line

joining ,1 3-^ h and ,2 4-^ h makes with the x -axis in the positive direction.

10. Find the equation of the line that passes through the point of intersection of lines 2 5 19 0x y+ + = and 4 3 1 0x y- - = that is perpendicular to the line 3 2 1 0.x y- + =

11. Prove , ,,A B2 5 4 5-^ ^h h and ,C 1 2-^ h are the vertices of a right-angled isosceles triangle.

12. Find the coordinates of the centre of a circle that passes through points , ,,7 2 2 3^ ^h h and , .4 1- -^ h

13. If 2 0ax y- - = and 5 11 0bx y- + = intersect at the point , ,3 4^ h fi nd the values of a and b .

14. Find the equation of the straight line through ,3 4-^ h that is perpendicular to the line with x -intercept and y -intercept −2 and 5 respectively.

15. Find the exact equation of the straight line through the midpoint of , ,0 5-^ h and ,4 1-^ h that is perpendicular to the line that makes an angle of 30c with the x -axis.

16. Point ,x yP ^ h moves so that it is equidistant from points ,A 1 4^ h and , .B 2 7-^ h By fi nding the distances AP and BP , fi nd the equation of the locus of P .

ch7.indd 390 7/18/09 11:23:05 AM

TERMINOLOGY

8 Introduction to Calculus

Composite function: A function of a function. One function, f (x), is a composite of one function to another function, for example g(x)

Continuity: Describing a line or curve that is unbroken over its domain

Continuous function: A function is continuous over an interval if it has no break in its graph. For every x value on the graph the limit exists and equals the function value

Derivative at a point: This is the gradient of a curve at a particular point

Derivative function: The gradient function of a curve obtained through differentiation

Differentiable function: A function which is continuous and where the gradient exists at all points on the function

Differentiation: The process of fi nding the gradient of a tangent to a curve which is called the derivative

Differentiation from fi rst principles: The process of fi nding the gradient of a tangent to a curve by fi nding the gradient of the secant between two points and fi nding the limit as the secant becomes a tangent

Gradient of a secant: The gradient (slope) of the line between two points that lie close together on a function

Gradient of a tangent: The gradient (slope) of a line that is a tangent to the curve at a point on a function. It is the derivative of the function

Rate of change: The rate at which the dependent variable changes as the independent variable changes

ch8.indd 392 8/1/09 11:43:59 AM

393Chapter 8 Introduction to Calculus

INTRODUCTION

CALCULUS IS A VERY IMPORTANT part of mathematics and involves the measurement of change. It can be applied to many areas such as science, economics, engineering, astronomy, sociology and medicine. We also see articles in newspapers every day that involve change: the spread of infectious diseases, population growth, infl ation, unemployment, fi lling of our water reservoirs.

For example, this graph shows the change in crude oil production in Iran over the years. Notice that while the graph shows that production is increasing over recent years, the rate at which it is being produced seems to be slowing down. Calculus is used to look at these trends and predict what will happen in the future.

There are two main branches of calculus. Differentiation is used to calculate the rate at which two variables change in relation to one another. Anti-differentiation, or integration, is the inverse of differentiation and uses information about rates of change to go back and examine the original variables. Integration can also be used to fi nd areas of curved objects.

DID YOU KNOW?

‘Calculus’ comes from the Latin meaning pebble or small stone. In ancient civilisations, stones were used for counting. However, the mathematics practised by these early people was quite sophisticated. For example, the ancient Greeks used sums of rectangles to estimate areas of curved fi gures.

However, it wasn’t until the 17 th century that there was a breakthrough in calculus when scientists were searching for ways of measuring motion of objects such as planets, pendulums and projectiles.

Isaac Newton, an Englishman, discovered the main principles of calculus when he was 23 years old. At this time an epidemic of bubonic plague closed Cambridge University where he was studying, so many of his discoveries were made at home.

He fi rst wrote about his calculus methods, which he called fl uxions, in 1671, but his Method of fl uxions was not published until 1704.

Gottfried Leibniz (1646–1716), in Germany, was also studying the same methods and there was intense rivalry between the two countries over who was fi rst!

Search the Internet for further details on these two famous mathematicians. You can fi nd out about the history of calculus and why it was necessary for mathematicians all those years ago to invent it.

7,000

Crude Oil Production (Mbbl/d)Iran

6,000

5,000

4,000

2,000

3,000

1,000

073

7475 77

76 7879 81 83 85 87 89

80 82 84 86 8891 93 95 97 99

90 92 94 96 9801 03 05 07

00 02 04 06

Tho

usan

d B

arre

ls p

er D

ay

January 1973–May 2007

You will learn about integration in the

HSC Course.

Isaac Newton

ch8.indd 393 8/1/09 11:44:08 AM


Gradient

Gradient of a straight line

The gradient of a straight line measures its slope. You studied gradient in the last chapter.

runrisem =

Class Discussion

Remember that an increasing line has a positive gradient and a decreasing line has a negative gradient.

positive negative

Notice also that a horizontal line has zero gradient.Can you see why?

Can you fi nd the gradient of a vertical line? Why?

Gradient plays an important part, not just in mathematics, but in many areas including science, business, medicine and engineering. It is used everywhere we want to fi nd rates.

On a graph, the gradient measures the rate of change of the dependent variable with respect to the change in the independent variable.

In this chapter you will learn about differentiation, which measures the rate of change of one variable with respect to another.

ch8.indd 394 8/1/09 11:44:10 AM


EXAMPLES

1. The graph shows the average distance travelled by a car over time. Find the gradient and describe it as a rate.

Hours

km

400

5t

d

Solution

The line is increasing so it will have a positive gradient.

runrisem

5400

180

80

=

=

=

=

This means that the car is travelling at the rate of 80 km/hour.

2. The graph shows the number of cases of fl u reported in a town over several weeks.

Weeks

Num

ber

ofca

ses

(100

s)

15

10t

N

Find the gradient and describe it as a rate.

CONTINUED

ch8.indd 395 8/1/09 11:44:10 AM


Solution

The line is decreasing so it will have a negative gradient. m

101500

1150

150

runrise

=

= -

= -

= -

This means that the rate is 150- cases/week, or the number of cases reported is decreasing by 150 cases/week.

When fi nding the gradient of a straight line in the number plane, we think of a change in y values as x changes. The gradients in the examples above show rates of change .

However, in most examples in real life, the rate of change will vary. For example, a car would speed up and slow down depending on where it is in relation to other cars, traffi c light signals and changing speed limits.

Class Discussion

The two graphs show the distance that a bicycle travels over time. One is a straight line and the other is a curve.

d

t

Hours

km

20

15

10

5

4321

t

Hours

km

20

15

10

5

4321

d

Is the average speed of the bicycle the same in both cases? What is different about the speed in the two graphs?

How could you measure the speed in the second graph at any one time? Does it change? If so, how does it change?

Gradient of a curve

ch8.indd 396 8/1/09 11:44:12 AM


Here is a more general curve. What could you say about its gradient? How does it change along the curve?

y

x

Copy the graph and mark on it where the gradient is positive, negative and zero.

Using what we know about the gradient of a straight line, we can see where the gradient of a curve is positive, negative or zero by drawing tangents to the curve in different places around the curve.

0

+-

y

x

Notice that when the curve increases it has a positive gradient, when it decreases it has a negative gradient and when it turns around the gradient is zero.

Investigation

There are some excellent computer programs that will draw tangents to a curve and then sketch the gradient curve. One of these is Geometer Sketchpad.

Explore how to sketch gradient functions using this or a similar program as you look at the examples below.

ch8.indd 397 8/1/09 11:44:12 AM


EXAMPLES

Describe the gradient of each curve.

1.

Solution

Where the curve increases, the gradient is positive. Where it decreases, it is negative. Where it turns around, it has a zero gradient.

2.

Solution

ch8.indd 398 8/1/09 11:44:13 AM


There are computer programs that will

draw these tangents.

EXAMPLE

Make an accurate sketch of (a) y x2= on graph paper. Draw tangents to this curve at the points where (b)

3, 2, 1, 0, 1, 2x x x x x x= - = - = - = = = and 3x = . Find the gradient of each of these tangents. (c) Draw the graph of the gradients (the gradient function) on a (d)

number plane.

Solution

(a) and (b)

9

8

7

6

5

4

3

2

1

1 2 3-3 -2x

y

(c)

3, 6

2, 4

1, 2

0, 0

1, 2

2, 4

3, 6

x m

x m

x m

x m

x m

x m

x m

At

At

At

At

At

At

At

= - = -

= - = -

= - = -

= =

= =

= =

= =

(d)

Since we have a formula for fi nding the gradient of a straight line, we fi nd the gradient of a curve by measuring the gradient of a tangent to the curve.

Use the ‘m’ values as the ‘y’ values on this

graph.

ch8.indd 399 8/1/09 11:44:13 AM


Drawing tangents to a curve is diffi cult. We can do a rough sketch of the gradient function of a curve without knowing the actual values of the gradients of the tangents.

To do this, notice in the example above that where m is positive, the gradient function is above the x -axis, where 0,m = the gradient function is on the x -axis and where m is negative, the gradient function is below the x -axis.

EXAMPLES

Sketch the gradient function of each curve.

1.

Solution

First we mark in where the gradient is positive, negative and zero.

Now on the gradient graph, place the points where 0m = on the x -axis. These are at ,x1 x2 and .x3

ch8.indd 400 8/1/09 11:44:14 AM


To the left of ,x1 the gradient is negative, so this part of the graph will be below the x -axis. Between x1 and ,x2 the gradient is positive, so the graph will be above the x -axis. Between x2 and ,x3 the gradient is negative, so the graph will be below the x -axis. To the right of ,x3 the gradient is positive, so this part of the graph will be above the x -axis.

2.

Solution

First mark in where the gradient is positive, negative and zero.

CONTINUED

ch8.indd 401 8/1/09 11:44:14 AM


8.1 Exercises

Sketch the gradient function for each graph.

The gradient is zero at x1 and .x2 These points will be on the x -axis. To the left of ,x1 the gradient is positive, so this part of the graph will be above the x -axis. Between x1 and ,x2 the gradient is negative, so the graph will be below the x -axis. To the right of ,x2 the gradient is positive, so this part of the graph will be above the x -axis.

1.

2.

3.

4.

5.

6.

ch8.indd 402 8/1/09 11:44:15 AM


7.

8.

9.

10.

Differentiation from First Principles

Seeing where the gradient of a curve is positive, negative or zero is a good fi rst step, but there are methods to fi nd a formula for the gradient of a tangent to a curve.

The process of fi nding the gradient of a tangent is called differentiation . The resulting function is called the derivative .

Differentiability

A function is called a differentiable function if the gradient of the tangent can be found.

There are some graphs that are not differentiable in places. Most functions are continuous , which means that they have a smooth

unbroken line or curve. However, some have a gap, or discontinuity, in the graph (e.g. hyperbola). This can be shown by an asymptote or a ‘hole’ in the graph. We cannot fi nd the gradient of a tangent to the curve at a point that doesn’t exist! So the function is not differentiable at the point of discontinuity.

a

y

x

This function is not differentiable at a since the curve is discontinuous at this point.

ch8.indd 403 8/1/09 11:44:15 AM


A function may be continuous but not smooth. It may have a sharp corner. Can you see why curves are not differentiable at the point where there is a corner?

A function ( )y f x= is differentiable at the point x a= if the derivative exists at that point. This can only happen if the function is continuous and smooth at .x a=

b

y

x

This function is not differentiable at b as the curve is discontinuous at this point.

c

y

x

The curve is not differentiable at point c since it is not smooth at that point.

ch8.indd 404 8/1/09 11:44:16 AM


EXAMPLES

1. Find all points where the function below is not differentiable.

C

B

A

y

x

Solution

The function is not differentiable at points A and B since there are sharp corners and the curve is not smooth at these points.

It is not differentiable at point C since the function is discontinuous at this point.

2. Is the function ( )f xx x

x x1

3 2 1for

for 2

1$

=-

) differentiable at all points?

Solution

The functions ( ) ( ) 3 2f x x f x xand2= = - are both differentiable at all points. However, we need to look at where one fi nishes and the other starts, at f (1).

f

f

( )

( )

f x x

f x x

1 11

3 2

1 3 1 21

For

For

2

2

=

=

=

= -

= -

=

]

] ]

g

g g

This means that both pieces of this function join up (the function is continuous). However, to be differentiable, the curve must be smooth at this point.

CONTINUED

ch8.indd 405 8/1/09 11:44:16 AM


8.2 Exercises

For each function, state whether it has any points at which it is not differentiable.

Sketching this function shows that it is not smooth (it has a sharp corner) so it is not differentiable at 1x = .

1

1

-2

y = x2

y = 3x - 2

y

x

1. y

x

2.

x1

y

x

3.

x1

y

x

4. y

x

ch8.indd 406 8/1/09 11:44:17 AM


5.

x1 x2

y

x

6. ( ) 4f x x=

7. 3

1yx

= -+

8. ( )f xx xx x

21 2

if if

3 2#

=+

)

9. ( )f x

x x

x

x x

2 3

3 2 3

1 2

for

for

for

2

2

1

# #= -

- -

Z

[

\

]]

]

10.

-4

-4

-5

-3

-3

-2

-2

-1-1

2

1

3

4

5

1 2 3 4x

y

11. tany x= for x0 360c c# #

12. ( )f x xx

=

13. ( ) cosf 3 2i i= -

14. ( ) sing 2z z=

15. 93y

xx

2=

-

-

Limits

To differentiate from fi rst principles, we need to look more closely at the concept of a limit.

A limit is used when we want to move as close as we can to something. Often this is to fi nd out where a function is near a gap or discontinuous point. You saw this in Chapter 5 when looking at discontinuous graphs. In this topic, it is used when we want to move from a gradient of a line between two points to a gradient of a tangent.

EXAMPLES

1. Find .limx

x x2

2x 2

2

-

- -"

Solution

( )

( ) ( )

( 1)

lim lim

limx

x xx

x x

x2

22

1 2

2 1

3

x x

x

2

2

2

2

-

- -=

-

+ -

= +

= +

=

" "

"

You did this in Chapter 5.

CONTINUED

ch8.indd 407 8/1/09 11:44:18 AM


2. Find an expression in terms of x for 2 3limh

xh h h0

2

h

- -"

.

Solution

( )

( )

lim lim

limh

xh h hh

h x h

x h

x

2 3 2 3

2 3

2 3

h h

h

0

2

0

0

- -=

- -

= - -

= -

" "

"

3. Find an expression in terms of x for 3 5limx

x x x x0

2 2

x dd d d+ -

"d .

Solution

( )

( )

lim lim

limx

x x x xx

x x x

x x

x

3 5 3 5

3 5

3 5

x x

x

0

2 2

0

2

0

2

2

dd d d

d

d d

d

+ -=

+ -

= + -

= -

" "

"

d d

d

1. Evaluate

(a) 3lim xx x

0

2

x

+"

(b) 5 2 7lim xx x x

0

3 2

x

- -"

(c) 33lim

xx x

3

2

x -

-"

(d) 416lim

tt

4

2

t -

-"

(e) 11

limgg

1

2

g -

-

"

(f) 2

2limx

x x2

2

x +

+ -"-

(g) 2limh

h h0

5

h

+"

(h) 3

7 12limx

x x3

2

x -

- +"

(i) 525lim

nn

5

2

n -

-"

(j) 1

4 3limx

x x1 2

2

x -

+ +"-

2. Find as an expression in terms of x

(a) 2 4limh

x h xh h0

2

h

- -"

(b) 2limh

x h xh h0

3

h

+ -"

(c) 3 7 4limh

x h xh h h0

2 2 2

h

- + -"

(d) 4 4limh

x h x h xh0

4 2 2

h

- -"

(e) 3 4 3limh

x h xh xh h0

2 2 2

h

+ - +"

(f) 2 5 6limh

x h xh h0

2 2

h

+ +"

(g) 2limx

x x x x0

2 2

x dd d-

"d

(h) 4 2limx

x x x0

2 2

x dd d-

"d

(i) 3limx

x x x x x0

3 2

x dd d d+ -

"d

(j) 2 9limx

x x x x x0

2

x dd d d- +

"d

8.3 Exercises

ch8.indd 408 8/1/09 11:44:18 AM


Differentiation as a limit

The formula m x xy y

2 1

2 1=

-

- is used to fi nd the gradient of a straight line when we

know two points on the line. However, when the line is a tangent to a curve, we only know one point on the line—the point of contact with the curve.

To differentiate from fi rst principles , we fi rst use the point of contact and another point close to it on the curve (this line is called a secant ) and then we move the second point closer and closer to the point of contact until they overlap and the line is at single point (the tangent ). To do this, we use a limit.

If you look at a close up of a graph, you can get some idea of this concept. When the curve is magnifi ed, two points appear to be joined by a straight line. We say the curve is locally straight .

Investigation

Use a graphics calculator or a computer program to sketch a curve and then zoom in on a section of the curve to see that it is locally straight.

For example, here is a parabola.

-20

-10

2 20

10

2

y

xf1(x) = x2

Notice how it looks straight when we zoom in on a point on the parabola?

f 1(x) = x22.99

7.99 y

x

Use technology to sketch other curves and zoom in to show that they are locally straight.

ch8.indd 409 8/1/09 11:44:18 AM


Before using limits to fi nd different formulae for differentiating from fi rst principles, here are some examples of how we can calculate an approximate value for the gradient of the tangent to a curve. By taking two points close together, as in the example below, we fi nd the gradient of the secant and then estimate the gradient of the tangent.

(3, f (3))

(3.01, f (3.01))

y

x

EXAMPLES

1. For the function f x x3=] g , fi nd the gradient of the secant PQ where P is the point on the function where 2x = and Q is another point on the curve close to P . Choose different values for Q and use these results to estimate the gradient of the curve at P .

(2, f(2))

(2.1, f(2.1))y

Q

P

x

ch8.indd 410 8/1/09 5:41:51 PM


Solution

2, (2)P f= ^ h Take different values of x for point Q , for example 2.1x = Using different values of x for point Q gives the results in the table.

Point Q Gradient of secant PQ

. , .f2 1 2 1]_ gi

.( . ) ( )

..

.

mf f

2 1 22 1 2

2 1 22 1 2

12 61

3 3

=-

-

=-

-

=

. , .f2 01 2 01]_ gi

.( . ) ( )

..

.

mf f

2 01 22 01 2

2 01 22 01 2

12 0601

3 3

=-

-

=-

-

=

. , .f2 001 2 001]_ gi

.( . ) ( )

..

.

mf f

2 001 22 001 2

2 001 22 001 2

12 006001

3 3

=-

-

=-

-

=

. , .f1 9 1 9]_ gi

.( . ) ( )

..

.

mf f

1 9 21 9 2

1 9 21 9 2

11 41

3 3

=-

-

=-

-

=

. , .f1 99 1 99]_ gi

.( . ) ( )

..

.

mf f

1 99 21 99 2

1 99 21 99 2

11 9401

3 3

=-

-

=-

-

=

. , .f1 999 1 999]_ gi

.( . ) ( )

..

.

mf f

1 999 21 999 2

1 999 21 999 2

11 994001

3 3

=-

-

=-

-

=

From these results, a good estimate for the gradient at P is 12. We can say that as x approaches 2, the gradient approaches 12.

We can write 2

( ) (2)12lim

x

f x f2x -

-=

" .

Use y

mx x

y

2 1

12

-

-= to fi nd

the gradient of the secant.

CONTINUED

ch8.indd 411 8/1/09 11:44:19 AM


2. For the curve y x2= , fi nd the gradient of the secant AB where A is the point on the curve where 5x = and point B is close to A . Find an estimate of the gradient of the curve at A by using three different values for B .

Solution

5, (5)A f= ^ h Take three different values of x for point B , for example . , .x x4 9 5 1= = and 5.01x = .

(a) . , ( . )

.( . ) ( )

..

.

B f

m x xy y

f f

4 9 4 9

4 9 54 9 5

4 9 54 9 5

9 9

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

(b) . , ( . )

.( . ) ( )

..

.

B f

m x xy y

f f

5 1 5 1

5 1 55 1 5

5 1 55 1 5

10 1

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

(c) . , ( . )

.( . ) ( )

..

.

B f

m x xy y

f f

5 01 5 01

5 01 55 01 5

5 01 55 01 5

10 01

2 1

2 1

2 2

=

=-

-

=-

-

=-

-

=

^ h

From these results, a good estimate for the gradient at A is 10. We can say that as x approaches 5, the gradient approaches 10.

We can write 5

( ) (5)10lim

xf x f

5x -

-=

" .

We can fi nd a general formula for differentiating from fi rst principles by using c rather than any particular number. We use general points , ( )P c f c^ h and , ( )Q x f x^ h where x is close to c .

The gradient of the secant PQ is given by

( ) ( )

m x xy y

x cf x f c

2 1

2 1=

-

-

=-

-

ch8.indd 412 8/1/09 11:44:19 AM


The gradient of the tangent at P is found when x approaches c . We call this ( )cfl .

( )( ) ( )

limf c x cf x f c

x c=

-

-

"l

There are other versions of this formula. We can call the points , ( )P x f x^ h and , ( )Q x h f x h+ +^ h where h is small.

(x+h, f(x+h))

(x , f(x))P

Q

y

x

Secant PQ has gradient

( ) ( )

( ) ( )

m x xy y

x h x

f x h f x

h

f x h f x

2 1

2 1=

-

-

=+ -

+ -

=+ -

To fi nd the gradient of the secant, we make h smaller as shown, so that Q becomes closer and closer to P .

P

Q

Q

Q

Q

y

x

(x+h, f(x+h))

(x, f (x))

Search the Internet using keywords ‘differentiation from

fi rst principles’, gradient of secant’ and ‘tangent’ to fi nd mathematical websites that

show this working.

ch8.indd 413 8/1/09 11:44:20 AM


As h approaches 0, the gradient of the tangent becomes ( ) ( )

limh

f x h f x0h

+ -

" .

We call this ( )xfl .

( )( ) ( )

limxh

f x h f xh 0

=+ -

"fl

If we use ,P x y^ h and ,Q x x y yd d+ +^ h close to P where x yandd d are small:

Gradient of secant PQ

m x xy

x x x

y y y

x

y

y

2 1

2 1

d

d

d

d

=-

-

=+ -

+ -

=

As xd approaches 0, the gradient of the tangent becomes limx

yx 0 d

d

"d . We

call this dx

dy .

The symbol d is a Greek letter called delta.

limdx

dy

x

yx 0d

d=

"d

All of these different notations stand for the derivative, or the gradient of the tangent:

, ( ), ( ) , ( ),dx

dy

dxd y

dxd f x f x yl l^ h

These occur because Newton, Leibniz and other mathematicians over the years have used different notation.

Investigation

Leibniz used dx

dy where d stood for ‘difference’. Can you see why he would

have used this?

Use the Internet to explore the different notations used in calculus and where they came from.

ch8.indd 414 8/1/09 11:44:20 AM


The three formulae for differentiating from fi rst principles all work in a similar way.

EXAMPLE

Differentiate from fi rst principles to fi nd the gradient of the tangent to the curve 3y x2= + at the point where 1.x =

Solution

Method 1:

( )( ) ( )

( )( ) ( )

( )( ) ( )

( )

( ) ( )

( )

lim

lim

lim

lim

lim

lim

lim

f c x cf x f c

f x x

f

f c x cf x f c

fx

f x f

xx

xx

xx x

x

3

1 1 34

11

1

13 4

11

11 1

1

1 1

2

x c

x c

x

x

x

x

x

2

2

1

1

2

1

2

1

1

=-

-

= +

= +

=

=-

-

=-

-

=-

+ -

=-

-

=-

+ -

= +

= +

=

"

"

"

"

"

"

"

l

l

l

]

]

g

g

Method 2:

( )( ) ( )

limf xh

f x h f x

f x x

f

f x h x hx

f h h

h h

h h

3

1 1 34

31

1 1 3

1 2 3

2 4

When

h 0

2

2

2

2

2

2

=+ -

= +

= +

=

+ = + +

=

+ = + +

= + + +

= + +

"l

]

]

] ]

] ]

g

g

g g

g g

Remember that 3y x2= - is the same as ( ) .f x x 32= -

CONTINUED

ch8.indd 415 8/1/09 11:44:20 AM


( )( ) ( )

( )( ) ( )

( )

( )

( )

lim

lim

lim

lim

lim

lim

f xh

f x h f x

fh

f h f

hh h

hh h

hh h

h

11 1

2 4 4

2

2

2

2 0

2

h

h

h

h

h

h

0

0

0

2

0

2

0

0

=+ -

=+ -

=+ + -

=+

=+

= +

= +

=

"

"

"

"

"

"

l

l

Method 3:

lim

dx

dy

x

y

y x 3

x 0

2

d

d=

= +

"d

1 3

x

y

1

4

When2

=

= +

=

So point ,1 4^ h lies on the curve. Substitute point ( , )x y1 4d d+ + :

( )

( )

( )

lim

lim

y x

x x

x x

y x x

x

y

xx x

x

x x

x

dx

dy

x

y

x

4 1 3

1 2 3

2 4

2

2

2

2

2

2 0

2

x

x

2

2

2

2

2

0

0

d d

d d

d d

d d d

d

d

dd d

d

d d

d

d

d

d

+ = + +

= + + +

= + +

= +

=+

=+

= +

=

= +

= +

=

"

"

d

d

We can also use these formulae to fi nd the derivative function generally.

ch8.indd 416 8/1/09 11:44:20 AM


EXAMPLE

Differentiate f x 2x x2 7 3= + -] g from fi rst principles.

Solution

f x x x

f x h x h x h

x xh h x hx xh h x h

2 7 3

2 7 3

2 2 7 7 32 4 2 7 7 3

2

2

2 2

2 2

= + -

+ = + + + -

= + + + + -

= + + + + -

]

] ] ]

^

g

g g g

h

f x h f x x xh h x h x xx xh h x h x x

xh h h

2 4 2 7 7 3 2 7 32 4 2 7 7 3 2 7 3

4 2 7

2 2 2

2 2 2

2

+ - = + + + + - - + -

= + + + + - - - +

= + +

] ] ^ ^g g h h

( )( ) ( )

( )

( )

lim

lim

lim

lim

f xh

f x h f x

hxh h h

hh x h

x h

x

x

4 2 7

4 2 7

4 2 7

4 0 7

4 7

h

h

h

h

0

0

2

0

0

=+ -

=+ +

=+ +

= + +

= + +

= +

"

"

"

"

l

Try this example using the other two formulae.

1. (a) Find the gradient of the secant between the point ,1 2^ h and the point where 1.01x = , on the curve 1.y x4= +

Find the gradient of the (b) secant between ,1 2^ h and the point where 0.999x = on the curve.

Use these results to fi nd the (c) gradient of the tangent to the curve 1y x4= + at the point ,1 2^ h .

2. A function f x x x3= +] g has a tangent at the point ,2 10^ h .

Find the value of (a) 2

( ) (2)x

f x f

-

-

when 2.1x = .

Find the value of (b) ( ) ( )

x

f x f

22

-

-

when 2.01x = .

Evaluate (c) 2

( ) (2)x

f x f

-

- when

.x 1 99= . Hence fi nd the gradient of the (d)

tangent at the point ,2 10^ h .

3. For the function ,f x x 42= -] g fi nd the derivative at point P where 3x = by selecting points near P and fi nding the gradient of the secant.

4. If ( )f x x2= , fi nd (a) ( )f x h+

show that (b) ( ) ( )f x h f x+ - xh h2 2= +

8.4 Exercises

ch8.indd 417 8/1/09 11:44:21 AM


show that (c)

( ) ( )

h

f x h f x+ - x h2= +

show that (d) ( )x x2=fl .

5. A function is given by ( ) 2 7 3f x x x2= - + .

Show that (a) ( )f x h+ =

2 4 2 7 7 3x xh h x h2 2+ + - - + . (b) Show that

( ) ( ) 4 2 7f x h f x xh h h2+ - = + - . Show that (c)

( ) ( )

4 2 7h

f x h f xx h

+ -= + - .

Find (d) ( )xfl .

6. A function is given by ( ) 5f x x x2= + + .

Find (a) f 2] g . Find (b) f h2 +] g . Find (c) f h f2 2+ -] ]g g . Show that (d)

( ) ( )

5h

f h fh

2 2+ -= + .

Find (e) (2)fl .

7. Given the curve ( ) 4 3f x x3= - fi nd (a) f 1-] g fi nd (b) f h1 1- + -f-] ]g g fi nd the gradient of the (c)

tangent to the curve at the point where x 1= - .

8. For the parabola 1y x2= - fi nd (a) f 3] g fi nd (b) f h f3 3+ -] ]g g fi nd (c) (3)fl .

9. For the function ( )f x x x4 3 5 2= - -

fi nd (a) ( )f 1l similarly, fi nd the gradient (b)

of the tangent at the point , 2 10- -^ h .

10. For the parabola y x x22= + show that (a)

2 2y x x x x2d d d d= + +

by substituting the point ,x x y yd d+ +^ h

show that (b) 2 2x

yx x

d

dd= + +

fi nd (c) dx

dy .

11. Differentiate from fi rst principles to fi nd the gradient of the tangent to the curve

(a) f x x2=] g at the point where 1x =

(b) y x x2= + at the point ,2 6^ h (c) f x x2 52= -] g at the point

where x 3= - (d) 3 3 1y x x2= + + at the point

where 2x = (e) f x x x7 42= - -] g at the

point , 1 6-^ h .

12. Find the derivative function for each curve by differentiating from fi rst principles

(a) f x x2=] g (b) y x x52= + (c) f x x x4 4 32= - -] g (d) y x x5 12= - - (e) y x3= (f) f x x x2 53= +] g (g) 2 3 1y x x x3 2= - + - (h) ( )f x x2 3= - .

13. The curve y x= has a tangent drawn at the point ,4 2^ h .

Evaluate (a) 4

( ) (4)x

f x f-

- when

.x 3 9= . Evaluate (b)

4( ) (4)

xf x f

-

- when

.x 3 999= . Evaluate (c)

4( ) (4)

xf x f

-

- when

4.01x = .

14. For the function ( )f x x 1= - ,

evaluate (a) 5

( ) (5)x

f x f-

- when

.x 4 99= .

Remember that

xx

11=

-

ch8.indd 418 8/1/09 11:44:21 AM


evaluate (b) ( ) ( )

xf x f

55

-

- when

5.01x = . Use these results to fi nd the (c)

derivative of the function at the point where x 5= .

15. Find the gradient of the tangent

to the curve 4yx2

= at point

,P 2 1^ h by fi nding the gradient of the secant between P and a point close to P .

Short Methods of Differentiation

The basic rule

Remember that the gradient of a straight line y mx b= + is m . The tangent to the line is the line itself, so the gradient of the tangent is m everywhere along the line.

y

x

y=mx+b

So if ,y mx= dx

dym=

For a horizontal line in the form y k= , the gradient is zero.

y

x

y= k

So if ,y kdx

dy0= =

dxd kx k=] g

dxd k 0=] g

ch8.indd 419 8/1/09 5:42:01 PM


Proof

Investigation

Differentiate from fi rst principles:

y x

y x

y x

2

3

4

=

=

=

Can you fi nd a pattern? Could you predict what the result would be for x n ?

Alternatively, you could fi nd an approximation to the derivative of a

function at any point by drawing the graph of .

( . ) ( )y

f x f x

0 010 01

=+ -

.

Use a graphics calculator or graphing computer software to sketch the derivative for these functions and fi nd the equation of the derivative.

Mathematicians working with differentiation from fi rst principles discovered this pattern that enabled them to shorten differentiation considerably!

For example: When ,y x y x22= =l When ,y x y x33 2= =l When ,y x y x44 3= =l

dxd x nxn n 1= -^ h

You do not need to know this proof.

( )

( ) ( )

( ) ( ) ( )

( ) [( ) ( ) ( ) ( )

. . . ( ) ]

[( ) ( ) ( ) ( )

. . . ( ) ]

f x x

f x h x h

f x h f x x h x

x h x x h x h x x h x x h x

x h x x

h x h x h x x h x x h x

x h x x

n

n

n n

n n n n

n n

n n n n

n n

1 2 3 2 4 3

2 1

1 2 3 2 4 3

2 1

=

+ = +

+ - = + -

= + - + + + + + + +

+ + + +

= + + + + + + +

+ + + +

- - - -

- -

- - - -

- -

^ h

( )( ) ( )

[( ) ( ) ( ) ( ) . . . ( ) ]

[( ) ( ) ( ) ( ) . . . ( ) ]

( ) ( ) ( ) ( ) . . . ( )

lim

lim

lim

f xh

f x h f x

hh x h x h x x h x x h x x h x x

x h x h x x h x x h x x h x x

x x x x x x x x x x

nx

h

h

n n n n n n

h

n n n n n n

n n n n n n

n

0

0

1 2 3 2 4 3 2 1

0

1 2 3 2 4 3 2 1

1 2 3 2 4 3 2 1

1

=+ -

=+ + + + + + + + + + +

= + + + + + + + + + + +

= + + + + + +

=

"

"

"

- - - - - -

- - - - - -

- - - - - -

-

l

ch8.indd 420 8/1/09 11:44:22 AM



EXAMPLE

Differentiate ( )f x x7= .

Solution

( ) 7f x x6=l

There are some more rules that give us short ways to differentiate functions. The fi rst one says that if there is a constant in front of the x (we call this a

coeffi cient), then it is just multiplied with the derivative.

dxd kx knxn n 1= -^ h

( ) ( )dxd kf x kf x= l^ h

A more general way of writing this rule is:

Proof

( )( ) ( )

[ ( ) ( )]

( ) ( )

( )

lim

lim

lim

dxd kf x

h

kf x h kf x

h

k f x h f x

kh

f x h f x

kf x

h

h

h

0

0

0

=+ -

=+ -

=+ -

=

"

"

"

l

^ h

EXAMPLE

Find the derivative of 3 x 8 .

Solution

3y x

dx

dyx

x

3 8

24

If 8

7

7

#

=

=

=

ch8.indd 421 8/1/09 11:44:22 AM


Many functions use a combination of these rules.

Also, if there are several terms in an expression, we differentiate each one separately. We can write this as a rule:

( ) ( ) ( ) ( )dxd f x g x f x g x+ = + ’l^ h

Proof

( ) ( )[ ( ) ( )] [ ( ) ( )]

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

lim

lim

lim

lim

lim lim

dxd f x g x

h

f x h g x h f x g x

h

f x h g x h f x g x

h

f x h f x g x h g x

h

f x h f x

h

g x h g x

h

f x h f x

h

g x h g x

f x g x

h

h

h

h

h h

0

0

0

0

0 0

+ =+ + + - +

=+ + + - -

=+ - + + -

=+ -

++ -

=+ -

++ -

= +

"

"

"

"

" "

l l

^ h

= G


EXAMPLE

Differentiate x x3 4+ .

Solution

( )dxd x x x x3 43 4 2 3+ = +

EXAMPLES

Differentiate 1. 7 x

Solution

dxd x7 7=] g

ch8.indd 422 8/1/09 11:44:23 AM


2. ( ) 5x x xf 4 3= - +

Solution

( )f x x x

x x

4 3 0

4 3

3 2

3 2

= - +

= -

l

3. y x4 7=

Solution

dx

dyx

x

4 7

28

6

6

#=

=

4. If ( )f x x x x2 7 5 45 3= - + - , evaluate ( )f 1-l

Solution

( )

( ) ( ) ( )

f x x x

f

10 21 5

1 10 1 21 1 56

4 2

4 2

= - +

- = - - - +

= -

l

l

5. Differentiate 2

3 5x

x x2 +

Solution

Divide by 2 x before differentiating.

xx x

xx

xx

x

dx

dy

23 5

23

25

23

25

23

121

2 2+= +

= +

=

=

6. Differentiate S r rh2 22r r= + with respect to r .

Solution

We are differentiating with respect to r , so r is the variable and r and h are constants.

( )

drdS r h

r h

2 2 2

4 2

r r

r r

= +

= +

ch8.indd 423 8/1/09 11:44:23 AM


1. Differentiate (a) 2x + (b) 5 9x - (c) 3 4x x2 + + (d) 5 8x x2 - - (e) 2 7 3x x x3 2+ - - (f) 2 7 7 1x x x3 2- + - (g) 3 2 5x x x4 2- + (h) 5 2x x x6 5 4- - (i) 2 4 2 4x x x x5 3 2- + - + (j) 4 7x x10 9-

2. Find the derivative of (a) 2 1x x +] g (b) 2 3x 2-] g (c) 4 4x x+ -] ]g g (d) 2 3x2 2

-^ h (e) 2 5 1x x x2+ - +] ^g h

3. Differentiate

(a) x x6

2

-

(b) 2 3

4x x4 3

- +

(c) ( )x x31 36 2 -

(d) xx x2 53 +

(e) 4

2x

x x2 +

(f) 3

2 3 6 2x

x x x x2

5 4 3 2- + -

4. Find ( )f xl when ( )f x x8 2= - x7 4+ .

5. If y x x2 54 3= - + , fi nd dx

dy when

.x 2= -

6. Find dx

dy if

y x x x6 5 710 8 5= - + - .x3 8+

7. If s t t5 202= - , fi nd dtds .

8. Find ( )g xl given ( )g x x5 4= - .

9. Find dtdv when v t15 92= - .

10. If h t t40 2 2= - , fi nd dtdh .

11. Given 34V r3r= , fi nd

drdV .

12. If ( )f x x x2 3 43= - + , evaluate (1)fl .

13. Given ( ) 5f x x x2= - + , evaluate (a) ( )f 3l (b) ( )f 2-l (c) x when ( )f x 7=l

14. If y x 73= - , evaluate

(a) dx

dy when x 2=

(b) x when dx

dy12=

15. Evaluate ( )g 2l when

( )g t t t t3 4 2 13 2= - - + .

8.5 Exercises

Expand brackets before differentiating.

Simplify by dividing before differentiating.

ch8.indd 424 8/1/09 11:44:23 AM


DID YOU KNOW?

• The word tangent comes from the Latin ‘tangens’, meaning ‘touching’. A tangent to a circle intersects it only once.

• However, a tangent to a curve could intersect the curve more than once.

• A line may only intersect a curve once but not be a tangent.

• So a tangent to a curve is best described as the limiting position of the secant PQ as Q approaches P .

This line is a tangent to the curve at point P.

Remember from earlier in the chapter that the derivative is the gradient of the tangent to a curve.

dx

dy is the gradient of the tangent to a curve

Tangents and Normals

ch8.indd 425 8/1/09 11:44:24 AM


u EXAMPLES

1. Find the gradient of the tangent to the parabola y x 12= + at the point ,1 2^ h .

Solution

, ( )

dx

dyx

x

dx

dy

2 0

2

1 2 2 1

2

At

= +

=

=

=

^ h

So the gradient of the tangent at ,1 2^ h is 2.

2. Find values of x for which the gradient of the tangent to the curve 2 6 1y x x3 2= - + is equal to 18.

Solution

dx

dyx x6 122= -

dx

dy is the gradient of the tangent, so substitute

dx

dy18= .

x 1= -

,

,

x x

x x

x xx x

x x

x

18 6 12

0 6 12 18

2 33 1

3 0 1 0

3

2

2

2

`

= -

= - -

= - -

= - +

- = + =

=

] ]g g

3. Find the equation of the tangent to the curve y x x x3 7 24 3= - + - at the point ,2 4^ h .

Solution

,

dx

dyx x

dx

dy

4 9 7

2 4 4 2 9 2 7

3

At

3 2

3 2

= - +

= - +

=

^ ] ]h g g

So the gradient of the tangent at ,2 4^ h is 3. Equation of the tangent:

y y m x x

y x4 3 21 1- = -

- = -

_

]

i

g

ch8.indd 426 8/1/09 11:44:24 AM


x

y x

x y

3 6

3 2

0 3 2or

= -

= -

= - -

The normal is a straight line perpendicular to the tangent at the same point of contact with the curve.

Normal

Tangent

y

x

If lines with gradients m 1 and m 2 are perpendicular, then m m 11 2 = - You used this rule in the

previous chapter.

EXAMPLES

1. Find the gradient of the normal to the curve y x x2 3 52= - + at the point where x 4= .

Solution

dx

dy is the gradient of the tangent .

13=

x4 3= -dx

dy

x

dx

dy

m

4

4 4 3

13

When

So 1

#

=

= -

=

The normal is perpendicular to the tangent . m m 1So 1 2 = -

CONTINUED

ch8.indd 427 8/1/09 11:44:24 AM


m

m

13 1

131

2

2

= -

= -

So the gradient of the normal is 131

- .

2. Find the equation of the normal to the curve y x x x3 2 13 2= + - - at the point , .1 3-^ h

Solution

dx

dy is the gradient of the tangent .

dx

dyx x

x

dx

dy

m

3 6 2

1

3 1 6 1 2

5

5

When

So

2

2

1

= + -

= -

= - + - -

= -

= -

] ]g g

The normal is perpendicular to the tangent .

m m

m

m

1

5 1

51

So 1 2

2

2

= -

- = -

=

So the gradient of the normal is 51 .

Equation of the normal:

y y m x x

y x

y x

x y

351 1

5 15 1

0 5 16

1 1- = -

- = - -

- = +

= - +

_

]]

i

gg

1. Find the gradient of the tangent to the curve

(a) y x x33= - at the point where 5x =

(b) f x 2x x 4= + -] g at the point ,7 38-^ h

(c) f x 3x x5 4 1= - -] g at the point where x 1= -

(d) 5 2 3y x x2= + + at the point ,2 19-^ h

(e) y x2 9= at the point where 1x =

(f) f x 3x 7= -] g at the point where 3x =

(g) v t t2 3 52= + - at the point where t 2=

(h) 3 2 8 4Q r r r3 2= - + - at the point where 4r =

(i) 4h t t4= - where t 0= (j) f t t t t3 8 55 3= - +] g at the

point where 2t = .

8.6 Exercise s

ch8.indd 428 8/1/09 11:44:25 AM


2. Find the gradient of the normal to the curve

(a) f x x x2 2 13= + -] g at the point where x 2= -

(b) 3 5 2y x x2= + - at the point ,5 48-^ h

(c) f x x x2 72= - -] g at the point where 9x = -

(d) 3 2y x x x3 2= + + - at the point ,4 62- -^ h

(e) f x x10=] g at the point where 1x = -

(f) 7 5y x x2= + - at the point ,7 5- -^ h

(g) 2 3 1A x x x3 2= + - + at the point where 3x =

(h) f a a a3 2 62= - -] g at the point where 3a = -

(i) 4 9V h h3= - + at the point ,2 9^ h

(j) 2 5 3g x x x x4 2= - + -] g at the point where x 1= - .

3. Find the gradient of the (i) tangent and (ii) normal to the curve

(a) 1y x2= + at the point ,3 10^ h (b) f x x5 2= -] g at the point where x 4= - (c) 2 7 4y x x5 2= - + at the point where x 1= - (d) 3 2 8p x x x x6 4= - - +] g where 1x = (e) f x x x4 2= - -] g at the point ,6 26-^ h .

4. Find the equation of the tangent to the curve

(a) 5 1y x x4= - + at the point ,2 7^ h

(b) ( ) 5 3 2 6f x x x x3 2= - - + at the point ,1 6^ h

(c) 2 8y x x2= + - at the point ,3 5- -^ h

(d) 3 1y x3= + at the point where 2x =

(e) 4 7 2v t t4 3= - - at the point where 2t =

5. Find the equation of the normal to the curve

(a) f x x x3 53= - +] g at the point ,3 23^ h

(b) 4 5y x x2= - - at the point ,2 7-^ h

(c) f x x x7 2 2= -] g at the point where 6x =

(d) 7 3 2y x x2= - - at the point ,3 70-^ h

(e) 2 4 1y x x x4 3= - + + at the point where 1x = .

6. Find the equation of the (i) tangent and (ii) normal to the curve

(a) f x x x4 82= - +] g at the point ,1 11^ h (b) 2 5y x x x3 2= + - at the point ,3 6-^ h (c) 5F x x x5 3= -] g at the point where 1x = (d) 8 7y x x2= - + at the point ,3 8-^ h (e) 2 4 1y x x x4 3= - + + at the point where 1x = .

7. For the curve 27 5,y x x3= - -

fi nd values of x for which 0dx

dy= .

8. Find the coordinates of the point at which the curve 1y x3= + has a tangent with a gradient of 3.

9. A function ( ) 4 12f x x x2= + - has a tangent with a gradient of 6- at point P on the curve. Find the coordinates of the point P .

10. The tangent at point P on the curve 4 1y x2= + is parallel to the x -axis. Find the coordinates of P .

11. Find the coordinates of point Q where the tangent to the curve 5 3y x x2= - is parallel to the line 7 3 0x y- + = .

ch8.indd 429 8/1/09 11:44:25 AM


12. Find the coordinates of point S where the tangent to the curve 4 1y x x2= + - is perpendicular to the line 4 2 7 0x y+ + = .

13. The curve 3 4y x2= - has a gradient of 6 at point A .

Find the coordinates of (a) A . Find the equation of the (b)

tangent to the curve at A .

14. A function 3 2 5h t t2= - + has a tangent at the point where t 2= . Find the equation of the tangent.

15. A function f x x x2 8 32= - +] g has a tangent parallel to the line 4 2 1 0x y- + = at point P . Find the equation of the tangent at P .

Further Differentiation and Indices

The basic rule for differentiating x n works for any rational number n .

Investigation

(a) Show that 1. 1 1( )x h x x x h

h+

- =+

- .

Hence differentiate (b) 1y x= from fi rst principles .

Differentiate (c) y x 1= - using a short method. Do you get the same answer as 1(b)? (a) Show that 2. ( )( )x h x x h x h+ - + + = .

Hence differentiate (b) y x= from fi rst principles .

Differentiate (c) 2y x=1

and show that this gives the same answer as 2(b).

EXAMPLES

1. Differentiate 7 x3 .

Solution

3

x x

dx

dyx

x

x

7 7

731

37

37 1

3 3

31

3

2

$

#

=

=

=

=

-

-

1

1

2

We sometimes need to change a function into index form before differentiating.

ch8.indd 430 8/1/09 11:44:26 AM


x

x

37 1

3

7

23

23

#=

=

2. Find the equation of the tangent to the curve 4yx2

= at the point where 2.x =

Solution

yxx

dx

dyx

x

4

4

8

8

2

2

3

3

=

=

= -

= -

-

-

2xWhen =

y24

1

2=

=

Gradient of the tangent at 2,1 :^ h

dx

dy

28

1

3= -

= -

Equation of the tangent:

y y m x x

y xx

y x

x y

1 1 22

3

3 0or

1 1- = -

- = - -

= - +

= - +

+ - =

_

]

i

g

8.7 Exercises

1. Differentiate

(a) x 3- (b) x1.4 (c) 6x0.2

(d) 2x1

(e) 22 3x x 1- -1

(f) 3x 31

(g) 8x 4

3

(h) 22x--

1

2. Find the derivative function, writing the answer without negative or fractional indices .

(a) 1x

(b) 5 x (c) x6

(d) 2x5

(e) 5x3

-

(f) 1x

ch8.indd 431 8/1/09 11:44:26 AM


(g) x21

6

(h) x x

(i) x3

2

(j) x x41 3

2 4+

3. Find the gradient of the tangent to the curve y x3= at the point where 27.x =

4. If 12xt

= , fi nd dtdx when .t 2=

5. A function is given by ( )f x x4= . Evaluate .( )f 16l

6. Find the gradient of the tangent

to the curve 23yx2

= at the point

1, 121

c m .

7. Find dx

dy if y x x

2= +^ h .

8. A function ( )2

f xx

= has a

tangent at , .4 1^ h Find the gradient of the tangent.

9. Find the equation of the tangent

to the curve 1yx3

= at the point

2,81

c m .

10. Find the equation of the tangent to ( ) 6f x x= at the point where 9.x =

11. (a) Differentiate xx

.

H(b) ence fi nd the gradient of the

tangent to the curve y xx

= at

the point where 4.x =


to the curve 4y x= at the point

8,21

c m .

13. If the gradient of the tangent to

y x= is 61 at point A , fi nd the

coordinates of A .

14. The function ( ) 3f x x= has

( )f x43

=l . Evaluate x .

15. The hyperbola 2y x= has two

tangents with gradient 252

- . Find

the coordinates of the points of contact of these tangents.

Note that 1

.x x2 2

1 16 6

#=

Use index laws to simplify fi rst.

Expand brackets fi rst.

Composite Function Rule

A composite function is a function composed of two or more other functions. For example, 3 4x2 5

-^ h is made up of a function u 5 where 3 4u x2= - . To differentiate a composite function, we need to use the result. .

This rule is also called the function of a function rule or chain rule.

dx

dy

du

dy

dxdu

#=

ch8.indd 432 8/1/09 11:44:26 AM


Proof

Let ,x yd d and ud be small changes in x , y and u where , , .x y u0 0 0" " "d d d

0, 0

lim lim lim

x

y

u

y

xu

x u

x

y

u

y

xu

Then

As

Sox u x0 0 0

" "

#

#

d

d

d

d

dd

d d

d

d

d

d

dd

=

=" " "d d d

Using the defi nition of the derivative from fi rst principles, this gives

dx

dy

du

dy

dxdu

#= .

You do not need to learn this proof.

EXAMPLES

Differentiate 1. (5 4)x 7+

Solution

Let 5 4u x= +

` 7

( )

dxdu

y u

du

dyu

dx

dy

du

dy

dxdu

u

x

5

7 5

35 5 4

Then

7

6

6

6

#

#

=

=

=

=

=

= +

2. (3 2 1)x x2 9+ -

Solution

( )

( ) ( )

u x x

dxdu x

y u

du

dyu

dx

dy

du

dy

dxdu

u x

x x x

3 2 1

6 2

9

9 6 2

9 6 2 3 2 1

Let

Then

2

9

8

8

2 8

#

= + -

= +

=

=

=

= +

= + + -

Can you see a quick way of doing this

question?

CONTINUED

ch8.indd 433 8/1/09 11:44:27 AM


3. 3 x-

Solution

2

2

2

2

2

( )

( 1)

(3 )

x xu x

dxdu

y u

du

dyu

dx

dy

du

dy

dxdu

u

x

x

3 33

1

21

21

21

2 31

Let

#

- = -

= -

= -

=

=

=

= -

= - -

= --

-

-

-

1

1

1

1

1

The derivative of a composite function is the product of two derivatives. One is the derivative of the function inside the brackets. The other is the derivative of the whole function.

[dxd n n 1-( )] ( ) [f x f x n= ( )]f xl

Proof

( )

( )

( )

( ) [ ( )]

u f x

dxdu f x

y u

du

dynu

dx

dy

du

dy

dxdu

nu f x

f x n f x

Let

Then

n

n

n

n

1

1

1

`

#

#

=

=

=

=

=

=

=

-

-

-

l

l

l


ch8.indd 434 8/1/09 11:44:27 AM


EXAMPLES

Differentiate 1. (8 1)x3 5-

Solution

( ) [ ( )]

( )

( )

dx

dyf x n f x

x x

x x

24 5 8 1

120 8 1

n 1

2 3 4

2 3 4

$

$

=

= -

= -

-l

2. (3 8)x 11+

Solution

.( ) [ ( )]

( )

( )

y f x n f x

x

x

3 11 3 8

33 3 8

n 1

10

10

#

=

= +

= +

-l l

3. (6 1)

1x 2+

Solution

( )( )

( ) [ ( )]

( )

( )

( )

xx

y f x n f x

x

x

x

6 11 6 1

6 2 6 1

12 6 1

6 112

n

22

1

3

3

3

$

#

+= +

=

= - +

= - +

= -+

-

-

-

-

l l

1. Differentiate (a) ( )x 3 4+ (b) ( )x2 1 3- (c) ( )x5 42 7- (d) ( )x8 3 6+ (e) ( )x1 5-

(f) 3(5 9)x 9+ (g) ( )x2 4 2- (h) ( )x x2 33 4+ (i) ( 5 1)x x2 8+ - (j) ( 2 3)x x6 2 6- +

(k) 2( )x3 1-1

8.8 Exercises

ch8.indd 435 8/1/09 11:44:27 AM


(l) (4 )x 2- -

(m) ( 9)x2 3- -

(n) 3( )x5 4+1

(o) 4( )x x x73 2- +3

(p) 3 4x +

(q) 5 2

1x -

(r) ( 1)

1x2 4+

(s) ( )x7 3 2-3

(t) 4

5x+

(u) 2 3 1

1x -

(v) 4(2 7)

3x 9+

(w) 3 31

x x x4 3- +

(x) ( )x4 1 43 +

(y) ( )x7

154 -

2. Find the gradient of the tangent to the curve 3 2y x 3= -] g at the point ., 1 1^ h

3. If ( ) 2( 3)f x x2 5= - , evaluate (2)fl .

4. The curve 3y x= - has a

tangent with gradient 21 at point

N . Find the coordinates of N .

5. For what values of x does the

function ( )4 1

1f xx

=-

have

( )f x494

= -l ?

6. Find the equation of the tangent to (2 1)y x 4= + at the point where 1.x = -

Product Rule

Differentiating the product of two functions y uv= gives the result

dx

dyu

dxdv v

dxdu

= +

Proof

y uv= Given that , y u vand d d d are small changes in y , u and v .

( ) ( )y y u u v v

uv u v v u u v

y u v v u u v y uv

x

yu

xv v

xu u

xv

since`

d d d

d d d d

d d d d d

d

d

dd

dd

ddd

+ = + +

= + + +

= + + =

= + +

^ h

ch8.indd 436 8/1/09 11:44:27 AM


0, 0

lim lim

lim lim lim

x u

x

yu

xv v

xu u

xv

uxv v

xu u

xv

dx

dyu

dxdv v

dxdu

As

x x

x x x

0 0

0 0 0

" "d d

d

d

dd

dd

ddd

dd

dd

ddd

= + +

= + +

= +

" "

" " "

d d

d d d

<< < <

FF F F

It is easier to remember this rule as y uv vu= +l l l . We can also write this the other way around which helps when learning the quotient rule in the next section.


If ,y uv y u v v u= = +l l l

EXAMPLES

Differentiate 1. 3 1 5x x+ -] ]g g

Solution

You could expand the brackets and then differentiate:

x x x x x

x x

dx

dyx

3 1 5 3 15 5

3 14 5

6 14

2

2

+ - = - + -

= - -

= -

] ]g g

Using the product rule:

y uv u x v x

u v

3 1 5

3 1

where and= = + = -

= =l l

y u v v u

x xx x

x

3 5 1 3 13 15 3 1

6 14

= +

= - + +

= - + +

= -

l l l

] ]g g

2. 2 5 3x x5 3+] g

Solution

.

y uv u x v x

u x v x

2 5 3

10 5 3 5 3

where and5 3

4 2

= = = +

= = +l l

]

]

g

g

CONTINUED

ch8.indd 437 8/1/09 11:44:28 AM


.

y u v v u

x x x x

x x x x

x x x x

x x x

10 5 3 5 3 5 3 2

10 5 3 30 5 3

10 5 3 5 3 3

10 5 3 8 3

4 3 2 5

4 3 5 2

4 2

4 2

$

= +

= + + +

= + + +

= + + +

= + +

l l l

] ]

] ]

] ]

] ]

g g

g g

g g

g g

6 @

3. (3 4) 5 2x x- -

Solution

2

2

2

( )

x x

y uv u x v x

u v x

5 2 5 2

3 4 5 2

3 221 5 2

Remember

where and

$

- = -

= = - = -

= = - --

1

1

1

l l

]

]

g

g

-

2

2

2

2

( )

( )

( )

( ) ( )

y u v v u

x x x

x x x

xx

x

xx

x

x

x x x

x

x x

xx x

xx

3 5 2 221 5 2 3 4

3 5 2 3 4 5 2

3 5 25 2

3 4

3 5 25 2

3 4

5 2

3 5 2 5 2 3 4

5 2

3 5 2 3 4

5 215 6 3 4

5 219 9

$

$

= +

= - +- - -

= - - - -

= - -

-

-

= - --

-

=-

- - - -

=-

- - -

=-

- - +

=-

-

-1 1

1

1

l l l

] ] ]

]

g g g

g

We can simplify this further by factorising.

1. Differentiate

(a) 2 3x x3 +] g (b) 3 2 2 1x x- +] ]g g (c) 3 5 7x x +] g (d) 4 3 1x x4 2 -^ h (e) 2 3x x x4 -^ h (f) 1x x2 3+] g

(g) 4 3 2x x 5-] g (h) x x3 44 3-] g (i) 1 2 5x x 4+ +] ]g g (j) 5 3 1x x x3 2 2 5

+ - +^ ^h h (k) 2x x-

(l) 2 15 3

xx-

+

8.9 Exercises

Change this into a product before differentiating.

ch8.indd 438 8/1/09 11:44:28 AM


2. Find the gradient of the tangent to the curve 2 3 2y x x 4= -] g at the point 1, 2^ h .

3. If ( ) (2 3)(3 1)f x x x 5= + - , evaluate ( )1fl .

4. Find the exact gradient of the tangent to the curve y x x2 5= + at the point where 1x = .

5. Find the gradient of the tangent where 3,t = given 2 5 1x t t 3= - +] ]g g .

6. Find the equation of the tangent to the curve 2 1y x x2 4= -] g at the point , .1 1^ h

7. Find the equation of the tangent to ( 1) ( 1)h t t2 7= + - at the point , .2 9^ h

8. Find exact values of x for which the gradient of the tangent to the curve 2 3y x x 2= +] g is 14.

9. Given ( ) (4 1)(3 2)f x x x 2= - + , fi nd the equation of the tangent at the point where x 1= - .

Quotient Rule

Differentiating the quotient of two functions y vu

= gives the result.

dx

dy

v

vdxdu u

dxdv

2=

-

Proof

y vu

=

Given that , y u vandd d d are small changes in y , u and v .

`

( )

( )

( )

( )

( )

( ) ( )

( )

( )

( ),

y yv vu u

yv vu u

vu y v

u

v v v

v u u

v v v

u v v

v v v

v u u u v v

v v vvu v u uv u v

v v vv u u v

x

y

v v v

vxu u

xv

x v0 0

since

As " "

ddd

ddd

d

d

d

d

d

d d

dd d

dd d

d

d

ddd

dd

d d

+ =+

+

=+

+- =

=+

+-

+

+

=+

+ - +

=+

+ - -

=+

-

=+

-

a k

ch8.indd 439 8/1/09 11:44:29 AM


( )lim lim

x

y

v v v

vxu u

xv

dx

dy

v

vdxdu u

dxdv

x x0 0

2

d

d

ddd

dd

=+

-

=

-

" "d d

R

T

SSSS

V

X

WWWW

It is easier to remember this rule as yv

u v v u2

=-

ll l

.


If ,y vu y

vu v v u

2= =

-l

l l

EXAMPLES

Differentiate

1. 5 23 5

xx+

-

Solution

y vu u x v x

u v

3 5 5 2

3 5

where and= = - = +

= =l l

( )

( ) ( )

( )

( )

yv

u v v u

x

x x

xx x

x

5 2

3 5 2 5 3 5

5 215 6 15 25

5 231

2

2

2

2

=-

=+

+ - -

=+

+ - +

=+

ll l

2. 1

4 5 2x

x x3

3

-

- +

Solution

=

y vu u x x v x

v x

4 5 2 1

3

where and3 3

2 2

= = - + = -

=u x12 5-l l

( )

( ) ( ) ( )

( )

( )

yv

u v v u

x

x x x x x

xx x x x x x

xx x

1

12 5 1 3 4 5 2

112 12 5 5 12 15 6

110 18 5

2

3 2

2 3 2 3

3 2

5 2 3 5 3 2

3 2

3 2

=-

=-

- - - - +

=-

- - + - + -

=-

- +

ll l

ch8.indd 440 8/1/09 11:44:29 AM


8.10 Exercises

1. Differentiate

(a) 2 1

1x -

(b) 5

3x

x+

(c) 4x

x2

3

-

(d) 5 1

3x

x+

-

(e) 7x

x2

-

(f) 3

5 4xx+

+

(g) 2x x

x2 -

(h) 24

xx-

+

(i) 4 32 7

xx-

+

(j) 3 1

5x

x+

+

(k) 3 7

1xx

2 -

+

(l) 2 3

2x

x2

-

(m) 54

xx

2

2

-

+

(n) 4x

x3

+

(o) 3

2 1x

x x3

+

+ -

(p) 3 4

2 1x

x x2

+

- -

(q) 1x x

x x2

3

- -

+

(r) 2( )x

x

5

2

+1

(s) 5 1

(2 9)xx 3

+

-

(t) (7 2)

1xx

4+

-

(u) (2 5)

(3 4)

x

x3

5

-

+

(v) 1

3 1xx+

+

(w) 2 3

1xx-

-

(x) ( 9)

1xx

2

2

-

+

2. Find the gradient of the tangent to

the curve 3 1

2yx

x=

+ at the point

1,21

c m .

3. If ( )f xxx

2 14 5

=-

+ evaluate ( )f 2l .

4. Find any values of x for which the gradient of the tangent to the

curve 2 14 1y

xx

=-

- is equal to .2-

5. Given ( )f xx

x3

2=

+ fi nd x if

( )f x61

=l .


to the curve 2

yx

x=

+ at the

point 4,32

c m .


to the curve 31y

xx2

=+

- at the

point where 2x = .

ch8.indd 441 8/1/09 11:44:29 AM


Test Yourself 8 1. Sketch the derivative function of

each graph (a)

(b)

2. Differentiate 5 3 2y x x2= - + from fi rst principles .

3. Differentiate (a) 7 3 8 4x x x x6 3 2- + - - (b) 2 13 4

xx+

-

(c) ( 4 2)x x2 9+ - (d) ( )x x5 2 1 4- (e) x x2

(f) 5x2

4. Find dtdv if 2 3 4v t t2= - - .

5. Given ( ) (4 3) ,f x x 5= - fi nd the value of (a) (1)f (b) ( )f 11 .

6. Find the gradient of the tangent to the curve 3 5y x x x3 2= - + - at the point ( , ) .1 10- -

7. If 60 3 ,h t t2= - fi nd dtdh when .t 3=

8. Find all x-values that are not differentiable on the following curves.(a)

(b)

(c)

9. Differentiate (a) f x x2 4 9 4= +] ]g g

(b) 3

5yx

=-

(c) 3 1y x x 2= -] g

(d) 4y x=

(e) ( )f x x5=

-4

-5

-3 -2 -1 2 3 4

21

345

-3-4

-2-1 1

y

x

-5

2 3 4

21

345

-3-4

-2-1

y

x11-44 -3 -2 -1

ch8.indd 442 8/1/09 11:44:30 AM


10. Sketch the derivative function of the following curve.

y

x

11. Find the equation of the tangent to the curve 5 3y x x2= + - at the point 2, 11 .^ h

12. Find the point on the curve 1y x x2= - + at which the tangent has a gradient of 3.

13. Find drdS if 4S r2r= .

14. At which points on the curve 2 9 60 3y x x x3 2= - - + are the tangents horizontal?

15. Find the equation of the tangent to the curve 2 5y x x2= + - that is parallel to the line 4 1.y x= -

16. Find the gradient of the tangent to the curve 3 1 2 1y x x3 2= - -] ]g g at the point where 2.x =

17. Find ( )f 4l when .f x x 3 9= -] ]g g

18. Find the equation of the tangent to the

curve 31yx

= at the point where .x61

=

19. Differentiate 21s ut at2= + with respect

to t and fi nd the value of t for which

5, 7dtds u= = and .a 10= -

20. Find the x -intercept of the tangent to

the curve 2 14 3y

xx

=+

- at the point where .x 1=

1. If ( ) 3 (1 2 ) ,f x x x2 5= - fi nd the value of (1)f and (1) .fl

2. If 7 15 3,Ahh

=-

+ fi nd dhdA when 1.h =

3. Given 2 100 ,x t t4 3= + fi nd dtdx and fi nd

values of t when 0.dtdx

=

4. Find the equations of the tangents to the curve ( 1)( 2)y x x x= - + at the points where the curve cuts the x -axis.

5. Find the points on the curve 6y x3= - where the tangents are parallel to the line 12 1.y x= - Hence fi nd the equations of the normals to the curve at those points.

6. Find ( )f 2l if ( ) 3 2 .f x x= -

7. Differentiate (5 1) ( 9) .x x3 5+ -

8. Find the derivative of (4 9)

2 1 .yxx

4=

-

+

9. If ( ) 2 3 4,f x x x3 2= + + for what exact values of x is ( ) ?x 7=fl


ch8.indd 443 8/3/09 5:54:27 PM


10. Find the equation of the normal to the curve 3 1y x= + at the point where 8.x =

11. The tangent to the curve 2y ax3= + at the point where 3x = is inclined at 135c to the x -axis. Find the value of a .

12. The normal to the curve 1y x2= + at the point where 2,x = cuts the curve again at point P . Find the coordinates of P .

13. Find the exact values of the x - coordinates of the points on the curve (3 2 4)y x x2 3= - - where the tangent is horizontal.

14. Find the gradient of the normal to the curve 2 5y x x= - at the point (4, 8) .

15. Find the equation of the tangent to the curve 2 6y x x x3 2= - + + at point (1, 8) .P Find the coordinates of point Q where this tangent meets the y -axis and calculate the exact length of PQ .

16. The equation of the tangent to the curve 3 2y x nx x4 2= - + - at the point where 2x = - is given by 3 2 0.x y- - = Evaluate n .

17. The function ( ) 3 1f x x= + has a tangent that makes an angle of 30c with the x -axis. Find the coordinates of the point of contact for this tangent and fi nd its equation in exact form.

18. Find all x values of the function ( ) ( 3)(2 1)f x x x2 8= - - for which ( )f x 0=l .

19. (a) Find any points at which the graph below is not differentiable.

Sketch the derivative function for (b) the graph.

y

x90c 180c 270c 360c

20. Find the point of intersection between the tangents to the curve 2 5 3y x x x3 2= - - + at the points where 2x = and 1.x = -

21. Find the equation of the tangent to the

parabola 2

3y x2

=- at the point where

the tangent is perpendicular to the line .x y3 3 0+ - =

22. Differentiate .x

x2

3 23

-

23. (a) Find the equations of the tangents to the parabola 2y x2= at the points where the line 6 8 1 0x y- + = intersects with the parabola.

Show that the tangents are (b) perpendicular.

24. Find any x values of the function

( )8 12

2f xx x x3 2

=- +

where it is not

differentiable.

25. The equation of the tangent to the curve 7 6 9y x x x3 2= + - - is y ax b= + at the point where .x 4= - Evaluate a and b .

ch8.indd 444 8/3/09 5:54:42 PM


26. Find the exact gradient with rational denominator of the tangent to the curve 3y x2= - at the point where 5.x =

27. The tangent to the curve y xp

= has a

gradient of 61

- at the point where 3.x =

Evaluate p .

28. Find drdV when

32r r

= and 6h = given

31V r h3r= .

29. Evaluate k if the function ( ) 2 1f x x kx3 2= - + has ( ) .f 2 8=l

30. Find the equation of the chord joining the points of contact of the tangents to the curve 4y x x2= - - with gradients 3 and .1-

31. Find the equation of the straight line passing through 4, 3^ h and parallel to the tangent to the curve y x4= at the point 1, 1 .^ h

32. Find ( )7fl as a fraction, given

( )1

1f xx3

=+

.

33. For the function ( ) , ( ) ,f x ax bx c f 2 42= + + =

( ) ( )x1 0 8and= =f fl l when .x 3= - Evaluate a , b and c .

34. Find the equation of the tangent to the curve 2 2S r rh2r r= + at the point where 2r = ( h is a constant) .

35. Differentiate (a) 2 3 5x x x3 4- -] g

(b) ( 3)

2 1x

x3-

+

36. The tangents to the curve 2 3y x x3 2= - + at points A and B are perpendicular to the tangent at , .2 3^ h Find the exact values of x at A and B .

37. (a) Find the equation of the normal to the curve 1y x x2= + - at the point P where 3.x =

Find the coordinates of (b) Q , the point where the normal intersects the parabola again.

ch8.indd 445 8/3/09 5:54:57 PM

TERMINOLOGY

9 The Quadratic Function

Axis of symmetry: A line about which two parts of a graph are symmetrical. One half of the graph is a refl ection of the other

Coeffi cient: A constant multiplied by a pronumeral in an algebraic term e.g. in ax 3 the a is the coeffi cient

Discriminant: Part of the quadratic formula, the algebraic expression 4b ac2

- is called the discriminant as its value determines the number and nature of the roots of a quadratic equation

Equations reducible to quadratics: Equations that can be reduced to the form: 0ax bx c2

=+ +

Indefi nite: A quadratic function where f ( x ) can be both positive and negative for varying values of x

Maximum value: The maximum or greatest y-value of a graph for a given domain

Minimum value: The minimum or smallest y-value of a graph for a given domain

Negative defi nite: A quadratic function where f(x) is always negative for all values of x

Positive defi nite: A quadratic function where f(x) is always positive for all values of x

Root of an equation: The solution of an equation

ch9.indd 450 7/18/09 2:40:31 PM

451Chapter 9 The Quadratic Function

INTRODUCTION THE SOLUTION OF QUADRATIC equations is important in many fi elds, such as engineering, architecture and astronomy. In this chapter you will study quadratic equations in detail, and look at the relationship between quadratic equations and the graphs of quadratic functions (the parabola). You will study the axis of symmetry and maximum and minimum values of the quadratic function. You will also look at the quadratic formula in detail, and at the relationships between the roots (solutions) of quadratic equations, the formula and the quadratic function.

DID YOU KNOW?

Thousands of clay tablets from ancient Babylonia have been discovered by archaeologists. These tablets are from as far back as 2000 BC. They show that the Babylonians had mastered many mathematical skills. Geometry, including Pythagoras’ theorem, was well developed, and geometric problems were often worked out by using algebra.

Quadratic equations were used in solving geometry problems. The word ‘quadratic’ comes from the Latin ‘ quadratum ’, meaning ‘four-sided fi gure’. Completing the square and the quadratic formula were both used to solve quadratic equations.

The Babylonians also had some interesting approximations for square roots. For example,

122 .=17 An approximation for 2 that is very accurate was found on a tablet dating back to 1600 BC:

2 32 1 1.4142136024

6051

6010

= + + + =

Graph of a Quadratic Function

Axis of symmetry

EXAMPLE

Sketch the parabola (a ) 4y x x2= - on the number plane. Find the equation of the axis of symmetry of the parabola. (b) Find the minimum value of the parabola. (c )

Solution

For the (a) y -intercept, 0x =

( )y 0 4 0

0i.e. 2= -

=

For the x -intercept, y 0=

`

( )x xx x

x x

x

0 44

0 4 0

4

i.e.

or

2= -

= -

= - =

=

CONTINUED

The axis of symmetry lies halfway between x 0= and x 4.=

ch9.indd 451 7/18/09 2:40:34 PM


The axis of symmetry has equation (b) 2.x = Since the parabola is symmetrical about the line (c) 2,x = the

minimum value is on this line. Substitute 2x = into the equation of the parabola

( )y 2 4 2

4i.e. 2= -

= -

So the minimum value is .4-

Class Investigation

How would you fi nd the axis of symmetry for a graph with no 1. x -intercepts?

How would you fi nd the axis of symmetry of a graph where the 2. x -intercepts are irrational numbers?

The axis of symmetry of the quadratic function y ax bx c2= + + has the equation

2

xab

= -

ch9.indd 452 7/18/09 2:40:37 PM


Proof

The axis of symmetry lies midway between the x -intercepts. For the x -intercepts, 0y =

ax bx c

xa

b b ac

0

24

i.e. 2

2!

+ + =

=- -

The x -coordinate of the axis of symmetry is the average of the x -intercepts.

x ab b ac

ab b ac

ab

ab

ab

22

42

4

222

42

2

i.e.

2 2

=

- - -+

- + -

=

-

=-

= -

The parabola has a minimum value if 0.a2 The shape of the parabola is concave upwards .

Minimum value

The parabola has a maximum value if 0.a1 The shape of the parabola is concave downwards.

Maximum value

ch9.indd 453 7/18/09 2:40:37 PM


EXAMPLES

1. Find the equation of the axis of symmetry and the minimum value of the quadratic function 5 1.y x x2= - +

Solution

The equation of the axis of symmetry is given by

( )( )

xab

x

x

y

2

2 15

25

221

25 5

25 1

425

225 1

541

i.e.

Equation is

Minimum value:2

= -

= --

=

=

= - +

= - +

= -

c cm m

So minimum value is 4

.5 1-

2. Find the equation of the axis of symmetry and the maximum value of the quadratic function .y x x3 52= - + -

Solution

The equation of the axis of symmetry is given by

( )

xab

x

x

y

2

2 31

61

61

361

61 5

121

61 5

41211

i.e.

Equation is

Maximum value:2

`

= -

= --

=

=

= - + -

= - + -

= -

c cm m

So maximum value is .41211

-

a 02 gives a minimum value.

a 01 gives a maximum value.

The minimum or maximum value is fab

2-c m

ch9.indd 454 7/18/09 2:40:38 PM


Class Investigation

Examine the graph of y x x3 52= - + - from the above example. Are there any solutions for the quadratic equation ?x x3 5 02- + - =

The minimum or maximum point of the parabola is called the vertex.

EXAMPLE

Find the equation of the axis of symmetry and the coordinates of the (a) vertex of the parabola y x x2 12 72= +- .

Find the (b) y -intercept and sketch the graph .

Solution

Axis of symmetry: (a)

xa

b2

2 212

3#

= -

= --

=

When 3x =

y 2 3 3

11

2=

= -

12 7- +] ]g g

So the vertex is ( , ) .3 11-

For (b) y -intercept, 0x =

y 2 0 12 0 7

7= - +

=

2] ]g g

The vertex is the minimum point of the parabola since a 02 .

1. By fi nding the intercepts on the axes, sketch the parabola 2 .y x x2= + Find the equation of its axis of symmetry, and the minimum value.

2. Find the equation of the axis of symmetry and the minimum value of the parabola 2 6 3.y x x2= + -

3. Find the equation of the axis of symmetry and the minimum value of the parabola 3 2.y x x2= + +

4. Find the equation of the axis of symmetry and the minimum value of the parabola 4.y x2= -

9.1 Exercises

y

x3

(3, -11)

7

-11

ch9.indd 455 7/31/09 4:56:57 PM


5. Find the equation of the axis of symmetry and the minimum point of the parabola 4 3 1.y x x2= - +

6. Find the equation of the axis of symmetry and the maximum value of the parabola 2 7.y x x2= - + -

7. Find the equation of the axis of symmetry and the maximum point of the parabola 2 4 5.y x x2= - - +

8. Find the minimum value of 4 3.y x x2= + + How many solutions does the equation 4 3 0x x2 + + = have?

9. Find the minimum value of 4.y x x2= + + How many solutions does the equation 4 0x x2 + + = have?

10. Find the minimum value of 4 4.y x x2= + + How many solutions does the equation 4 4 0x x2 + + = have?

11. Find the equation of the axis of symmetry and the coordinates of the vertex for each parabola .

(a) y x x6 32= + - (b) y x x8 12= - - + (c) 2 5y x x2= - + (d) y x x4 10 72= + - (e) 3 18 4y x x2= + +

12. Find the equation of the axis of (i)

symmetrythe minimum or maximum (ii)

value andthe vertex of the parabola. (iii)

(a) y x x2 22= + - (b) y x x2 4 12= - + -

13. Find the maximum or minimum point for each function.

(a) 2 1y x x2= + + (b) y x x8 72= --

(c) f x x x4 32= + -] g (d) y x x22= - (e) f x x x4 72= - -] g (f) f x x x2 32= + -] g (g) y x x2 52= - +- (h) 2 8 3y x x2= - + + (i) f x x x3 3 72= - + +] g (j) f x x x2 42= - + -] g

14. For each quadratic function fi nd any (i) x -intercepts using the

quadratic formula . state whether the function (ii)

has a maximum or minimum value and fi nd this value .

sketch the function on a (iii) number plane .

(a) f x x x4 42= + +] g (b) f x x x2 32= - -] g (c) y x x6 12= - + (d) f x x x22= +] g (e) y x2 182= - (f) y x x3 22= + - (g) f x x x2 62= - - +] g (h) f x x x 32= - - +] g (i) y x x3 22= - - + (j) 2 4 5y x x2= - + +

15. (a) Find the minimum value of the parabola 2 5y x x2= - + . (b) How many solutions does the quadratic equation x x2 5 02 + =- have? (c) Sketch the parabola .

16. (a) How many x -intercepts has the quadratic function f x x x3 92= - +] g ? (b) Find the minimum point of the function . (c) Sketch the function.

17. (a) Find the maximum value of the quadratic function f x x x2 42= - + -] g . (b) How many solutions has the quadratic equation x x2 4 02- + =- ? (c) Sketch the graph of the quadratic function .

ch9.indd 456 7/18/09 2:40:40 PM


Investigation

Could you tell without sketching the function y x x 52= +- if x x 5 02 2- + for all x ? How could you do this?

How could you know that x x2 7 02 1- + - for all x without sketching the graph of f x x x2 72= - + -] g ?

18. (a) Sketch the parabola .y x x5 62= - + (b) From the graph, fi nd values of x for which .x x5 6 02 2- + (c) Find the domain over which .x x5 6 02 #- +

19. Sketch y x x3 2 42= - + and hence show that x x3 2 4 02 2- + for all x.

20. By sketching f x x x 22= + +] g , show that x x 2 02 2+ + for all x.

21. Show by a sketch that x x2 7 02 1- + - for all x .

22. Sketch y x x5 4 12= - + - and show that x x5 4 1 02 1- + - for all x .

Quadratic Inequalities

You looked at solving quadratic inequations in Chapter 3 using the number line. You can also solve them using the graph of a parabola.

For any curve on a number plane 0y = is on the x -axis (all values of y are zero on the x -axis) y 02 is above the x -axis (all positive values of y lie above the x -axis) y 01 is below the x -axis (all negative values of y lie below the x -axis)

For the parabola y ax bx c2= + + 0ax bx c2 + + = on the x -axis

ax bx c 02 2+ + above the x -axis ax bx c 02 1+ + below the x -axis

Substituting ax bx c2 + + for y in the general parabola y ax bx c2= + + gives the following results:

You will look at this later on in the chapter.

ch9.indd 457 7/18/09 2:40:40 PM


EXAMPLES

1. Solve .x x3 2 02 $- +

Solution

First sketch y x x3 22= +- showing x -intercepts ( a 02 so it is concave upwards) . For x -intercepts, 0y =

x x

x x0 3 2

2 1

2= - +

= - -] ]g g

,

,

x x

x x

2 0 1 0

2 1

- = - =

= =

ax2 + bx + c 2 0

ax2 + bx + c = 0

ax2 + bx + c 1 0

a 2 0

x

y

ax2 + bx + c 2 0

ax2 + bx + c = 0

ax2 + bx + c 1 0

a 1 0

x

y

ch9.indd 458 7/18/09 2:40:41 PM


1 2

y

x

y 0$ on and above the x -axis So x x3 2 02 $- + on and above the x -axis. 1, 2x x` # $

2. Solve .x x4 022-

Solution

First sketch y x x4 2= - showing x -intercepts ( a 01 so it is concave downwards). For x -intercepts, 0y =

,

,

x xx x

x x

x

0 44

0 4 0

0

2= -

= -

= - =

= x4 =

] g

y

x

y 02 above the x -axis So 4 0x x22- above the x -axis. .x0 4` 1 1

CONTINUED

ch9.indd 459 7/18/09 2:40:44 PM


3. Solve .x 25 02 1-

Solution

First sketch y x 252= - showing x -intercepts ( a 02 so it is concave upwards). For x -intercepts, 0y =

,0=

,

xx x

x x

x

0 255 5

5 5 0

5

2= -

= + -

+ - =

= - x 5=

] ]g g

y 01 below the x -axis So x 25 02 1- below the x -axis. x5 5` 1 1-

1. x 9 02 2-

2. n n 02 #+

3. a a2 02 $-

4. x4 021-

5. y y6 02 #-

6. t t2 02- 2

7. x x2 8 02 2+ -

8. p p4 3 02 $+ +

9. m m6 8 02 2- +

10. x x6 02#- -

9.2 Exercises

Solve

5-5x

y

ch9.indd 460 7/31/09 4:57:02 PM


11. h h2 7 6 02 1- +

12. x x 20 02 #- -

13. k k35 9 2 02$+ -

14. q q9 18 02 2+-

15. x 2 02$+] g

16. n n12 02#- -

17. x x2 152 1-

18. t t4 122$- -

19. y y3 14 522 +

20. x x3 1 5$+-] ]g g

The Discriminant

The values of x that satisfy a quadratic equation are called the roots of the equation.

The roots of 0ax bx c2 + + = are the x -intercepts of the graph y ax bx c2= + +

If 1. y ax bx c2= + + has 2 x -intercepts, then the quadratic equation 0ax bx c2 + + = has 2 real roots.

y

x

a 2 0

y

x

a 1 0

Since the graph can be both positive and negative, it is called an indefi nite function .

If 2. y ax bx c2= + + has 1 x -intercept, then the quadratic equation 0ax bx c2 + + = has 1 real root

y

x

a 2 0

y

x

a 1 0

ch9.indd 461 7/18/09 2:41:22 PM


If 3. y ax bx c2= + + has no x -intercepts, then the quadratic equation 0ax bx c2 + + = has no real roots

y

x

a 2 0

y

x

a 1 0

Since this graph is always positive, Since this graph is always it is called a positive defi nite negative, it is called a negative function. defi nite function.

This information can be found without sketching the graph.

In the quadratic formula ,xa

b b ac2

42!

=- -

the expression b ac42 - is called

the discriminant . It gives us information about the roots of the quadratic equation 0ax bx x2 + + = .

Investigation

Solve the following quadratic equations using the quadratic formula 1. (a) x x3 2 02 + =- (b) x x4 7 02 + =- (c) 5 0x x2 + + = (d) x x6 9 02 - + =

Without solving a quadratic equation, can you predict how many 2. roots it has by looking at the quadratic formula?

ch9.indd 462 7/18/09 2:41:22 PM


EXAMPLES

Use the quadratic formula to fi nd how many real roots each quadratic equation has.

1. x x5 3 02 + - =

Solution

xa

b b ac2

4

2 15 5 4 1 3

25 25 12

25 37

2

2

!

#

! # #

!

!

=- -

=- - -

=- +

=-

There are 2 real roots:

,x2

5 372

5 37=

- + - -

2. x x 4 02 + =-

Solution

( ) ( )

xa

b b ac2

4

2 11 1 4 1 4

21 15

2

2

!

#

! # #

!

=- -

=- - - -

=-

There are no real roots since 15- has no real value.

3. x x2 1 02 + =-

Solution

( ) ( )

xa

b b ac2

4

2 12 2 4 1 1

22 0

2

2

!

#

! # #

!

=- -

=- - - -

=

CONTINUED

ch9.indd 463 7/18/09 2:41:23 PM


There are 2 real roots:

,

,

x2

2 02

2 0

1 1

=+ -

=

However, these are equal roots.

Notice that when there are 2 real roots, the discriminant .b ac4 02 2- When there are 2 equal roots (or just 1 real root), .b ac4 02 - = When there are no real roots, .b ac4 02 1- We often use .b ac42D = -

If 1. ,0T2 then the quadratic equation 0ax bx c2 + + = has 2 real unequal (different) roots.

y

x

a 2 0

y

x

a 1 0

If T is a perfect square, the roots are rational. If T is not a perfect square, the roots are irrational.

If 2. 0T= , then the quadratic equation 0ax bx c2 + + = has 1 real root or 2 equal roots.

y

x

a 2 0

y

x

a 1 0

Tis the Greek letter 'delta'.

ch9.indd 464 7/18/09 2:41:23 PM


If 3. 0T1 , then the quadratic equation 0ax bx c2 + + = has no real roots.

y

x

y

x

a 1 0a 2 0

If 0T1 and a 02 , it is positive If 0T1 and a 01 , it is defi nite and ax bx c 02 2+ + negative defi nite and for all x. ax bx c 02 1+ + for all x.

We can examine the roots of the quadratic equation by using the discriminant rather than the whole quadratic formula.

EXAMPLES

1. Show that the equation 2 4 0x x2 + + = has no real roots .

Solution

b ac4

1 4 2 41 32

31

0

2

2

T

1

= -

= -

= -

= -

] ]g g

So the equation has no real roots.

2. Find the values of k for which the quadratic equation x x k5 2 02 + =- has real roots .

Solution

For real unequal roots, .0T 2 For real equal roots, .0T = So for real roots, .0T $

CONTINUED

ch9.indd 465 7/18/09 2:41:24 PM


b ac

kk

k

k

0

4 0

2 4 5 04 20 0

4 20

51

2

2

T $

$

$

$

$

$

-

- -

-

] ] ]g g g

3. Show that x x2 4 02 2+- for all x .

Solution

If a 02 and 0T1 , then ax bx c 02 2+ + for all x .

y

x

a 2 0

a

b ac

1

0

4

2 1 44 16

12

0

2

2

T

1

=

= -

= -

= -

= -

42 -] ] ]g g g

Since a 02 and , x x0 2 4 02T1 2- + for all x.

4. Show that the line 4 4 0x y+ + = is a tangent to the parabola y x2= .

Solution

For the line to be a tangent, it must intersect with the curve in only 1 point.

ch9.indd 466 7/18/09 2:41:25 PM


It is too hard to tell from the graph if the line is a tangent, so we solve simultaneous equations to fi nd any points of intersection.

y x

x y1

4 4 0 2

2=

+ + =

]

]

g

g


x x

x x

4 4 0

4 4 0

2

2

+ + =

+ + =

We don’t need to fi nd the roots of the equation as the question only asks how many roots there are. We fi nd the discriminant.

b ac4

4 4 1 416 16

0

2

2

D =

=

=

=

-

-

-

] ]g g

the equation has 1 real root (equal roots) so there is only one point of intersection. So the line is a tangent to the parabola.

1. Find the discriminant of each quadratic equation.

(a) x x4 1 02 =- - (b) 2 3 7 0x x2 + + = (c) x x4 2 1 02- + - = (d) x x6 2 02 =- - (e) x x3 02 =- - (f) 4 0x2 + = (g) x x2 1 02 - + = (h) x x3 2 5 02- - + =

(i) 2 2 0x x2- + + = (j) x x4 4 02- + =-

2. Find the discriminant and state whether the roots of the quadratic equation are real or imaginary (not real), and if they are real, whether they are equal or unequal, rational or irrational.

9.3 Exercises

y

x

1

-1

-3

-4

-5

2

3

4

5

-2

- 521 3 44 -21

-3

-6

ch9.indd 467 7/31/09 6:48:58 PM


(a) x x 4 02 =- - (b) 2 3 6 0x x2 + + = (c) x x9 20 02 + =- (d) 6 9 0x x2 + + = (e) x x2 5 1 02 =- - (f) x x2 5 02- + - = (g) x x2 5 3 02- - + = (h) x x5 2 6 02- + - = (i) x x 02- + = (j) x x2 8 2 02- + - =

3. Find the value of p for which the quadratic equation 2 0x x p2 + + = has equal roots.

4. Find any values of k for which the quadratic equation 1 0x kx2 + + = has equal roots .

5. Find all the values of b for which 2 1 0x x b2 + + + = has real roots .

6. Evaluate p if 4 2 0px x2 + + = has no real roots .

7. Find all values of k for which k x x2 3 02+ + - =] g has 2 real unequal roots .

8. Prove that x x3 7 02 2+- for all real x .

9. Find the values of k for which 1 4 0x k x2 + + + =] g has real roots .

10. Find values of k for which the expression 3 9kx kx2 + + is positive defi nite .

11. Find the values of m for which the quadratic equation x mx2 9 02 + =- has real and different roots .

12. If x kx2 1 02 + =- has real roots, evaluate k .

13. Find exact values of p if px x p2 3 02 + =- is negative defi nite .

14. Evaluate b if 2 5 0b x bx b2 2- - + =] g has real roots .

15. Find values of p for which the quadratic equation 3 0x px p2 + + + = has real roots.

16. Show that the line 2 6y x= + cuts the parabola 3y x2= + in 2 points .

17. Show that the line x y3 4 0+ =- cuts the parabola 5 3y x x2= + + in 2 points .

18. Show that the line y x 4= - - does not touch the parabola y x2= .

19. Show that the line y x5 2= - is a tangent to the parabola y x x3 12= + - .

20. The line y x p3 1= +- is a tangent to the parabola y x2= . Evaluate p .

21. Which of these lines is a tangent to the circle 4x y2 2+ = ?

(a) x y3 1 0- - = (b) x y5 3 0+ =- (c) x y4 3 10 0+ =- (d) x y5 12 26 0+ =- (e) x y2 7 0+ =-

Quadratic Identities

When you use the quadratic formula to solve an equation, you compare a quadratic, say, 3 2 5 0x x2 - + = with the general quadratic 0.ax bx c2 + + =

ch9.indd 468 7/18/09 2:45:19 PM


You are assuming when you do this that x x3 2 52 - + and ax bx c2 + + are equivalent expressions.

We can state this as a general rule: If two quadratic expressions are equivalent to each other then the

corresponding coeffi cients must be equal .

If a x b x c a x b x c12

1 1 22

2 2/+ + + + for all real x then ,a a b b1 2 1 2= = and c c1 2=

Proof

If a x b x c a x b x c12

1 1 22

2 2+ + = + + for more than two values of x , then ( ) ( ) ( ) 0.a a x b b x c c1 2

21 2 1 2- + - + - =

That is, ,a a b b1 2 1 2= = and .c c1 2=

EXAMPLES

1. Write 2 3 5x x2 - + in the form ( ) ( ) .A x B x C1 12- + - +

Solution

( ) ( )

( )

( )

A x B x C A x x Bx B C

Ax Ax A Bx B C

Ax A B x A B C

x x Ax A B x A B C

1 1 2 1

2

2

2 3 5 2For

2 2

2

2

2 2/

- + - + = - + + - +

= - + + - +

= + - + + - +

- + + - + + - +

] g

( )

( )

( )

( ) ( ):

( )

( ):

( ) ( )

A

A B

A B C

B

B

B

A B

C

C

C

x x x x

2 1

2 3 2

5 3

1 2

2 2 3

4 3

1

2 1 3

2 1 5

1 5

4

2 3 5 2 1 1 4

Substitute into

Substitute and into

2 2` /

=

- + = -

- + =

- + = -

- + = -

=

= =

- + =

+ =

=

- + - + - +

You learnt how to solve simultaneous equations

with 3 unknowns in Chapter 3.

CONTINUED

ch9.indd 469 7/18/09 2:45:20 PM


2. Find values for a , b and c if 2( ) .x x a x bx c3 12 /- + + + -

Solution

( )

( )

( )

a x bx c a x x bx c

ax ax a bx c

ax a b x a c

x x ax a b x a c

3 1 6 9 1

6 9 1

6 9 1

6 9 1For

2 2

2

2

2 2/

+ + + - = + + + + -

= + + + + -

= + + + + -

- + + + + -

] g

( )

( )

( )

( ) ( ):

( )

( ) ( ):

( )

a

a b

a c

b

b

b

c

c

c

1 1

6 1 2

9 1 0 3

1 2

6 1 1

6 1

7

1 3

9 1 1 0

8 0

8

Substitute into

Substitute into

=

+ = -

+ - =

+ = -

+ = -

= -

+ - =

+ =

= -

, ,a b c1 7 8` = = - = -

3. Find the equation of the parabola that passes through the points ( , ), ( , )1 3 0 3- - and (2,21).

Solution

The parabola has equation in the form .y ax bx c2= + + Substitute the points into the equation:

, :

, :

, :

a b ca b c

a b c

a b cc

c

a b ca b c

a b c

1 3 3 1 1

3 1

0 3 3 0 0

3 2

2 21 21 2 24 2

4 2 21 3

2

2

`

`

`

- - - = - + - +

= - +

- + = -

= + +

=

=

= + +

= + +

+ + =

2^ ] ]

]

^ ] ]

]

^ ] ]

]

h g g

g

h g g

g

h g g

g

Solve simultaneous equations to fi nd a , b and c . Substitute (2) into (1):

( )

a b

a b

3 3

6 4

- + = -

- = -

ch9.indd 470 7/18/09 2:45:21 PM



a6

( )

( ) :

( )

( ) ( ):

a b

a b

a b

a b

a b

a

4 2 3 21

4 2 18 5

4 2

2 2 12 6

6 5

2 2 12

4 2 18

6

1

#

+ + =

+ =

- = -

+

- = -

+ =

=

=

a 1Substitute into(5):=

( ) b

b

b

b

4 1 2 18

4 2 18

2 14

7

+ =

+ =

=

=

, ,a b c1 7 3= = = Thus the parabola has equation .y x x7 32= + +

1. Find values of a , b and c for which

(a) x x4 32 + - a x b x c1 12/ + + + +] ]g g

(b) x x2 3 12 - +

a x b x c2 22/ + + + +] ]g g (c) x x 22 - -

a x b x c1 12/ - + - +] ]g g

(d) x x 62 + +

a x b x c3 3/ - + - +2] ]g g (e) x x3 5 22 - -

a x b x c1 12/ + + - +] ]g g

(f) x x4 72 + - a x b x c2 22/ - + - +] ]g g

(g) x x2 4 12 + - a x b x c4 22/ + + + +] ]g g

(h) x x3 2 52 - + a x bx c1 2/ + + +] g

(i) x x4 32- + - a x b x c3 32/ + + + +] ]g g

(j) x x2 4 32- + - a x b x c1 12/ - + + +] ]g g

2. Find values of m , p and q for which x x2 12 - -

.m x p x q1 12/ + + + +] ]g g

3. Express x x4 52 +- in the form .Ax x B x C2 1 4- + + + +] ]g g

4. Show that 2 9x x2 + + can be written in the form a x x b x c2 3 2+ + +- -] ] ]g g g where 1a = , 1b = and .c 17=

5. Find values of A , B and C if .x x A x Bx C2 22 2/+ - - + +] g

6. Find values of a , b and c for which x x3 5 12 + -

.ax x bx c x3 12/ + + + +] ]g g

7. Evaluate K , L and M if .x K x L x M3 1 22 / - + + -2] ]g g

9.4 Exercises

ch9.indd 471 7/31/09 6:49:16 PM


8. Express 4 2x2 + in the form .a x b x c5 2 3 22+ + - + -] ]g g

9. Find the values of a , b and c if .x a x b x c20 17 4 5 12/- - - + +] ]g g

10. Find the equation of the parabola that passes through the points

(0, (a) 5- ), (2, 3- ) and ( 3- , 7) (1, (b) 2- ), (3, 0) and ( 2- , 10) ((c) 2- , 21), (1, 6) and ( 1- , 12) (2, 3), (1, (d) 4- ) and ( 1- , 12- ) (0, 1), ((e) 2- , 1) and (2, 7- )

Sum and Product of Roots

When you solve a quadratic equation, you may notice a relationship between the roots. You also used this to factorise trinomials in Chapter 2.

EXAMPLE

Solve (a) .x x9 20 02 - + = Find the sum of the roots. (b) Find the product of the roots. (c)

Solution

(a) ( ) ( )

,

,

x xx x

x x

x

9 20 04 5 0

4 0 5 0

4

2

`

- + =

- - =

- = - =

= x 5=

(b) 4 5

9

Sum = +

=

(c) 4 5

20

Product #=

=

This relationship with the sum and product of the roots works for any quadratic equation.

The general quadratic equation can be written in the form ( )x x 02 a b ab- + + =

where a and b are the roots of the equation.

Proof

Suppose the general quadratic equation 0ax bx c2 + + = has roots a and .b Then this equation can be written in the form

Notice 9- is the coeffi cient of x and 20 is the constant term in the equation.

ch9.indd 472 7/18/09 2:45:23 PM


( ) ( ) 0

0

( ) 0

x x

x x x

x x

i.e. 2

2

a b

b a ab

a b ab

- - =

- - + =

- + + =

EXAMPLES

1. Find the quadratic equation that has roots 6 and .1-

Solution

Method 1: Using the general formula ( )x x 02 a b ab- + + = where 6a = and 1b = -

6 15

6 16#

a b

ab

+ = + -

=

= -

= -

Substituting into ( )x x 02 a b ab- + + = gives x x5 6 02 =- -

Method 2: If 6 and 1- are the roots of the equation then it can be written as

x x

x x x

x x

6 1 0

6 6 0

5 6 0

2

2

- + =

+ - - =

- - =

] ]g g

2. Find the quadratic equation that has roots 3 2+ and .3 2-

Solution

Method 1: Using the general formula

2

( ) ( )

( )

3 2 3 26

3 2 3 2

3 29 2

7

2

#

a b

ab

+ = + + -

=

= + -

= -

= -

=

Substituting into ( )x x 02 a b ab- + + = gives x x6 7 02 + =-

Method 2: If 3 2+ and 3 2- are the roots of the equation then it can be written as

3 2 3 2

x x

x xx x x x x

x x

0

3 2 3 2 03 2 3 9 3 2 2 3 2 2 0

6 7 0

2

2

- + - - =

- - - + =

- + - + - - + - =

- + =

_ _

^ ^

i i

h h

" ", ,

It doesn’t matter which way around we

name these roots.

ch9.indd 473 7/18/09 2:45:24 PM


We can fi nd a more general relationship between the sum and product of roots of a quadratic equation.

If a and b are the roots of the quadratic equation 0ax bx c2 + + = :

Sum of roots: ab

a b+ = -

Product of roots: ac

ab =

Proof

If an equation has roots a and b , it can be written as ( )x x 02 a b ab- + + = . But we know that a and b are the roots of the quadratic equation

0ax bx c2 + + = . Using quadratic identities, we can compare the two forms of the

equation.

( )

( )

ax bx c

ax bx c

x ab x a

c

x x x ab x a

c

ab

ab

ac

a a a a

0

0

0

For

Also

2

2

2

2 2

`

/a b ab

a b

a b

ab

+ + =

+ + =

+ + =

- + + + +

- + =

+ = -

=

EXAMPLES

1. Find (a) a b+ (b) ab (c) 2 2a b+ if a and b are the roots of .x x2 6 1 02 - + =

Solution

(a)

( )ab

26

3

a b+ = -

= --

=

ch9.indd 474 7/18/09 2:45:25 PM


(b) ac

21

ab =

=

(c)

2( )

2

2

3 221

9 1

8

2 2

2 2 2

2 2 2

2 2 2

2 2

2 2

!a b a b

a b a ab b

a b ab a b

a b

a b

a b

+ +

+ = + +

+ - = +

- = +

- = +

= +

^

^

] c

h

h

g m

2. Find the value of k if one root of kx x k7 1 02 - + + = is .2-

Solution

If 2- is a root of the equation then x 2= - satisfi es the equation. Substitute 2x = - into the equation:

5 15 0

k kk k

k

k

k

2 7 2 1 04 14 1 0

5 15

3

- - - + + =

+ + + =

+ =

= -

= -

2] ]g g

3. Evaluate p if one root of x x p2 5 02 + =- is double the other root.

Solution

If one root is a then the other root is .2a Sum of roots:

ab

212

3 2

32

a b

a a

a

a

+ = -

+ = -

= -

= -

CONTINUED

You could use b and 2b instead.

ch9.indd 475 7/18/09 2:45:26 PM


Product of roots:

ac

p

p

p

p

p

p

215

2 5

232 5

294 5

98 5

458

2

2

#

ab

a a

a

=

=-

= -

- = -

= -

= -

- =

c

c

m

m

1. Find a b+ and ab if a and b are the roots of

(a) x x2 1 02 + + = (b) 2 3 6 0x x2 - - = (c) 5 9 0x x2 - - = (d) x x7 1 02 + + = (e) 3 8 3 0y y2 - + =

2. If a and b are the roots of the quadratic equation 3 6 0,x x2 - - = fi nd the value of

(a) a b+ (b) ab

(c) 1 1a b

+

(d) 2 2a b+

3. Find the quadratic equation whose roots are

2 and (a) 5- (b) 3- and 7 (c) 1- and 4- (d) 4 5+ and 4 5- (e) 1 2 7+ and 1 2 7-

4. Find the value of m in 2 6 0x mx2 + - = if one of the roots is 2.

5. If one of the roots of the quadratic equation x x k2 5 1 02 - + - = is 3,- fi nd the value of k .

6. One root of 3 2(3 1) 4 0x b x b2 - + + = is 8. Find the value of b .

7. In the quadratic equation 2 3 0,x x k2 - + = one root is double the other. Find the value of k .

8. In the quadratic equation ,x x p8 1 02 - + - = one root is triple the other. Find the value of p .

9. In the quadratic equation ( ) ,k x x k2 2 3 0502- + + =+ the roots are reciprocals of each other. Find the value of k .

9.5 Exercises

Reciprocals are n and 1n .

ch9.indd 476 7/31/09 6:49:34 PM


10. In the quadratic equation 2 0,x mx2 + + = the roots are consecutive. Find the values of m .

11. In the quadratic equation ( ) ,x k x3 1 5 02- - + + = the roots are equal in magnitude but opposite in sign. Find the value of k .

12. Find values of n in the equation ( )x n x2 5 1 12 02 - - + = if the two roots are consecutive.

13. If the sum of the roots of 0x px r2 + + = is 2- and the product of roots is 7,- fi nd the values of p and r .

14. One root of the quadratic equation 0x bx c2 + + = is 4 and the product of the roots is 8. Find the values of b and c .

15. The roots of the quadratic equation 4 0x x a2 + - = are 1b + and 3.b - Find the values of a and b .

16. Show that the roots of the quadratic equation 3 2 3 0mx x m2 + + = are always reciprocals of one another.

17. Find values of k in the equation

( )x k x k14

1 02 + + ++

=c m if:

roots are equal in magnitude (a) but opposite in sign

roots are equal (b) one root is 1 (c) roots are reciprocals of one (d)

another roots are real .(e)

18. Find exact values of p in the equation 3 0x px2 + + = if

the roots are equal (a) it has real roots (b) one root is double the other .(c)

19. Find values of k in the equation x kx k 1 02 + + =- if

the roots are equal (a) one root is 4 (b) the roots are reciprocals of (c)

one another .

20. Find values of m in the equation mx x m 3 02 + + =- if

one root is (a) 2- it has no real roots (b) the product of the roots is 2. (c)

Equations Reducible to Quadratics

To solve a quadratic equation such as x x3 3 2 02- - - - =] ]g g , you could expand the brackets and then solve the equation. However, in this section you will learn a different way to solve this.

There are other equations that do not look like quadratic equations that can also be solved this way.

Consecutive numbers are numbers that follow each other in order, such as 3 and 4.

ch9.indd 477 7/18/09 2:45:29 PM


EXAMPLES

1. Solve .x x2 3 2 4 02+ - + - =] ]g g

Solution

Let 2u x= +

,

,

u uu u

u u

u u

3 4 04 1 0

4 0 1 0

4 1

Then 2 - - =

- + =

- = + =

= = -

] ]g g

But 2u x= +

,

,

x x

x x

2 4 2 1

2 3

So + = + = -

= = -

2. Solve 2 3x x+ = where .x 0!

Solution

,

,

x x

x xx x

x xx x

x x

x x

x x x

2 3

2 3

2 3

3 2 02 1 0

2 0 1 0

2 1

2

2

# # #

+ =

+ =

+ =

- + =

- - =

- = - =

= =

] ]g g

3. Solve . .9 4 3 3 0x x- + =

Solution

9 3 32 2x x x= =^ ^h h So .9 4 3 3 0x x + =- can be written as .3 4 3 3 0x x2

+ =-^ h Let 3k x=

,

,

k kk k

k k

k k

4 3 03 1 0

3 0 1 0

1 3

2 - + =

- - =

- = - =

= =

] ]g g

But k 3x=

,,x x

3 1 3 30 1

So x x= =

= =

ch9.indd 478 7/18/09 2:45:30 PM


4. Solve sin sinx x2 1 02 + - = for .x0 360c c# #

Solution

Let sinx u=

u uu u

u u

u u

u

2 1 02 1 1 0

2 1 0 1 0

2 1 1

21

Then

or

2 + - =

- + =

- = + =

= = -

=

] ]g g

But sinu x=

So sin x21

= or sin x 1= -

21sinx = has solutions in the 1 st and 2 nd quadrants

sin3021

c =

,

,

x 30 180 30

30 150

So c c c

c c

= -

=

For 1sinx = - , we use the graph of siny x=

From the graph: x 270c= So solutions to sin sinx x2 1 02 + - = are , ,x 30 150 270c c c=

See Chapter 6 if you have forgotten how to solve a trigonometric equation.1

2 3

60c00

30c

90c 180c 270c 360cx

y

1

-1

ch9.indd 479 7/31/09 6:49:46 PM


1. Solve (a) x x1 7 1 10 0- + - + =2] ]g g (b) y y3 3 2 02- - - - =^ ^h h (c) x x2 2 2 8 02+ - + - =] ]g g (d) n n5 7 5 6 0- + - + =2] ]g g (e) a a4 6 4 7 0- + - - =2] ]g g (f) p p1 9 1 20 02+ - + + =^ ^h h (g) x x3 4 3 5 02+ - + - =] ]g g (h) k k8 8 12 02- - - - =] ]g g (i) t t2 2 2 24 02- + - - =] ]g g (j) b b9 2 9 15 02+ - + - =] ]g g

2. Solve ( ) .x 0!

(a) 6 1x x- =

(b) 6 5x x+ =

(c) 20 9 0x x+ - =

(d) 15 8x x+ =

(e) 2 12 11x x+ =

3. Solve (a) x x7 18 04 2 =- - (b) y y6 8 04 2 + =- , giving exact

values (c) x x x x 2 02 2 2+ =- - -^ ^h h

giving exact values (d) x x x x3 1 7 3 1 10 02 2 2+ - - + - + =^ ^h h correct to 2 decimal places

(e) a a a a4 2 4 8 02 2 2+ + + =-^ ^h h giving exact values .

4. Solve (a) .2 9 2 8 0x x2 + =- (b) 3 3 12 0p p2 + - =

(c) 5 5 20 0x x2 =- - (d) 9 3 12 0x x+ - = (e) .4 10 2 16 0x x- + =

5. Solve ( ) .xx

x4 5 022

!+ =

6. Solve x x x x1 1 2 0

2

+ + + - =b bl l

( ) .x 0!

7. Solve

xx

xx

1 9 1 20 022

22

2+ - + + =d dn n

correct to 2 decimal places ( )x 0! .

8. Solve for .x0 360c c# # (a) sin sinx x 02 - = (b) cos cosx x 02 + = (c) sin sinx x2 1 02 - - = (d) cos cosx x2 2 = (e) sin cosx x 12= -

9. Solve for .x0 360c c# # (a) tan tanx x 02 - = (b) cos x 1 02 - = (c) sin sinx x2 02 - = (d) sin sinx x8 10 3 04 2- + = (e) tan tanx x3 10 3 04 2- + =

10. Show that the equation

33

2 5xx

+ ++

= has 2 real

irrational roots ( ) .x 3! -

9.6 Exercises

ch9.indd 480 7/31/09 6:51:05 PM


Test Yourself 9 1. Solve

(a) 3 0x x2 #- (b) n 9 02 2- (c) 4 0y2$-

2. Evaluate a , b and c if x x2 5 72 - +

( ) ( ) .a x b x c2 1 12= + + + +

3. Findthe equation of the axis of (a)

symmetry andthe minimum value of the parabola (b)

.y x x4 12= - +

4. Show that 2 7y x x2= - + is a positive defi nite quadratic function.

5. If a and b are roots of the quadratic equation 6 3 0,x x2 - + = fi nd

(a) a b+ (b) ab

(c) 1 1a b

+

(d) 2 2ab a b+ (e) 2 2a b+

6. Solve (3 2) 2(3 2) 3 0.x x2- - - - =

7. Describe the roots of each quadratic equation as

real, different and rational (i) real, different and irrational (ii) equal or (iii) unreal. (iv)

(a) 2 3 0x x2 - + = (b) 10 25 0x x2 - - = (c) 10 25 0x x2 - + = (d) 3 7 2 0x x2 + - = (e) 6 2 0x x2 - - =

8. Show that x x4 3 021- + - for all x .

9. Findthe equation of the axis of (a)

symmetry andthe maximum value of the quadratic (b)

function .y x x2 62= - - +

10. Write 3 7x2 + in the form ( ) ( ) .a x b x c2 32- + + +

11. Solve sin sinx x2 1 02 + - = for .x0 360c c# #

12. Find the value of k in x x k3 1 02 + + - = if the quadratic equation has

equal roots (a) one root (b) 3- one root double the other (c) consecutive roots (d) reciprocal roots .(e)

13. Solve ( ) .x x x2 5 3 0!= +

14. Find values of m such that mx x3 4 02 1+ - for all x .

15. Solve 5 26.5 25 0.2x x- + =

16. For each set of graphs, state whether they have

2 points (i) 1 point (ii) no points of intersection. (iii)

(a) 7xy = and x y3 5 1 0- - = (b) 9x y2 2+ = and 3 3y x= - (c) x y 12 2+ = and 2 3 0x y- - =

(d) 2y x= and y x3 1= +

(e) y x2= and 4 4y x= -

ch9.indd 481 7/18/09 2:49:37 PM



1. Show that the quadratic equation 2 2 0x kx k2 - + - = has real rational roots.

2. Find the equation of a quadratic function that passes through the points ( , ), ( , )2 18 3 2- - and ( , ) .1 0

3. Find the value of a , b and c if ( ) ( ) .x x ax x b x cx5 3 1 12 2/+ - + + + +

4. Solve .xx

11

25 1022

+ ++

=

5. Find the maximum value of the function ( ) 2 4 9.f x x x2= - - +

6. Find the value of n for which the equation ( 2) 3 5 0n x x2+ + - = has one root triple the other.

17. State if each quadratic function isindefi nite (i) positive defi nite or (ii) negative defi nite. (iii)

(a)

(b)

(c)

(d)

18. Show that kx px k 02 - + = has reciprocal roots for all x .

19. Find the quadratic equation that has roots

4 and (a) 7- (b) 5 7+ and 5 7-

20. Solve . .2 10 2 16 0x x2 - + =

ch9.indd 482 7/18/09 2:49:39 PM


7. Find the values of p for which x x p3 2 02 2- + - for all x .

8. Show that the quadratic equation 2 0x px p2 2- + = has equal roots.

9. Solve 2 5.2 2 0.2 1x x- + =+

10. Find values of A , B and C if ( ) ( ) .x x Ax B x C4 3 7 4 42 2/- + + + + +

11. Express 2

4 1x x

x2 - -

+ in the form

2 1

.x

ax

b-

++

12. Find exact values of k for which 2 5 0x kx k2 + + + = has real roots.

13. Solve cos sinx x3 2 3 02- - = for 0 360 .xc c# #

14. Solve .x x x x1 5 1 6 0

2

+ - + + =b bl l

15. Solve sin cosx x2 2 02 + - = for 0 360 .xc c# #

16. If a and b are the roots of the quadratic equation x x2 4 5 02 + =- , evaluate 3 3a b+ .

ch9.indd 483 7/18/09 2:49:40 PM

TERMINOLOGY

10 Locus and the Parabola

Axis: A line around which a curve is refl ected e.g. the axis of symmetry of a parabola

Chord: An interval joining any two points on a curve. In this chapter, any two points on a parabola

Circle: The locus of a point moving so that it is equidistant from a fi xed point on a plane surface

Directrix: A fi xed line from which all points equidistant from this line and a fi xed point called the focus form a parabola

Focal chord: A chord that passes through the focus

Focal length: The distance between the focus and the vertex of a parabola or the shortest distance between the vertex and the directrix

Focus: A fi xed point from which all points equidistant from this point and the directrix form a parabola

Latus rectum: A focal chord that is perpendicular to the axis of the parabola

Locus: The path traced out by a point that moves according to a particular pattern or rule. Locus can be described algebraically or geometrically

Parabola: The locus of a point moving so that it is equidistant from a fi xed point called the focus and a fi xed line called the directrix

Tangent: A straight line that touches a curve at a single point only

Vertex: The turning point (maximum or minimum point) of a parabola. It is the point where the parabola meets the axis of symmetry

ch10.indd 484 8/4/09 10:39:57 AM

485Chapter 10 Locus and the Parabola

INTRODUCTION

THIS CHAPTER EXPANDS THE work on functions that you have already learned. It shows a method of fi nding the equation of a locus. In particular, you will study the circle and the parabola , defi ned as a locus.

DID YOU KNOW?

Locus problems have been studied since very early times. Apollonius of Perga (262–190 BC), a contemporary (and rival) of Archimedes , studied the locus of various fi gures. In his Plane Loci , he described the locus points whose ratio from two fi xed points is constant. This locus is called the ‘Circle of Apollonius’.

Apollonius also used the equation y lx2= for the parabola.

René Descartes (1596–1650) was another mathematician who tried to solve locus problems. His study of these led him to develop analytical (coordinate) geometry.

Locus

A relation can be described in two different ways. It can be a set of points that obey certain conditions, or a single point that moves along a path according to certain conditions.

A locus is the term used to describe the path of a single moving point that obeys certain conditions.

Circle

ch10.indd 485 8/7/09 1:52:05 PM


EXAMPLES

Describe the locus of the following. 1. A pencil on the end of compasses.

Solution

The path of the pencil is a circle with centre at the point of the compasses.

2. A person going up an escalator (standing still on one step).

Solution

The body travels along a straight line parallel to the escalator.

3. A doorknob on a closing door.

What would the locus be if the person walks up the escalator?

ch10.indd 486 7/18/09 3:44:48 PM


Solution

If the door could swing right around it would follow a circle. So a door closing swings through an arc of a circle.

4. A point on the number line that is 3 units from 0.

Solution

The locus is .3!

5. A point in the number plane that moves so that it is always 3 units from the y -axis.

Solution

The locus is 2 vertical lines with equations .x 3!=

Class Discussion

Describe the path of a person abseiling down a cliff.

1. a racing car driving around a track

2. a person climbing a ladder

3. a child on a swing

4. a ball’s fl ight when thrown

5. a person driving up to the 5th fl oor of a car park

10.1 Exercises

Describe the locus of the following:

ch10.indd 487 7/18/09 3:45:01 PM


6. a point that moves along the number line such that it is always less than 2 units from 0

7. a point on the number plane that moves so that it is always 2 units from the origin

8. a point that moves so that it is always 1 unit from the x -axis

9. a point that moves so that it is always 5 units from the y -axis

10. a point that moves so that it is always 2 units above the x -axis

11. a point that moves so that it is always 1 unit from the origin

12. a point that moves so that it is always 4 units from the point ,1 2-^ h

13. a point that is always 5 units below the x -axis

14. a point that is always 3 units away from the point (1, 1)

15. a point that is always 7 units to the left of the y -axis

16. a point that is always 3 units to the right of the y -axis

17. a point that is always 8 units from the x -axis

18. a point that is always 4 units from the y -axis

19. a point that is always 6 units from the point ( 2- , 4)

20. a point that is always 1 unit from the point ( 4- , 5).

A locus describes a single point ,P x y^ h that moves along a certain path. The equation of a locus can often be found by using ,P x y^ h together with the information given about the locus.

EXAMPLES

1. Find the equation of the locus of a point ,P x y^ h that moves so that it is always 3 units from the origin.

Solution

You may recognise this locus as a circle, centre ,0 0^ h radius 3 units. Its equation is given by 9.x y2 2+ =

Alternatively, use the distance formula.

d x x y y

d x x y yor2 1

22 1

2

22 1

22 1

2

= - + -

= - + -

_ _

_ _

i i

i i

You studied this formula in Chapter 7. It is easier to use d

2

than d to fi nd the equation of the locus.

ch10.indd 488 7/18/09 3:45:03 PM


Let ,P x y^ h be a point of the locus.

0

PO

PO

x y

x y

3

9

0 9

9

We want

i.e. 2

2 2

=

=

- + - =

+ =

22^ ^h h

2. Find the equation of the locus of point ,x yP ^ h that moves so that distance PA to distance PB is in the ratio 2:1 where , , .A B3 1 2 2and= - = -^ ^h h

Solution


: :

[ 3 ] 2 [ 2 ]3 [ 2 ]

4 4 4 4

PA PB

PBPA

PA PB

PA PB

PB

x y x yx y x y

x x y y x x y x

x x y y

x x y y

x x y y

2 1

12

2

2

4

1 41 4 2

6 9 2 1 4

4 16 16 4 16 16

0 3 22 3 18 22

3 22 3 18 22 0

i.e.

i.e.

or

2

2

2 2 2

2

2 2 2 2

2 2

2 2

2 2

`

=

=

=

=

=

- - + - = - + - -

+ + - = - + +

+ + + - + = - + + + +

= - + + + +

= - + + +

- + + + =

2

2

2 2 2

]

^ ^ ^ ^

^ ^ ^ ^

^

g

h h h h

h h h h

h

$ .

3. Find the equation of the locus of a point ,x yP ^ h that moves so that the line PA is perpendicular to line PB , where ,A 1 2= ^ h and , .B 3 1= - -^ h

Place P anywhere on the number plane.

This is the equation of a circle.

Use the distance formula as in

Example 1.

CONTINUED

ch10.indd 489 7/18/09 3:45:05 PM


Solution

Let ,x yP ^ h be a point of the locus. For perpendicular lines, m m 11 2 = -

:

:

m x xy y

PA mx

y

PB m

x

yx

y1

2

3

1

3

1

Using

1

2

2 1

2 1=

-

-

=-

-

=-

-

=+

+

-

-

]

]

g

g

For PA perpendicular to PB

x

y

x

y

x x

y y

y y x x

x x

x x y y

1

2

3

11

2 3

21

2 2 3

2 3

2 5 0i.e.

2

2

2 2

2

2 2

#-

-

+

+= -

+ -

- -= -

- - = - + -

= - - +

+ + - - =

^ h

4. Find the equation of the locus of point ,P x y^ h that is equidistant from fi xed point ,A 1 2-^ h and fi xed line with equation 5.y =

Solution

Let ,P x y^ h be a point of the locus. B has coordinates , .x 5^ h

[ ]

PA PB

PA PB

x y x x y

x y y

x x y y y y

x x y

1 2 5

1 2 5

2 1 4 4 10 25

2 14 20 0

We want

i.e. 2 2

2

2 2 2

2

=

=

- + - - = - + -

- + + = -

- + + + + = - +

- + - =

2 2 2

2 2 2

^ ^ ^ ^

^ ^ ^

h h h h

h h h

These results come from Chapter 7.

The locus is a circle with diameter AB .

This is the equation of a parabola. Can you see where the parabola lies?

ch10.indd 490 7/18/09 3:45:07 PM


1. Find the equation of the locus of point ,P x y^ h that moves so that it is always 1 unit from the origin.

2. Find the equation of the locus of point ,P x y^ h that moves so that it is always 9 units from the point , .1 1- -^ h

3. Find the equation of the locus of a point that moves so that it is always 2 units from the point , .5 2-^ h

4. Find the equation of the locus of point ,P x y^ h that moves so that it is equidistant from the points ,3 2^ h and , .1 5-^ h

5. Find the equation of the locus of a point that moves so that it is equidistant from the points ,4 6-^ h and , .2 7-^ h

6. Find the equation of the locus of point ,P x y^ h that moves so that it is equidistant from the x -axis and the y -axis.

7. Find the equation of the locus of a point P that moves so that PA is twice the distance of PB where ,A 0 3= ^ h and , .B 4 7= ^ h

8. Find the equation of the locus of point ,P x y^ h that moves so that the ratio of PA to PB is :3 2 where ,A 6 5= -^ h and , .B 3 1= -^ h

9. Find the equation of the locus of a point that moves so that it is equidistant from the point ,2 3-^ h and the line 7.y =

10. Find the equation of the locus of a point that moves so that it is equidistant from the point ,0 5^ h and the line 5.y = -

11. Find the equation of the locus of a point that moves so that it is equidistant from the point ,2 0^ h and the line 6.x =



14. Find the equation of the locus of a point ,P x y^ h that moves so that the line PA is perpendicular to line PB where ,A 1 3= -^ h and , .B 4 5= ^ h

15. Find the equation of the locus of a point ,P x y^ h that moves so that the line PA is perpendicular to line PB , where ,A 4 0= -^ h and , .B 1 1= ^ h

16. Find the equation of the locus of a point ,P x y^ h that moves so that the line PA is perpendicular to line PB where ,A 1 5= ^ h and , .B 2 3= - -^ h

17. Point P moves so that 4PA PB2 2+ = where ,A 3 1= -^ h and , .B 5 4= -^ h Find the equation of the locus of P .

18. Point P moves so that 12PA PB2 2+ = where ,A 2 5= - -^ h and , .B 1 3= ^ h Find the equation of the locus of P .

19. Find the equation of the locus of a point that moves so that its distance from the line 3 4 5 0x y+ + = is always 4 units.

10.2 Exercises

ch10.indd 491 7/18/09 3:45:08 PM


20. Find the equation of the locus of a point that moves so that its distance from the line 12 5 1 0x y- - = is always 1 unit.

21. Find the equation, in exact form, of the locus of a point that moves so that its distance from the line 2 3 0x y- - = is always 5 units.

22. Find the equation of the locus of a point that moves so that it is equidistant from the line 4 3 2 0x y- + = and the line 3 4 7 0.x y+ - =

23. Find the equation of the locus of a point that moves so that it is equidistant from the line 3 4 5 0x y+ - = and the line 5 12 1 0.x y+ - =

24. Given two points ,A 3 2-^ h and , ,B 1 7-^ h fi nd the equation of the locus of ,P x y^ h if the gradient of PA is twice the gradient of PB .

25. If R is the fi xed point ,3 2^ h and P is a movable point , ,x y^ h fi nd the equation of the locus of P if the distance PR is twice the distance from P to the line .y 1= -

PROBLEM

Can you see 2 mistakes in the solution to this question ? Find the locus of point ,P x y^ h that moves so that its perpendicular distance from the line 12 5 1 0x y+ - = is always 3 units.

Solution


| |

| |

| |

| |

| |

da b

ax by c

x y

x y

x y

x y

x y

x y

35 12

5 12 1

25 144

5 12 1

169

5 12 1

13

5 12 1

39 5 12 1

0 5 12 40

2 2

1 1

2 2

`

=+

+ +

=+

+ -

=+

+ -

=+ -

=+ -

= + -

= + -

Can you fi nd the correct locus?

ch10.indd 492 7/18/09 3:45:10 PM


Circle as a Locus

The locus of point P (x, y) that is always a constant distance from a fi xed point is a circle.

The circle, centre ,0 0^ h and radius r , has the equation x y r2 2 2+ =

Proof

Find the equation of the locus of point ,P x y^ h that is always r units from the origin.

Let ,x yP ^ h be a point of the locus.

0

OP r

OP r

x y r

x y r

0

i.e. 2 2

2

2 2 2

=

=

- + - =

+ =

22^ ^h h

So x y r2 2 2+ = is the equation of the locus. It describes a circle with radius r and centre , .0 0^ h

The circle, centre ,a b^ h and radius r , has the equation

x a y b r2- + - =22^ ^h h

Proof

Find the equation of the locus of point ,P x y^ h that is always r units from point , .A a b^ h

ch10.indd 493 7/18/09 3:45:11 PM



AP r

AP r

x a y b r

i.e. 2 2

2

=

=

- + - =22^ ^h h

So x a y b r2 2 2- + - =] ^g h is the equation of the locus. It describes a circle with radius r and centre , .a b^ h

EXAMPLES

1. Find the equation of the locus of a point that is always 2 units from the point , .1 0-^ h

Solution

This is a circle with radius 2 and centre , .1 0-^ h Its equation is in the form

[ 1 ]

1

x a y b r

x y

x y

x x y

x x y

0 2

4

2 1 4

2 3 0

i.e.

2

2 2 2

2

2 2

2 2

- + - =

- - + - =

+ + =

+ + + =

+ + - =

2 2

2

^ ^

^ ^

^

h h

h h

h

2. Find the radius and the coordinates of the centre of the circle .x x y y2 6 15 02 2+ + - - =

Solution

We put the equation into the form .x a y b r2- + - =2 2^ ^h h

To do this we complete the square. In general, to complete the square on ,x bx2 + add

2b 2

c m to give:

2 2

x bx b x b22 2

+ + = +c cm m

First we move any constants to the other side of the equation, then complete the square. To complete the square on 2 ,x x2 + we add .

22 1

2

=c m

You could fi nd this equation by using (x, y)P and treating the question as a locus problem.

You learned how to complete the square in Chapter 3.

ch10.indd 494 7/18/09 3:45:14 PM


To complete the square on ,y y62 - we add .26 9

2

=c m x x y y2 6 15 02 2+ + =- -

2x x2 + y y62+ - 15= 2 6 151 9 1 9x x y y2 2+ + - =+ + + +

1 25x y 3+ + - =22^ ^h h

x y1 3 52- - + - =22]^ ^gh h The equation is in the form .x a y b r2- + - =2 2

^ ^h h This is a circle, centre ,1 3-^ h and radius 5.

1. Find the length of the radius and the coordinates of the centre of each circle.

(a) 100x y2 2+ = (b) 5x y2 2+ = (c) x y4 5 16- + - =22^ ^h h (d) x y5 6 49- + + =22^ ^h h (e) x y 3 812 2+ - =^ h

2. Find the equation of each circle in expanded form (without grouping symbols) .

Centre (0, 0) and radius 4 (a) Centre (3, 2) and radius 5 (b) Centre (c) ,1 5-^ h and radius 3 Centre (2, 3) and radius 6 (d) Centre (e) ,4 2-^ h and radius 5 Centre (f) ,0 2-^ h and radius 1 Centre (4, 2) and radius 7 (g) Centre (h) ,3 4- -^ h and radius 9 Centre (i) ,2 0-^ h and radius 5 Centre (j) ,4 7- -^ h and

radius .3

3. Find the equation of the locus of a point moving so that it is 1 unit from the point , .9 4-^ h

4. Find the equation of the locus of a point moving so that it is 4 units from the point , .2 2- -^ h

5. Find the equation of the locus of a point moving so that it is 7 units from the point , .1 0^ h

6. Find the equation of the locus of a point moving so that it is 2 units from the point , .3 8-^ h

7. Find the equation of the locus of a point moving so that it is 2 units from the point , .5 2-^ h

8. Find the equation of a circle with centre ,0 0^ h and radius 3 units.

9. Find the equation of a circle with centre ,1 5^ h and radius 1 unit.

10. Find the equation of a circle with centre ,6 1-^ h and radius 6 units.

11. Find the equation of a circle with centre ,4 3^ h and radius 3 units.

12. Find the equation of a circle with centre ,0 3-^ h and radius 2 2 units.

13. Find the coordinates of the centre and the length of the radius of each circle.

(a) x x y y4 2 4 02 2+ =- - - (b) x x y y8 4 5 02 2+ + =- - (c) x y y2 02 2+ =-

10.3 Exercises

ch10.indd 495 7/18/09 3:45:15 PM


Concentric circles have the same centre.

(d) x x y y10 6 2 02 2+ + =- - (e) x x y y2 2 1 02 2+ + + =- (f) x x y12 02 2+ =- (g) x x y y6 8 02 2+ + =- (h) x x y y20 4 40 02 2+ + + =- (i) x x y y14 2 25 02 2+ + + =- (j) x x y y2 4 5 02 2+ + + - =

14. Find the centre and radius of the circle with equation given by 6 2 6 0.x x y y2 2- + + - =

15. Find the centre and radius of the circle with equation given by 4 10 4 0.x x y y2 2- + - + =

16. Find the centre and radius of the circle with equation given by .x x y y2 12 12 02 2+ + + - =

17. Find the centre and radius of the circle with equation given by 8 14 1 0.x x y y2 2- + - + =

18. Find the centre and radius of the circle with equation given by 3 2 3 0.x x y y2 2+ + - - =

19. Sketch the circle whose equation is given by 4 2 1 0.x x y y2 2+ + - + =

20. Prove that the line 3 4 21 0x y+ + = is a tangent to the circle 8 4 5 0.x x y y2 2- + + - =

21. (a) Show that 2 4 1 0x x y y2 2- + + + = and 2 4 4 0x x y y2 2- + + - = are concentric .

Find the difference between (b) their radii.

22. Given two points ,A 2 5-^ h and , ,B 4 3-^ h fi nd the equation of the circle with diameter AB .

23. Find the exact length of AB where A and B are the centres of the circles 6 0x x y2 2- + = and 4 6 3 0x x y y2 2+ + + - = respectively.

24. (a) Find the length of XY where X and Y are the centres of the circles 6 2 1 0x x y y2 2+ + - + = and 4 2 1 0x x y y2 2- + - + = respectively.

Find the radius of each circle. (b) What conclusion can you draw (c)

from the results for (a) and (b)?

25. Show that the circles 4x y2 2+ = and 2 4 4 0x x y y2 2+ + - - = both have 3 4 10 0x y+ + = as a tangent.

26. A circle has centre ,C 1 3-^ h and radius 5 units.

Find the equation of the (a) circle.

The line (b) 3 1 0x y- + = meets the circle at two points. Find their coordinates.

Let the coordinates be (c) X and Y , where Y is the coordinate directly below the centre C . Find the coordinates of point Z , where YZ is a diameter of the circle.

Hence show (d) .ZXY 90c+ =

27. (a) Find the perpendicular distance from ,P 2 5-^ h to the line 5 12 2 0.x y+ - =

Hence fi nd the equation (b) of the circle with centre P and tangent .x y5 12 2 0+ - =

ch10.indd 496 7/18/09 3:45:17 PM


Parabola as a Locus

The locus of a point that is equidistant from a fi xed point and a fi xed line is always a parabola. The fi xed point is called the focus and the fi xed line is called the directrix.

Work on the parabola as a locus is very important, as the properties of the parabola are useful to us. The parabola is used in lenses of glasses and cameras, in car headlights, and for bridges and radio telescope dishes.

DID YOU KNOW?

Any rope or chain supporting a load (e.g. a suspension bridge) is in the shape of a parabola.

Find some examples of suspension bridges that have a parabola shaped chain.

Other bridges have ropes or chains hanging freely. These are not in the shape of a parabola, but are in a shape called a catenary. Can you fi nd some bridges with this shape?

More recent bridges are cable-stayed, where ropes or chains are attached to towers, or pylons, and fan out along the sides of the bridge. An example is the Anzac Bridge in Sydney.

There are many different bridge designs. One famous bridge in Australia is the Sydney Harbour Bridge.

Research different bridge designs and see if you can fi nd some with parabolic shapes.

Parabola with vertex at the origin

Just as the circle has a special equation when its centre is at the origin, the parabola has a special equation when its vertex is at the origin. Both also have a more general formula.

ch10.indd 497 7/18/09 3:45:18 PM


The locus of a point that is equidistant from a fi xed point and a fi xed line is always in the shape of a parabola.

If the fi xed point is (0, a ) and the fi xed line is y a= - (where a 02 ), then one of the equidistant points is the origin (0, 0). The distance between the points (0, 0) and (0, a ) is a units.

The point on y a= - directly below the origin is , a0 -^ h and the distance from (0, 0) to , a0 -^ h is also a units.

(0, a)

(0,-a) y=-a

a

a

y

x

To fi nd the equation of the parabola, we use the general process to fi nd the equation of any locus. The features of the parabola have special names.

A parabola is equidistant from a fi xed point and a fi xed line. The fi xed point is called the • focus. The fi xed line is called the • directrix. The turning point of the parabola is called the • vertex. The axis of symmetry of the parabola is called its • axis. The distance between the vertex and the focus is called the • focal length. An interval joining any two points on the parabola is called a • chord. A chord that passes through the focus is called a • focal chord. The focal chord that is perpendicular to the axis is called the • latus rectum. A • tangent is a straight line that touches the parabola at a single point.

ch10.indd 498 7/18/09 3:45:24 PM


The locus of point ,P x y^ h moving so that it is equidistant from the point , a0^ h and the line y a= - is a parabola with equation

4x ay2 =

PARABOLA x2 4= ay

The parabola 4x ay2 = has • focus at , a0^ h • directrix with equation y a= - • vertex at ,0 0^ h • axis with equation 0x = focal length• the distance from the vertex to the focus with length a latus rectum• that is a horizontal focal chord with length 4 a

Since the focal length is a , a is always a positive number.

Proof

Let ,P x y^ h be a point of the locus. Taking the perpendicular distance from P to the line ,y a= - point , .B x a= -^ h

0 [ ]

PA PB

PA PB

x y a x x y a

x y a y a

x y ay a y ay a

x ay

2 2

4

2 2

2

2

2 2 2 2 2

2

`

=

=

- + - = - + - -

+ - = +

+ - + = + +

=

2 2 2

2 2

^ ^ ^ ^

^ ^

h h h h

h h

Class Investigation

Find the equation of the locus if point ,P x y^ h is equidistant from , a0 -^ h and .y a=

ch10.indd 499 7/18/09 3:45:26 PM


EXAMPLES

1. Find the equation of the parabola whose focus has coordinates ,0 2^ h and whose directrix has equation 2.y = -

Solution

The focus has coordinates in the form , a0^ h and the directrix has equation in the form ,y a= - where 2.a = ` the parabola is in the form 4x ay2 = where 2a =

( )x y

x y

4 2

8

i.e. 2

2

=

=

2. Find the coordinates of the focus and the equation of the directrix of (a)

the parabola .x y202 = Find the points on the parabola at the endpoints of the latus rectum (b)

and fi nd its length.

Solution

The parabola (a) 20x y2 = is in the form 4x ay2 =

4 20

5

a

a`

=

=

The focal length is 5 units. We can fi nd the coordinates of the focus and the equation of the directrix in two ways.

Method 1: Draw the graph 20x y2 = and count 5 units up and down from the origin as shown.

y

x

(0, 5)

(0, -5) y=-5

x2=20y

5

5

The focus is (0, 5) and the directrix has equation y = -5.

ch10.indd 500 7/18/09 3:45:28 PM


Method 2: The focus is in the form (0, a ) where .a 5= So the focus is (0, 5).

The directrix is in the form y a= - where .a 5= So the directrix is .y 5= -

The latus rectum is a focal chord that is perpendicular to the axis of (b) the parabola as shown

(0, 5)

x2= 20 y

y

x

The endpoints of the latus rectum will be where the line 5y = and the parabola intersect. Substitute 5y = into the parabola.

x y

x

2020 5100

10010

2

!

!

=

=

=

=

=

] g

So the endpoints are ( 10- , 5) and (10, 5).

(0, 5) (10, 5)(-10, 5)

y

x

x2=20 y

From the graph, the length of the latus rectum is 20 units.

CONTINUED

The latus rectum is 4 a units long which gives

20 units.

ch10.indd 501 7/31/09 5:02:19 PM


3. Find the equation of the focal chord to the parabola x y42 = that passes through ( 4- , 4).

Solution

The parabola 4x y2 = is in the form .x ay42 =

4 4

1

a

a`

=

=

The focal length is 1 unit. The focus is 1 unit up from the origin at (0, 1) and the focal chord also passes through ( 4- , 4).

(0, 1)

(-4, 4)

y

x

x2=4y

We can fi nd the equation of the line between (0, 1) and ( 4- , 4) by using either formula

y y m x x x xy y

x xy y

or1 11

1

2 1

2 1- = -

-

-=

-

-_ i

x xy y

x xy y

x

y

xy

y xy x

x y

0

1

4 04 1

14

3

4 1 34 4 3

0 3 4 4

1

1

2 1

2 1

-

-=

-

-

-

-=

- -

-

-=

-

- - =

- + =

= + -

^ h

You used these formulae in Chapter 7.

As you saw in the previous chapter, a parabola can be concave downwards. Can you guess what the equation of this parabola might be?

PARABOLA x2 4= - ay

The locus of a point P ( x , y ) moving so that it is equidistant from the point , a0 -^ h and the line y a= is a parabola with equation 4x ay2 = -

ch10.indd 502 7/18/09 3:50:07 PM


Proof

A(0, -a)

P(x, y)

B(x, a) y=a

y

x

Let P ( x , y ) be a point of the locus. Taking the perpendicular distance from P to the line ,y a= point , .B x a= ^ h

PA PB

PA PB

x y a x x y ax y a y a

x y ay a y ay a

x ay

0

2 2

4

2 2

2 2

2 2 2

2 2 2 2 2

2

`

=

=

- + - - = - + -

+ + = -

+ + + = - +

= -

22^ ^ ^ ^

^ ^

h h h h

h h

7 A

The parabola 4x ay2 = - has • focus at , a0 -^ h • directrix with equation y a= vertex • at (0, 0) axis• with equation 0x = focal length• a latus rectum• a horizontal focal chord with length 4 a

ch10.indd 503 7/18/09 3:51:28 PM


EXAMPLES

1. Find the equation of the parabola with focus ,0 4-^ h and directrix .y 4=

Solution

If we draw this information, the focus is below the directrix as shown. So the parabola will be concave downwards (the parabola always turns away from the directrix).

y

x

(0, -4)

y=4

4

4

The focal length is 4 so .a 4= The parabola is in the form 4x ay2 = - where .a 4=

4416

x ayy

y4

2 = -

= -

= -

] g

2. Find the coordinates of the vertex, the coordinates of the focus and the equation of the directrix of the parabola .x y122 = -

Solution

The parabola x y122 = - is in the form .x ay42 = -

a 3` =

a4 12=

The focal length is 3 units. The vertex is (0, 0). We can fi nd the coordinates of the focus and the equation of the directrix in two ways. Method 1: Draw the graph x y122 = - and count 3 units up and down from the origin as shown. (The parabola is concave downward.)

ch10.indd 504 7/18/09 3:51:30 PM


y

x

(0, -3)

y=3

3

3

x2=-12y

Counting down 3 units, the focus is , .0 3-^ h Counting up 3 units, the directrix has equation .y 3= Method 2: The focus is in the form , a0 -^ h where .a 3= So the focus is , .0 3-^ h The directrix is in the form y a= where .a 3= So the directrix is .y 3=

3. Find the equation of the parabola with focal length 5 and whose vertex is ,0 0^ h and equation of the axis is 0.x =

Solution

Vertex ,0 0^ h and axis given by 0x = give a parabola in the form ,x ay42

!= since there is not enough information to tell whether it is concave upwards or downwards. This gives two possible parabolas.

CONTINUED

ch10.indd 505 7/18/09 3:51:31 PM


( )

a

x y

x y

5

4 5

20

Focal length of 5 means

The equation is

i.e.

2

2

!

!

=

=

=

1. Find the equation of each parabola.

focus (0, 5), directrix (a) 5y = - focus (0, 9), directrix (b) 9y = - focus (0, 1), directrix (c) y 1= - focus (0, 4), directrix (d) 4y = - focus (0, 10), directrix (e)

y 10= - focus (0, 3), directrix (f) 3y = - focus (0, 6), directrix (g) 6y = - focus (0, 11), directrix (h)

y 11= - focus (0, 2), directrix (i) 2y = - focus (0, 12), directrix (j)

y 12= -


focus (0, (a) 1- ), directrix 1y = focus (0, (b) 3- ), directrix 3y = focus (0, (c) 4- ), directrix 4y = focus (0, (d) 7- ), directrix 7y = focus (0, (e) 6- ), directrix 6y = focus (0, (f) 9- ), directrix 9y = focus (0, (g) 8- ), directrix 8y = focus (0, (h) 2- ), directrix 2y = focus (0, (i) 15- ), directrix

15y = focus (0, (j) 13- ), directrix

y 13=

3. Find(i) the coordinates of the focus and(ii) the equation of the directrix of

(a) x y42 = (b) 28x y2 = (c) x y162 =

(d) 36x y2 = (e) 40x y2 = (f) 44x y2 = (g) x y122 = (h) 6x y2 = (i) x y102 = (j) x y152 =

4. Find (i) the coordinates of the focus and (ii) the equation of the directrix of

(a) 4x y2 = - (b) 24x y2 = - (c) 8x y2 = - (d) 48x y2 = - (e) 20x y2 = - (f) 16x y2 = - (g) 32x y2 = - (h) 40x y2 = - (i) 2x y2 = - (j) x y222 = -

5. Find the equation of the parabola with

coordinates of the focus (a) ,0 7^ h and equation of the directrix y 7= -

coordinates of the focus (b) ,0 11^ h and equation of the directrix y 11= -

coordinates of the focus (c) ,0 6-^ h and equation of the directrix y 6=

coordinates of the focus (d) ,0 2^ h and coordinates of the vertex , .0 0^ h

10.4 Exercises

ch10.indd 506 8/4/09 11:42:11 AM


coordinates of the vertex (e) , ,0 0^ h equation of the axis 0x = and focal length 3

coordinates of the vertex (f) , ,0 0^ h equation of the axis 0x = and focal length 8

coordinates of the vertex (g) ,0 0^ h and equation of the axis 0,x = and passing through the point ,8 2-^ h

coordinates of the vertex (h) ,0 0^ h and equation of the axis 0,x = and passing through the point , .1 7-^ h

6. Find the coordinates of the focus, the equation of the directrix and the focal length of the parabola

(a) 8x y2 = (b) 24x y2 = (c) x y122 = - (d) 2x y2 = (e) 7x y2 = - (f) x y2 2 =

7. Find the equation of the focal chord that cuts the curve 8x y2 = at , .4 2-^ h

8. The tangent with equation 2 4 0x y- - = touches the parabola 4x y2 = at A . Find the coordinates of A .

9. The focal chord that cuts the parabola x y62 = - at ,6 6-^ h cuts the parabola again at X . Find the coordinates of X .

10. Find the coordinates of the endpoints of the latus rectum of the parabola 8 .x y2 = - What is the length of the latus rectum?

11. The equation of the latus rectum of a parabola is given by 3.y = - The axis of the parabola is 0,x = and its vertex is , .0 0^ h

Find the equation of the (a) parabola.

Find the equation of the (b) directrix.

Find the length of the focal (c) chord that meets the parabola at

, .231

-c m

12. (a) Show that the point ,3 3-^ h lies on the parabola with equation 3 .x y2 =

Find the equation of the line (b) passing through P and the focus F of the parabola.

Find the coordinates of the (c) point R where the line PF meets the directrix.

13. (a) Find the equation of chord

PQ where ,P 141

-c m and ,Q 2 1^ h

lie on the parabola 4 .x y2 = Show that (b) PQ is not a focal

chord. Find the equation of the circle (c)

with centre Q and radius 2 units. Show that this circle passes (d)

through the focus of the parabola.

14. (a) Show that ,Q aq aq2 2_ i lies on the parabola 4 .x ay2 =

Find the equation of the focal (b) chord through Q .

Prove that the length of the (c) latus rectum is 4 a .

ch10.indd 507 7/18/09 3:51:32 PM


Investigation

Sketch the parabola .x y2= You may like to complete the table below to help you with its sketch.

x

y 3- 2- 1- 0 1 2 3

Is this parabola a function? What is its axis of symmetry?

The parabola that has y 2 rather than x 2 in its equation is a sideways parabola. It still has the same properties, but generally the x and y values are swapped around.

PARABOLA y2 4= ax

The locus of point ,P x y^ h moving so that it is equidistant from the point ,a 0^ h and the line x a= - is a parabola with equation

4y ax2 =

Proof

Find the equation of the locus of point , ,P x y^ h which moves so that it is equidistant from the point ,a 0^ h and the line .x a= -

Coordinates of B are , .a y-^ h

[ ]

PA PB

PA PB

x a y x a y y

x a y x a

x ax a y x ax a

y ax

0

2 2

4

We want

i.e. 2 2

2

2

2 2 2 2 2

2

=

=

- + - = - - + -

- + =

- + + = + +

=

+

2 2 2 2

2 2

^ ^ ^ ^

^ ^

h h h h

h h

ch10.indd 508 7/18/09 3:51:33 PM


The parabola y ax42 = has • focus at ,a 0^ h equation of • directrix x a= - vertex• at ,0 0^ h axis• with equation 0y = focal length• the distance from the vertex to the focus with length a • latus rectum that is a vertical focal chord with length 4 a

EXAMPLES

1. Find the equation of the parabola with focus (7, 0) and directrix .x 7= -

Solution

If we draw this information, the focus is to the right of the directrix as shown (the parabola always turns away from the directrix). So the parabola turns to the right.

y

x

x=-7

77

(7, 0)

CONTINUED

ch10.indd 509 7/18/09 3:51:34 PM


The focal length is 7 so .a 7= The parabola is in the form 4y ax2 = where .a 7=

.

y axx

x

44 728

2 =

=

=

^ h

2. Find the coordinates of the focus and the equation of the directrix of the parabola .y x322 =

Solution

The parabola 32y x2 = is in the form .y ax42 =

4 32

8

a

a`

=

=

The focal length is 8 units. Method 1: Draw the graph 32y x2 = and count 8 units to the left and right from the origin as shown. (The parabola turns to the right.)

y

x

x=-8

88

(8, 0)

y2=32x22

Counting 8 units to the right, the focus is (8, 0). Counting 8 units to the left, the directrix has equation .x 8= - Method 2: The focus is in the form ( a , 0) where .a 8= So the focus is (8, 0). The directrix is in the form x a= - where .a 8= So the directrix is .x 8= -

A parabola can also turn to the left.

ch10.indd 510 7/18/09 3:51:35 PM


PARABOLA y2 4= - ax

The locus of a point P ( x , y ) moving so that it is equidistant from the point ,a 0-^ h and the line x a= is a parabola with equation 4y ax2 = -

Proof

y

x

P (x, y)B (a, y)

A (- a, 0)

x=a

Let P ( x , y ) be a point of the locus. Taking the perpendicular distance from P to the line ,x a= point , .B a y= ^ h

PA PB

PA PB

x a y x a y yx a y x a

x ax a y x ax a

y ax

0

2 2

4

2 2

2 2

2

2 2 2 2 2

2

`

=

=

- - + - = - + -

+ + = -

+ + + = - +

= -

2 2

2 2

^ ^ ^ ^

^ ^

h h h h

h h

7 A

ch10.indd 511 7/18/09 4:13:17 PM


The parabola 4y ax2 = - has • focus at ( a- , 0) • directrix with equation x a= vertex • at (0, 0) axis• with equation 0y = focal length• a latus rectum• a vertical focal chord with length 4 a

EXAMPLES

1. Find the equation of the parabola with focus ( 4- , 0) and directrix .x 4=

Solution

Drawing this information shows that the parabola turns to the left.

y

x(-4, 0)

x=4

4 4

The focal length is 4 so .a 4= The parabola is in the form 4y ax2 = - where .a 4=

.

y axx

x

44 416

2 = -

= -

= -

^ h

2. Find the coordinates of the focus and the equation of the directrix of the parabola .y x22 = -

Solution

The parabola 2y x2 = - is in the form .y ax42 = -

4 2a

a21

`

=

=

The focal length is 21 unit.

ch10.indd 512 7/18/09 3:51:37 PM


Method 1: Draw the graph 2y x2 = - and count

21 unit to the left and right from the

origin as shown. (The parabola turns to the left.)

12

12

(-12

, 0)

12

x =

y

x

Counting 21 units to the left, the focus is , .

21 0-c m

Counting 21 units to the right, the directrix has equation .x

21

=

Method 2: The focus is in the form ( a- , 0) where .a

21

=

So the focus is , .21 0-c m

The directrix is in the form x a= where .a21

=

So the directrix is .x21

=


focus (2, 0), directrix (a) 2x = - focus (5, 0), directrix (b) 5x = - focus (14, 0), directrix (c)

x 14= - focus (9, 0), directrix (d) 9x = - focus (8, 0), directrix (e) 8x = - focus (6, 0), directrix (f) 6x = - focus (7, 0), directrix (g) 7x = - focus (3, 0), directrix (h) 3x = - focus (4, 0), directrix (i) 4x = - focus (1, 0), directrix (j) x 1= -


focus ((a) 9- , 0), directrix 9x = focus ((b) 4- , 0), directrix 4x = focus ((c) 10- , 0), directrix x 10= focus ((d) 6- , 0), directrix 6x = focus ((e) 2- , 0), directrix 2x = focus ((f) 12- , 0), directrix x 12= focus ((g) 11- , 0), directrix x 11= focus ((h) 5- , 0), directrix 5x = focus ((i) 3- , 0), directrix 3x = focus ((j) 7- , 0), directrix x 7=

10.5 Exercises

ch10.indd 513 7/18/09 3:51:38 PM



(a) 8y x2 = (b) y x122 = (c) y x162 = (d) 4y x2 = (e) 28y x2 = (f) 32y x2 = (g) 24y x2 = (h) 36y x2 = (i) y x2 = (j) y x182 =


(a) 8y x2 = - (b) y x122 = - (c) 28y x2 = - (d) 4y x2 = - (e) 24y x2 = - (f) 52y x2 = - (g) 60y x2 = - (h) 2y x2 = - (i) 26y x2 = - (j) y x52 = -

5. Find the equation of the parabola with

coordinates of the focus (a) ,5 0^ h and equation of the directrix x 5= -

coordinates of the focus (b) ,1 0^ h and equation of the directrix x 1= -

coordinates of the focus (c) ,4 0-^ h and equation of the directrix x 4=

coordinates of the focus (d) ,3 0^ h and coordinates of the vertex ,0 0^ h

coordinates of the vertex (e) ,0 0^ h equation of the axis 0y = and focal length 9

coordinates of the vertex (f) , ,0 0^ h equation of the axis 0y = and focal length 2

coordinates of the vertex (g) ,0 0^ h and equation of the axis 0y = and passing through the point ,3 6^ h

coordinates of the vertex (h) ,0 0^ h and equation of the axis 0y = and passing through the point , .2 1^ h

6. Find the coordinates of the focus, the equation of the directrix and the focal length of the parabola

(a) 8y x2 = (b) 4y x2 = (c) y x122 = - (d) 6y x2 = (e) 5y x2 = - (f) y x3 2 =

7. Find the equation of the focal chord that cuts the curve y x162 = at , .4 8^ h

8. Find the length of the latus rectum of the parabola .y x122 = What are the coordinates of its endpoints?

9. The line with equation 3 27 0x y- - = meets the parabola 4y x2 = at two points. Find their coordinates.

10. Let ,R51 2-c m be a point on the

parabola 20 .y x2 = Find the equation of the focal (a)

chord passing through R . Find the coordinates of the (b)

point Q where this chord cuts the directrix.

Find the area of (c) D OFQ where O is the origin and F is the focus.

Find the perpendicular (d) distance from the chord to the point , .P 1 7- -^ h

Hence fi nd the area of (e) D PQR .

ch10.indd 514 7/31/09 5:02:24 PM


Application

A parabolic satellite dish receives its signals through the focus. If the dish has height 12 m and a span of 20 m, fi nd where the focus should be placed, to the nearest mm.

SOLUTION

The parabola is of the form 4x ay2= and passes through (10, 12) and ( 10, 12)-

Substituting (10, 12) gives

10 4 (12)

100 48

2.083

a

a

a

2=

=

=

So the focus should be placed 2.083 m from the vertex. This is 2083 mm to

the nearest millimetre.

1. 4x ay2 =

y

x

x2=4ay

Focus (0, a)

Directrix y=-a

Here is a summary of the 4 different types of parabola with the vertex at the origin.

ch10.indd 515 7/18/09 3:51:40 PM


2. 4x ay2 = -

y

x2=-4ay

Focus (0, -a)

Directrix y=a

x

3. 4y ax2 =

y

y2=4ax

Directrixx=-a

xFocus(a, 0)

4. 4y ax2 = -

y2=-4ax

Directrixx=a

y

xFocus(-a, 0)

General Parabola

When the parabola does not have its vertex at the origin, there is a more general formula.

Since we use a to mean the focal length, we cannot use ( a , b ) as the vertex. We use ( h , k ) instead.

ch10.indd 516 7/18/09 3:51:41 PM


PARABOLA (x - h)2 = 4a(y - k)

The concave upwards parabola with vertex ( h , k ) and focal length a has equation x h a y k4- = -2

^ ^h h

Proof

Find the equation of the parabola with vertex ,h k^ h and focal length a .

Counting up a units from vertex V gives the focus , .F h k a= +^ h Counting down a units from V gives the point on the directrix , .D h k a= -^ h So the equation of the directrix is given by .y k a= - We fi nd the equation of the locus of ,x yP ^ h that is equidistant from point ,h k aF +^ h and line .y k a= -

B has coordinates , .x k a-^ h

i.e.

We want

[ ] [ ]

[ ] [ ]

PF PB

PF PB

x h y k a x x y k a

x h y k a y k a

x h y k a y k ay k a y k a y k a y k a

y k aay ak

a y k

2 2 24 4

4

difference of two squares

2 2

2 2

#

=

=

- + - + = - + - +

- + - - = - +

- = - + - - -

= - + + - - - + - - -

= -

= -

= -

2 2

2 2 2

2 2 2

^ ^ ^ ^

^ ^ ^

^ ^ ^

^ ^ ^ ^

^

^ ^

^

h h h h

h h h

h h h

h h h h

h

h h

h

ch10.indd 517 7/18/09 4:03:50 PM


The parabola x h y ka42- -=^ ^h h has • axis parallel to the y -axis • vertex at ,h k^ h focus• at ,h k a+^ h • directrix with equation y k a= -

EXAMPLES

1. Find the equation of the parabola with focus ,2 3^ h and directrix with equation 7.y = -

Solution

Coordinates of B are , .2 7-^ h The vertex is the midpoint of ,2 3^ h and , .2 7-^ h ,2 2vertex` = -^ h Focal length is the distance from the focus to the vertex. a 5` = From the diagram the parabola is concave upwards. The equation is in the form

[ ]

x h a y k

x yy

x x y

x x y

4

2 4 5 220 2

4 4 20 40

4 20 36 0

i.e.

2

2

- = -

- = - -

= +

- + = +

- - - =

2

2

^ ^

^ ^ ^

^

h h

h h h

h

2. Find the coordinates of the vertex and the focus, and the equation of the directrix, of the parabola with equation .x x y6 12 3 02 + - - =

Draw a diagram to fi nd the vertex and to fi nd a .

ch10.indd 518 7/18/09 4:03:51 PM


PARABOLA (x - h)2 = - 4a(y - k)

Solution

Complete the square on x .

3( )

x x y

x x y

x x y

x yy

6 12 3 0

6 12 3

6 12 3

12 1212 1

9 9

2

2

2

2

+ - - =

+ = +

+ = +

+ = +

= +

+ +

^ h

So the parabola has equation .x y3 12 12+ = +^ ^h h Its vertex has coordinates , .3 1- -^ h

4 12

3

a

a`

=

=

The parabola is concave upwards as it is in the form .x h a y k4- = -2^ ^h h

Count up 3 units to the focus ,3 2focus` = -^ h Count down 3 units to the directrix directrix has equation .y 4= -

It is easy to fi nd the focus and the

directrix by counting along the y -axis.

The concave downwards parabola with vertex ( h , k ) and focal length a has equation x h a y k4- = - -2

^ ^h h

Proof

Find the equation of the concave downwards parabola with vertex ( h , k ) and focal length a.

ch10.indd 519 7/18/09 4:03:52 PM


Counting down a units from the vertex V gives the focus , .F h k a= -^ h Counting up a units from the vertex V gives the point on the directrix , .D h k a= +^ h So the equation of the directrix is given by .y k a= + We fi nd the equation of the locus of P ( x , y ) that is equidistant from point ,F h k a-^ h and line .y k a= +

y

x

F (h , k- a)

P (x, y)

B y= k+a

B has coordinates , .x k a+^ h

( )

PF PB

PF PB

x h y k a x x y k ax h y k a y k a

x h y k a y k ay k a y k a y k a y k a

y k aay ak

a y k

2 2 24 4

4

We want


2 2

2 2

=

=

- + - - = - + - +

- + - + = - -

- = - - - - +

= - - + - + - - - - +

= - -

= - +

= - -

2 2

2 2 2

2 2 2

^ ^ ^ ^

^ ^ ^

^ ^ ^

^ ^ ^ ^

^ ^

^

h h h h

h h h

h h h

h h h h

h h

h

7 7

7 7

A A

A A

ch10.indd 520 7/18/09 4:03:53 PM


The parabola x h a y k4- = - -2^ ^h h has

• axis parallel to the y -axis vertex• at ( h , k ) focus• at ,h k a-^ h • directrix with equation y k a= +

EXAMPLES

1. Find the equation of the parabola with focus ( 2- , 1) and directrix .y 3=

Solution

x(-2, 1)

y=3

-2 -1

1

1

B

2

1

y

3

Coordinates of B are ( 2- , 3) . The vertex is the midpoint of ( 2- , 1) and ( 2- , 3). vertex = ( 2- , 2) Focal length .a 1= From the diagram the curve is concave downwards. The equation is in the form

.

x h a y k

x yx y

x x y

x x y

4

2 4 1 22 4 2

4 4 4 8

4 4 4 0

i.e. 2

2

2

- = - -

- - = - -

+ = - -

+ + = - +

+ + - =

2

2

^ ^

^ ] ^

^ ^

h h

h g h

h h

7 A

2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola .x x y8 8 16 02 - + - =

CONTINUED

ch10.indd 521 7/18/09 4:03:54 PM


PARABOLA ( y - k)2 = 4a(x - h)

Solution

Complete the square on x.

x x y

x x y

x x y

x yy

8 8 16 0

8 8 16

8 8 16

4 8 328 4

16 16

2

2

2

2

- + - =

- = - +

- = - +

- = - +

= - -

+ +

^

^

h

h

So the parabola has equation .x y4 8 4- = - -2^ ^h h

Its vertex has coordinates (4, 4).

4 8

2

a

a`

=

=

The parabola is concave downwards as it is in the form .x h a y k4- = - -2^ ^h h

(4, 4)

(4, 2)

1

1

2

3

4

5

2 3 4

2

2

y

y=6

Count down 2 units to the focus ,4 2focus` = ^ h Count up 2 units to the directrix directrix has equation .y 6=

The parabola with vertex ( h , k ) and focal length a that turns to the right has equation y k a x h4- = -2

^ ^h h

Proof

Find the equation of the parabola that turns to the right with vertex ( h , k ) and focal length a.

ch10.indd 522 7/18/09 4:03:55 PM


Counting a units to the right from the vertex V gives the focus , .F h a k= +^ h Counting a units to the left from the vertex V gives the point on the directrix , .D h a k= -^ h So the equation of the directrix is given by .x h a= - We fi nd the equation of the locus of P ( x , y ) that is equidistant from point ,F h a k+^ h and line .x h a= -

y

F (h+a, k)

x

P (x, y)

x=h-a

B

B has coordinates , .h a y-^ h

( )

PF PB

PF PB

x h a y k x h a y yx h a y k x h a

y k x h a x h ax h a x h a x h a x h a

x h aax ah

a x h

2 2 24 4

4

We want


2 2

2 2 2

2

=

=

- + + - = - - + -

- - + - = - -

- = - + - - -

= - + + - - - + - - -

= -

= -

= -

2

2 2

2 2 2

^ ^ ^ ^

^ ^ ^

^ ^ ]

^ ^ ^ ^

^ ^

^

h h h h

h h h

h h g

h h h h

h h

h

7 7

7 7

A A

A A

ch10.indd 523 7/18/09 4:03:55 PM


The parabola y k a x h4- = -2^ ^h h has

• axis parallel to the x -axis • vertex at ,h k^ h focus• at ,h a k+^ h directrix• with equation x h a= -

EXAMPLES

1. Find the equation of the parabola with focus (1, 1- ) and directrix .x 5= -

Solution

x

(1, -1)

4

2

3

-4 -3 -2-1 1 2 3

33

B4 5

5x=-5

1

y

-5

-2

-3

Coordinates of B are ( 5- , 1- ). The vertex is the midpoint of ( 5- , 1- ) and (1, 1- ). ,2 1vertex` = - -^ h

Focal length 3a = From the diagram the parabola curves to the right. The equation is in the form

1 2

y k a x h

y xy x

y y x

y y x

4

4 31 12 2

2 1 12 24

2 12 23 0

i.e.

2

2

2

2

- = -

- - = - -

+ = +

+ + = +

+ - - =

2

^ ]

^ ] ^

^ ]

h g

h g h

h g

7 7A A

ch10.indd 524 7/18/09 4:03:56 PM


2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola .y y x12 4 8 02 + - - =

Solution

Complete the square on y.

y y x

y y x

y y x

y xx

12 4 8 0

12 4 8

12 4 8

6 4 444 11

36 36

2

2

2

+ - - =

+ = +

+ = +

+ = +

= +

+ +2

^

^

h

h

So the parabola has equation 11y x6 42+ = +^ ^h h .y x6 4 11or 2

- - = - -] ]g g7 6A @ Its vertex has coordinates ( 11- , 6- ).

4 4

1

a

a`

=

=

The parabola turns to the right as it is in the form .y k a x h4- = -2^ ^h h

y

x

(-10,-6)(-11,-6)

x=-12

11

Count 1 unit to the right for the focus , .10 6focus` = - -^ h Count 1 unit to the left for the directrix directrix has equation .x 12= -

PARABOLA (y – k)2 = – 4a(x – h)

The parabola with vertex ( h , k ) and focal length a that turns to the left has equation y k a x h4- = - -2

^ ^h h

ch10.indd 525 7/18/09 4:03:57 PM


Proof

Find the equation of the parabola that turns to the left with vertex ( h , k ) and focal length a.

Counting a units to the left from the vertex V gives the focus ,F h a k= -^ h . Counting a units to the right from the vertex V gives the point on the directrix ,D h a k= +^ h . So the equation of the directrix is given by x h a= + . We fi nd the equation of the locus of P ( x , y ) that is equidistant from point ,F h a k-^ h and line x h a= + .

x=h+a

B

F (h-a, k)

P (x, y)

y

x

B has coordinates , .h a y+^ h

PF PB

PF PB

x h a y k x h a y yx h a y k x h a

y k x h a x h ax h a x h a x h a x h a

We want2 2

=

=

- + - = - + -

- + - = -

- = - - -

= - + - - - -

- +

+ -

- +

- + - +

2 2 2 2

2 2 2

2 2 2

^ ^ ^ ^

^ ^ ^

^ ^ ^

^ ^ ^ ^

h h h h

h h h

h h h

h h h h

7 7

7 7

A A

A A

(difference of two squares)

2 2 2x h aax ah

a x h

a y k

4 4

4

4

= - -

= - +

= - -

= - -

^ ^

^

^

h h

h

h

ch10.indd 526 7/18/09 4:03:58 PM


The parabola y k a x h4- = - -2^ ]h g has

• axis parallel to the x -axis • vertex at ( h , k ) focus• at ,h a k-^ h directrix• with equation x h a= +

EXAMPLES

1. Find the equation of the parabola with focus (2, 1) and directrix 3x = .

Solution

x

y

B

x=3

1

1 2

(2, 1)

12

12

12(2 , 1)

Coordinates of B are (3, 1). The vertex is the midpoint of (3, 1) and (2, 1).

,221 1vertex` = c m

Focal length 21a =

From the diagram the parabola curves to the left. The equation is in the form

4

1

2

y a x h

y x

y x

y y x

y y x

k

421 2

21

1 221

2 1 2 5

2 2 4 0

i.e.

2

2

- = - -

- = - -

- = - -

- + = - +

- + - =

2

2

2

^ ^

^ c c

^ c

h h

h m m

h m

2. Find the coordinates of the vertex and focus, and the equation of the directrix of the parabola .y y x4 8 4 02 + + - =

CONTINUED

ch10.indd 527 7/18/09 4:03:59 PM


Solution

Complete the square on y.

y y x

y y x

y y x

y xx

4 8 4 0

4 8 4

4 8 4

2 8 88 1

4 4

2

2

2

2

+ + - =

+ = - +

+ = - +

+ = - +

= - -

+ +

^

]

h

g

So the parabola has equation y x2 8 12+ = - -^ ]h g .y x2 8 1or 2

- - = - -] ]g g7 A Its vertex has coordinates , .1 2-^ h

4 8

2

a

a`

=

=

The parabola turns to the left as it is in the form y k a x h4- = - -2^ ^h h

Count 2 units to the left for the focus , .1 2focus` = - -^ h Count 2 units to the right for the directrix directrix has equation .x 3=

1. Complete the square on x to write each equation in the form .x h a y k42

!- = -] ^g h (a) x x y6 8 15 02 - - - = (b) x x y10 4 1 02 - - + = (c) x x y2 4 11 02 - - - = (d) x x y8 12 20 02 - + - =

(e) x x y12 8 20 02 =- - - (f) x x y14 16 1 02 + + + = (g) x x y4 4 16 02 - + - = (h) x x y18 12 9 02 + - + = (i) x x y2 8 7 02 + - - = (j) x x y6 4 1 02 - + + =

10.6 Exercises

x=3

y

x

(1, -2)

1

-1-2-3 -1

(-1, -2)

2 2

-222222

21 3

ch10.indd 528 7/18/09 4:04:00 PM


2. Complete the square on y to write each equation in the form y k a x h42

!- = -^ ]h g (a) y y x8 4 02 =- - (b) y y x2 8 15 02 =- - - (c) y y x4 12 8 02 + =- - (d) y y x20 4 16 02 + =- - (e) y y x6 16 7 02 + + =- (f) y y x12 8 4 02 + =- - (g) y y x10 24 23 02 + + =- (h) y y x24 4 02 + =- (i) y y x4 20 16 02 + =- - (j) y y x8 8 02 + + =

3. Find the equation of each parabola focus (a) , ,1 3-^ h directrix 1y = - focus (b) 4, 1 ,-^ h directrix y 1= - focus (2, 0), directrix (c) 4y = - focus (3, 6), directrix (d) 2y = focus (e) 2, 5 ,-^ h directrix

3y = - focus (f) , ,1 4- -^ h directrix 4y = focus (g) 3),( ,4 - directrix 7y = focus (h) 5, 1 ,-^ h directrix 5y = focus (i) , ,3 6- -^ h directrix 0y = focus (j) , ,0 7-^ h directrix 5y = - focus (2, 3), directrix (k) 4x = - focus (l) , ,1 4-^ h directrix 3x = - focus (6, 0), directrix (m) 2x = focus (n) 2( ),,3 - directrix

5x = - focus (o) , ,1 1-^ h directrix 3x = - focus (p) , ,2 4- -^ h directrix 4x = focus (2, 1), directrix (q) 4x = focus (r) 5, 3 ,-^ h directrix 3x = focus (s) 1, 2 ,-^ h directrix 0x = focus (3, 1), directrix (t) x 4=

4. Find (i) the coordinates of the focus and

(ii) the equation of the directrix of (a) x x y6 4 3 02 =- - - (b) x x y2 8 7 02 =- - - (c) x x y4 4 02 + =- (d) x x y8 12 4 02 + =- - (e) x x y10 8 1 02 + + =- (f) x x y6 4 1 02 + + =-

(g) x x y2 8 15 02 + + =- (h) x x y4 4 02 + =- (i) x x y8 12 4 02 + + =- (j) x x y4 16 12 02 + + =-

5. Find (i) the coordinates of the focus and

(ii) the equation of the directrix of (a) y y x2 4 3 02 + =- - (b) y y x8 12 4 02 + =- - (c) y y x6 8 7 02 =- - - (d) y y x4 16 12 02 + =- - (e) y y x2 24 25 02 + =- - (f) y y x10 8 1 02 + + + = (g) y y x14 4 1 02 + + + = (h) y y x12 20 4 02 - + - = (i) y y x4 32 28 02 + =- - (j) y y x6 40 29 02 + + + =

6. Find the equation of the parabola with vertex ,0 3^ h if it is concave upwards and 3.a =

7. Find the equation of the parabola with vertex , ,2 1- -^ h focal length 2, and axis parallel to the y -axis.

8. A parabola has its vertex at ,1 5-^ h and its focal length as 1. If the parabola is concave upwards, fi nd its equation.

9. A parabola has its axis parallel to the x -axis. If its vertex has coordinates ,2 6^ h and ,a 3= fi nd its equation if it turns to the left.

10. Find the equation of the parabola with vertex at ,1 0^ h and focus at , .1 4^ h

11. Find the equation of the parabola that has vertex ,1 1^ h and focus , .1 8^ h

12. A parabola has its vertex at ,2 2-^ h and focus at , .4 2- -^ h Find its equation.

ch10.indd 529 7/18/09 4:07:37 PM


13. Find the equation of the parabola with vertex ,0 3^ h and focus , .8 3^ h

14. Find the equation of the parabola with vertex ,3 3^ h and equation of directrix .y 5=

15. Find the equation of the parabola with vertex ,3 1-^ h and directrix .x 1= -

16. A parabola has directrix y 5= and focus , .3 3-^ h Find its equation.

17. Find the equation of the locus of a point moving so that it is equidistant from the point ,2 2^ h and the line .y 4= -

18. Find the equation of the parabola with focus ,2 1-^ h and directrix .x 10=

19. Find the coordinates of the vertex and focus and the equation of the directrix for the parabola

(a) x x y4 8 12 02 + - + = (b) x x y6 12 33 02 - - + = (c) x x y2 4 5 02 - + + = (d) y y x8 16 64 02 - - + = (e) y y x4 24 4 02 + - + = (f) .y x8 40 02 + + =

20. For the parabola ,x x y2 28 111 02 + + - = fi nd the coordinates of its vertex and focus, and the equations of its directrix and axis. What is its maximum value?

21. The latus rectum of a parabola has endpoints ,2 3-^ h and , .6 3^ h Find two possible equations for the parabola.

22.

Find the equation of the arch (a) above.

Find the coordinates of its (b) focus and the equation of its directrix.

23. (a) Sketch ,y x x2 82= + - showing intercepts and the minimum point.

Find the coordinates of the (b) focus and the equation of the directrix of the parabola.

24. Find the equation of the parabola with vertex ,2 3-^ h that also passes through ,2 1^ h and is concave downwards.

25. A parabolic satellite dish has a diameter of 4 m at a depth of 0.4 m. Find the depth at which its diameter is 3.5 m, correct to 1 decimal place.

DID YOU KNOW?

The word ‘directrix’ is due to the Dutch mathematician Jan De Witt (1629–72). He published a work called Elementa curvarum , in which he defi ned the properties of the parabola, ellipse, circle and hyperbola. These curves are all called conic sections .

ch10.indd 530 7/18/09 4:08:24 PM


De Witt was well known as the ‘Grand Pensionary of Holland’. He took part in the politics and wars of his time, opposing Louis XIV. When the French invaded Holland in 1672, De Witt was seized and killed.

Tangents and Normals

Remember that the gradient of the tangent to a curve is given by the derivative.

The normal to the curve is perpendicular to its tangent at that point. That is, 1m m1 2 = - for perpendicular lines.

EXAMPLES

1. Find the gradient of the tangent to the parabola 8x y2 = at the point , .4 2^ h

Solution

x y

y x

dx

dy x

x

8

8

82

4

2

2

`

=

=

=

=

CONTINUED

ch10.indd 531 7/18/09 4:08:24 PM


, ,dx

dy4 2

44

1

At =

=

^ h

So the gradient of the tangent at ,4 2^ h is 1.

2. Find the equation of the normal to the parabola 4x y2 = at the point , .8 16-^ h

Solution

( , ):

the .

x

dx

dy x

x

dx

dy

m

4

42

2

8 1628

So

At

So gradient of the tangent

2

2

1

--

x y

y

4

4

4

=

=

=

=

=

= -

= -

The normal is perpendicular to the tangent.

m m

m

m

1

4 1

41

So1 2

2

2

`

`

= -

- = -

=

] g

The equation of the normal is given by

( )

[ ( )]

( )

.

y y m x x

y x

x

y x

x y

1641 8

41 8

4 64 8

0 4 72

i.e.

1 1- = -

- = - -

= +

- = +

= - +

1. Find the gradient of the tangent to the parabola 12x y2 = at the point where 2.x =

2. Find the gradient of the tangent to the parabola 3x y2 = - at the point , .6 12-^ h

3. Find the gradient of the normal to the parabola 4x y2 = at the point where 2.x =

4. Find the gradient of the tangent to the parabola 16x y2 = at the point , .4 1^ h

10.7 Exercises

ch10.indd 532 7/18/09 4:08:26 PM


5. Show that the gradient of the tangent to the curve 2x y2 = at any point is its x -coordinate.

6. Find the equation of the tangent to the curve 8x y2 = at the point , .4 2^ h

7. Find the equation of the normal to the curve 4x y2 = at the point where 4.x = -

8. Find the equations of the tangent and normal to the parabola 24x y2 = - at the point , .12 6-^ h

9. Find the equations of the tangent and normal to the parabola 16x y2 = at the point where 4.x =

10. Find the equation of the tangent to the curve 2x y2 = - at the point , .4 8-^ h This tangent meets the directrix at point M . Find the coordinates of M .

11. Find the equation of the normal to the curve 12x y2 = at the point , .6 3^ h This normal meets the parabola again at point P . Find the coordinates of P .

12. The normal of the parabola 18x y2 = at ,6 2-^ h cuts the parabola again at Q . Find the coordinates of Q .

13. Find the equations of the normals to the curve 8x y2 = - at the

points ,16 32- -^ h and , .221

- -c m

Find their point of intersection and show that this point lies on the parabola.

14. Find the equation of the tangent at ,8 4^ h on the parabola 16 .x y2 = This tangent meets the tangent at the vertex of the parabola at point R . Find the coordinates of R .

15. (a) Show that the point ,P p p2 2_ i lies on the parabola 4 .x y2 =

Find the equation of the (b) normal to the parabola at P .

Show that (c) p 1 02 + = if the normal passes through the focus of the parabola .p 0!^ h

ch10.indd 533 7/18/09 4:08:26 PM


Test Yourself 10 1. Find the equation of the locus of a point

moving so that it is equidistant from ,A 1 2-^ h and ,B 3 5^ h .

2. Find the equation of the parabola with focus ,2 1^ h and directrix .y 3= -

3. Find the radius and centre of the circle 6 2 6 0.x x y y2 2- + - - =

4. Find the coordinates of the vertex and (a) the focus of the parabola (b)

( ) .y x3 12 12+ = -] g

5. Find the equation of the locus of a point that is always 5 units from the origin.

6. Find the equation of the directrix and(a) the coordinates of the focus of the (b)

parabola .x y82 = -

7. A point ,P x y^ h moves so that AP and BP are perpendicular, given ,A 3 2= ^ h and , .B 4 1= -^ h Find the equation of the locus of P .

8. Point ,x yP ^ h is equidistant from the point ,A 4 2-^ h and the line 6.y = Find the equation of the locus.

9. Find (a) the coordinates of the (i) vertex and (ii) focus and (b) the equation of the directrix of the parabola .x x y2 4 5 02 - - + =

10. Find the equation of the tangent to the parabola 18x y2 = at the point , .6 2-^ h

11. Find the length of the diameter of the circle 8 12 3 0.x x y y2 2+ + - + =

12. Find the equation of the parabola with directrix x 6= and focus , .6 0-^ h

13. A parabola has a focus at ,0 4^ h and its vertex is at ,0 2^ h . Find the equation of the parabola.

14. Find the equation of the locus of a point that is always 3 units from the line .x y4 3 1 0- - =

15. A point is equidistant from the x - and y -axis. Find the equation of its locus.

16. Find the equation of the parabola with vertex at the origin, axis 0y = and

passing through the point , .141 5c m

17. Find the gradient of the tangent and (a) the normal to the parabola (b) 12x y2 = -

at the point where 3.x =

18. (a) Find the equation of the normal to the parabola 4x y2 = at the point , .8 16-^ h

This normal cuts the parabola again (b) at Q . Find the coordinates of Q .

19. Show that 7 3 12 0x y- + = is a focal chord of the parabola 16 .x y2 =

20. Find the point of intersection of the normals to the parabola x y122 = - at the

points ,4 131

-c m and , .231

- -c m

21. (a) Find the equation of the tangent to the parabola x y122 = at the point P (6, 3).

Find (b) R , the y- intercept of the tangent. Show that (c) FP FR= where F is the

focus.

ch10.indd 534 7/18/09 4:08:26 PM


1. (a) Find the equation of the locus of point P , which is equidistant from fi xed points ,A 3 5^ h and , .B 1 2-^ h

Show that this locus is the (b) perpendicular bisector of line AB .

2. (a) Find the equation of the circle with centre ,1 3^ h and radius 5 units.

Show that the circle cuts the (b) x -axis at the points ,5 0^ h and , .3 0-^ h

3. The line with equation x y5 12 36 0- + = is a chord of the parabola .x y122 = Find the point of intersection of the tangents to the parabola from the endpoints of the chord.

4. (a) Find the equation of the normals to the parabola x y82 = at the points

,M 221

-c m and , .N 8 8^ h

Show that these normals are (b) perpendicular.

Find the point of intersection (c) X of the normals.

Find the equation of line (d) MN and show that it is a focal chord.

5. From which point on the parabola x ay42 = does the normal pass through the focus?

6. (a) Find the equation of the tangents to

the parabola x y42 = at the points

,A 141

c m and , .B 4 4-^ h

Show that the point of intersection of (b) these tangents lies on the directrix.

7. Find the equation of the parabola with axis parallel to the y -axis and passing through points , ,0 2-^ h ,1 0^ h and , .3 8-^ h

8. Find the equation of the straight line through the centres of the circles with equations x x y y4 8 5 02 2+ + - - = and .x x y y2 10 10 02 2- + + + =

9. Sketch the region 2 4 4 0.x x y y2 2 #+ + - -

10. (a) Find the equation of the locus of a point P moving so that PA is perpendicular to PB where ,A 4 3-= ^ h and , .B 0 7= ^ h

Show that this locus is a circle with (b) centre ,2 5-^ h and radius .2 2

11. Find the exact gradient, with rational denominator, of the normal to the parabola y x122 = at the point where x 4= in the fi rst quadrant.

12. (a) Find the equation of the parabola with vertex ,3 2-^ h and focus , .7 2-^ h

Find the equation of the tangent to (b) the parabola at the point where x 4= in the fi rst quadrant.

13. Find the exact length of the line from ,2 7^ h to the centre of the circle .x x y y4 6 3 02 2+ + - - =

14. Find the equation of the locus of midpoints of all chords of length 2 units in the circle with equation .x y y2 3 02 2+ - - =

15. A satellite dish is to be 3.5 m wide and 1.1 m deep. Find the position of the focus in millimetres, correct to the nearest millimetre.

3.5 m

1.1 m

16. Find the equation of the locus of point P that moves such that the distance from P to the lines 3 4 1 0x y- + = and 12 5 3 0x y+ + = is in the ratio : .3 1


ch10.indd 535 7/20/09 10:38:50 AM

195PRACTICE ASSESSMENT TASK 1

1. Solve for : .pp p

23

51

1-

-+

=

2. Factorise fully: .x xy y y10 2 10 2 2+ - -

3. Write in index form

(a) x1

(b) x43

4. Simplify the expression .y y8 2 5- +^ h

5. Rationalise the denominator of .5 2

5-

6. Expand and simplify .x x x3 5 12- + -] ^g h

7. Solve the equation .x

x x1

332 1!

+= -^ h

8. Simplify .x x x5

1102 32

'+ - -

9. Show that ABCT and EDCT are congruent triangles. Hence, or otherwise, show that ACET is an isosceles triangle.

10. Evaluate . ,3 9 4] g correct to 1 decimal place.

11. Simplify .2 3 27-

12. Find the size of each interior angle in a regular octagon.

13. Evaluate . .0 72 9 82' in scientifi c notation, correct to 3 signifi cant fi gures.

14. Expand and simplify .2 3 5 2 2-^ h

15. Find, correct to 2 decimal places, the

value of . .

..

6 94 3 722 14 3

-

] g

16. Simplify .x2

2 6+

17. Solve x x2 3 1 02 - - = correct to 3 signifi cant fi gures.

18. The radius r of a circle with area A is

given by .r Ar= Find r , correct to

2 decimal places, if 7.59.A =

19. Solve x5 2 31- and sketch the solution on a number line.

20. Evaluate 203

151

32

125 .+ + -

21. Solve the equation 4 1 0,x x2 - + = giving exact solutions in simplest surd form.

22. Write 7 2- as a rational number.

23. Solve simultaneous equations 3 1y x= - and 5.y x2= -

24. Find integers x and y such that

2 3 3

33 .x y

+= +

25. Evaluate 2| | | | | |.2 1 4- - - +

26. Find the value of x .

27. Factorise .x8 13 -


.3 5 2

2 3

-

Practice Assessment TaskSET 1

pat_1.indd 195 7/18/09 9:45:02 AM


29. Simplify | | | | | |.2 4 3 2- - + -

30. Find the sale price if a discount of 8% is given on a DVD player that usually sells for $699.

31. Evaluate . .. .

3 94 1 765 21 4 713

#

+ correct to


32. Rationalise the denominator of .2 2 3

5 1

+

+

33. The price of roller skates has increased by 6.5% to $89. Find the price before the increase.

34. Find the values of all pronumerals, giving reasons for each step of your calculations.

35. Find the area of this fi gure.

36. Simplify .a b

a b9 1 4

4 3 6

#

#

-

-

^

^

h

h

37. Solve .x5 9 212-

38. Evaluate 24-

3


39. Simplify .x x2 5 3 1- - -] ]g g

40. Solve .4 8x2 1 =+

41. Solve .x 92$

42. Write .0 16o as a fraction.

43. Prove that the diagonals in any rhombus bisect the angles they make with the sides of the rhombus.

44. Write x 3

1+

in index form.

45. Expand and simplify .x 2 3+] g

46. Find the value of a b3 2- in index form if

a21 3

= c m and .b54 2

= c m

47. Find the value of x , giving reasons for each step of your working out.

48. Find values of x and y .

49. Write 2x3 2+-

1

] g without an index.

50. Simplify

(a) x y y x8 7 4- - +

(b) 124

(c) xx

279

3

2

+

-

pat_1.indd 196 7/18/09 9:45:06 AM


(d) 2 1

12 1

2+

+-

(e) x x x1

31

21

42+

+-

--

(f) x x1

- when x 2 3=

(g) x y z

x y z4 3 1 4 2

2 5 4 3

- - -

- -

_ ^

^

i h

h

(h) a ab

a bb

a ab b5 20 3 6

22

2 2

'-

+

-

+ +

(i) 8 5 3 20 2 45- +

(j) ,a bc

a b c2 2 5

3 2 4 2

^

^

h

h if ,a b

21

322 3

= =c cm m

and c94 1

=-

c m

51. Find the values of x and y , correct to 1 decimal place.

52. Evaluate x .

53. The volume of a sphere is given by the

formula .V r34 3r= Find the exact radius r ,

if the volume V is 1032 cm .3

54. Find the perimeter of the fi gure below, correct to 3 signifi cant fi gures.

55. ABCD is a parallelogram with CD produced to E so that .ED AD=

Prove that .ABC DEA2+ +=

56. Find the exact value of .1

65

52

31

-

-

57. Tran spent 31 of her salary on rent,

41

on food, 51 on bus and taxi fares, and

61

on going out. If she puts the rest of her salary into savings, what percentage of her salary is savings?

58. The speed of light is about . .2 99 10 ms8 1

#- If a rocket travels at

one-fi fth the speed of light, fi nd its speed in kmh 1- (in scientifi c notation correct to 2 signifi cant fi gures).

59. Find the value of k if .x x kx2 5 4 252 2+ = + +] g

60. Simplify .x y81 2 3

61. The sum of the interior angles in a regular polygon is .1620c Find the size of each interior angle, to the nearest minute.

62. Find the area of the shaded region in this fi gure, correct to 2 decimal places.

pat_1.indd 197 7/18/09 9:45:07 AM


63. Factorise (a) a b5 2 403 3- +] g (b) a b c a b c2 52 2- + - + -] ]g g

64. Solve x25

8 1 91#-- .

65. ABCD and BCEF are parallelograms. Show that AFED is a parallelogram.

66. Find the value of b correct to 2 decimal places.

67. The diagonals of a rhombus are 6 cm and 10 cm long. Find the

exact length of the sides of the (a) rhombus

area of the rhombus .(b)

68. Write as a single fraction with a rational

denominator 3 3 2

22 5

1-

-+

.

69. The exterior angles of a regular polygon are .18c How many sides has the polygon?

70. A cable is used to support a 20 m tower as shown. If the cable is placed 25 m away from the base of the tower, how long must it be, to the nearest metre?

20 m

25 m

71. In the fi gure, BD is the perpendicular bisector of AC . Prove that triangle ABC is isosceles.

B

A D C

72. The diagonals of a rhombus are x and y . Find the length of its side.

73. Write x3 2

15-] g in index form .

(a) 3x 2--

5

] g

(b) 2x

32-

-5

] g

(c) 23 x 2--

5

] g

(d) 3x 2

1

-5

] g

74. Write the number 54 000 000 in scientifi c notation .

5.4 (a) # 10 8 54 (b) # 10 6 5.4 (c) # 10 7 54 (d) # 10 −8

pat_1.indd 198 7/18/09 9:45:09 AM


75. Simplify aba b2

2

3 3

]

^

g

h .

8 (a) a 7 b 8 (b) a 8 b 2 (c) a 7 b 2 (d) a 8 b

76. A computer costs $1850. If it has increased in cost by 4% since last week, how much did it cost last week?

$1924.00 (a) $1778.85 (b) $1867.80 (c) $1776.00 (d)

77. Evaluate 24-

3

. (a) 8-

(b) 81

(c) 61

−6 (d)

78. Convert 0.36o to a fraction .

(a) 259

(b) 3312

(c) 331

(d) 3011

79.

C

A

B

E

D

The triangles ABC and CDE can be proven congruent by using the test

(a) SSS (b) SAS (c) RHS (d) AAS .

pat_1.indd 199 7/18/09 9:45:09 AM



1. Evaluate sin309 41c l to 2 decimal places.

2. Simplify 38 52 .sin sin2 2c c+

3. Find the equation of the straight line through ,1 3-^ h that is perpendicular to the line .x y2 5 9- =

4. Calculate the point of intersection of the lines x y2 3 10 0- - = and .x y5 4 2 0+ - =

5. Evaluate cot 107 9c l to 3 decimal places.

6. ABCD is a parallelogram in which 2 ,BC cm= 150ABC c+ = and .AB 3 cm= Find

the exact area of the parallelogram (a) the exact lengths of both diagonals. (b)

7. The lines AB and AC have equations x y3 4 9 0- + = and x y8 6 1 0+ - = respectively. Show that the lines are perpendicular and fi nd the coordinates of A .

8. If 70 ,sin cosx 5 c c+ =^ h fi nd a possible value of x .

9. Simplify as a fraction the expression sin cos tan60 45 302 2 2c c c- + without using a calculator.

10. Sketch .x y5 2 10 0- - =

11. On the number plane, shade in the region given by x y 92 2#+ and .x y 2$-

12. A soccer goal is 8 m wide. A man shoots for goal when he is 9 m from one post and 11 m from the other. Within what

angle must a shot be made in order to score a goal?

13. Find the domain and range of .yx2 1

3=

-

14. (a) On a number plane, draw the line .x y2 4- =

(b) On your diagram, shade the region given by , .y x y0 2 4$ $-

15. Sketch .y x x5 42= - +

16. Simplify .sin sin cos2 2 2i i i-

17. Find the perpendicular distance from ,2 3-^ h to the line .x y5 12 20 0- + =

18. Find the equation of the straight line through ,1 2^ h parallel to the line through ,3 4-^ h and , .5 5^ h

19. Solve cos21

i = - for 0 360 .c c# #i

20. Find the gradient of the line through the origin and ,3 5-^ h .

21. Simplify .tan

tan cot20

20 70c

c c+

22. Find a in degrees and minutes.

23. Find the value of y correct to 3 signifi cant fi gures.


pat_2.indd 446 8/1/09 1:43:50 PM


24. Find the equation of the line passing through ,1 2^ h that passes through the intersection of lines x y2 11 0- - = and .x y5 19 0- - =

25. Solve .x3 7 21-

26. If ,f x x9 2 2= -] g fi nd the value of .f 1-] g

27. Find the value of x if xf 7=] g where .xf 2 1x= -] g

28. Find the exact value of .tan300c

29. Show that x y3 4 10 0- + = is a tangent to the circle .x y 42 2+ =

30. If g xx x

x x

3 1

2 1

if

if

2

#=

-] g )

fi nd (a) g 2] g and g 3-] g sketch the graph of (b) xg ] g .

31. Find .limh

x h xh h3 4 2h 0

2 2- +"

32. Find the exact value of 135 .cos c

33. A bird at the top of a 10 m tree sees a mouse on the ground. If the angle of depression is 34 51,c l how far, to 1 decimal place, does the bird need to fl y to reach the mouse?

34. If the point ,3 7-^ h is the midpoint of ,x 3^ h and , y8^ h , fi nd the values of x and y .

35.

Find (a) AB , correct to 1 decimal place. Find the area of (b) T ABC ,

to 3 signifi cant fi gures.

36. Simplify .sin9 9 2 i-

37. (a) Find the equation of the straight line l through ,1 2-^ h that is perpendicular to the line .x y3 6 7 0+ - =

(b) Line l cuts the x -axis at P and the y -axis at Q . Find the coordinates of P and Q .

Find the area of (c) T OPQ where O is the origin.

38. Two points A and B are 100 m apart on the same side of a tower. The angle of elevation of A to the top of the tower is 20c and the angle of elevation from B is 27 .c Find the height of the tower, to the nearest metre.

39. The lines ,x y3 6- = x y2 14+ = and y 0= enclose a triangle. Find the area of the triangle.

40. Show that f x x x 36 2= - -] g is an even function.

41. Differentiate .x x2 12 4+^ h

42. Solve .y3 7 20#-

43. Differentiate .x3

44. (a) Find the equation of the tangent to the curve y x 23= - at the point , .P 1 1-^ h

(b) The curve y x 23= - meets the y -axis at Q . Find the equation of PQ .

(c) Find the equation of the normal to y x 23= - at the point ,1 3- -^ h .

Find the point (d) R where this normal cuts the x -axis.

45. Find the distance between the parallel lines x y5 12 3 0- - = and .x y5 12 5 0- + =

46. What is the domain and range of the

function 3 ?f xx 4

=+

] g

pat_2.indd 447 8/1/09 1:43:51 PM


47. Differentiate .x2 7-

48. Justin walks for 3.1 km due west, then turns and walks for 2.7 km on a bearing of 205 .c How far is he from his starting point? Give your answer correct to 1 decimal place.

49. Find the derivative of .x x x4 7 3 52 2- + + -

50. Find the derivative of .x

x1

5+

51. Differentiate from fi rst principles .f x x x3 22= - +] g

52. Write as a single fraction

.x x x5

3 45

22 +

- ++x

53. Find the equation of the normal to the curve y x x3 6 92= - + at the point where .x 2=

54. If ,f x x x x2 5 4 13 2= - + -] g fi nd f 2-] g and .f 2-l ] g

55. At the point ,2 3-^ h on the curve ,y ax bx 72= + + the tangent is inclined at 135c to the x -axis. Find the values of a and b .

56. Find the equation of the straight line passing through ,3 6^ h that also passes through the intersection of the lines x y2 0- = and .x y3 7 0+ + =

57. Find the equation of the tangent to the curve f x x x2 7 13= - +] g at the point where .x 3= -

58. Solve for : .x 481x2 1 =-

59. Find the domain and range of (a) y x2 1= -

(b) yx 7

5=

+

(c) y x4 2= - -

60. For the interval AB where ,A 3 2= -^ h and ,B 1 4= -^ h fi nd

the midpoint (a) the exact distance (b) the gradient (c) the equation of line (d) AB .

61. Differentiate x x3 2 2 1 6- +] ]g g (a) x x x12 3 2 2 1 3 2 15 6- + + +] ] ]g g g

(b) x36 2 1 5+] g

(c) x x x6 3 2 2 1 3 2 15 6- + + +] ] ]g g g

(d) x18 2 1 5+] g .

62. Find an expression involving i for the triangle below (there may be more than one answer) .

7

i a

54

(a) 2 5 4

5 4 7cos2 2 2

# #i =

+ -

(b) 4 5

sin sini a=

(c) 5 4

sin sini a=

(d) 2 5 7

5 7 4cos2 2 2

# #i =

+ -

63. If f x

x x

x xx

8 3

3 2 0 39 0

if

ifif

3

2

2

1# #= -] g *

evaluate f f f3 1 1+ + -] ] ]g g g 35 (a) 226 (b) 233 (c) 53 (d)

pat_2.indd 448 8/1/09 1:43:52 PM


64. The equation of the normal to the parabola y x x3 5 12= - + at the point ,2 3^ h is

(a) 7 11 0x y- - = (b) 7 17 0x y- - = (c) 7 23 0x y+ - = (d) 7 19 0x y+ - =

65. The linear function with equation x y4 2 3 0- + = has

gradient (a) ,2- y -intercept 121

-

gradient (b) 21 , y -intercept

43

gradient 2, (c) y -intercept 121

gradient 4, (d) y -intercept 3 .

pat_2.indd 449 8/1/09 1:43:52 PM



1. Solve .m m5 6 02 $- +

2. Find the locus of point P that moves so that it is equidistant from the points ,A 3 1-^ h and ,B 5 7^ h .

3. Find the centre and radius of the circle with equation .x x y y6 10 15 02 2+ + - - =

4. If a and b are the roots of the quadratic equation ,x x3 2 1 02 - - = fi nd the value of

(a) a b+ (b) ab (c) 2 2a b+

5. Find the coordinates of the focus and the equation of the directrix of the parabola .x y82 = -

6. Solve ( ) .x x3 5 3 6 02+ + + + =] g

7. Find the value of k in the equation x k x k4 3 02 - - + =] g if the sum of the roots is .5-

8. Find the equation of the locus of a point whose distance from the line x y3 4 1 0- + = is 3 units.

9. Find the coordinates of the vertex and focus of the parabola .y x x8 12= + -

10. Solve . .2 9 2 8 0x x2 - + =

11. Find the equation of the tangent to the parabola 16x y2 = at the point , .4 1-^ h

12. For what values of b does the equation 4 2 0x x b2 + - = have real roots?

13. A and B are the points ,4 0-^ h and ,4 0^ h respectively. Point ,P x y^ h moves so that .PA PB 642 2+ = Find the equation of the locus of P and describe it geometrically.

14. Find the equation of the circle with centre ,2 3- -^ h and radius 5 units.

15. The lines PA and PB are perpendicular, where A is , ,2 7-^ h B is ,5 1-^ h and P is , .x y^ h Find the equation of the locus of P .

16. Find the gradient of the normal to the curve x y62 = - at the point where .x 4= -

17. Find the locus of a point moving so that the ratio of PA to PB is 2:3 where A is ,3 2^ h and B is , .0 7^ h

18. If x x2 3 12 - + ( ) ( ) ,a x b x c1 12/ - + - + fi nd the values of a , b and c .

19. Differentiate .x9 2-

20. Find the locus of the point that is equidistant from the point ,2 5^ h and the line .y 3= -

21. Show that ABCD is similar to CDED and hence fi nd y , correct to 1 decimal place.


pat_3.indd 536 8/1/09 4:33:54 PM


22. Find the equation of the tangent to the curve x y2 82- =] g at the point where .x 6=

23. Find the equation of the locus of point ,P x y^ h that moves so that it is always equidistant from the point ,1 3-^ h and the line .y 5= -

24. Solve . .2 5 2 4 0x x2 - + =

25. Show that x x 9 02 1- + - for all x .

26. Differentiate .x x3 1 2 5 4- +^ ^h h

27. Simplify .cot tanx x+

28. Find the centre and radius of the circle whose equation is .x x y y10 6 30 02 2+ + - + =

29. Show that x x 3 02 2- + for all x .

30. Find the value of k in the quadratic equation 3 1 0x x k2 - + + = if the roots are consecutive numbers.

31. Find the equation of the locus of the point that is equidistant from ,2 1-^ h and ,4 5^ h.

32. A ship sails from port due east for 150 km, then turns and sails on a bearing of 195c for 200 km.

How far from port is the ship, to the (a) nearest kilometre?

On what bearing, to the nearest (b) degree, is the ship from port?

33. Find the values of a , b and c if .x a x bx c3 7 32 2/- + + +] g

34. Solve .x2 7 12-

35. Find the value of i in degrees and minutes.

36. Show that the quadratic equation x x6 15 02 + - = has 2 real, rational roots.

37. Find the equation of the normal to the curve y x x2 5 14 2= - - at the point , .1 4- -^ h

38. Find values of k for which the quadratic equation x x k2 2 02 - + - = has real roots.

39. Find the equation of the straight line through ,5 4-^ h , that is parallel to the line through ,7 4^ h and , .3 1-^ h


.3 3 5

2 1

+

+

41. Find the values of x and y correct to 1 decimal place.

42. Given ,f x x8 3= -] g fi nd the value of x for which .f x 5=] g

43. Find the distance between ,0 7^ h and ,2 1- -^ h correct to 3 signifi cant fi gures.

44. Find the value of p correct to 1 decimal place.

45. Simplify a b

a b1 2 7

3 2 4

-^

^

h

h if a

32

= and .b94

=

46. Solve cos x221

= - for .x0 360c c# #

47. Find the equation of the straight line through ,3 1-^ h perpendicular to the line .x y3 2 7 0- - =

pat_3.indd 537 8/1/09 4:33:56 PM


48. Solve .y y5 3 5- = -

49. Find the size of each internal angle in a regular 20-sided polygon.

50. Solve cos x2 12 = for .x0 360c c# #

51. Solve equations x xy 1 02 + + = and x y3 5 0- + = simultaneously.

52. Factorise .a b83 3-

53. Solve .x x2

13

2 7+-

+=

54. Find the gradient of the normal to the curve y x x2 7 13= + + at the point where .x 2= -

55. Find the perpendicular distance from ,3 2-^ h to the line .x y4 3 9 0- - =

56. Simplify .sec sec1 1i i+ -] ]g g

57. Differentiate ( ) .x x2 5 12 4+ -] g

58. Find .limxx

42

x 2 2 -

-"

59. Find the equation of the locus of point P ( x , y ) if PA is perpendicular to PB , given 3, 2A = -^ h and , .B 5 5= -^ h

60. Find the coordinates of the focus and the equation of the directrix of the parabola .x x y4 8 20 02 - + - =

61. Find the equation of the normal to the parabola x y122 = - at the point where .x 12=

62. Prove that the line x y6 8 40 0- + = is a tangent to the circle with centre the origin and radius 4 units.

63. In the quadratic equation ( ) ,k x x k1 5 3 4 02- - + + = the roots are reciprocals of each other. Find the value of k .

64. Solve .x 4 02 2-

65. A circle with centre at the origin O passes through the point (2, )5 . Find the radius of the circle, and hence its equation.

66. Find values of a , b and c for which ( ) ( ) .x x a x b x c3 2 7 2 22 2/- - + + + +

67. What is the domain and range of ?y x 32= -

68. Prove that ABCT is congruent to CDET .


70. Find the equation of the straight line through the midpoint of ( , )5 7- and (1,3) and making an angle of 135c with the x -axis.

71. Complete the square on 12 .x x2 -

72. Solve 2 4 9.y $+

73. (a) Find the equation of the tangent to the curve y x x33= - at the point ( , ).P 2 2- -

pat_3.indd 538 8/1/09 4:33:57 PM


(b) Find the equation of the normal to y x x33= - at P .

Find the point (c) Q where this normal cuts the x -axis.

74. (a) Find the equation of the normal to the curve y x x6 92= - + at the point where .x 1= -

(b) This normal cuts the curve again at point R . Find the coordinates of R .

75. The function ( )f x ax bx c2= + + has a tangent at ( , )1 3- with a gradient of .1- It also passes through (4,3). Find the values of a , b and c .

76. The equation of the locus of point P ( x , y ) that moves so that it is always 4 units from ,1 3-^ h is

(a) x y1 3 422- + + =^ ^h h

(b) x y1 3 422+ + - =^ ^h h

(c) x y1 3 162 2+ + - =] ^g h

(d) x y1 3 1622- + + =^ ^h h

77. If a and b are the roots of the

quadratic equation ,x x5 2 02 - + =

evaluate ba

ab

+

11 (a) 21

12 (b) 21

2 (c) 21

10 (d) 21

78. The equation of the locus of point P ( x , y ) moving so that it is equidistant from (3, 2) and the line x 1= - is given by

(a) x x y2 8 15 02 - + - =

(b) y y x4 8 12 02 - - + =

(c) x x y2 8 17 02 - + =-

(d) y y x4 8 4 02 + =- -

79. The quadratic equation x k x k3 02 + + =-] g has real roots. Evaluate k

(a) ,k k1 9# $

(b) ,k 1 9=

(c) 1 9k# #

(d) ,k k1 91 2

80. Find the centre and radius of the circle .x x y y2 8 13 02 2+ + - + =

Centre (a) , ,1 4-^ h radius 4

Centre (b) , ,1 4-^ h radius 2

Centre (c) , ,1 4-^ h radius 2

Centre (d) , ,1 4-^ h radius 4

81. For the quadratic function y ax bx c2= + + to be positive defi nite

(a) ,a b ac0 4 022 2-

,a b ac0 4 02 21 -(b)

,a b ac0 4 022 1-(c)

(d) ,a b ac0 4 021 1-

pat_3.indd 539 8/1/09 5:45:29 PM


Answers Chapter 1: Basic arithmetic

Problem

5

Exercises 1.1

1. (a) Rational (b) Rational (c) Rational (d) Irrational (e) Rational (f) Irrational (g) Irrational (h) Rational (i) Rational (j) Irrational

2. (a) 18 (b) 11 (c) 6 (d) 11 (e) .4 3- (f) −1 (g) 2157

(h) 12019

(i) 2 (j) 331

3. (a) 16.36 (b) 21.87 (c) 8.80 (d) 22.71 (e) .13 20-

(f) 0.17 (g) 0.36 (h) 1.20 (i) .4 27- (j) 8.16

4. 1300 5. 950 6. 3000 7. 11 000

8. 600 9. $8 000 000 10. $34 600 000

11. 844 km 12. 0.73 13. 33 14. 3.248 15. 4.21

16. 1.7 17. 79 cents 18. 2.73 19. 1.1 20. 3.6 m

21. $281.93 22. 1.8 g 23. $3.20

24. (a) 7.95 (b) 30.03 (c) 0.37 (d) 5.74 (e) 0.52 25. 0.2

Exercises 1.2

1. 1 2. 11- 3. 56- 4. 10 5. 4-

6. .1 2- 7. .7 51- 8. .35 52- 9. 6.57

10. 2154

- 11. 7- 12. −23 13. 10 14. 1

15. 5 16. 3 17. 1 18. 60 19. −20 20. 9

Exercises 1.3

1. (a) 2516

(b) 100051

(c) 5201

(d) 1154

2. (a) 0.4 (b) 1.875 (c) .0 416o (d) .0 63oo

3. (a) 501

(b) 83

(c) 1000

1 (d) 1

100097

4. (a) 0.27 (b) 1.09 (c) 0.003 (d) 0.0623

5. (a) 35% (b) %3331

(c) %22632

(d) 0.1%

6. (a) 124% (b) 70% (c) 40.5% (d) 127.94%

7. (a) . ;0 522513

(b) . ;0 071007

(c) . ;0 16812521

(d) . ;1 09 11009

(e) . ;0 434500217

(f) . ;0 122540049

8. (a) .0 83o (b) .0 07oo (c) .0 13oo (d) .0 16o (e) .0 6o (f) .0 15oo

(g) .0 142857o o or 0.142857 (h) .1 18oo

9. (a) 98

(b) 92

(c) 195

(d) 397

(e) 9967

(f) 116

(g) 457

(h) 6013

(i) 990217

(j) 149537

10. (a) .0 5o (b) 7.4 (c) 0.73o (d) .0 68oo (e) .1 72oo

11. (a) 85

(b) 281

(c) 118

(d) 2187

(e) 454

12. 74% 13. 77.5% 14. 17.5% 15. 41.7%

Exercises 1.4

1. 203

2. 207

3. (a) 2017

(b) 107

(c) 1201

(d) 283

(e) 53

4. $547.56 5. 714.3 g 6. 247

7. $65

8. 179 cm 9. (a) 11.9 (b) 5.3 (c) 19 (d) 3.2 (e) 3.5 (f) 0.24 (g) 0.000 18 (h) 5720 (i) 0.0874 (j) 0.376

10. $52.50 11. 54.925 mL 12. 1152.125 g 13. $10.71

14. 5.9% 15. 402.5 g 16. 41.175 m 17. $30.92

18. 3.2 m 19. 573 20. $2898

Problem

5115 minutes after 1 o’clock.

Exercises 1.5

1. (a) 500 (b) 145 (c) 641

(d) 3 (e) 2

2. (a) 13.7 (b) 1.1 (c) 0.8 (d) 2.7 (e) .2 6- (f) 0.5

3. (a) a17 (b) y 10 = (c) a 4- (d) w (e) x5 (f) p10

(g) y6 (h) x21 (i) x4 10 (j) y81 8- (k) a (l) y

x45

10

(m) w10 (n) p5 (o) x 3- (p) a ba

bor2 3

2

3-

(q) x yx

yor5 2

5

2

-

4. (a) x 14 (b) a 7- (c) m 4 (d) k 10 (e) a 8- (f) x (g) mn 2

(h) p 1- (i) 9 x 22 (j) x 21

5. (a) p 5 q 15 (b) b

a8

8

(c) b

a6412

3

(d) 49 a 10 b 2 (e) 8 m 17

(f) x 4 y 10 (g) k27

2 23

(h) 16 y 47 (i) a 3 (j) x y125 21 18-

Answer S1-S5.indd 540 7/31/09 1:35:59 PM

541ANSWERS

6. 421

7. 324 8. 22710

9. (a) a 3 b (b) 251

10. (a) pq 2 r 2 (b) 327

11. 94

12. 181

13. 274

14. 811

15. 108

1 16.

121

17. 2

558

22

18. 388849

Exercises 1.6

1. (a) 271

(b) 41

(c) 3431

(d) 10 000

1 (e)

2561

(f) 1

(g) 321

(h) 811

(i) 71

(j) 811

(k) 641

(l) 91

(m) 1

(n) 361

(o) 125

1 (p)

100 0001

(q) 1281

(r) 1

(s) 641

(t) 641

2. (a) 1 (b) 16 (c) 121

(d) 12511

(e) 1 (f) 125 (g) 131

(h) 49 (i) 383

(j) 32 (k) 231

(l) 1 (m) 13613

(n) 18119

(o) 1 (p) 16 (q) 1585

- (r) 237

- (s) 1 (t) 2516

3. (a) m 3- (b) x 1- (c) p 7- (d) d −9 (e) k −5 (f) x 2-

(g) 2x 4- (h) 3 y −2 (i) 21

z 6- or 2

z 6-

(j) 5

3t 8-

(k) 7

2x 1-

(l) 2

5m 6-

(m) 3

2y 7-

(n) 3 4x 2+ -] g (o) a b 8+ -] g

(p) 2x 1- -] g (q) 5 1p 3+ -^ h (r) 2 4 9t 5- -] g

(s) 41x 11+ -] g

(t) 9

5 3a b 7+ -] g

4. (a) 1

t5 (b)

1

x6 (c)

1

y3 (d)

1

n8 (e)

1

w10 (f)

2x (g)

3

m4

(h) 5

x7 (i)

8

1

x3 (j)

41n

(k) 1

1

x 6+] g (l)

81

y z+

(m) 3

1

k 2-] g ( n)

3 2

1

x y 9+^ h (o) x 5 (p) y 10 (q)

2

p

(r) a b 2+] g (s) x y

x y

+

- (t)

2

3

w z

x y 7

-

+e o

Exercises 1.7

1. (a) 9 (b) 3 (c) 4 (d) 2 (e) 7 (f) 10 (g) 2 (h) 8 (i) 4 (j) 1 (k) 3 (l) 2 (m) 0 (n) 5 (o) 7 (p) 2

(q) 4 (r) 27 (s) 21

(t) 161

2. (a) 2.19 (b) 2.60 (c) 1.53 (d) 0.60 (e) 0.90 (f) 0.29

3. (a) y3 (b) y yor23 32

_ i (c) 1

x (d) 2 5x +

(e) 3 1

1

x - (f) 6q r3 + (g)

x x7

1

7

1or

25 5 2+ +] ^g h

4. (a) 2t1

(b) 5y1

(c) 2x3

(d) x9 31

-] g (e) 2s4 1+1

] g

(f) 2t2 3+-

1

] g (g) -

2x y5 -

3

^ h (h) 2x3 1+5

] g

(i) 3x 2--

2

] g (j) 2y21

7+-

1

^ h (k) -

3x5 4+1

] g

(l) y32 1 2

1

--

2a k (m) x53 2 4

3

+-

2_ i

5. (a) x 23

(b) x 21

-

(c) x 32

(d) x 35

(e) x 45

6. (a) x x x2 23

+ +2 (b) a b32

32

- (c) p p p2 21

+ +1-2

(d) 2x x 1+ +- (e) x x x321

23

25

- +- - -

7. (a) 2

1

a b3 - (b)

3

1

y 23 -^ h (c)

6 1

4

a 47 +] g

(d) 3

1

x y 54 +^ h (e)

7 3 8

6

x 29 +] g

Exercises 1.8

1. (a) .3 8 103# (b) .1 23 106

# (c) .6 19 104#

(d) 1.2 107# (e) .8 67 109

# (f) .4 16 105#

(g) 9 102# (h) .1 376 104

# (i) 2 107# (j) 8 104

#

2. (a) .5 7 10 2#

- (b) .5 5 10 5#

- (c) 4 10 3#

-

(d) 6.2 10 4#

- (e) 2 10 6#

- (f) 8 10 8#

-

(g) 7.6 10 6#

- (h) 2.3 10 1#

- (i) 8.5 10 3#

- (j) 7 10 11#

-

3. (a) 36 000 (b) 27 800 000 (c) 9 250 (d) 6 330 000 (e) 400 000 (f) 0.072 3 (g) 0.000 097 (h) 0.000 000 038 (i) 0.000 007 (j) 0.000 5

4. (a) 240 000 (b) 9 200 000 (c) 11 000 (d) 0.36 (e) 1.3 (f) 9.0 (g) 16 (h) 320 (i) 2900 (j) 9.1

5. (a) 6.61 (b) 0.686 (c) 8.25 (d) 1.30

6. 1.305 1010# 7. 6.51 10 10

#-

Exercises 1.9

1. (a) 7 (b) 5 (c) 6 (d) 0 (e) 2 (f) 11 (g) 6 (h) 24 (i) 25 (j) 125 2. (a) 5 (b) −1 (c) 2 (d) 14 (e) 4 (f) −67 (g) 7 (h) 12 (i) −6 (j) 10 3. (a) 3 (b) 3 (c) 1 (d) 3 (e) 1 4. (a) a (b) a- (c) 0 (d) 3 a (e) −3 a (f) 0 (g) 1a + (h) a 1- - (i) 2x - (j) 2 x-

5. (a) | | 6a b+ = | | | |a b 6+ = | | | | | |a b a b` #+ + (b) | | 3a b+ = | | | |a b 3+ = | | | | | |a b a b` #+ + (c) | | 1a b+ = | | | |a b 5+ = | | | | | |a b a b` #+ + (d) | | 1a b+ = | | | |a b 9+ = | | | | | |a b a b` #+ + (e) | |a b 10+ = | | | |a b 10+ = | | | | | |a b a b` #+ +

6. (a) | | 5x x2 = = (b) | | 2x x2 = = (c) | | 3x x2 = =

(d) | | 4x x2 = = (e) | | 9x x2 = =

7. (a) x x x x5 5 5 5for and for2 1+ - - - - (b) b b b x3 3 3 3for and for2 1- - (c) a a a a4 4 4 4for and for2 1+ - - - - (d) y y y y2 6 3 6 2 3for and for2 1- - (e) 3 9 3 3 9 3x x x xfor and for2 1+ - - - - (f) 4 4 4 4x x x xfor and for1 2- -

(g) 2 1 2 1k k k k21

21

for and for2 1+ - - - -

(h) 5 2 5 2x x x x52

52

for and for2 1- - +

(i) a b a b a b a bfor and for2 1+ - - - - (j) p q p q q p p qfor and for2 1- -

8. x 3!= 9. 1! 10. , x1 2! !

Answer S1-S5.indd 541 7/31/09 1:36:00 PM


Test yourself 1

1. (a) 209

(b) 0.14 (c) 0.625 (d) 200157

(e) 1.2%

(f) 73.3% 2. (a) 491

(b) 51

(c) 31

3. (a) 8.83 (b) 1.55 (c) 1.12 (d) 342 (e) 0.303 4. (a) 1

(b) 1 (c) 39 (d) 2 (e) 10- (f) 1- (g) 4 5. (a) x9

(b) 25y6 (c) a b11 6 (d) 27

8x18

(e) 1 6. (a) 4029

(b) 371

(c) 12 (d) 221

(e) 1221

7. (a) 4 (b) 6 (c) 19

(d) 641

(e) 4 (f) 3 (g) 71

(h) 2 (i) 1 (j) 4

8. (a) a 5 (b) x 30 y 18 (c) p 9 (d) 16 b 36 (e) 8 x 11 y 9. (a) 2n1

(b) x 5- (c) x y 1+ -^ h (d) 4x 1+

1

] g (e) 7a b+1

] g

(f) 2x 1- (g) 21

x 3- (h) 3x4

(i) 7x5 3+9

] g (j) 4m-

3

10. (a) 1

a5 (b) n4 (c) 1x + (d)

1x y-

(e) 4 7

1

t 4-] g

(f) a b5 + (g) 1

x3 (h) b34 (i) 2 3x 43 +] g (j)

1

x3

11. | | 2a b+ = | | | | 8a b+ = | | | | | |a b a b` #+ +

12. 1 13. 192

1 14. 689 mL 15. (a) 6 h (b)

127

(c) 81

(d) 33.3% 16. $38 640 17. 70% 18. 6.3 1023#

19. (a) 2x1

(b) y 1- (c) 6x 3+1

] g (d) 2 3x 11- -] g (e) 3y7

20. (a) 1.3 10 5#

- (b) 1.23 1011# 21. (a)

97

(b) 33041

22. (a) 1

x3 (b)

2 51

a + (c) a

b 5

c m 23. 14 500

24. | | ,2 5 7LHS = - + - = | | | | .2 5 7RHS = - + - = So | | | | | |a b a b#+ + since .7 7#

Challenge exercise 1

1. 4303278

2. 11811

3. . , %, , .0 502 519951

0 5o

4. 5331

% 5. 161

6. 3.04 1014# 7. 83% 8. 1

99903271

9. 18 h 10. 1.98

11.

2 2 1 22 2 22 2 22 2 1

2 2 1 2 2 2 1

LHS

RHS

k k

k k

k

k

k k k

1

1 1

1

1

1 1

:

`

= - +

= - +

= -

= -

=

- + = -

+

+ +

+

+

+ +

^

^

^ ^

h

h

h h

12. −2 4 .3 5 13. . , , . , ,0 34 2 1 5 073

- o 14. 632

%

15. ,x

xx

x1

11

11

1when when2 1-

--

- 16. 0.73

17. 0.6% 18 4.54 19. 4.14 10 20#

-

20.

| | | | | | , , ;| | | | | | , , ;a b a b a b a ba b a b b ba

0 0 0 00 0 0 0

when orwhen ora

2 2 1 11 2 1 1 2

+ = +

+ +

| | | |a b a b` #+ + for all a , b

Chapter 2: Algebra and surds

Exercises 2.1

1. 7 x 2. 3 a 3. z 4. 6 a 5. 3 b 6. −3 r

7. y- 8. −5 x 9. 0 10. 3 k 11. 9 t 12. 10 w

13. m- 14. x- 15. 0 16. 5 b 17. 11 b 18. 10x-

19. 6 6x y- 20. 3a b- 21. 4 2xy y+ 22. 6ab2-

23. m m6 122 - + 24. 2 6p p2 - - 25. 8 3x y+

26. 2 10ab b- + 27. 2bc ac- 28. 2 9 1a x5 3- +

29. 2 3 2x xy x y y3 2 2 3- + + 30. 3 7 6x x x3 2+ - -

Exercises 2.2

1. b10 2. xy8 3. p10 2 4. wz6-

5. ab15 6. xyz14 7. abc48 8. d12 2

9. a12 3 10. y27 3- 11. x32 10

12. a b6 2 3 13. a b10 3 2- 14. p q21 3 4

15. a b5 3 3 16. h8 10- 17. k p33 18. t81 12

19. 14m11- 20. x y24 6 3

Exercises 2.3

1. 6 x 2. 2 3. 4a2 4. 8 a 5. 4 a 6. 2

y 7. 3 p

8. 2ab

9. 34y

10. 3x3- 11. 3 a 12. 3

1

ab2 13.

2qs-

14. 3

2

c d2 15.

x

z

2 2

2

16. 6p q4 17. 4c

a b4 7

18. 2ab6

19. 3y

x z3 3

- 20. 2b

a6

13

Exercises 2.4

1. x2 8- 2. h6 9+ 3. a5 10- + 4. xy x2 3+

5. x x22 - 6. a ab6 162 - 7. a b ab2 2 2+ 8. n n5 202 -

9. 3x y x y63 2 2 3+ 10. k4 7+ 11. t2 17-

12. y y4 112 + 13. b5 6- - 14. x8 2-

15. m3 1- + 16. h8 19- 17. d 6- 18. a a2 42 - +

19. x x3 9 52 - - 20. ab a b b2 2 2- + 21. x4 1-

22. y7 4- + 23. b2 24. t5 6- 25. a2 26+

Exercises 2.5

1. 7 10a a2 + + 2. 2 3x x2 + - 3. 2 7 15y y2 + -

4. 6 8m m2 - + 5. 7 12x x2 + + 6. 3 10y y2 - -

7. 2 6x x2 + - 8. 10 21h h2 - + 9. 25x2 -

10. 15 17 4a a2 - + 11. 8 6 9y y2 + - 12. 7 4 28xy x y+ - -

13. 2 3 6x x x3 2- + - 14. 4n2 - 15. 4 9x2 -

16. 16 49y2- 17. 4a b2 2- 18. 9 16x y2 2- 19. 9x2 -

20. 36y2 - 21. 9 1a2 - 22. 4 49z2 -

Answer S1-S5.indd 542 7/31/09 1:36:01 PM

543ANSWERS

23. 2 11 18 18x xy x y2 - + - + 24. 2 2 7 6 3ab b b a2+ - - +

25. x 83 + 26. 27a3 - 27. 18 81a a2 + +

28. 8 16k k2 - + 29. 4 4x x2 + + 30. 14 49y y2 - +

31. 4 12 9x x2 + + 32. 4 4 1t t2 - +

33. 9 24 16a ab b2 2+ + 34. 10 25x xy y2 2- +

35. 4 4a ab b2 2+ + 36. a b2 2- 37. 2a ab b2 2+ +

38. 2a ab b2 2- + 39. a b3 3+ 40. a b3 3-

Exercises 2.6

1. 8 16t t2 + + 2. 12 36z z2 - + 3. 2 1x x2 - +

4. 16 64y y2 + + 5. 6 9q q2 + + 6. 14 49k k2 - +

7. 2 1n n2 + + 8. 4 20 25b b2 + + 9. 9 6x x2- +

10. 9 6 1y y2 - + 11. 2x xy y2 2+ + 12. 9 6a ab b2 2- +

13. 16 40 25d de e2 2+ + 14. 16t2 - 15. x 92 -

16. 1p2 - 17. 36r2 - 18. 100x2 - 19. 4 9a2 -

20. 25x y2 2- 21. 16 1a2 - 22. 49 9x2- 23. 4x4 -

24. 10 25x x4 2+ + 25. 9 16a b c2 2 2- 26. 44

xx

2

2+ +

27. 1

aa

2

2- 28. 2 4 4x y x y y2 2 2 2- - = - + -^ h

29. 2 2 2 2a b a b c c a ab b ac bc c2 2 2 2 2+ + + + = + + + + +] ]g g

30. 1 2 1 2 1 2 2x x y y x x xy y y2 2 2 2+ - + + = + + - - +] ]g g

31. 12a 32. 32 z2- 33. 9 8 3x x2 + -

34. 3 2x xy y x2 2+ + - 35. 14 4n2 -

36. 12 48 64x x x3 2- + - 37. x2 38. 2x x y y4 2 2 4- +

39. 8 60 150 125a a a3 2+ + +

40. 4 16 15 4 4x x x x4 3 2+ + - -

Problem

2,a = 7,b = 9,c = 4,d = 3,e = 8,f = 0,g = 6,h = 1i =

Exercises 2.7

1. y2 3+^ h 2. x5 2-] g 3. m3 3-] g 4. x2 4 1+] g

5. y6 4 3-^ h 6. xx 2+] g 7. m m 3-] g 8. y y2 2+^ h

9. a a3 5 -] g 10. ab b 1+] g 11. xy x2 2 1-] g

12. mn n3 32 +^ h 13. xy x z2 4 -] g 14. a b a6 3 2+ -] g

15. x x y5 2- +^ h 16. q q3 22 3 -_ i 17. b b5 32 +] g

18. a b ab3 22 2 -] g 19. 5)( 7)(m x+ + 20. 1 2y y- -^ ^h h

21. 7 )(4 3 )( y x+ - 22. 2 6 5a x- +] ]g g

23. 2 1t x y+ -] ^g h 24. 3 2 2 3x a b c- + -] ]g g

25. 3 2 3x x2 +] g 26. 3 2q pq3 2 -_ i 27. ab a b3 5 13 2 +^ h

28. 4 6x x2 -] g 29. 5 7 5m n mn2 3 -^ h 30. 4 6 4ab ab2 3 +^ h

31. r r h2r +] g 32. 3 2x x- +] ]g g 33. ( ) ( )x y4 22+ +

34. 1a- +] g 35. ( ) ( )a ab1 4 32 + -

Exercises 2.8

1. 4 2x b+ +] ]g g 2. 3y a b- +^ ]h g 3. 5 2x x+ +] ]g g

4. 2 3m m- +] ]g g 5. d c a b- +] ]g g 6. 1 3x x2+ +] ^g h

7. 5 3 2a b- +] ]g g 8. 2y x x y- +^ ^h h 9. 1 1y a+ +^ ]h g

10. 5 1x x+ -] ]g g 11. 3)(1 )(y a+ + 12. 2)(1 2 )(m y- -

13. 5 2 3x y x y+ -^ ^h h 14. 4a b ab2+ -^ ]h g

15. 5 3x x- +] ]g g 16. 7)( 4)(x x3+ - 17. 3 7x y- -] ^g h

18. 3 4d e+ -] ]g g 19. 4 3x y- +] ^g h 20. 3 2a b+ -] ]g g

21. 3)( 6)(x x2- + 22. 3q p q- +^ ^h h

23. 2 3 5x x2- -] ^g h 24. 3 4a b c- +] ]g g

25. 7 4y x+ -^ ]h g 26. 4)( 5)(x x3- -

27. (2 3)(2 4) (2 3)( )x x x x2 22 2- + = - +

28. ( ) ( )a b a3 2 3+ + 29. 5( 3)(1 2 )y x- +

30. r r2 3r+ -] ]g g

Exercises 2.9

1. 3 1x x+ +] ]g g 2. 4 3y y+ +^ ^h h 3. 1m 2+] g

4. 4t 2+] g 5. 3 2z z+ -] ]g g 6. 1 6x x+ -] ]g g

7. 3 5v v- -] ]g g 8. 3t 2-] g 9. 10 1x x+ -] ]g g

10. 7 3y y- -^ ^h h 11. 6 3m m- -] ]g g

12. 12 3y y+ -^ ^h h 13. 8 3x x- +] ]g g

14. a 2 2-] g 15. 2 16x x- +] ]g g

16. 4 9y y+ -^ ^h h 17. 6 4n n- -] ]g g 18. x 5 2-] g

19. 9 1p p+ -^ ^h h 20. 2 5k k- -] ]g g 21. 4 3x x+ -] ]g g

22. 7 1m m- +] ]g g 23. 10 2q q+ +^ ^h h

24. 5 1d d- +] ]g g 25. 9 2l l- -] ]g g

Exercises 2.10

1. 2 1)( 5)( a a+ + 2. 5 2 1y y+ +^ ^h h

3. 3 7)( 1)( x x+ + 4. 3 2)( 2)( x x+ + 5. 2 3)( 1)( b b- -

6. 7 2)( 1)( x x- - 7. 3 1 2y y- +^ ^h h 8. 2 3 4x x+ +] ]g g

9. 5 2 3p p- +^ ^h h 10. 3 5 2 1x x+ +] ]g g

11. 2 1)( 6)( y y+ - 12. 5 1 2 1x x- +] ]g g

13. 4 1)(2 3)( t t- - 14. 3 4)(2 3)( x x+ -

15. 6 1 8y y- +^ ^h h 16. 4 3 2n n- -] ]g g

17. 4 1 2 5t t- +] ]g g 18. 3 2 4 5q q+ +^ ^h h

19. r r r r4 1 2 6 4 12 3- + - +=] ] ] ]g g g g

20. 2 5 2 3x x- +] ]g g 21. 6 1 2y y- -^ ^h h

22. 2 3 3 2p p- +^ ^h h 23. 8 7)( 3)( x x+ +

24. 3 4 4 9b b- -] ]g g 25. 6 1)( 9)( x x+ -

26. 3 5x 2+] g 27. 4 3y 2+^ h 28. 5 2k 2-] g

29. 6 1a 2-] g 30. 7 6m 2+] g

Answer S1-S5.indd 543 7/31/09 1:36:02 PM


Exercises 2.11

1. 1y 2-^ h 2. ( 3)x 2+ 3. ( 5)m 2+ 4. ( 2)t 2-

5. ( 6)x 2- 6. 2 3x 2+] g 7. 4 1b 2-] g 8. 3 2a 2+] g

9. 5 4x 2-] g 10. 7 1y 2+^ h 11. 3 5y 2-^ h 12. 4 3k 2-] g

13. 5 1x 2+] g 14. 9 2a 2-] g 15. 7 6m 2+] g 16. 21

t2

+d n

17. 32

x2

-d n 18. 351

y2

+d n 19. 1

x x

2

+c m 20. 52

kk

2

-d n

Exercises 2.12

1. 2)( 2)(a a+ - 2. 3)( 3)(x x+ - 3. 1)( 1)(y y+ -

4. 5 5x x+ -] ]g g 5. 2 7)(2 7)( x x+ - 6. 4 3)(4 3)( y y+ -

7. 1 2 )(1 2 )( z z+ - 8. 5 1 5 1t t+ -] ]g g 9. 3 2 3 2t t+ -] ]g g

10. 3 4 3 4x x+ -] ]g g 11. 2 )( 2 )(x y x y+ -

12. 6 6x y x y+ -^ ^h h 13. 2 3 2 3a b a b+ -] ]g g

14. 10 10x y x y+ -^ ^h h 15. 2 9 2 9a b a b+ -] ]g g

16. 2 2x y x y+ + + -^ ^h h 17. 3)( 1)(a b a b+ - - +

18. 1 1z w z w+ + - -] ]g g 19. 21

21

x x+ -d dn n

20. 3

13

1y y+ -e eo o 21. 2 3 2 1x y x y+ + - +^ ^h h

22. ( )( ) ( )( )( )x x x x x1 1 1 1 12 2 2+ - = + + -

23. 3 2 3 2x y x y3 3+ -_ _i i 24. 4 2 2x y x y x y2 2+ + -_ ^ ^i h h

25. 1)( 1)( 1)( 1)(a a a a4 2+ + + -

Exercises 2.13

1. 2)( 2 4)(b b b2- + + 2. 3 3 9x x x2+ - +] ^g h

3. 1 1t t t2+ - +] ^g h 4. 4)( 4 16)(a a a2- + +

5. 1 )(1 )( x x x2- + + 6. 2 3 4 6 9y y y2+ - +^ _h i

7. ( ) ( )y z y yz z2 2 42 2+ - + 8. 5 )( 5 25 )(x y x xy y2 2- + +

9. 2 3 4 6 9x y x xy y2 2+ - +^ _h i 10. 1 1ab a b ab2 2- + +] ^g h

11. 10 2 )(100 20 4 )( t t t+ - + 2 12. 2

34 2

39

x x x2

- + +d en o

13. 10 1 100 10 1a b a ab b2 2

+ - +d en o

14. 1 2 1x y x x xy y y2 2+ - + + + + +^ _h i

15. xy z x y xyz z5 25 306 362 2 2+ - +^ _h i 16. a a 19 2 - +- ^ h

17. 13

13 9

x x x2

- + +d en o

18. 3 3 9 6x y y y xy x x2 2+ + - - + + +^ _h i

19. 1 4 5 7x y x x xy y y2 2+ - + - + - +^ _h i

20. 2 6 )(4 24 2 6 36)( a b a a ab b b2 2+ - + + + + +

Exercises 2.14

1. x x2 3 3+ -] ]g g 2. p p3 3 4+ -^ ^h h

3. y y y5 1 12- + +^ _h i 4. ) (ab a b a2 2 2 1+ -^ h

5. 5 1a 2-] g 6. x x2 3 4- -- ] ]g g 7. z z z3 5 4+ +] ]g g

8. ab ab ab3 2 3 2+ -] ]g g 9. x xx 1 1+ -] ]g g

10. x x2 3 2 2- +] ]g g 11. 5 3m n- +] ]g g 12. x7 2 1- +] g

13. 5 4 4y y y+ + -^ ^ ^h h h 14. 1 2 2 4x x x x2- + - +] ] ^g g h

15. x x x x x x1 1 1 12 2+ - + - + +] ^ ] ^g h g h

16. x x x2 5+ -] ]g g 17. ( )x x3 3 2+ -] g

18. ( ) ( )xy xyy 2 1 2 1+ - 19. b b b3 2 4 2 2- + +] ^g h

20. x x3 3 2 2 5- +] ]g g 21. x3 1 2-] g

22. 2)( 5)( 5)(x x x+ + - 23. 3z z 2+] g

24. 1 1 2 3 2 3x x x x+ - + -] ] ] ]g g g g

25. x x x y x xy y2 2 2 2 2+ - + - +] ] ^ _g g h i

26. ( ) ( )a a a4 3 3+ - 27. x x xx 2 4 25 2- + +] ^g h

28. 2)( 2)( 3)( 3)(a a a a+ - + - 29. 4 ( 5)k k 2+

30. 3( 1) 1) 3)( (x x x+ - +

Exercises 2.15

1. 4 4 2x x x2 2+ + = +] g 2. 6 9 3b b b2 2- + = -] g

3. 10 25 5x x x2 2- + = -] g 4. 8 16 4y y y2 2+ + = +^ h

5. 14 49 7m m m2 2- + = -] g 6. 18 81 9q x q2 2+ + = +^ h

7. 2 1 1x x x2 2+ + = +] g 8. 16 64 8t t t2 2- + = -] g

9. 20 100 10x x x2 2- + = -] g

10. 44 484 22w w w2 2+ + = +] g

11. 32 256 16x x x2 2- + = -] g 12. 349

23

y y y22

+ + = +d n

13. 74

4927

x x x22

- + = -d n 14. 41

21

a a a22

+ + = +d n

15. 94

8129

x x x22

+ + = +d n 16. 5

yy

y1625

45

22

2

- + = -d n

17. 11

kk

k16

1214

112

22

- + = -d n

18. 6 9 3x xy y x y2 2 2+ + = +^ h 19. 4 4 2a ab b a b2 2 2- + = -] g

20. 8 16 4p pq q p q2 2 2- + = -^ h

Exercises 2.16

1. 2a + 2. 2 1t - 3. 3

4 1y + 4.

2 14

d - 5.

5 2xx-

6. 4

1y -

7. ab a

322

-

-] g 8.

31

ss+

- 9.

11

bb b2

+

+ +

10. 3

5p + 11.

31

aa+

+ 12.

2 4

3

x x

y2 + +

+ 13. 3x -

14. 4 2 1

2

p p

p2 - +

- 15.

2a ba b

-

+

Answer S1-S5.indd 544 7/31/09 1:36:03 PM

545ANSWERS

Exercises 2.17

1. (a) 45x

(b) 15

13 3y + (c)

128a +

(d) 6

4 3p + (e)

613x -

2. (a) 2 1a

b-

(b) 1

2 1

q

p q q2

+

- - +^ _h i (c)

b

x yb

2 1

2

10

2

-

+

]

^

g

h

(d) ab

x xy y2 2- + (e)

5 2

3 1

x x

x x

- -

- -

] ]

] ]

g g

g g

3. (a) 5x (b)

xx

x 12

-

- +

] g (c)

3a b

a b+

+ + (d)

22

xx+

(e) p q

p q p q

p q

p q 11 2 2

+

+ -=

+

- ++^ ^h h (f)

x x

x

1 3

12

+ -

-

] ]

]

g g

g

(g) 2 23 8

x xx

+ -

- +

] ]g g (h)

1

2

a

a2+

+

] g

(i) y y y

y y

2 3 1

3 14 132 2

+ + -

+ +

^ ^ ^

_

h h h

i (j)

x x x

x

4 4 3

5 22

+ - +

+-

] ] ]

]

g g g

g

4. (a) y y

xx

3 9

2

8

2

2 - +

+

_

]

i

g (b)

15

2 1

y

y y+ +^ ^h h

(c) x x

x x2 3 4

210 42

- -

-+

] ]g g (d)

bb b

b 13 5 10

2

2

+

- -

] g (e) x

5. (a) 5 2 3

3 13x x x

x- - +

-

] ] ]g g g (b)

2 23 5

x xx

+ -

-

] ]g g

(c) p q p q

p pq q

pq

3 5 22 2

+ -

+ -

^ ^h h (d)

2 1a b a b

a ab b2 2

+ -

- - +

] ]g g

(e) x y x y

x yy 1

+ -

+ +

^ ^

^

h h

h

Exercises 2.18

1. (a) 7.1- (b) 6.9- (c) 48.1 (d) 37.7- (e) 0.6

(f) 2.3 (g) 5.3- 2. 47 3. 7- 4. 375 5. 196-

6. 5.5 7. 377 8. 284 9. 40- 10. 51.935 11. 143

-

12. 22.4 13. 1838.8 14. 43

15. 15 16. 10

17. 2 312 = 18. 23.987 19. 352.47 20. 93 21. 4

Exercises 2.19

1. (a) 2 3 (b) 3 7 (c) 2 6 (d) 5 2 (e) 6 2

(f) 10 2 (g) 4 3 (h) 5 3 (i) 4 2 (j) 3 6

(k) 4 7 (l) 10 3 (m) 8 2 (n) 9 3 (o) 7 5

(p) 6 3 (q) 3 11 (r) 5 5

2. (a) 6 3 (b) 20 5 (c) 28 2 (d) 4 7 (e) 16 5

(f) 8 14 (g) 72 5 (h) 30 2 (i) 14 10 (j) 24 5

3. (a) 18 (b) 20 (c) 176 (d) 128 (e) 75

(f) 160 (g) 117 (h) 98 (i) 363 (j) 1008

4. (a) 45x = (b) 12x = (c) 63x = (d) 50x =

(e) 44x = (f) 147x = (g) 304x = (h) 828x =

(i) 775x = (j) 960x =

Exercises 2.20

1. 3 5 2. 2 3. 6 3 4. 3 3 5. 3 5- 6. 3 6

7. 7 2- 8. 8 5 9. 4 2- 10. 4 5 11. 2 12. 5 3

13. 3- 14. 2 15. 5 7 16. 2 17. 13 6

18. 9 10- 19. 47 3 20. 2 2 35 - 21. 7 5 2-

22. 2 3 4 5- - 23. 7 6 3 5+ 24. 2 2 3- -

25. 17 5 10 2- +

Exercise 2.21

1. 21 2. 15 3. 3 6 4. 10 14 5. 6 6- 6. 30

7. 12 55- 8. 14 9. 60 10. 12 2 3=

11. 2 48 8 3= 12. 15 28 30 7=

13. 2 20 4 5= 14. 84- 15. 2

16. 28 17. 30 18. 2 105- 19. 18

20 . 30 50 150 2= 21. 2 6 22. 4 3 23. 1 24. 6

8

25. 2 3 26. 3 10

1 27.

2 5

1 28.

3 5

1 29.

21

30. 2 2

3 31.

2

3 32.

2 5

9 33.

2 2

5 34.

32

35. 75

Exercises 2.22

1. (a) 10 6+ (b) 2 6 15- (c) 12 8 15+

(d) 5 14 2 21- (e) 6 4 18 6 12 2- + = - +

(f) 5 33 3 21+ (g) 6 12 6- - (h) 5 5 15-

(i) 6 30+ (j) 2 54 6 6 6 6+ = +

(k) 8 12 12 8 24 3- + = - + (l) 210 14 15-

(m) 10 6 120- (n) 10 2 2- - (o) 4 3 12-

2. (a) 10 3 6 3 5 9 3+ + +

(b) 10 35 2 14- - +

(c) 2 10 6 10 15 15 6- + -

(d) 12 18 8 1224 36 8 12

20 60 10 305 15 10 30

+ - - =

+ - -

(e) 52 13 10- (f) 15 15 18 10 6 6- + -

(g) 4 (h) 1- (i) 12- (j) 43 (k) 3 (l) 241-

(m) 6- (n) 7 2 10+ (o) 11 4 6- (p) 25 6 14+

(q) 57 12 15+ (r) 27 4 35-

(s) 77 12 40 77 24 10- = - (t) 53 12 10+

3. (a) 18 (b) 108 2 (c) 432 2 (d) 19 6 2+ (e) 9

4. (a) 21, 80a b= = (b) 19, 7a b= = -

5. (a) 1a - (b) p pp2 1 2 1- - -^ h

6. 25k = 7. 2 3 5x y xy- - 8. 17, 240a b= =

9. 107, 42a b= = - 10. 9 5 units2+

Answer S1-S5.indd 545 8/7/09 12:28:47 PM


Exercises 2.23

1. (a) 77

(b) 46

(c) 5

2 15 (d)

106 14

53 14

=

(e) 3

3 6+ (f)

212 5 2-

(g) 5

5 2 10+

(h) 14

3 14 4 7- (i)

208 5 3 10+

(j) 35

4 15 2 10-

2. (a) 4 4 3 243 2- -= ^ h (b) 47

6 7 3+- ^ h

(c) 19

2 15 4 1819

15 6 22-=

-- -^ ^h h

(d) 13

19 8 313

8 3 19-=

-- ^ h (e) 6 2 5 3 5 2+ + +

(f) 2

6 15 9 6 2 10 6- + -

3. (a) 2 2

(b) 2 3 32 6 3 2 3 3 6 2 3- + - + = - - + -^ h

(c) 39

22 5 14 2+

(d)

106 6 16 3 84 8 14

6 21 145

3 8 3 4

- - - +

=- + + -

^ h

(e) 4- (f) 4 2

(g) 15

20 12 19 6 25 3 615

19 6 65 3 6+ + -=

+ -

(h) 6

6 9 2 2 3+ + (i)

214 6 9 3+

(j) 415

30330 30 5- -

(k) 13

28 2 6 7 3- -

(l) 2

2 15 2 10 2 6 3 5+ - - -

4. (a) 45, 10a b= = (b) 1, 8a b= = (c) 21

,21

a b= - =

(d) 195

,98

a b= - = - (e) 5, 32a b= =

5.

2

2

3 2 23

2 1

2 1

2

4

2 1

2 1

2 1

2 1

2

4

2

2

2 1

2 1 2 12

4 2

2 12 2 2 1

2

13 2 2

2

2 2

2 2

# #

+

-+

=+

-

-

-+

=-

- -+

=-

- - ++

=-

+

= - +

=

^

^ ^

h

h h

So rational

6. (a) 4 (b) 14 (c) 16

7. 3

3 5 2 15 3- - -

8. 3 2 2

2

2

8

3 2 2

2

3 2 2

3 2 2

2

8

2

2

3 2 2

2 3 2 22

8 2

9 4 26 4 2

4 2

16 4 2

4 2

6 4 2 4 26

2 2

# #

#

++

=+ -

-+

=-

-+

=-

-+

=-

+

= - +

=

^

^

h

h

So rational

9. x 3 2= - +^ h 10. 4

4 4b

b b-

+ +

Test yourself 2

1. (a) 2y- (b) 4a + (c) 6k5- (d) 15

5 3x y+ (e) 3 8a b-

(f) 6 2 (g) 4 5

2. (a) 6 6x x+ -] ]g g (b) 3 1a a+ -] ]g g (c) ab b4 2-] g

(d) 3)(5 )(y x- + (e) n p2 32 - +^ h

(f) 2 )(4 2 )( x x x2- + +

3. (a) 4 6b - (b) 2 5 3x x2 + - (c) 4 17m +

(d) 16 24 9x x2 - + (e) 25p2 - (f) 1 7a- -

(g) 2 6 5 3- (h) 3 3 6 21 2 7- + -

4. (a) a ab 3 9

822 + +^ h

(b) 2

15

m 2-] g

5. 157.464V = 6. (a) 17 (b) 17

6 15 9-

7. 3 2

4 5x x

x+ -

+

] ]g g 8. (a) 36 (b) 2- (c) 2 (d) 216 (e) 2

9. (a) 5

1 (b) 8 10. 11.25d =

11. (a) 15

2 3 (b)

22 6+

12. (a) 3 6 6 4 3 4 2- - + (b) 11 4 7+

13. (a) 3( 3)( 3)x x- + (b) x x6 3 1- +] ]g g

(c) y y y5 2 2 42+ - +^ _h i

14. (a) 3y

x4

3

(b) 3 1

1x -

15. (a) 99 (b) 24 3

16. (a) a b2 2- (b) 2a ab b2 2+ + (c) 2a ab b2 2- +

17. (a) a b 2-] g (b) a b a ab b2 2- + +] ^g h

18. 2

3 3 1+ 19. (a)

4 3ab

b a+ (b)

103 11x -

20. 7

21 5 46 2- -

Answer S1-S5.indd 546 7/31/09 1:36:05 PM

547ANSWERS

21. (a) 6 2 (b) 8 6- (c) 2 3 (d) 3

4 (e) 30a b2

(f) 3n

m4 (g) 2 3x y-

22. (a) 2 6 4+ (b) 10 14 5 21 6 10 3 15- - +

(c) 7 (d) 43 (e) 65 6 14-

23. (a) 7

3 7 (b)

156

(c) 5 1

2+

(d) 15

12 2 6- (e)

5320 3 15 4 10 3 6+ + +

24. (a) 10

10x + (b)

2117 15a -

(c) ( 1)( 1)x x

x3 2+

-

-

(d) 1

1k -

(e) 3

15 6 15 3 15 2- - -

25. (a) 48n = (b) 175n = (c) 392n =

(d) n 5547= (e) 1445n =

26. 312171

27. (b), (c) 28. (d) 29. (a), (d) 30. (c)

31. (c) 32. (b) 33. (a) 34. (d) 35. (b)


1. (a) 2 8 6a b ab a2 2 3- + (b) 4y4 -

(c) 8 60 150 125x x x3 2- + -

2. 17

17 3 2 5 20+ + 3.

2 2

142

or

4. ab

xa

bx

ab

x4 2

2

2

2 2

+ + += d n

5. (a) x x4 9+ +] ]g g

(b) ( ) ( )x y x y x y x y x y3 2 3 3 22 2 2- + = + - +_ _ _i i i

(c) 5 7 25 35 49x x x2+ - +] ^g h (d) 2 2 2b a a- + -] ] ]g g g

6. 4 12 9 2 3x x x2 2+ + = +] g 7. x

y

1

1

2 -

+

] g 8. 2 5

9. 1

1

a a

a2

2

- +

+] g 10.

2 2x b

ax b

a+ -d dn n

11. x x x

x x x xy y

3 3 2

3 6 3 643 2

- + -

- + + -

] ] ]g g g

12. (a) 8 12 6 1x x x3 2- + - (b) 2 1

3 4

x

x2-

+

] g

13. x x x 97 153 2 + --

14. 13

66 6 4 2 15 4 5 65 3- + - + -

15. xx

x32

91

312

2

+ + = +d n 16. 2x =

17. 10

400 59 5- 18. (a) 3

12171

(b) ,a b2317

2314

= = -

19. 121

i = 20. 4

r4

3 3

r r

r= =

21. 2 6 3s = +

Chapter 3: Equations

Exercises 3.1

1. 5t = - 2. 5.6z = - 3. 1y = 4. 6.7w = 5. 12x =

6. 4x = 7. 151

y = 8. 35b = 9. 16n = - 10. 4r =

11. 9y = 12. 6k = 13. 2d = 14. 5x = 15. 15y =

16. 20x = 17. 20m = 18. 4x = 19. 7a = - 20. 3y =

21. 4b = - 22. 3x = 23. 132

a = - 24. 4t = -

25. 1.2x = 26. 1.6a = 27. 81

b = 28. 39t =

29. 5p = 30. 4.41x Z

Exercises 3.2

1. 331

b = 2. 35x = 3. 494

y = 4. 1359

x = 5. 585

k =

6. 36x = 7. 0.6t = 8. 3x = - 9. 1.2y = - 10. 69x =

11. 13w = 12. 30t = 13. 14x = 14. 1x = -

15. 0.4x = - 16. 3p = 17. 8.2t = 18. 9.5x = -

19. 22q = 20. 3x = - 21. 0.8b = 22. 0.375a = -

23. 3x = 24. 1y = 25. 132

t = -

Exercises 3.3

1. 8.5t = 2. 122l = 3. 8b = 4. 41a = 5. 4y =

6. 6.68r = 7. 6.44x = 8. 15n = 9. 332

y1 =

10. 3.7h = 11. (a) 25.39BMI = (b) 69.66w =

(c) 1.94h = 12. 0.072r = 13. 9x1 = - 14. 2.14t =

15. x 2!= 16. 2.12r = 17. 10.46r = 18. 1.19x =

19. 5.5x = 20. 3.3r =

Exercises 3.4

1. (a) x 32

-4 -3 -2 -1 0 1 2 3 4

(b) y 4#

-4 -3 -2 -1 0 1 2 3 4

2. (a) 7t 2 (b) x 3$ (c) 1p 2 - (d) x 2$ - (e) 9y 2 -

(f) a 1$ - (g) y 221

$ - (h) x 21 - (i) a 6# -

(j) y 121 (k) b 181 - (l) 30x 2 (m) x 343

#

(n) m 1432

2 (o) b 1641

$ (p) r 9# - (q) 8z 2

(r) w 254

1 (s) x 35$ (t) t 9$ - (u) 6q52

2 -

(v) 1x32

2 - (w) b 1141

# -

Answer S1-S5.indd 547 7/31/09 1:36:06 PM


3. (a) x1 71 1

0 1 2 3 4 5 6 7 8

(b) p2 51#-

-3 -2 -1 0 1 2 3 4 5

(c) x1 41 1

-3 -2 -1 0 1 2 3 4 5

(d) y3 5# #-

-3 -2 -1 0 1 2 3 4 5

(e) y61

132

1 1

-3 -2 -1 0 1 2 3 4 5

Exercises 3.5

1. (a) x 5!= (b) y 8!= (c) 4 4a1 1-

(d) ,k k1 1$ # - (e) 6, 6x x 12 -

(f) p10 10# #- (g) 0x = (h) ,a a14 142 1 -

(i) 12 12y1 1- (j) ,b b20 20$ # -

2. (a) ,x 5 9= - (b) ,n 4 2= - (c) ,a a2 212 -

(d) x4 6# # (e) ,x 3 6= - (f) ,x 5 475

= -

(g) 3 2y211 1- (h) ,x x9 6$ # - (i) x 12!=

(j) a2 10# #

3. (a) 141

x = (b) ,a 331

= - (c) 231

b =

(d) No solutions (e) 272

y = - (f) 7x = (g) ,m 5 132

=

(h) ,d 221

143

= - (i) ,y54

2= - (j) No solutions

4. (a) ,x 221

= - (b) 3, 231

y = (c) 10, 153

a = -

(d) ,x 4 731

= - (e) ,d 4 5= -

5. (a) ,t 3 152

= - (b) 1 3t521 1-

-3 -2 -1 0 1 2 3 4 5

Exercises 3.6

1. (a) 3x = (b) y 8!= (c) 2n != (d) 2x 5!=

(e) 10p = (f) x 5!= (g) y 3!= (h) 2w =

(i) n 4!= (j) 2q = -

2. (a) 6.71p != (b) 4.64x = (c) 2.99n = (d) 5.92x !=

(e) 1.89y = (f) .d 2 55!= (g) 4.47k != (h) 2.22x =

(i) .y 3 81!= (j) 3.01y =

3. (a) 27n = (b) 16t = (c) 32x = (d) 8t =

(e) 243p = (f) 625m = (g) 216b = (h) 27y =

(i) 128a = (j) 81t =

4. (a) 51

x = (b) 21

a = (c) 21

y = (d) x71

!=

(e) 32

n = (f) 2a = (g) 2x != (h) 9b =

(i) x32

!= (j) b 121

!=

5. (a) x512

1= (b) 6

41

x = (c) 811

a = (d) 625

1k =

(e) x81

!= (f) 4x = (g) y 8!= (h) n 73219

=

(i) 8b = (j) 1216127

m =

6. (a) 4n = (b) 5y = (c) 9m = (d) 5x = (e) 0m =

(f) 3x = (g) 2x = (h) 2x = (i) 1x = (j) k 2=

7. (a) 2x = (b) 1x = (c) 2x = - (d) 2n = (e) 0x =

(f) 6x = (g) y31

= (h) 2x = (i) 2x = (j) a 0=

8. (a) 21

m = (b) 31

x = (c) 31

x = (d) 21

k = -

(e) 32

k = - (f) 43

n = (g) 121

x = (h) 32

n =

(i) 61

k = - (j) 132

x =

9. (a) x 1= - (b) x 131

= - (c) k 4= - (d) n 3=

(e) x 212

= - (f) x32

= - (g) x 421

= - (h) x 1117

= -

(i) x 154

= (j) x 18=

10. (a) 41

m = (b) 243

k = - (c) 283

x = (d) 121

k =

(e) 181

n = (f) 21

n = - (g) 54

x = (h) 361

b = -

(i) 171

x = - (j) 5m =

Puzzle

1. All months have 28 days. Some months have more days as well. 2. 10 3. Bottle $1.05; cork 5 cents

4. 16 each time 5. Friday

Exercises 3.7

1. ,y 0 1= - 2. ,b 2 1= - 3. ,p 3 5= - 4. ,t 0 5=

5. ,x 2 7= - - 6. q 3!= 7. x 1!= 8. ,a 0 3= -

Answer S1-S5.indd 548 7/31/09 1:36:07 PM

549ANSWERS

9. ,x 0 4= - 10. x21

!= 11. ,x 1 131

= - -

12. ,y 1 121

= - 13. ,b43

21

= 14. ,x 5 2= - 15. ,x 032

=

16. ,x 1 221

= 17. 0, 5x = 18. 1, 2y = - 19. ,n 53=

20. 3, 4x = 21. 6, 1m = - 22. , ,x 0 1 2= - -

23. , ,y 1 5 2= - - 24. ,x 5 7= - 25. ,m 8 1= -

Exercises 3.8

1. (a) x 5 2!= - (b) 3a 7!= + (c) 4y 23!= +

(d) 1x 13!= - (e) p 44 7 2 11 7! != - = -

(f) x 28 5 2 7 5! != + = +

(g) 510y 88 2 22 2210 2! ! ! -= - = - = ^ h

(h) 1x 2!= + (i) 12n 137!= -

(j) 3

y25!

=+

2. (a) . , .x 3 45 1 45= - (b) . , .x 4 59 7 41= - -

(c) . , .q 0 0554 18 1= - (d) . , .x 4 45 0 449= -

(e) . , .b 4 26 11 7= - - (f) . , .x 17 7 6 34=

(g) . , .r 22 3 0 314= - (h) . , .x 0 683 7 32= - -

(i) . , .a 0 162 6 16= - (j) . , .y 40 1 0 0749= -

Exercises 3.9

1. (a) . , .y 0 354 5 65= - - (b) , .x 1 1 5=

(c) . , .b 3 54 2 54= - (d) , .x 1 0 5= -

(e) . , .x 0 553 0 678= - (f) . , .n 0 243 8 24= -

(g) ,m 2 5= - - (h) ,x 0 7= (i) ,x 1 6= -

(j) . , .y 2 62 0 382=

2. (a) x2

1 17!=

- (b) x

65 13!

=

(c) q2

4 282 7

!!= =

(d) h8

12 1282

3 2 2! !=

-=

-

(e) s6

8 403

4 10! != =

(f) x2

11 133!=

- (g) d

125 73!

=-

(h) x2

2 321 2 2

!!= = (i) t

21 5!

=

(j) x4

7 41!=

Exercises 3.10

1. 3 0x 11- 2. 0 4y1 1 3. ,n n0 1# $

4. ,x x2 2# $- 5. ,n n1 11 2- 6. n5 3# #-

7. ,c c1 21 2- 8. x4 2# #- - 9. 4 5x1 1

10. ,b b221

# $- - 11. ,a a131

1 2-

12. ,y y121

21 2- 13. ,x x32

1# $

14. ,b b352

1 2- 15. x121

31

# #- -

16. y4 3# #- 17. ,x x4 41 2- 18. a1 1# #-

19. 2 3x 11- 20. ,x x1 3# $- 21. 0 2x1 1

22. a1 121

# # 23. ,y y254

# $-

24. ,m m132

121

1 2- 25. x1 131

# #

Exercises 3.11

1. ,a b1 3= = 2. ,x y2 1= = 3. ,p q2 1= = -

4. ,x y6 17= = 5. ,x y10 2= - = 6. ,t v3 1= =

7. ,x y3 2= - = 8. ,x y64 39= - = - 9. ,x y3 4= = -

10. ,m n2 3= = 11. ,w w1 51 2= - = 12. ,a b0 4= =

13. ,p q4 1= - = 14. ,x x1 11 2= = -

15. ,x y1 4= - = - 16. ,s t2 1= = -

17. ,a b2 0= - = 18. ,k h4 1= - =

19. ,v v2 41 2= - = 20. , .x y2 1 41Z=

Problem

23 adults and 16 children.

Exercises 3.12

1. ,x y0 0= = and ,x y1 1= =

2. ,x y0 0= = and ,x y2 4= - =

3. ,x y0 3= = and ,x y3 0= =

4. ,x y4 3= = - and ,x y3 4= = - 5. ,x y1 3= - = -

6. ,x y3 9= = 7. ,t x2 4= - = and ,t x1 1= =

8. ,m n4 0= - = and ,m n0 4= = -

9. ,x y1 2= = and ,x y1 2= - = -

10. ,x y0 0= = and ,x y1 1= =

11. ,x y2 1= = and ,x y1 2= - = - 12. ,x y0 1= =

13. ,x y1 5= = and ,x y4 11= =

14. ,x y41

4= = and ,x y1 1= - = - 15. ,t h21

41

= - =

16. ,x y2 0= =

17. ,x y0 0= = and ,x y2 8= - = - and ,x y3 27= =

18. ,x y0 0= = and ,x y1 1= = and ,x y1 1= - =

19. ,x y21

243

= = 20. 135

,1312

x y= - = -

Exercises 3.13

1. , ,x y z2 8 1= - = - = - 2. , ,a b c2 1 2= - = - =

3. , ,a b c4 2 7= - = = 4. , ,a b c1 2 3= = = -

5. , ,x y z5 0 2= = = - 6. , ,x y z0 5 4= = - =

7. , ,p q r3 7 4= - = = 8. , ,x y z1 1 2= = - =

9. , ,h j k3 2 4= - = = - 10. , ,a b c3 1 2= = - = -

Answer S1-S5.indd 549 7/31/09 1:36:08 PM


Test yourself 3

1. (a) 10b = (b) 116a = - (c) 7x = - (d) p 4#

2. (a) 1262.48A = (b) 8558.59P =

3. (a) x x x8 16 42 2- + = -] g (b) k k k4 4 22 2+ + = +] g

4. (a) ,x y2 5= - = (b) ,x y4 1= = and ,x y21

8= - = -

5. (a) 2x = (b) 41

y =

6. (a) ,b 2 131

= - (b) ,g 241

= (c) ,x x4 3$ #

7. (a) 36A = (b) 12b = 8. ,x21

1=

9. 1 3y1 #-

10. (a) . , .x 0 298 6 70= - - (b) . , .y 4 16 2 16= -

(c) . , .n 0 869 1 54= -

11. (a) 764.5V = (b) 2.9r = 12. x 7141

2

13. ,x x2 91 2 14. . , .x y2 4 3 2= = 15. (a) 2100V =

(b) 3.9r = 16. (a) ii (b) i (c) ii (d) iii (e) iii

17. , ,a b c3 2 4= = = -

18. ,n n0 331

2 1 -

19. 4x = - 20. 2x = - 21. (a) 3y 2 (b) n3 0# #-

(c) 2x = (d) 2x = (e) ,x 3 152

= - (f) 1, 2t t$ # -

(g) 4 2x# #- (h) 3x = - (i) ,y y2 22 1 -

(j) 1, 1x x# $- (k) 65

x = (l) 21

2b# #-

(m) No solutions (n) 231

,53

t = (o) 1 3x1 1-

(p) ,m m3 2# $-


1. 1y = 2. ,x a x a1 2-

3. ,a b3 2!= = 4. . , .x 2 56 1 56= -

5. ; ,x x x x x x3 3 2 2 4 3 22!+ - - + + =] ] ] ^g g g h

6. ,x y1 2= = and ,x y1 0= - =

7. ; . , .b x4 17 4 8 12 0 123! Z= = + - 8. x 1!=

9. 1 1t1 1- 10. x3 8# #- 11. 41

x =

12. 2.31r = 13. No solutions 14. x b a a2!= + +

15. 2247.36P = 16. x3

4 102 !=^ h

17. ,y y153

1 2-

Chapter 4 : Geometry 1

Exercises 4.1

1. (a) 47y c= (b) 39x c= (c) 145m c= (d) 60y c= (e) 101b c= (f) 36x c= (g) 60a c= (h) 45x c=

(i) 40y c= (j) 80x c= 2. (a) 121c (b) 72 29c l (c) 134 48c l 3. (a) 42c (b) 55 37c l (c) 73 3c l

4. (a) (i) 47c (ii) 137c (b) (i) 9c (ii) 99c (c) (i) 63c (ii) 153c (d) (i) 35c (ii) 125c (e) (i) 52c (ii) 142c (f) (i) 15c7l (ii) 105c7l (g) (i) 47c36l (ii) 137c36l (h) (i) 72c21l (ii) 162c21l (i) (i) 26c 11 l (ii) 116c 11 l (j) (i) 38c 15 l (ii) 128c 15 l 5. (a) 49x c= (b) 41c (c) 131c 6. (a) ,y x z15 165c c= = =

(b) , ,x y z142 48 28c c c= = =

(c) , ,a b c43 137 101c c c= = =

(d) ,,a b d c97 41 42c c c= = = =

(e) , ,a b c68 152 28c c c= = = (f) ,a b10 150c c= =

7. 0x x x x

xx

8 10 2 10 10 7 10 36

18 36020

- + - + + + + =

=

=

(angleof revolution)

( )

( )

ABE x

EBC x

ABE EBC

8 108 20 101502 102 20 1030150 30180

c

cc cc

+

+

+ +

= -= -

== -= -

=+ = +

=

ABC`+ is a straight angle

( )

DBC x

DBC EBC

7 107 20 10150150 30180

cc cc

+

+ +

= +

= +

=

+ = +

=

DBE`+ is a straight angle AC and DE are straight lines

8.

AFC x

CD bisects

`

`

+ =

AFE+

( )

( )

( )

( )

DFB x

x

CFE x x

xAFC CFE

AFB

AFB

180 180

180 180 2

is a straight angle


is a straight angle

`

c c

c c

+

+

+ +

+

+

= - -

=

= - + -

=

=

9. ABD DBC+ ++

110 3 3 70180

x xc

= - + +

=

So ABC+ is a straight angle. AC is a straight line.

10. AEB BEC CED+ + ++ +

y y y50 8 5 20 3 60

90c= - + - + +

=

So AED+ is a right angle.

Exercises 4.2

1. (a) ,a b e f c d g148 32c c= = = = = = =

(b) ,x z y70 110c c= = =

(c) , ,x y z55 36 89c c c= = = (d) ,y x z125 55c c= = =

(e) ,n e g a c z x 98c= = = = = = = 82o m h f b d y w c= = = = = = = =

Answer S1-S5.indd 550 8/1/09 6:50:34 PM

551ANSWERS

(f) , ,a b c95 85 32c c c= = =

(g) , ,a b c27 72 81c c c= = =

(h) , , ,x y z a b56 124 116 64c c c c= = = = =

(i) 61x c= (j) 37y c=

2. (a) CGF

BFG CGF

180 121

5959

( is a straight angle)FGH

`

c c

cc

+

+ +

= -

=

= =

These are equal alternate angles. AB CD` < (b) BAC 360 292 68

(angle of revolution)c c c+ = - =

BAC DCA 68 112180

` c cc

+ ++ = +

=

These are supplementary cointerior angles.

AB CD` <

(c) 180 76104

104

BCD

ABC BCDc

c

+

+ +

= -

=

= =

( BCE+ is a straight angle)

These are equal alternate angles.

AB CD` ;

(d) 180 12852

52

CEF

CEF ABEc

c

+

+ +

= -

=

= =

( CED+ is a straight angle)

These are equal corresponding angles.

AB CD` ;

(e) 180 23 115CFH+ = - +] g ( EFG+ is a straight angle)

42c=

42BFD` c+ = (vertically opposite angles)

ABF BFD 138 42

180c cc

+ ++ = +

=

These are supplementary cointerior angles. AB CD` ;

Exercises 4.3

1. (a) 60x c= (b) 36y c= (c) 71m c= (d) 37x c=

(e) 30x c= (f) 20x c= (g) 67x c= (h) 73a c=

(i) , ,a b c75 27 46c c c= = =

(j) , ,a b c36 126 23c c c= = =

(k) , ,x y z w67 59 121c c c= = = =

2. All angles are equal. Let them be x . x x x 180Then (angle sum of )D+ + =

xx

3 18060

=

=

So all angles in an equilateral triangle are 60 .c

3. x90 c-] g

4. 50

180 (50 45 )ACBABC

DEC ABC85

85

(vertically opposite angles)(angle sum of )

`

cc c cc

c

++

+ +

D

=

= - +

=

= =


AB DE` <

5.

124 68

ACB

CBACBA

CBA ACBABC

180 12456

68 124

5656

is isosceles

( is a straight angle)

(exterior angle of )

DCB

`

`

c cc

c cc cc

c

+

++

+ +D

D

= -

=

+ =

= -

=

= =

6. 38y c=

7. (a) x 64c= (b) ,x y64 57c c= = (c) 63x c=

(d) ,a b29 70c c= =

8. 180 (35 25 )120180 12060180 (90 30 )60180 (60 60 )60

HJI

IJL

JIL

ILJ

(angle sum of )

( is a straight angle)

(angle sum of )

(angle sum of )

HJI

HJL

IKL

JIL

c c ccc ccc c cc

c cc

+

+

+

+

D

D

D

= - +

=

= -

=

= - +

=

= - +

=

Since 60 ,IJL JIL ILJ c+ + += = = IJLD is equilateral

( )

( )

( )

KJL

JLK

KJI

JKL

180 60120180 30 12030

is a straight angle

angle sum of

c cc

c c cc

+

+ D

= -

=

= - +

=

°JLK JKL 30`+ += =

JKL` D is isosceles

9. BC BD=

BDC 46` c+ = (base angles of isosceles triangle)

CBD 180 2 4688

#c

+ = -=

CBD BDE 88` c+ += = These are equal alternate angles.

AB ED` ;

10. 18032

OQP 75 73c

+ = - +

=

] g (angle sum of triangle)

MNO OQP 32` c+ += =


MN QP` ;

Exercises 4.4

1. (a) Yes

5AB EF cm= = (given)

6BC DF cm= = (given)

8AC DE cm= = (given)

ABC DEF` /D D ( SSS )

(b)Yes

4.7XY BC m= = (given)

XYZ BCA 110c+ += = (given)

2.3YZ AC m= = (given)

XYZ ABC` /D D ( SAS )

(c) No

Answer S1-S5.indd 551 7/31/09 1:36:09 PM


(d) Yes

PQR SUT 49c+ += = (given)

PRQ STU 52c+ += = (given)

8QR TU cm= = (given)

PQR STU` /D D ( AAS )

(e) No

2. (a)

,

AB KLB L

BC JLABC JKL

438

5by SAS

(given)(given)(given)

`

c+ +

/D D

= =

= =

= =

(b)

,

Z BXY ACYZ BC

RHS XYZ ABC

9072

by


`

c+ +

/D D

= =

= =

= =

(c)

,

MN QRNO PRMO PQ

MNO PQR

885

by SSS


` /D D

= =

= =

= =

(d)

.

Y TZ S

XY TRXYZ STR

90351 3

by AAS,


`

cc

+ ++ +

/D D

= =

= =

= =

(e)

,

BC DEC E

AC EFABC DEF

490

7by SAS


`

c+ +

/D D

= =

= =

= =

3. (a) B CBDA CDA

ADABD ACD

90is common

by AAS,

(base angles of isosceles )(given)

`

c+ +

+ +

/D D

D=

= =

(b) BD DCAD BCbisects

(corresponding sides in congruent s)`

`

D=

4. , )AB CDABD BDC ernate angles+ + <= (alt

( , )ADB DBCBD

ABD CDBAD BC

AD BCis common

by AAS,

alternate angles

(corresponding sides in congruent s)

`

`

+ +

/

<

D D

D

=

=

5. (a) OA OC= (equal radii)

OB OD= (similarly)

AOB COD+ += (vertically opposite angles)

AOB COD` /D D ( SAS )

(b) AB CD= (corresponding sides in congruent

triangles)

6. (a) AB AD= (given)

BC DC= (given)

AC is common

ABC ADC` /D D ( SSS )

(b) ABC ADC+ += (corresponding angles in congruent

triangles)


OB is common

AOB COB 90c+ += = (given)

OAB OBC` /D D ( SAS )

(b) OCB OBC+ += (base angles of OBC, an isosceles

right angled triangle)

But OCB OBC 90c+ ++ = (angle sum of triangle)

So OCB OBC 45c+ += =

Similarly 45OBA c+ =

45 45 90OBA OBC` c c c+ ++ = + =

So ABC+ is right angled

8. (a) 90AEF BDC c+ += = (given)

AF BC= (given)

FE CD= (given)

AFE BCD` /D D ( RHS )

(b) AFE BCD+ += (corresponding angles in

congruent triangles)


OB is common

AB BC= (given)

OAB OBC` /D D ( SSS )

(b) OBA OBC+ += (corresponding angles in

congruent triangles)

But 180OBA OBC c+ ++ = ( ABC is a straight angle)

So 90OBA OBC c+ += =

OB is perpendicular to AC.

10. (a) AD BC= (given)

ADC BCD 90c+ += = (given) DC is common ADC BCD` /D D ( SAS )

(b) AC BD= (corresponding sides in congruent

triangles)

Exercises 4.5

1. (a) .x 15 1= (b) 4.4x = (c) 6.6m =

(d) , ,76 23 81c c ca i b= = = (e) 4.5b =

(f) , , .x y115 19 3 2c ca = = = (g) 9.7p =

2. . , .a b1 81 5 83= =

3. ( , )

( )

BAC EDCABC DECACB ECD

AB EDalternate angles(similarly)vertically opposite angles

+ ++ ++ +

<=

=

=

since 3 pairs of angles are equal, | CDED||ABCD

Answer S1-S5.indd 552 7/31/09 1:36:10 PM

553ANSWERS

4.

.

..

..

.

GFE EFD

EFGF

DFEF

EFGF

DFEF

2 71 5

0 5

4 862 7

0 5

(given)

`

+ +=

= =

= =

=

o

o

Since two pairs of sides are in proportion and their included angles are equal, then | FGED||DEFD

5. ..

.

..

.

..

.

DEAB

DFAC

EFBC

DEAB

DFAC

EFBC

1 821 3

0 714

5 884 2

0 714

6 864 9

0 714

`

= =

= =

= =

= =

Since three pairs of sides are in proportion, | DEFD||ABCD

y 41c=

6. (a) OA OBOC OD

ODOA

OCOB

AOB COD

(equal radii)(similarly)


`

+ +

=

=

=

=

Since two pairs of sides are in proportion and their included angles are equal, | OCD3||OAB3

(b) 5.21AB cm=

7. (a) A+ is common

( , )ABC ADE

ACB AEDBC DEcorresponding angles

(similarly)+ ++ +

<=

=

since 3 pairs of angles are equal, | ADED||ABCD

(b) . , .x y2 17 2 25= =

8. ( , )( , )( )

ABF BECCBE BFA

C A

s AB CDBC AD

s

alternate anglesimilarlyangle sum of`

+ ++ ++ +

z

z

D

=

=

=

since 3 pairs of angles are equal, | CEBD||ABFD

9. A+ is common

..

..

ABAD

ACAE

ABAD

ACAE

31 2

0 4

20 8

0 4

`

= =

= =

=

Since two pairs of sides are in proportion and their included angles are equal, | , .ABC m 4 25D =||AEDD

10. .

.

..

..

.

CDAB

ACBC

ADAC

CDAB

ACBC

ADAC

2 62

0 769

3 93

0 769

5 073 9

0 769

`

= =

= =

= =

= =

Since three pairs of sides are in proportion,

,c| ,ACD x y109 47cD = =||ABCD

11. (a) 7.8x = (b) . , .m p4 0 7 2= = (c) 6.5x =

(d) . , .x y6 2 4 4= = (e) . , .x y1 4 9 2= =

12. (a) BCAB

DEAD

DEAD

FGAF

BCAB

FGAF

Also

`

=

=

=

(b)ACAB

AEAD

AEAD

AGAF

ACAB

AGAF

Also

`

=

=

=

(c) CEBD

AEAD

AEAD

EGDF

CEBD

EGDF

Also

`

=

=

=

13. . , .a b4 8 6 9= = 14. 0.98y = 15. . , .x y3 19 1 64= =

Exercises 4.6

1. (a) 6.4x = (b) 6.6y = (c) 5.7b = (d) 6.6m =

2. (a) 61p = (b) 58t = (c) 65x = (d) 33y =

3. .s 6 2 m= 4. .CE 15 3 cm=

5. 81, 144, 225AB CB CA2 2 2= = =

AB CB

CA

81 144225

2 2

2

+ = +

=

=

ABC` D is right angled

6. 1XY YZ= = XYZ` D is isosceles

,YZ XY XZYZ XY

XZ

1 21 12

2 2 2

2 2

2

= = =

+ = +

=

=

XYZ` D is right angled

7. AC AB BC

BCBC

BCBC

AC

BC

2 34 311

22 12

2 2 2

2 2 2

2

2

`

#

= +

= +

= +

=

=

=

=

=

^ h

8. (a) 5AC =

(b) , ,AC CDAD

25 144169

2 2

2

= =

=

25 144169

AC CD

AD

2 2

2

+ = +

=

=

ACD` D has a right angle at ACD+ AC` is perpendicular to DC

Answer S1-S5.indd 553 7/31/09 1:36:10 PM


9. AB b3= 10. xx y2 2+

11. d t tt t t t

t t

20 3 15 2400 120 9 225 60 413 180 625

2 2 2

2 2

2

= - + -

= - + + - +

= - +

] ]g g

12. 1471 mm 13. 683 m 14. 12.6 m 15. 134.6 cm

16. 4.3 m 17. 42.7 cm

18. 1.3 1.1 2.9 1.5 2.25and2 2 2+ = =

. . .1 3 1 1 1 52 2 2!+ so the triangle is not right angled the property is not a rectangle

19. No. The diagonal of the boot is the longest available space and it is only 1.4 m.

20. (a) 6 4BC2 2 2= - 20= 20BC = 6AO cm= (equal radii) So 6 4AC2 2 2= - 20= 20AC = Since ,BC AC= OC bisects AB

(b) OCA OCB 90c+ += = (given) OA OB= (equal radii) OC is common OAC OBC` /D D ( RHS ) So AC BC= (corresponding sides in congruent triangles) OC bisects AB

Exercises 4.7

1. (a) x 94c= (b) y 104c= (c) x 111c= (d) x 60c= (e) y 72c= (f) °, °x y102 51= = (g) °, °x y43 47= =

2. ABED is isosceles.

( s )

( )

B ECBE DEB

76180 76104

base equal

straight s

` cc cc

+ ++ +

+

+

= =

= = -

=

D

DD

62 104 104 360270 360

90

(angle sum of quadrilateral)c c c cc c

c

++

+

+ + + =

+ =

=

CD is perpendicular to AD

3. (a)

( )( , )

( , )

( , )

D x

C x

xx

A C xB xB D x

A D AB DC

C D AD BC

B C AB DC

180

180 180

180 180

180180

and cointerior angles

and cointerior angles

and cointerior angles`

`

c

c c

c c

cc

+

+

+ +++ +

+ +

+ +

+ +

<

<

<

= -

= - -

= - +

=

= =

= -

= = -

(b) x x x x180 180360

Angle sum c cc

= + + - + -

=

4. ,a b150 74c c= =

5. (a) 5 , 3 , 108 , 72a b x z ym m c c= = = = = (b) , ,x y z53 56 71c c c= = = (c) 5 , 68x y cm ca b= = = =

(d) , ,121 52 77c c ca b i= = = (e) 60x c= (f) ,x y3 7= =

6. ( ), ),

ADB CDBCDB ABDADB DBCABD DBC

BD ABC

BD ADCAB DCAD BC

bisects

bisects(alternate angles(alternate angles )

`

`

+ ++ ++ ++ +

+

+

<

<

=

=

=

=

7. (a) ..

AD BCAB DC

3 85 3

cmcm

(given)(given)

= =

= =

Since two pairs of opposite sides are equal, ABCD is a parallelogram.

(b) AB DCAB DC

7cm (given)

(given)<

= =

Since one pair of opposite sides is both equal and parallel, ABCD is a parallelogram.

(c) 54 126180

X M c cc

+ ++ = +

=

These are supplementary cointerior angles. XY MN` <

XM YNAlso, (given)<

XMNY is a parallelogram

(d) AE ECDE EB

56

cmcm

(given)(given)

= =

= =

Since the diagonals bisect each other, ABCD is a parallelogram.

8. (a) ,x 5 66cm ci= = (b) , ,90 25 65c c ca b c= = = (c) ,x y3 5cm cm= = (d) ,x y58 39c c= = (e) x 12 cm=

9. 6.4 cm 10. 59 , 31 , 59ECB EDC ADEc c c+ + += = =

11. 4 2 cm 12. 57x y c= =

Exercises 4.8

1. (a) 540c (b) 720c (c) 1080c (d) 1440c (e) 1800c (f) 2880c 2. (a) 108c (b) 135c (c) 150c (d) 162c (e) 156c 3. (a) 60c (b) 36c (c) 45c (d) 24c

4. 128 34c l 5. (a) 13 (b) 152 18c l 6. 16 7. 3240c

8. 2340c 9. 168 23c l

10. ( )

.

n nn n

nn

145145 180 360

3510 3

2 180Sum # c= = -

= -

=

=

But n must be a positive integer. no polygon has interior angles of 145 .c

11. (a) 9 (b) 12 (c) 8 (d) 10 (e) 30

12. (a) ABCDEF is a regular hexagon. AF BC= (equal sides) FE CD= (equal sides) AFE BCD+ += (equal interior angles) AFE BCD` /D D ( SAS )

Answer S1-S5.indd 554 7/31/09 1:36:11 PM

555ANSWERS

(b) ( )

S n

AFE

6720

6720

120

2 1802 180#

#

cc

cc

c

+

= -

= -

=

=

=

] g

Since ,AF FE= triangle AFE is isosceles. So FEA FAE+ += (base angles in isosceles triangle)

FEA2

180 120

30

`c

c

+ =-

=

(angle sum of triangle)

EDA 120 3090

cc

+ = -

=

Similarly, DEB 90c+ =

So ED DEA B 180c+ ++ = These are supplementary cointerior angles AE BD` <

13. A regular octagon has equal sides and angles. AH AB= (equal sides)

GH BC= (equal sides) AHG ABC+ += (equal interior angles)

AHG ABC` /D D ( SAS )

So AG AC= (corresponding sides in congruent triangles)

( )S n

81080

2 1802 180#

#

cc

c

= -

= -

=

] g

AHG

81080

135

`c

c

+ =

=

HGA HAG+ += (base angles in isosceles triangle)

HAG2

180 135

22 30

`c

c

+ =-

= l


GAC 135 2 22 30

90# c

c+ = -

=

l

We can similarly prove all interior angles are 90c and adjacent sides equal . So ACEG is a square .

14. EDC5

5

108

2 180# c

c

+ =-

=

] g

ED CD= (equal sides in regular pentagon)

So EDC is an isosceles triangle. DEC ECD`+ += (base angles in isosceles triangle)

36

DEC2

180 108c

c

+ =-

=


108 3672

AEC cc

+ = -

=

Similarly, using triangle ABC , we can prove that 72EAC c+ = So EAC is an isosceles triangle. (Alternatively you could prove EDC and ABC congruent triangles and then AC EC= are corresponding sides in congruent triangles.)

15. (a) p

360

(b) Each interior angle:

180360

180 360

180 360

180 2

p

p

p

p

p

p

p

p

-

= -

=-

=-^ h

Exercises 4.9

1. (a) .26 35 m2 (b) .21 855 cm2 (c) .18 75 mm2 (d) 45 m2 (e) 57 cm2 (f) 81 m2 (g) .28 27 cm2 2. .4 83 m2

3. (a) .42 88 cm2 (b) .29 5 m2 (c) .32 5 cm2 (d) .14 32 m2 (e) .100 53 cm2 4. (a) 25 m2 (b) .101 85 cm2 (c) .29 4 m2 (d) .10 39 cm2 (e) 45 cm2

5. 7 51 98 7 51 14 cm2+ = +^ h 6. .22 97 cm2

7. $621.08 8. (a) .161 665 m2 (b) 89 m2 (c) 10.5 m

9. (a) 48 cm (b) 27 cm 10. w12 units2

Test yourself 4

1. (a) , ,x y z43 137 147c c c= = = (b) 36x c= (c) , ,a b c79 101 48c c c= = = (d) 120x c= (e) 7.2r cm= (f) 5.6 , 8.5x ycm cm= = (g) 45ci =

2. )AGF HGB(vertically opposite+ +i=

AGF CFESo+ + i= =

These are equal corresponding .s+ AB CD` <

3. 118.28 cm 2

4. (a)

( )

DAE BACADE ABCAED ACB

ABC ADE AAAand are similar

(common)(corresponding angles, DE BC)(similarly)

`

+ ++ ++ +

<

D D

=

=

=

(b) 3.1 , 5.2x ycm cm= =

5. 162c 6. 1020.7 cm 3 7. 36 m

8. (a) AB ADBC DC

(adjacent sides in kite)(similarly)

=

=

AC is common Δ ABC and Δ ADC are congruent (SSS)

(b) AO COBO DO

AOB COD

(equal radii)(similarly)(vertically opposite angles)+ +

=

=

=

Δ AOB and Δ COD are congruent (SAS)

9. 73.5 cm 2

10. 6 2 7 36 28 64 82 2 2+ = + = =^ h ` Δ ABC is right angled (Pythagoras)

Answer S1-S5.indd 555 7/31/09 1:36:11 PM


11. AGAF

AEAD

AEAD

ACAB

AGAF

ACAB

(equal ratios on intercepts)

(similarly)

`

=

=

=

12. (a) (base s of isosceles+ D)( , )

AB ACB C

BD DC AD BC

(given)

bisects given+ +

=

=

=

ABD ACD SAS` /D D ] g

(b) 180

ADB ADCADB ADCBut

(corresponding s in congruent s)(straight )c

+ ++ +

+

+

D=

+ =

So 90ADB ADC c+ += =

So AD and BC are perpendicular.

13.

34˚ 34

( )( )

ACBCAD

CAD ADC

6868 34

base s of isoscelesexterior of

`

cc cc

c

++

+ +

+

+

D

D

=

= -

=

= =

So Δ ACD is isosceles base s equal+^ h

14.

( , )

, )DAC ACBBAC ACD

AD BCs AB DC

alternate s(alternate

+ ++ +

+

+

<

<

=

=

AC is common

ABC ADC

AB DC(AAS)


`

/D D

D=

Similarly, AD BC= opposite sides are equal

15. (a) 24 cm 2 (b) 5 cm 16. 9

17. BFG FGD x x109 3 3 71180c cc

+ ++ = - + +

=

These are supplementary cointerior .s+ AB CD` <

18. 57 cm 2

19. (

(( )

)

)

ACB A Bx y

ACD ACBz x y

x yx y

180180180180 180180 180

sum of

straight

cccc cc c

+ + +

+ +

+

+

D= - +

= - -

= -

= - - -

= - + +

= +

] g

20. (a)

..

.

..

.

A E

EFAC

DEAB

EFAC

DEAB

2 72 97

1 1

3 63 96

1 1

given

`

+ +=

= =

= =

=

^ h

So Δ ABC and Δ DEF are similar (two sides in proportion, included s+ equal).

(b) 4.3x cm=


1. 94c 2. , ,x y z75 46 29c c c= = = 3. ,1620 32 44c c l

4. , )

( )

BAD DBCABD BDCADB DCB

AB DC(given)(alternate anglesangle sum of`

+ ++ ++ +

<

D

=

=

=

since 3 pairs of angles are equal, BCDD;ABD <D

6.74d cm=

5. AB DCA D 131 49

180

(given)c cc

+ +=

+ = +

=

A+ and D+ are supplementary cointerior angles AB DC` <

Since one pair of opposite sides are both parallel and equal, ABCD is a parallelogram.

6. .27 36 m2

7.

Let ABCD be a square with diagonals AC and BD and

D

AD DC90

(adjacent sides of square)c+ =

=

°

°°

ADCDAC DCA

DAC DCADAC DCABAC BCA

904545

is isosceles

Similarly,

(base angles of isosceles )(angle sum of )

(other angles can be proved similarly)

`

`

`

+ ++ +

+ ++ +

D

D

D

=

+ =

= =

= =

8.

Let ABCD be a kite

AD ABDC BC

(given)(given)

=

=

AC is common

, ADC ABCDAC BAC

AD ABDAE BAE

by SSS

(corresponding angles in congruent s)(given)(found)

`

` + +

+ +

/D D

D

=

=

=

AE is common

Answer S1-S5.indd 556 8/1/09 8:50:48 PM

557ANSWERS

,(

( )

ADE ABEDEA BEADEA BEADEA BEA

DEB18090

by SAS

But

the diagonals are perpendicular

corresponding angles in congruent s)is a straight angle

`

`

`

`

cc

+ ++ ++ +

/D D

D=

+ =

= =

9. 84 (15 112 ) )

( )

MNYMNY

XYZXYZMNY XYZ

MNZ

XYZ43

69 11243

43

(exterior angle of

exterior angle of`

`

`

c c cc

c cc

c

++

+++ +

D

D

+ = +

=

+ =

=

= =

These are equal corresponding angles. MN XY` <

10. .x 2 12 m= 11. (a) 6 m2 (b) 10 2 5 2 5 5 m+ = +^ h

12. . , .x y28 7 3 8cm cm== 13. 7.40 , 4.19x ym m= =

14. (a) AB BCABE CBE 45

(adjacent sides in square)

(diagonals in square make 45 with sides)c

+ +=

= =

EB is common.

, ABE CBE

AE CEby SAS


`

/D D

D=

Since AB BC= and ,AE CE= ABCE is a kite.

(b) BD x x

xx

DE BD

x

22

21

22

units

2 2

2

= +

=

=

=

=

Practice assessment task set 1

1. 9p = 2. 2 5 y x y+ -^ ^h h 3. (a) x 1- (b) 3x4

4. 6 10y - 5. 23

25 5 2+ 6. 2 16 3x x x3 2+ - +

7. 72

x = 8. 3

2x -

9. °ABC EDCACB ECD

AB EDABC EDC

AC ECACE

90

by AAS

is isosceles

(given)(vertically opposite angles)(given)

(corresponding sides in congruent triangles)`

`

`

+ ++ +

/D D

D

= =

=

=

=

10. 231.3 11. 3- 12. 135c 13. 7.33 10 2#

-

14. 3 10 4- 15. 3.04 16. 3x + 17. . , .x 1 78 0 281= -

18. 1.55r = 19. x 12

20. 157

21. x2

42 3

12!!= = 22.

491

23. 4, 11 1, 4x y x yor= = = - = - 24. ,x y2 1= = -

25. 7 26. 7.02 cm 27. 2 1 4 2 1x x x2- + +] ^g h

28. 43

6 15 2 6+ 29. 7 30. $643.08 31. 1.1

32. 2 10 3 5 2 2 3- + - + 33. $83.57

34. , ,x y w z22 29 90c c c= = = = 35. 56.7 cm2

36. a ba

b21 10

21

10

=- 37. ,x x6 252

2 1 - 38. 81

39. x 7- - 40. 41

x = 41. ,x x3 3# $- 42. 61

43. Given diagonal AC in rhombus ABCD :

)

)

AB BCDAC ACBBAC ACBDAC BAC

AD BCABC

(adjacent sides in rhombus)(alternate s,(base s of isosceles

`

+ ++ ++ +

+

+

<

D

=

=

=

=

` diagonal AC bisects the angle it meets. Similarly, diagonal BD bisects the angle it meets.

44. x 3 1+ -] g 45. 6 12 8x x x3 2+ + + 46. 2

517

4

47. 53x c=

48. ,x y98 41c c= = 49. 3 2

1

x +

50. (a) 12 8x y- (b) 2 31 (c) 3 9

3

x x

x2 - +

- (d) 3 2 1+

(e) 1 1

5

x x

x

+ -

- +

] ]

]

g g

g (f)

611 3

(g) x y zx z

y14 7 11

14 11

7

=- -

(h) 5 1 2

3a a b b+ +] ]g g

(i) 8 5 (j) 1321

51. . , .x y2 7 3 1= = 52. 25x = 53. r2

cm3 r

=

54. 17.3 cm

55. DEA xEAD xCD x x

xABC xABC DEA

A222

LetThen (base s of isosceles )

(exterior of )

(opposite s of gram are equal)

EAD

`

`

+++

++ +

+

+

+ <

D

D

=

=

= +

=

=

=

56. 52

57. 5% 58. 2.2 10 kmh8 1#

- 59. 20k =

60. 9xy y 61. 147 16c l 62. 5.57 m2

63. (a) a b a a ab b b5 2 2 4 2 4 4 42 2+ - - - + + +] ^g h

(b) 3 4 6 2a b a b c+ - +] ]g g

Answer S1-S5.indd 557 7/31/09 1:36:13 PM


64. x181

543

1#-

65. (BCEF is a gram)<

(BC AD ABCDBC FEAD FE

is a gram)

`

< <

<

<

BC ADBC FEAD FE

Also opposite sides of gram

similarly`

<=

=

=

^

^

h

h

Since AD and FE are both parallel and equal, AFED is a parallelogram.

66. 11.95b m= 67. (a) 34 cm (b) 30 cm 2

68. 75

18 3 31 2 25 5+ - 69. 20 70. 32 m

71. BD bisects AC So AD DC= 90BDC BDA c+ += = (given) BD is common BAD BCD` /D D ( SAS ) AB CB` = (corresponding sides in congruent

triangles) So triangle ABC is isosceles

72. 2

x y2 2+ 73. (b) 74. (c) 75. (a) 76. (b) 77. (b)

78. (d) 79. (d)

Chapter 5 : Functions and graphs

Exercises 5.1

1. Yes 2. No 3. No 4. Yes 5. Yes 6. Yes 7. No

8. Yes 9. Yes 10. No 11. Yes 12. No 13. Yes

14. No 15. Yes

Exercises 5.2

1. 4, 0f f1 3= - =] ]g g 2. , ,h h h0 2 2 2 4 14= - = - =] ] ]g g g

3. 25, 1, 9, 4f f f f5 1 3 2= - - = - = - - = -] ] ] ]g g g g 4. 14

5. 35- 6. 9x = 7. x 5!= 8. x 3= - 9. ,z 1 4= -

10. 2 9, 2 2 9f p p f x h x h= - + = + -^ ]h g

11. 1 2g x x2- = +] g 12. f k k k k1 12= - + +] ] ^g g h

13. ; ,t t1 2 4= - = - 14. 0

15. 125, 1, 1f f f5 1 1= = - = -] ] ]g g g

16. 0 4 1 3f f f2 2 1- - + - = - + = -] ] ]g g g

17. 10 18. 7 19. 28-

20. (a) 3 (b) 3 3 3 0x - = - = Denominator cannot be 0 so the function doesn’t exist for .x 3= (c) 4

21. 2 5f x h f x xh h h2+ - = + -] ]g g 22. 4 2 1x h+ +

23. x c5 -] g 24. 3 5k2 + 25. (a) 2 (b) 0 (c) 2n n4 2+ +

Exercises 5.3

1. (a) x -intercept 32

, y -intercept -2

(b) x -intercept -10, y -intercept 4 (c) x -intercept 12, y -intercept 4 (d) x -intercepts 0, -3, y -intercept 0 (e) x -intercepts 2! , y -intercept -4 (f) x -intercepts -2, -3, y -intercept 6 (g) x -intercepts 3, 5, y -intercept 15

(h) x -intercept 53- , y -intercept 5 (i) x -intercept -3, no y -intercept (j) x -intercept ,3! y -intercept 9

2. 2

( )

f x xxf x

2

even function

2

`

- = - -

= -

=

2] ]g g

3. (a) 1f x x2 6= +^ h (b) f x x x2 12 6 3= + +] g7 A

(c) 1f x x3- = - +] g (d) Neither odd nor even

4.

( )

g x x x xx x xg x

3 23 2

even function

8 4 2

8 4 2

`

- = - + - - -

= + -

=

] ] ] ]g g g g

5. f x x f x- = - = -] ]g g odd function

6. 1

( )

f x xxf x

1

even function

2

2

`

- = - -

= -

=

] ]g g 7. f x x xx xx x

f x

444

odd function

3

3

3

`

- = - - -

= - +

= - -

= -

] ] ]

^

]

g g g

h

g

8. f x x xx xf x

even function

4 2

4 2

`

- = - + -

= +

=

] ] ]

]

g g g

g

0f x f x- - =] ]g g

9. (a) Odd (b) Neither (c) Even (d) Neither (e) Neither

10. (a) Even values i.e. , , ,n 2 4 6 f=

(b) Odd values i.e. , , ,n 1 3 5 f=

11. (a) No value of n (b) Yes, when n is odd (1, 3, 5, …)

12. (a) (i) x 02 (ii) x 01 (iii) Even

(b) (i) x 21 (ii) x 22 (iii) Neither

(c) (i) x2 21 1- (ii) ,x x2 21 2- (iii) Neither

(d) (i) All real x 0! (ii) None (iii) Odd

(e) (i) None (ii) All real x (iii) Neither

Exercises 5.4

1. (a) x -intercept 2, y -intercept -2

(b) x -intercept 121

- , y -intercept 3

(c) x -intercept 21

, y -intercept 1

(d) x -intercept -3, y -intercept 3

(e) x -intercept 32

, y -intercept 31

-

Answer S1-S5.indd 558 7/31/09 1:36:13 PM

559ANSWERS

2. (a)

-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

(b) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

(c) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

(d) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2-1

1

(e) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2-1

112

(f) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2-1

1

(g) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2-1

1

23

-

(h) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

Answer S1-S5.indd 559 7/31/09 1:36:14 PM


(i) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

(j) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-1111

2

3. (a) ,x yall real all real" ", , (b) :,x y y 2all real =" ", , (c) : ,x x y4 all real= -! "+ , (d) : ,x x y2 all real=! "+ , (e) , :x y y 3all real =! "+ ,

4. (a) Odd (b) Even (c) Neither (d) Odd (e) Odd

5. y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-1111

2

(3, -1)

Exercises 5.5

1. (a) x -intercepts 0, -2, y -intercept 0 (b) x -intercepts 0, 3, y -intercept 0 (c) x -intercepts ! 1, y -intercept -1 (d) x -intercepts -1, 2, y -intercept -2 (e) x -intercepts 1, 8, y -intercept 8

2. (a) y

x-4

-5

-3 -2-1 2 3 4 5

2

1

3

4

5

6

-3-4

-2-1

1

(b) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

6

-3

-4

-2

-11

(c) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

6

-3

-4

-2-1

1

Answer S1-S5.indd 560 7/31/09 1:36:15 PM

561ANSWERS

(d) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

6

-3

-4

-2

-1 1

(e) y

x-4

-5

-3 -2 -1 2 3 4 5

21

3

4

5

6

-3

-4

-2

-11

(f) y

x-4

-10

-3 -2 -1 2 3 4 5

4

6

8

2

10

12

-6

-8

-4-2

1

(g) y

x-4

-5

-3 -2 -1 2 3 4 5

21

3

4

5

-3

-4

-6

-2

-11

(h) y

x

-5

-3-4 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-6

-2

-1 1

(i) y

x-4

-5

-3 -2 -1 3 4 5

2

1

3

4

5

-3

-4

-6

-2-1 2111

2

(j) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3-4

-6

-2-1

1

3. (a) (i) x -intercepts 3, 4, y -intercept 12 (ii) {all real x },

:y y41

$ -( 2

(b) (i) x -intercepts 0, -4, y -intercept 0 (ii) {all real x }, :y y 4$ -" ,

(c) (i) x -intercepts -2, 4, y -intercept -8 (ii) {all real x }, : 9y y $ -" ,

(d) (i) x -intercept 3, y -intercept 9 (ii) {all real x }, :y y 0$" ,

(e) (i) x -intercepts ,2! y -intercept 4 (ii) {all real x }, :y y 4#" ,

4. (a) {all real x }, :y y 5$ -" , (b) {all real x }, :y y 9$ -" ,

Answer S1-S5.indd 561 7/31/09 1:36:16 PM


(c) {all real x }, :y y 241

$ -( 2 (d) {all real x }, :y y 0#" ,

(e) {all real x }, : 0y y $" ,

5. (a) y0 9# # (b) y0 4# # (c) y1 24# #-

(d) y4 21# #- (e) y18 241

# #-

6. (a) (i) x 02 (ii) x 01 (b) (i) x 01 (ii) x 02

(c) (i) x 02 (ii) x 01 (d) (i) x 21 (ii) x 22 (e) (i) x 52 - (ii) x 51 -

7.

( )

f x xx

f xeven

2

2

`

- = - -

= -

=

] ]g g

8. (a) Even (b) Even (c) Even (d) Neither (e) Neither (f) Even (g) Neither (h) Neither (i) Neither (j) Neither

Exercises 5.6

1. (a) x -intercept 0, y -intercept 0 (b) No x -intercepts, y -intercept 7 (c) x -intercepts ,2! y -intercept -2 (d) x -intercept 0, y -intercept 0 (e) x -intercepts ,3! y -intercept 3 (f) x -intercept -6, y -intercept 6

(g) x -intercept 32

, y -intercept 2

(h) x -intercept 54

- , y -intercept 4

(i) x -intercept 71

, y -intercept 1

(j) No x -intercepts, y -intercept 9

2. (a) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(b) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(c) y

-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(d) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(e) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(f) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

Answer S1-S5.indd 562 7/31/09 1:36:16 PM

563ANSWERS

(g) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(h) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(i) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(j) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

3. (a) {all real x }, :y y 0$" ,

(b) {all real x }, :y y 8$ -" ,

(c) {all real x }, :y y 0$" ,

(d) {all real x }, :y y 3$ -" ,

(e) {all real x }, :y y 0#" ,

4. (a) (i) x 22 (ii) x 21 (b) (i) x 02 (ii) x 01

(c) (i) x 121

2 (ii) x 121

1 (d) (i) x 02 (ii) x 01

(e) (i) x 01 (ii) x 02

5. (a) 0 2y# # (b) y8 4# #- - (c) 0 6y# #

(d) 0 11y# # (e) y1 0# #-

6. (a) x 32 - (b) x 01 (c) x 92 (d) x 22 (e) x 21 -

7. (a) x 3!= (b) ,x x1 12 1 - (c) x2 2# #-

(d) ,x 1 3= - - (e) 3x = (f) ,x 1 2= (g) x3 51 1-

(h) x4 2# #- (i) ,x x4 02 1 (j) ,x x2 4# $

(k) x4 1# #- (l) ,x x0 1# $ (m) ,x 221

= -

(n) No solutions (o) 0x = (p) 1x = (q) ,x 0 2= -

(r) No solutions (s) 31

x = ( t) 0, 6x =

Exercises 5.7

1. (a) (i) {all real x : x ! 0}, {all real y : y ! 0} (ii) no y -intercept

(iii) y

x-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

(b) (i) {all real : },x x 0! {all real :y y 0! } (ii) no y -intercept

(iii) y

x-2 -1 2

2

1

-2

-1

1

Answer S1-S5.indd 563 7/31/09 1:36:17 PM


(c) (i) {all real :x x 1! - }, {all real : 0y y ! } (ii) 1

(iii) y

x-2 -1 2

2

1

-2

-1

1

(d) (i) {all real :x x 2! }, {all real : 0y y ! } (ii) 121

-

(iii) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(e) (i) {all real :x x 2! - }, {all real : 0y y ! } (ii) 61

(iii) y

x-2 -1 2

2

1

-2

-1

1

(f) (i) {all real :x x 3! }, {all real :y y 0! } (ii) 32

(iii) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(g) (i) {all real : 1x x ! }, {all real : 0y y ! } (ii) -4

(iii) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

(h) (i) {all real : 1x x ! - }, {all real : 0y y ! } (ii) -2

(iii) y

x-4

-5

-3 -2-1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

Answer S1-S5.indd 564 7/31/09 1:36:18 PM

565ANSWERS

(i) (i) :x x21

all real !' 1 , {all real : 0y y ! } (ii) 32

-

(iii) y

x-2 -1 2

2

1

-2

-1

1

23

-

12

(j) (i) {all real :x x 2! - }, {all real :y y 0! } (ii) -3

(iii) y

x-4

-5

-3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-2

-11

2.

( )

f x x

xf x

2

2

odd function`

- =-

= -

= -

] g

3. (a) 1y91

## (b) 1y31# # (c) y2

21

21

# #- -

(d) 3y73

## (e) 2 y81

# #- -

4. (a) 1 3x# # (b) 1 4x# # (c) 6 0x# #-

(d) 1 4x# # (e) 1 2x# #

Exercises 5.8

1. (a) (i) y

x-3

3

3

-3

(ii) : , :x x y y3 3 3 3# # # #- -! "+ ,

(b) (i) y

x-4

4

4

-4

(ii) : , :x x y y4 4 4 4# # # #- -! "+ ,

(c) (i)

(2, 1)

-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

y

x

Answer S1-S5.indd 565 7/31/09 1:36:18 PM


(ii) : 0 4 , : 1 3x x y y# # ## -! "+ ,

(d) (i)

-4

-5

-3 -2 -1 2 3 4

2

1

3

4

5

-3

-4

-2

-11

y

x

(ii) : , :x x y y4 2 3 3# # # #- -! "+ ,

(e) (i)

-4 -3 -2 -1 2 3 4

2

1

3

4

5

-2

-1

(-2, 1)

1

y

x

(ii) : , :x x y y3 1 0 2# # # #- -! "+ ,

2. (a) (i) Below x -axis

(ii) y

x-5 5

-5

(iii) : , :x x y y5 5 5 0# # # #- -! "+ ,

(b) (i) Above x -axis

(ii) y

x-1

1

1

(iii) : , :x x y y1 1 0 1# # # #-! "+ ,

(c) (i) Above x -axis

(ii) y

x-6

6

6

(iii) : , :x x y y6 6 0 6# # # #-! "+ ,

(d) (i) Below x -axis

(ii) y

x-8 8

-8

(iii) : , :x x y y8 8 8 0# # # #- -! "+ ,

Answer S1-S5.indd 566 7/31/09 1:36:19 PM

567ANSWERS

(e) (i) Below x -axis

(ii) y

x- 7

- 7

7

(iii) : , :x x y y7 7 7 0# # # #- -" #, -

3. (a) Radius 10, centre (0, 0) (b) Radius 5 , centre (0, 0)

(c) Radius 4, centre (4, 5) (d) Radius 7, centre (5, -6) (e) Radius 9, centre (0, 3)

4. (a) 16x y2 2+ =

(b) 6 4 12 0x x y y2 2- + - - =

(c) 2 10 17 0x x y y2 2+ + - + =

(d) 4 6 23 0x x y y2 2- + - - =

(e) 8 4 5 0x x y y2 2+ + - - =

(f) 4 3 0x y y2 2+ + + =

(g) 8 4 29 0x x y y2 2- + - - =

(h) 6 8 56 0x x y y2 2+ + + - =

(i) 4 1 0x x y2 2+ + - =

(j) 8 14 62 0x x y y2 2+ + + + =

Exercises 5.9

1. (a) {all real x }, {all real y } (b) {all real x }, {y: y = -4} (c) {x: x = 3}, {all real y } (d) {all real x }, { y : y $ -1 }

(e) {all real x }, {all real y } (f) {all real x }, : 1241

y y #' 1

(g) { : 8 8}, { : 8 8}x x y y# # # #- -

(h) {all real :t t 4! }, {all real ( ): ( )f t f t 0! }

(i) {all real : 0!z z }, {all real :g g 5!zz^ ^h h }

(j) {all real x }, { :y y 0$ }

2. (a) { x : 0x $ }, { y : y 0$ } (b) { x : x 2$ }, { y : y 0$ } (c) {all real x }, { y : y 0$ } (d) {all real x }, { y : y 2$ - }

(e) : 221

, { : }x x y y 0$ #-' 1

(f) {all real x }, { :y y 5# } (g) {all real x }, { : }y y 02

(h) {all real x }, { : }y y 01

(i) {all real :x x 0! }, {all real :y y 1! } (j) {all real :x x 0! }, {all real :y y 2! }

3. (a) ,x 0 5= (b) , ,x 3 1 2= - (c) , ,x 0 2 4=

(d) ,x 0 4!= (e) x 7!= 4. (a) x1 1# #-

(b) { : }x x1 1# #-

5. (a) { : , }x x x1 2# $- (b) { : , }t t t6 0# $-

6. (a) { y : y9 3# #- }

(b) { y : y0 9# # } (c) { y : y8 1# #- }

(d) :51

1y y# #' 1 (e) { y : 0 4y# # }

(f) { y : y1 15# #- } (g) { y : y1 0# #- }

(h) :y y1 8# #-" , (i) { y : 4 21y# #- }

(j) :y y61

64

# #-' 1

7. (a) {all real :x x 1! - }

(b) x -intercept: 0y =

01

3x

=+

0 3= This is impossible so there is no x -intercept (c) {all real :y y 0! }

8. (a) {all real :x x 0! } (b) {all real :y y 1!! }

9. (a) y

x-4 -3 -2 -1 2 3 4 5

10

5

15

20

25

-15

-10

-51

(b) y

x-4 -3 -2 -1 2 3 4

4

2

6

8

-6

-8

-4

-21

(c) y

x-4 -3 -2 -1 2 3 4 5

10

5

15

20

25

-15

-10

-51

Answer S1-S5.indd 567 7/31/09 1:36:20 PM


(d) y

x-4 -3 -2 -1 2 3 4

4

2

6

8

-6

-8

-4

-21

(e) y

x-4 -3 -2 -1 2 3 4

4

2

6

8

-6

-8

-4

-21

(f) y

x-10 10

10

-10

(g) y

x-1

1

2

3

-1

1

10. (a) : : 0x x y y1$ $" ", ,

(b) y

x2 3

2

1

-11

11. y

x-1

4

3

2

1

5

6

-1 1

12. (a) (i) {all real x }, {all real y } (ii) All x (iii) None (b) (i) {all real x }, :y y 22 -" , (ii) x 02 (iii) x 01 (c) (i) {all real :x x 0! }, {all real : 0y y ! } (ii) None (iii) All 0x ! (d) (i) {all real x }, {all real y } (ii) All x (iii) None (e) (i) {all real x }, :y y 02" , (ii) All x (iii) None

13. (a) 2 2x ##- (b) (i) { x : 2 2x# #- }, { y: 0 2y# # } (ii) { x : 2 2x# #- }, { y: 2 0y# #- }

Exercises 5.10

1. (a) 21 (b) 10- (c) 8 (d) 3 (e) 3 (f) 75 (g) 0

(h) 6- (i) 41

(j) 1 (k) 7- (l) 3x x2 -

(m) 2 3 5x x3 + - (n) 3c2

Answer S1-S5.indd 568 7/31/09 1:36:21 PM

569ANSWERS

2. (a) Continuous (b) Discontinuous at 1x = - (c) Continuous (d) Continuous (e) Discontinuous at x 2!=

3. (a)

(b)

(c)

Exercises 5.11

1. (a) y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-4

-2

-11

(b) y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-4

-2

-11

(c) y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-4

-2

-11

(d) y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-4

-2

-11

Answer S1-S5.indd 569 7/31/09 1:36:21 PM


(e)

y = x +1

y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-4

-2

-11

(f)

y = 2x-3

y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-2

-11

(g)

x + y = 1

y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-2

-11

-4

(h)

3x - y - 6 = 0

y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-2

-11

-4

-5

-6

(i)

x + 2y - 2 = 0

y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

6

-3

-2

-11

-4

-5

-6

(j)

x-4 -3 -2 -1 2 3 41

y

2

1

3

4

5

6

-3

-2

-1

-4

-5

-6

x =12

Answer S1-S5.indd 570 7/31/09 1:36:22 PM

571ANSWERS

2. (a) x 32 - (b) y 2$ - (c) y x 1$ + (d) y x 422 -

(e) y 2x$

3. (a) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-2

-11

-4

-5

y = x2 - 1

(b)

-3 3

3

-3

y

x

(c) y

x-1 1

1

-1

(d)

x-3-4 -2 -1 2 3 4 51

y = x 2

y

1

2

3

4

5

-3

-2

-1

-4

-5

(e) y

x-4 -3 -2 -1 2 3 4

4

2

6

8

-6

-8

-4

-21

y = x3

4. (a) y x3 21 - (b) y x 222 +

(c) x y 492 21+

(d) x y 812 22+

(e) ,x y5 21 2

5. (a) y

x-4 -3 -2 -1 2 3 4

3

1

2

4

5

-2

-11

Answer S1-S5.indd 571 7/31/09 1:36:23 PM


(b) y

x-4 -3 -2 -1 2 3 4

3

1

2

4

5

-2

-11

(c) y

x-4-5 -3 -2 -1 2 3 4

3

1

2

4

5

-2

-11

6. (a) y

x-4 -3 -1-2 2 3 4

3

1

2

4

5

6

-2

-3

-4

-11

(b) y

x-4 -3 -1-2 2 3 4

3

1

2

4

5

6

-2

-3

-4

-5

-6

-11

y = x - 3

(c) y

x-4 -3 -1-2 2 3 4

3

1

2

4

5

6

-2

-3

-4

-5

-11

y = 3x – 5

-6

(d) y

x-4 -3 -1-2 2 3 4

3

1

2

4

5

6

-2

-3

-4

-5

-6

-11

y = x + 1

y = 3 – x

(e) y

x-3 3

3

-3

y = 1

Answer S1-S5.indd 572 8/1/09 8:13:02 PM

573ANSWERS

(f) y

x-1-2 2

1

2

-2

x = – 1

(g) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-2

-11

-4

-5

y = x2

y = 4

(h) y

x-4 -3 -2 -1 2 3 4

4

2

6

8

-6

-4

-21

-8

y = x3

y = 3

x = -2

(i) y

x-1 1

1

1

-1

(j) y

x-4 -3 -1-2 2 3 4

3

1

2

4

5

6

-2

-3

-4

-5

-6

-11

x - y = 2

x - y = -1

7. (a) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-2

-11

-4

-5

y = x2

Answer S1-S5.indd 573 7/31/09 1:36:25 PM


(b) y

x-4 -3 -2 -1 2 3 4

4

2

6

8

-6

-4

-21

-8

y = x3

y = 1

(c) y

x1-2 2

2

-2

x = 1

(d)

1-1 2 3 4

1

2

-2

y

x

y =2x

(e)

-1 2 3-2-3-4 1 4

1

2

-1

-2y =

1x + 2

x

y

8. (a)

x2 3 4 51-1-3-4 -2

y

y = x2

y = 5

x = 2

3

2

1

4

5

-2

-1

-3

-4

-5

(b)

x2 3 41-1-3-4 -2

y

x = 3

y = -1

y = x - 2

3

2

1

4

5

6

-2

-1

-3

-4

-5

-6

Answer S1-S5.indd 574 7/31/09 1:36:26 PM

575ANSWERS

(c)

x2 3 41-1-3-4 -2

y

y = 2x + 1

2x - 3y = 6

3

2

1

4

5

6

-2

-1

-3

-4

-5

-6

(d)

-3 3

3

-3

x

x = -3

y = 2

y

(e)

x2 3 41-1-3-4 -2

y

y = 3

y = |x |

x = 2

3

2

1

4

5

6

-2

-1

-3

Test yourself 5

1. (a) f 2 6- =] g (b) f a a a3 42= - -] g (c) ,x 4 1= -

2. (a)

(b)

(c)

(d)

(e)

(f)

Answer S1-S5.indd 575 7/31/09 1:36:26 PM


(g)

(h)

3. (a) Domain: all real x ; range: y 641

$ -

(b) Domain: all real x ; range: all real y (c) Domain: 1 1;x# #- range: 1 1y# #- (d) Domain: 1 1;x# #- range: 0 1y# # (e) Domain: 1 1;x# #- range: 1 0y ##- (f) Domain: all real ;x 0! range: all real y 0! (g) Domain: all real x ; range: all real y (h) Domain: all real x ; range: y 0$

4. 15 5. (a) 4 (b) 5 (c) 9 (d) 3 (e) 2

6.

7.

8.

9.

10.

11. (a) y 3# (b) y x 22 + (c) ,y x y 02$ #-

12. (a) Domain: all real ,x 3! range: all real y 0!

(b)

13. (a)

(b) (i) ,x 2 4= - (ii) 4 2x 11- (iii) ,x x2 42 1 -

14. (a) 2 (b) 332

x = (c) 131

15. (a) x -intercept ,10- y -intercept 4 (b) x -intercepts , ,2 7- y -intercept 14-

16. (a) i (b) iii (c) ii (d) i (e) iii

Answer S1-S5.indd 576 7/31/09 1:36:28 PM

577ANSWERS

17. (a) 4 (b) 52

(c) 121

- (d) 3

18.

19. (a) Domain: 2,x $ range: 0y $

(b)

20. (a) ( ) 3 1( )

( )

f x x xf x x x

x xf x

3 13 1

4 2

4 2

4 2

= + -

- = - + - -

= + -

=

] ]g g

So f x] g is even.

(b) ( )( ) ( )

( )( )

f x x xf x x x

x xx x

f x

3

3

3

3

= -

- = - - -

= - +

= - -

= -

] g

So f x] g is odd.


1. ,b32

3= -

2.

3.

4.

5. , ,f f f3 9 4 16 0 1= - = =] ] ]g g g

6. Domain: all real ;x 1!! range: ,y y1 02# -

7.

Answer S1-S5.indd 577 7/31/09 1:36:29 PM


8. Domain: ;x 0$ range: y 0$ 9. , ,x 0 3 2= -

10.

11. h h h2 1 0 3 0 1 2+ - - = - + - - = -] ] ] ]g g g g

12.

13.

14. ( ) ( )

( )

f a aa

f a

2 12 1

2 2

2

2

- = - -

= -

=

^ h

15. x4

1 41!=

16. (a) 2

2

x

xx

x

xx

xx

xx

x

31

32 3

31

32 6 1

32 7

32 7

31

RHS

LHS

`

= ++

=+

++

+

=+

+ +

=+

+

=

+

+= +

+

] g

(b) Domain: all real ;x 3! - range: all real y 2!

(c)

17.

18.

19. Domain: ;x 3$ range: 0y $ 20. Domain: x2 2# #-

21.

Answer S1-S5.indd 578 7/31/09 1:36:30 PM

579ANSWERS

Chapter 6: Trigonometry

Exercises 6.1

1. , ,cos sin tan135

1312

512

i i i= = =

2. , ,sin cot sec54

43

35

b b b= = =

3. , ,sin tan cos74

757

74

5b b b= = =

4. , ,cos tan cosecx x x95

556

56

9= = =

5. ,cos sin53

54

i i= =

6. , ,tan sec sin25

23

35

i i i= = =

7. ,cos tan635

35

1i i= =

8. ,tan sin751

1051

i i= =

9. (a) 2 (b) 45c

(c) , ,sin cos tan452

145

2

145 1c c c= = =

10. (a) 3 (b) , ,sin cos tan3021

3023

303

1c c c= = =

(c) , ,sin cos tan6023

6021

60 3c c c= = =

11. .sin cos67 23 0 92c c= = 12. .sec cosec82 8 7 19c c= =

13. .tan cot48 42 1 11c c= = 14. (a) 2 61 2 29cos sinorc c

(b) 0 (c) 0 (d) 1 (e) 2

15. 80x c= 16. 22y c= 17. 31p c= 18. 25b c=

19. 20t c= 20. 15k c=

Exercises 6.2

1. (a) 47c (b) 82c (c) 19c (d) 77c (e) 52c

2. (a) 47 13c l (b) 81 46c l (c) 19 26c l

(d) 76 37c l (e) 52 30c l

3. (a) 77.75c (b) 65.5c (c) 24.85c

(d) 68.35c (e) 82.517c

4. (a) 59 32c l (b) 72 14c l (c) 85 53c l

(d) 46 54c l (e) 73 13c l

5. (a) 0.635 (b) 0.697 (c) 0.339 (d) 0.928 (e) 1.393

6. (a) 17 20c l (b) 34 20c l (c) 34 12c l

(d) 46 34c l (e) 79 10c l

Exercises 6.3

1. (a) 6.3x = (b) 5.6y = (c) 3.9b = (d) 5.6x = (e) 2.9m = (f) 13.5x = (g) 10.0y = (h) 3.3p = (i) 5.1x = (j) 28.3t = (k) 3.3x cm= (l) 2.9x cm= (m) 20.7x cm= (n) 20.5x mm= (o) 4.4y m= (p) 20.6k cm= (q) 17.3h m= (r) 1.2d m= (s) 17.4x cm= (t) 163.2b m=

2. 1.6 m 3. 20.3 cm 4. 13.9 m

5. (a) 18.4 cm (b) 13.8 cm 6. 10 cm and 10.5 cm

7. 47.4 mm 8. 20.3 m 9. (a) 7.4 cm (b) 6.6 cm (c) 9.0 cm

10. (a) 6.8 cm (b) 6.5 cm 11. 38 cm

Exercises 6.4

1. (a) x 39 48c= l (b) 35 06ca = l (c) 37 59ci = l (d) 50 37ca = l (e) 38 54ca = l (f) 50 42cb = l (g) x 44 50c= l (h) 3 10 5ci = l (i) 29 43ca = l (j) 45 37ci = l (k) 57 43ca = l (l) 43 22ci = l (m) 37 38ci = l (n) 64 37ci = l (o) 66 16cb = l (p) 29 56ca = l (q) 54 37ci = l (r) 35 58ca = l (s) °59 2i = l (t) 56 59cc = l

2. 37 57c l 3. 22 14c l 4. 36 52c l 5. 50c

6. (a) 11.4 cm (b) 37 52c l 7. ,31 58 45 44c ca b= =l l

8. (a) 13 m (b) 65 17c l 9. (a) 11 19c l (b) 26 cm

10. 4.96 cm and 17.3 cm 11. (a) 12.9 m (b) 56 34c l

Exercises 6.5

1. (a)

100c

Boat

Beachhouse

North

AnswerS6.indd 579 7/31/09 11:07:53 AM


(b)

320c

Campsite

Jamie

North

(c)

200c

Seagull

Jetty

North

(d)

50c

Alistair

Bus stop

North

(e)

B Hill285c

Plane

North

(f)

12c

Dam

FarmhouseNorth

(g)

160cHouse

Mohammed

North

(h)

80c

Town

Mine shaft

North

(i)

349cSchool

YvonneNorth

AnswerS6.indd 580 7/31/09 11:07:54 AM

581ANSWERS

(j)

Island

Boat ramp

280c

North

2. (a) 248c (b) 145c (c) 080c (d) 337c (e) 180c

3. 080c 4. 210c 5. 160c 6. 10.4 m

7. 21 m 8. 126.9 m 9. 72 48c l

10. (a) 1056.5 km (b) 2265.8 km (c) 245c

11. 83.1 m 12. 1.8 km 13. 12 m 14. 242c 15. 035c

16. 9.2 m 17. 171 m 18. 9.8 km 19. 51 41c l 20. 2.6 m

21. 9 21c l 22. 1931.9 km 23. 34.6 m 24. 149c

25. 198 m 26. 4.8 km 27. 9.2 m 28. 217c

29. (a) 1.2 km (b) 7.2 km 30. (a) 13.1 m (b) 50 26c l

Exercises 6.6

1. (a) 2

3 1+ (b) 1 (c) 2 (d) 4 (e)

34 3

(f) 3

2 3

(g) 141

(h) 4

6 24

2 3 1+=

+^ h (i) 3

(j) 2 3- +^ h (k) 0 (l) 1 (m) 2 2 1-^ h (n) 6

(o) 131

(p) 3 2 2- (q) 2 3 (r) 21

- (s) 632

(t) 2 3

2-

2. (a) 2

3 2x = (b)

29 3

y = (c) 2 3p =

3. 60c 4. 2 m 5. 3 m 6. 3

10 3m

7. (a) 6 2 m (b) 4 m 8. 0.9 m 9. 3

5 3 3m

+^ h

10. 100 3 m

Exercises 6.7

1. (a) 1 st , 4 th (b) 1 st , 3 rd (c) 1 st , 2 nd (d) 2 nd , 4 th (e) 3 rd , 4 th (f) 2 nd , 3 rd (g) 3 rd (h) 3 rd (i) 2 nd (j) 4 th

2. (a) 3 rd (b) 21

- 3. (a) 4 th (b) 2

1-

4. (a) 2 nd (b) 3- 5. (a) 2 nd (b) 2

1

6. (a) 1 st (b) 23

7. (a) 1 (b) 2

1 (c) 3- (d)

21

(e) 21

- (f) 21

- (g) 23

(h) 3

1- (i)

23

- (j) 2

1-

8. (a) 2

1- (b)

23

- (c) 3 (d) 23

- (e) 23

-

(f) 3- (g) 21

(h) 3

1- (i)

2

1 (j)

2

1-

9. (a) 23

- (b) 3 (c) 23

(d) 21

(e) 21

- (f) 3

(g) 2

1 (h)

2

1 (i) −1 (j)

21

10. ,sin cos53

54

i i= - = -

11. ,cos tan733

33

4i i= - = -

12. ,cos cosecx x89

8589

= = -

13. , ,cosec cot tanx x x21

5

21

2221

= - = - = -

14. ,cos sinx x74

7 7474

5 74= - = -

15. ,tan sec65

4

65

9i i= - =

16. , ,tan sec cosecx x x355

38

55

8= = - = -

17. (a) 103

sinx = (b) 1091

,91

3cos tanx x= - = -

18. , ,cot sec cosec65

561

661

a a a= - = = -

19. ,sin cot1051

51

7i i= = -

20. (a) sin i (b) cos x (c) tan b (d) sin a- (e) tan i-

(f) sin i- (g) cos a (h) tan x-

Exercises 6.8

1. (a) ,20 29 159 31c ci = l l (b) ,120 240c ci = (c) ,135 315c ci = (d) ,60 120c ci = (e) ,150 330c ci = (f) ,30 330c ci =

(g) , , ,30 120 210 300 0 2 720c c c c c c# #i i= ] g (h) 70 , 110 , 190 , 230 , 310 , 350

0 3 1080c c c c c c

c c# #

i

i

=

] g

(i) , , ,30 150 210 330c c c ci = (j) , , , , , , , ,

, , ,15 45 75 105 135 165 195 225255 285 315 345c c c c c c c c

c c c c

i =

2. (a) 79 13! ci = l (b) ,30 150c ci = (c) ,45 135c ci = -

(d) ,60 120c ci = - - (e) ,150 30c ci = -

(f) ,30 150! !c ci =

(g) , , ,22 30 112 30 67 30 157 30c c c ci = - -l l l l

AnswerS6.indd 581 7/31/09 11:07:55 AM


(h) , , , , ,15 45 75 105 135 165! ! ! ! ! !c c c c c ci =

(i) ,135 45c ci = - (j) , , ,30 60 120 150! ! ! !c c c ci =

3.

4. 1-

5.

6. , ,x 0 180 360c c c= 7. 1- 8. 1

9. ,x 0 360c c=

10.

11. 0 12. 270x c= 13. , ,x 0 180 360c c c=

14. , ,x 0 180 360c c c= 15. ,x 270 90c c= -

16.

17.

Exercises 6.9

1. (a) cos i (b) tan i- (c) cos i (d) tan i (e) sec a-

2. (a) sin i (b) sec i (c) cosec x (d) cos 2 x (e) sin a

(f) cosec 2 x (g) sec 2 x (h) tan2 i (i) cosec5 2 i

(j) sin 2 x (k) 1 (l) sin cosi i

3. (a) 1cos xLHS 2= -

sin

sinx

x1 1

RHS

2

2

= - -

= -

=

So cos sinx x12 2- = -

(b) sec tanLHS i i= +

cos cossin

cossin

1

1

RHS

i i

i

i

i

= +

=+

=

So sec tancos

sin1i i

i

i+ =

+

(c) 3 3 tanLHS 2 a= +

( )tansec

cos

sin

3 13

3

1

3

RHS

2

2

2

2

a

a

a

a

= +

=

=

=-

=

So tansin

3 31

32

2a

a+ =

-

(d) sec tantan tan

cosec cot

x xx x

x x

11

LHS

RHS

2 2

2 2

2 2

= -

= + -

=

= -

=

So sec tan cosec cotx x x x2 2 2 2- = -

(e) sin cossin cos sin cossin cos sin sin cos cossin cos sin cos

sin sin cos cos sin cos

x xx x x xx x x x x xx x x x

x x x x x x

21 2

2 2

LHS

RHS

2 2

2 2

= -

= - -

= - - +

= - -

= - - +

=

3

2

]

] ]

] ^

] ]

g

g g

g h

g g

So sin cos sin sin cos cossin cos

x x x x x xx x2

2

2

2

- = - -

+

3] g

AnswerS6.indd 582 7/31/09 11:07:55 AM

583ANSWERS

(f) sin cossin sin

sin coscos sin

sin coscos

sin cossin

sincos

coscot sec

1 2

2

2

2

2

RHS

LHS

2

2

2

i i

i i

i i

i i

i i

i

i i

i

i

i

ii i

=- +

=+

= +

= +

= +

=

So cot secsin cossin sin

21 22

i ii i

i i+ =

- +

(g) cos cotsin cot

sinsincos

sin cos

90LHS

RHS

2

2

2#

c i i

i i

ii

i

i i

= -

=

=

=

=

] g

So 90cos cot sin cos2 c i i i i- =] g

(h) cosec cot cosec cotcosec cot

cot cot

x x x xx x

x x11

LHS

RHS

2 2

2 2

= + -

= -

= + -

=

=

] ]g g

So cosec cot cosec cotx x x x 1+ - =] ]g g

(i)

( )

cos

sin cos

cos cos

sin cos

sec sintan costan costan cos

1

1

1 11 1

LHS

RHS

2

2 2

2 2

2 2

2 2

2 2

2 2

2 2

i

i i

i i

i i

i i

i i

i i

i i

=-

= -

= -

= + - -

= + - +

= +

=

So cos

sin costan cos

12

2 22 2

i

i ii i

-= +

(j) cosec

cotcos

cosec

cot cos cosec

cosec

cot cossin

cosec

cot cot

cosecsin

1

1

11

1

1

LHS

#

b

bb

b

b b b

b

b bb

b

b b

b

b

=+

-

=+ -

=

+ -

=+ -

=

=

tan cot

sec

cos

sin

sin

cos

sec

sin cos

sin cos

sec

sin cos

sec

seccos sin

cos

cos sin

sin

1

1

11

RHS

2 2

#

#

b b

b

b

b

b

b

b

b b

b b

b

b b

b

bb b

b

b b

b

=+

=

+

=+

=

=

=

=

LHS RHS=

So cosec

cotcos sin

1

b

bb b

+- =

4.

( )

cos sincos sincos sin

x y2 2

4 444 14

LHS

RHS

2 2

2 2

2 2

2 2

i i

i i

i i

= +

= +

= +

= +

=

=

=

] ]

]

g g

g

So 4x y2 2+ =

5.

( )

cos sincos sincos sin

x y9 9

81 818181 181

LHS

RHS

2 2

2 2

2 2

i i

i i

i i

= +

= +

= +

= +

=

=

=

2 2] ]

]

g g

g

So 81x y2 2+ =

Exercises 6.10

1. (a) 8.9x = (b) 9.4y cm= (c) 10.0a =

(d) 10.7b m= (e) 8.0d =

2. (a) 51 50ci = l (b) c61 23a = l (c) x 43 03c= l

(d) 87 04ca = l (e) 150 56ci = l

3. 126 56c l 4. (a) 13.5 mm (b) 25 mm

5. (a) 1.8 m (b) 2.7 m 6. 5.7 cm

7. (a) 10.3 m (b) 9.4 m 8. (a) 60 22c l (b) 57 9c l

9. (a) 14.1 cm (b) 15.6 cm

10. (a) 54.7 mm (b) 35.1 mm

AnswerS6.indd 583 8/7/09 12:40:20 PM


Exercises 6.11

1. (a) 5.8m = (b) 10.4b m= (c) 7.4h cm=

(d) 16.4n = (e) 9.3y =

2. (a) 54 19ci = l (b) 60 27ci = l (c) x 57 42c= l

(d) 131 31cb = l (e) 73 49ci = l

3. 32.94 mm 4. 11.2 cm and 12.9 cm

5. (a) 11.9 cm (b) 44 11c l (c) 82 13c l

6. ,XYZ XZY YXZ66 10 47 40c c+ + += = =l l

7. (a) 18.1 mm (b) 80 49c l 8. (a) 6.2 cm (b) 12.7 cm

9. 12.9 cm 10. (i) 11 cm (ii) 30c

Exercises 6.12

1. 12.5 cm and 4.7 cm 2. (a) 040c (b) 305c 3. 16.4 m

4. 103c 5. 1.97 m 6. 11c

7. (a) 1.21 km (b) 1 minute 8. 32 m 9. 107 m

10. (a) .sin

sinAC

101 365 8 42 29

c

c=

l

l (b) 74 50ci = l

11. 8.5h = 12. 7.7 km 13. 5.7 km and 5.4 km

14. 1841 km 15. 35.8 m 16. 89 52c l 17. 9.9 km

18. 163.5 km 19. 64.1 m 20. 3269 km

21. (a) 11.3 cm (b) 444 0c l 22. 141c

23. (a) 11.6 cm (b) 73 14c l

24. (a) 265.5 km (b) 346 33c l

25. (a) 35 5c l (b) (i) 4.5 m (ii) 0.55 m

Exercises 6.13

1. (a) 7.5 cm2 (b) 32.3 units2 (c) 9.9 mm2 (d) 30.2 units2 (e) 6.3 cm2

2. 2

15 3m2 3. 7.5 cm2 4. 15.5 cm2 5. 34.8 cm2

6. 1.2 m2 7. 42 cm2 8. 247.7 mm2

9. (a) 7.8 cm (b) 180.8 cm2

10. (a) 5.6 cm (b) 18.5 cm2 (c) 19.1 cm2

Test yourself 6

1. ,cos sin34

5

34

3i i= =

2. (a) cos x (b) 2 (c) cosec A

3. (a) 0.64 (b) 1.84 (c) 0.95

4. (a) 46 3ci = l (b) 73 23ci = l (c) 35 32ci = l

5.

( )

sincos

sinsin

sinsin sin

sinsin

12

12 1

12 1 1

2 12 2

LHS

RHS

2

2

i

i

i

i

i

i i

ii

=-

=-

-

=-

+ -

= +

= +

=

^

] ]

h

g g

2 2sin

cossin

12

So2

i

ii

-= +

6. b 40c= 7. (a) 2

1 (b)

23

- (c) 3-

8. ,x 120 240c c=

9.

,x 90 270c c=

10. 122 km 11. 5 3 12. (a) 6.3 cm (b) 8.7 m

13. (a) 65 5ci = l (b) 84 16ci = l (c) 39 47ci = l

14. 65.3 cm2 15. (a) ,x 60 120!! c c=

(b) , , ,x 15 105 75 165c c c c= - -

(c) , , ,x 0 180 30 150!c c c c= -

16. ,sin cot53

34

i i= - = 17. (a) 209c (b) 029c

18. (a) sin

sinAD

9920 39

c

c= (b) 8.5 m 19. 2951 km


1. 92 58c l 2. 50.2 km 3. 12.7x cm=

4. (a) .sin

sinAC

41 2125 3 39 53

c

c=

l

l (b) 25.2h cm= 5. 4.1 km

6. cos x- 7. 16 3 cm2 8. 2

1

9. , , ,x 22 30 112 30 202 30 292 30c c c c= l l l l 10. 75 45ci = l

11. 5.4 m 12. ,110 230c ci = 13. 6.43 km

14. 956

- 15. 31 m 16. sin

cos sin cos

coscos

sin cos

cossin cos

tan

1

1

LHS

RHS

2

2

i

i i i

ii

i i

i

i i

i

=-

+

=+

=+

= +

=

]

]

g

g

17. 4 5 0x y y2 2+ + - =

AnswerS6.indd 584 8/4/09 1:22:03 PM

585ANSWERS

Chapter 7: Linear functions

Exercises 7.1

1. (a) 5 (b) 10 (c) 13 2. (a) 13 (b) 65

(c) 85 (d) 52 2 13=

3. (a) 9.85 (b) 6.71 (c) 16.55 4. 12 units

5. , 134 128Two sides side= =

6. Show 85AB BC= =

7. Show points are 17 units from ,7 3-^ h

8. 3 , 9x yRadius units equation 2 2= + =

9. Distance of all points from ,0 0^ h is 11, equation

11x y2 2+ = 10. 3a = 11. a 6 2!= -

12. All 3 sides are 2 units. 13. ,a 10 2= -

14. , ,MQ NP QP MN37 20= = = = so parallelogram

15. 98BD AC= = 16. (a) ,AB AC BC40 4= = =

(b) OC OB 2= = 17. 2 101 18. 61 units

19. 29, 116, 145AB BC AC= = =

AB BC

AC

29 116145

2 2

2

+ = +

=

=

So triangle ABC is right angled (Pythagoras’ theorem)

20. , ,XY YZ XZ65 130 65= = =

Since XY YZ= , triangle XYZ is isosceles.

XY XZ

YZ

65 65130

2 2

2

+ = +=

=

So triangle XYZ is right angled. (Pythagoras’ theorem)

Problem

30.2

Exercises 7.2

1. (a) ,2 4^ h (b) ,1 1-^ h (c) ,2 1-^ h (d) ,3 2-^ h

(e) ,1 1-^ h (f) ,3 2-^ h (g) ,321

d n (h) ,121

1d n

(i) ,21

221

d n (j) ,0 521

d n

2. (a) ,a b9 3= = - (b) ,a b5 6= - =

(c) ,a b1 2= - = - (d) ,a b1 2= - = -

(e) ,a b6 1= =

3. ,2

3 30

24 4

0+ -

=- +

=] g

4. ,P Q 2 1= = -^ h

5. ,4 3^ h 6. 3x = is the vertical line through midpoint ,3 2^ h .

7. Midpoint of , .AC BD 221

321

midpoint of= = d n

Diagonals bisect each other

8. 125,AC BD= = midpoint AC midpoint=

,BD 421

= - ;d n rectangle 9. ,8 13-^ h

10. (a) , , ,X Y Z21

321

21

21

1 1= - = =, ,d d ^n n h

(b) , ; ,XY BC XZ10 40 2 10234

= = = =

; ,AC YZ AB3422

2= = =

11. 4x y2 2+ = 12. 1x y2 2+ =

Exercises 7.3

1. (a) 2 (b) 131

(c) 131

- (d) 252

- (e) 32

(f) 81

-

(g) 421

- (h) 32

- (i) 241

(j) 2- 2. 21y1 =

3. 1.8x = 4. 9x = 5. (a) Show 53

m m1 2= =

(b) Lines are parallel .

y

-3 -2 -1 3 4 5 6 7

1

3

4

-2

-1 2(2, -1)

(-2, 1)

(7, 2)

(3, 4)

1

2

6. Gradient of 1AB CD21

gradient of= =

Gradient of 0BC ADgradient of= =

7. Gradient of 1AB CD31

gradient of= = -

Gradient of BC AD43

gradient of= =

Gradient of ,AC 521

= -

gradient of 21

BD = -

8. Gradient of 1,AC = gradient of BD 1= -

9. (a) Show AB BC AC2 2 2+ =

(b) Gradient of 45

,AB =

gradient of 54

BC = -

10. (a) , , ,F G1 2 421

= - =^ dh n

(b) Gradient of FG BC65

gradient of= =

Answer S7-S8.indd 585 7/31/09 12:08:49 PM


11. 4 3 11 0x y- - = 12. Gradient of ,2 4-^ h and , ,3 1 3 1gradient of- = -^ ^h h and ,5 5 3=^ h

13. 1 14. 0.93 15. 21 16. 50 12c l 17. 108 26c l

18. (a) 3 (b) 3

1 (c) 3-

19.

tantan

m

m

7 45 2

33

1

1180 45 2135

nd quadrant` c c

c

ii

i

=-

- - -

=-

= -

=

- =

= -

=

]

^

g

h

20. 3

2 3 3x =

+^ h

Exercises 7.4

1. (a) (i) 3 (ii) 5 (b) (i) 2 (ii) 1 (c) (i) 6 (ii) 7-

(d) (i) 1- (ii) 0 (e) (i) 4- (ii) 3 (f) (i) 1 (ii) 2-

(g) (i) 2- (ii) 6 (h) (i) 1- (ii) 1 (i) (i) 9 (ii) 0

(j) (i) 5 (ii) 2- 2. (a) (i) 2- (ii) 3 (b) (i) 5- (ii) 6-

(c) (i) 6 (ii) 1- (d) (i) 1 (ii) 4 (e) (i) 2- (ii) 21

(f) (i) 3 (ii) 121

(g) (i) 31

- (ii) 2- (h) (i) 54

- (ii) 2

(i) (i) 321

(ii) 21

- (j) (i) 132

(ii) 32

3. (a) 4 (b) 2-

(c) 0 (d) 2- (e) 1- (f) 3- (g) 2 (h) 41

- (i) 121

(j) 141

(k) 32

(l) 21

(m) 51

(n) 72

(o) 53

-

(p) 141

- (q) 15 (r) 121

- (s) 61

(t) 83

-

Exercises 7.5

1. (a) 4 1y x= - (b) y x3 4= - + (c) 5y x=

(d) 4 20y x= + (e) 3 3 0x y+ - = (f) x y4 3 12 0- - =

(g) 1y x= - (h) 5y x= + 2. 8 0x y+ - =

3. (a) 4 3 7 0x y- + = (b) 3 4 4 0x y- + =

(c) 4 5 13 0x y- + = (d) 3 4 25 0x y+ - =

(e) 2 2 0x y- + = 4. 4 8 0x y+ - = 5. (a) 3y =

(b) x 1= - 6. y x2= - 7. 3 4 12 0x y- - =

8. 2 3 0x y+ - = 9. 4x = - 10. 3 8 15 0x y+ - =

Exercises 7.6

1. (a) 3- (b) 31

(c) 43

(d) 121

(e) 1 (f) 65

- (g) 3

1

(h) 31

(i) 3

1 (j)

51

2. (a) 1 0x y- + = (b) 3 16 0x y- + = (c) 5 0x y+ - =

(d) 2 5 0x y+ + = (e) 2 4 0x y- + =

(f) 3 1 0x y+ - = (g) 3 4 13 0x y+ + =

3. 3m m1 2= = so parallel

4. m m51

5 11 2 #= - = - so perpendicular

5. 151

m m1 2= =

6. m m37

73

11 2# #= - = - 7. 32

k = - 8. 4m m1 2= =

9. AB CD m m 31 2< = =_ i and BC AD m m85

1 2< = = -d n

10. Gradient of : ,AC m21

1 = gradient of BD : 2,m2 = -

m m21

2 11 2# #= - = -

11. (a) y x= - (b) 5 8 0x y- - = (c) 2 2 0x y+ + =

(d) 2 3 16 0x y- + = 12. 7 6 24 0x y+ - =

13. 3 0x y+ - = 14. 2 5 0x y- - =

15. 2 3 18 0x y- + =

Exercises 7.7

1. (a) ,2 4-^ h (b) ,1 3- -^ h (c) ,4 4^ h (d) ,0 2-^ h

(e) ,5 1-^ h (f) ,1 1-^ h (g) ,3 7^ h (h) ,4 0^ h (i) ,41 26^ h

(j) ,191

197

-d n 2. Substitute ,3 4-^ h into both lines

3. , , ,2 5 4 1^ ^h h and ,1 1- -^ h 4. All lines intersect

at ,2 3-^ h

5. All lines meet at ,5 0-^ h 6. 11 6 0x y+ =

7. 5 6 27 0x y+ - = 8. x y4 7 23 0+ =+

9. 1 0x y+ - = 10. 2 2 0x y+ - =

11. 3 0x y+ - = 12. 2 3 0x y- - =

13. x y 1 0- + = 14. 3 2 0x y- + =

15. 3 7 0x y+ - = 16. 5 13 0x y+ + =

17. 27 5 76 0x y- - = 18. 3 14 0x y- - =

19. 2 1 0x y- - = 20. 3 11 0x y- - =

21. 5 17 0x y- + =

Exercises 7.8

1. (a) 2.6 (b) 1133

(c) 2.5 (d) 2.4 (e) 138

2. (a) 3.48 (b) 1.30 (c) 0.384 (d) 5.09 (e) 1.66

3. (a) 13

7 13 (b) 5 (c)

2054 205

(d) 13

5 26 (e)

1314 13

4. d d d 11 2 3= = =

Answer S7-S8.indd 586 7/31/09 12:08:50 PM

587ANSWERS

5. : , :A d B d5

14

5

3= =

-

Opposite signs so points lie on opposite sides of the line

6. , : , , :d d2 310

139 2

10

5- = =^ ^h h

Same signs so points lie on the same side of the line

7. , : , , :d d3 2 4 4 1 251

- = - =^ ^h h


8. 2d d1 2= = so the point is equidistant from both lines

9. , : , , :d d8 337

551 1

37

9- = =^ ^h h

Same signs so points lie on same side of the line

10. , : , , :d d3 25

64 1

5

7- =

-=^ ^h h


11. 4d d1 2= = so same distance 12. 5

8 5 units

13. 1 14. 4.2 15. 9 17x32

or= - 16. 3 1b41

121

or= -

17. m 1 1832

31

or= - -

18. Show distance between ,0 0^ h and the line is 5

19. Show distance between ,0 0^ h and the line is greater than 1

20. (a) , , , , ,3 1 374

71

2 2- -^ d ^h n h (b) , ,5

2 105

13 5119

26 34

Test yourself 7

1. 6.4 units 2. ,221

2-d n

3. (a) 151

- (b) 2 (c) 3

1 (d)

53

4. (a) 7 11 0x y- - = (b) 5 6 0x y+ - = (c) 3 2 0x y+ =

(d) 3 5 14 0x y+ - = (e) 3 3 0x y- - =

5. 5

6 5units

6. ,m m41

41 2= - = so m m 11 2 = -

` lines are perpendicular.

7. x -intercept 5, y -intercept 2-

8. (a) 2 1 0x y+ - = (b) 21

(c) 25

units

9. 5,m m1 2= = so lines are parallel 10. 3 4 0x y- =

11. ,1 1-^ h 12. ,a b6 1= =

13. Solving simultaneously, 4 0x y- - = and

2 1 0x y+ + = have point of intersection , .1 3-^ h

Substitute ,1 3-^ h in 5 3 14 0:x y- - =

5 1 3 3 14 0LHS RHS# #= - - - = =

point lies on 5 3 14 0:x y- - =

Substitute ,1 3-^ h in 3 2 9 0:x y- - =

3 1 2 3 9 0LHS RHS# #= - - - = =

point lies on 3 2 9 0:x y- - =

lines are concurrent

14. 0.499- 15. ,c 13 65= - - 16. 3y = 17. 154

x =

18. , : , , :d d2 113

86 3

13

2- =

-=^ ^h h


19. 4 0x y- - = 20. 3 7 14 0x y- - =


1. 2k = - 2. 3 3 3 0x y- - = 3. 10 10 81x y2 2+ =

4. Show AC and BD have the same midpoint ,1 2^ h and m m 1AC BD# = -

5. Show distance of all points from ,0 0^ h is 3; radius 3; equation 9x y2 2+ =

6. 13

4 13 7. 45 ; ( )OBA a b sides of isoscelesc+ D= =

8. 13

12 13 9. 113 12c l 10. 2 3 13 0x y+ + =

11. .angled

, , ;,

BC AC ABm m

18 61

so is isoscelesso is rightBC AC#

D

D

= = =

= -

12. ,3 5-^ h

13. ,a b2 3= = 14. 2 5 14 0x y+ + =

15. 3 3 2 3 0x y+ + - = 16. 6 0x y- + =

Chapter 8: Introduction to calculus

Exercises 8.1

1.

Answer S7-S8.indd 587 7/31/09 12:08:51 PM


2.

3.

4.

5.

6.

7.

8.

9.

10.

Exercises 8.2

1. Yes, 0x = 2. Yes, x x1= 3. No 4. Yes, 0x =

5. Yes, ,x x x x1 2= = 6. Yes, 0x = 7. Yes, x 3= -

8. Yes, 2x = 9. Yes, ,x 2 3= - 10. Yes, x1 01#-

11. Yes, ,x 90 270c c= 12. Yes, 0x = 13. No 14. No

15. Yes, x 3!=

Exercises 8.3

1. (a) 3 (b) 7- (c) 3 (d) 8 (e) 2 (f) 3- (g) 2 (h) 1- (i) 10 (j) 1-

2. (a) 2 4x x2 - - (b) 2 1x x3 + - (c) 7 1x- - (d) 4x x4 2- (e) 4 3x- + (f) 2 6x2 + (g) 2x- (h) 4x2 (i) 3 1x - (j) 2 9x x2 - +

Exercises 8.4

1. (a) 4.06 (b) 3.994 (c) 4

2. (a) 13.61 (b) 13.0601 (c) 12.9401 (d) 13 3. 6

4. (a) 2f x h x xh h2 2+ = + +] g

(b) ( ) ( )f x h f x x xh h xxh h

22

2 2 2

2

+ - = + + -

= +

(c) h

f x h f x

hxh h

hh x h

x h

2

2

2

2+ -=

+

=+

= +

] ]

]

g g

g

Answer S7-S8.indd 588 7/31/09 12:08:51 PM

589ANSWERS

(d) ( )

( )

lim

lim

f xh

f x h f x

x h

x

2

2

h

h

0

0

=+ -

= +

=

"

"

l] ]g g

5. (a) ( ) ( )( )

f x h x h x hx xh h x h

x xh h x h

2 7 32 2 7 7 32 4 2 7 7 3

2

2 2

2 2

+ = + - + +

= + + - - +

= + + - - +

] g

(b) ( ) ( ) ( )( )

f x h f x x xh h x hx x

x xh h x hx x

xh h h

2 4 2 7 7 32 7 3

2 4 2 7 7 32 7 3

4 2 7

2 2

2

2 2

2

2

+ - = + + - - +

- - +

= + + - - +

- + -= + -

(c)

h

f x h f x

hxh h h

hh x h

x h

4 2 7

4 2 7

4 2 7

2+ -=

+ -

=+ -

= + -

] ]

]

g g

g

(d) f x x4 7= -l] g

6. (a) f 2 11=] g (b) 2 5 11f h h h2+ = + +] g

(c) f h f h h2 2 52+ - = +] ]g g

(d)

h

f h f

hh h

hh h

h

2 2 5

5

5

2+ -=

+

=+

= +

] ]

]

g g

g

(e) f 2 5=l] g

7. (a) f 1 7- = -] g

(b) f h f h h h1 1 4 12 123 2- + - - = - +] ]g g (c) 12

8. (a) f 3 8=] g (b) f h f h h3 3 6 2+ - = +] ]g g (c) f 3 6=l] g

9. (a) f 1 13= -l] g (b) 17

10. (a) 2y x x2= +

Substitute ,x x y yd d+ +_ i :

( )

2

y y x x x xx x x x x x

y x xy x x x x

22 2 2

2 2Since

2

2 2

2

2

d d d

d d d

d d d d

+ = + + +

= + + + += +

= + +

] g

(b) x

y

xx x x x

x

x x x

x x

2 2

2 2

2 2

2

d

d

d

d d d

d

d d

d

=+ +

=+ +

= + +

] g

(c) 2 2dx

dyx= +

11. (a) 2 (b) 5 (c) 12- (d) 15 (e) 9-

12. (a) f x x2=l] g (b) 2 5dx

dyx= +

(c) f x x8 4= -l] g (d) 10 1dx

dyx= -

(e) 3dx

dyx2= (f) f x x6 52= +l] g

(g) 3 4 3dx

dyx x2= - + (h) x xf 6 2= -l] g

13. (a) 0.252 (b) 0.25 (c) 0.2498

14. (a) 0.04008- (b) 0.03992- (c) 0.04- 15. 1-

Exercises 8.5

1. (a) 1 (b) 5 (c) 2 3x + (d) 10 1x - (e) 3 4 7x x2 + - (f) 6 14 7x x2 - + (g) 12 4 5x x3 - + (h) 6 25 8x x x5 4 3- - (i) 10 12 2 2x x x4 2- + - (j) 40 63x x9 8-

2. (a) 4 1x + (b) 8 12x - (c) 2 x (d) 16 24x x3 - (e) 6 6 3x x2 + -

3. (a) x3

1- (b) x x2 3 2- (c) 3

86

xx

75- (d) 4 x (e)

41

(f) 2 2 2x x2 - +

4. f x x16 7= -l] g 5. 56-

6. 60 40 35 3dx

dyx x x9 7 4= - + - 7. 10 20

dtds

t= -

8. g x x20 5= - -l] g 9. 30dtdv

t= 10. 40 4dtdh

t= -

11. drd

rV

4 2r= 12. 3 13. (a) 5 (b) 5- (c) 4x =

14. (a) 12 (b) x 2!= 15. 18

Exercises 8.6

1. (a) 72 (b) 13- (c) 11 (d) 18- (e) 18 (f) 27

(g) 11 (h) 136 (i) 4- (j) 149

2. (a) 261

- (b) 251

(c) 201

(d) 431

- (e) 101

(f) 71

(g) 711

- (h) 201

(i) 81

- (j) 51

-

3. (a) (i) 6 (ii) 61

- (b) (i) 8 (ii) 81

-

(c) (i) 24 (ii) 241

- (d) (i) 8- (ii) 81

(e) (i) 11 (ii) 111

-

4. (a) 27 47 0x y- - = (b) 7 1 0x y- - = (c) 4 17 0x y+ + = (d) 36 47 0x y- - = (e) 44 82 0t v- - =

5. (a) x y24 555 0+ - = (b) 8 58 0x y- + = (c) 17 516 0x y- - = (d) 45 3153 0x y- + = (e) 2 9 0x y+ - =

6. (a) (i) 7 4 0x y- + = (ii) 7 78 0x y+ - = (b) (i) 10 36 0x y- + = (ii) 10 57 0x y+ - = (c) (i) 10 6 0x y+ - = (ii) 10 41 0x y- - = (d) (i) 2 2 0x y+ + = (ii) 2 19 0x y- - = (e) (i) 2 2 0x y- + = (ii) 2 9 0x y+ - =

7. x 3!= 8. (1, 2) and ( 1- , 0) 9. ( 5- , 7- )

Answer S7-S8.indd 589 7/31/09 12:08:52 PM


10. (0, 1) 11. (1, 2) 12. ,143

41615

- -d n

13. (a) (1, 1- ) (b) 6 7 0x y- - =

14. 10 7 0t h- - = 15. x y4 2 19 0- - =

Exercises 8.7

1. (a) 3x 4- - (b) 1.4x0.4 (c) 1.2x 0.8- (d) 2x21 -

1

(e) 2x x3 2+-

-

1

(f) 3x-

2

(g) 4x6-

1

(h) 2x-

3

2. (a) x

12

- (b) 2

5

x (c)

6

1

x56 (d)

10

x6- (e)

15

x4

(f) 2

1

x3- (g)

3

x7- (h)

23 x

(i) 3

2

x2-

(j) 2

1 12

x x3 5- -

3. 271

4. −3 5. 321

6. −3 7. 2 3 1x x+ +

8. 81

9. 3 16 8 0x y+ - = 10. 9 0x y- + =

11. (a) 2

1

x3- (b)

161

- 12. x y16 016+ - = 13. (9, 3)

14. 4x = 15. , , ,552

552

- -d dn n

Exercises 8.8

1. (a) 4 3x 3+] g (b) 6 2 1x 2-] g (c) 70 5 4x x2 6-^ h

(d) 48 8 3x 5+] g (e) 5 1 x 4- -] g (f) 135 5 9x 8+] g (g) x4 4-] g (h) 4 6 3 2 3x x x2 3 3

+ +^ ^h h (i) 8 2 5 5 1x x x2 7

+ + -] ^g h (j) 6 6 4 2 3x x x x5 6 2 5

- - +^ ^h h (k) 2x23

3 1--

1

] g

(l) 2 4 x 3- -] g (m) 6 9x x2 4- -

-^ h (n) -

3x35

5 4+2

] g

(o) -

4x x x x x43

3 14 1 72 3 2- + - +

1

^ ^h h (p) 2 3 4

3

x +

(q) 5 2

5

x 2-

-] g (r)

1

8

x

x2 5

-+^ h

(s) 7 3

2

x3-

-

(t) 2 4

5

x 3-

+] g (u)

4 3 1

3

x 3-

-] g (v)

2 2 7

27

x 10-

+] g

(w) 3 3

4 9 3

x x x

x x4 3 2

3 2

-- +

- +

^

^

h

h (x)

316 4 1x3 +

(y) 4 7

5

x 94 -] g

2. 9 3. 40 4. (4, 1) 5. ,x 2 121

= - 6. 8 7 0x y+ + =

Exercises 8.9

1. (a) 8 9x x3 2+ (b) 12 1x - (c) 30 21x +

(d) 72 16x x5 3- (e) 30 4x x4 -

(f) 5 2 1x x x 2+ +] ]g g (g) 8 9 1 3 2x x 4- -] ]g g (h) x x x3 16 7 4 23 - -] ]g g (i) 10 13 2 5x x 3+ +] ]g g (j) x x x x x x x

x x x x x

10 5 3 1 3 10 1

13 60 3 20 1

3 2 2 4 2 2 5

3 2 2 4

+ - + + + +

= + + - +

^ ^ ^ ^

^ ^

h h h h

h h

(k) x

xx

x

x

2 22

2 2

4 3-

-+ - =

-

-

(l) x

xx x2 1

2 5 32 1

5

2 1

112 2-

- ++

-= -

-]

]

]g

g

g

2. 26 3. 1264 4. 77

1

7

8+ = 5. 176

6. 10 9 0x y- - = 7. 69 129 0x y- - =

8. x3

6 30!=

- 9. 34 29 0x y- + =

Exercises 8.10

1. (a) x2 1

22-

-

] g (b)

5

15

x 2+] g (c)

x

x x

x

x x

4

12

4

122 2

4 2

2 2

2 2

-

-=

-

-

^ ^

^

h h

h

(d) 5 1

16

x 2+] g (e)

14 14

x

x x

x

x4

2

3

- +=

- + (f)

3

11

x 2+] g

(g) 2

2

x x

x2 2

2

-

-

^ h (h)

2

6

x 2-

-

] g (i)

x4 3

342-

-

] g (j)

x3 1

142+

-

] g

(k) 3 7

3 6 7

x

x x2 2

2

-

- - -

^ h (l)

x

x x

x

xx

2 3

4 12

2 3

342

2

2-

-=

-

-

] ]

]

g g

g

(m) x

x

5

182 2-

-

^ h (n)

x

x x

x

x x

4

2 12

4

2 62

3 2

2

2

+

+=

+

+

] ]

]

g g

g

(o) x

x x

3

2 9 72

3 2

+

+ +

] g (p)

3 4

3 8 5

x

x x2

2

+

+ -

] g

(q) x x

x x x

1

2 4 12 2

4 3 2

- -

- - -

^ h (r)

-2 2

xx x x

52 5 5

+

+ - +

1 1

] ]g g

(s)

(t) 28

x

x x x

x

x

7 2

7 1 7

7 2

21 302 28 5

4 3

+

-=

+

- ++ - +

]

] ] ]

]g

g g g

g

(u) x

x x x x

x

x x2 5

15 2 5 3 4 6 3 4 2 5

2 5

3 3 4 4 33

6

3 4 5 2

4

4

-

- + - + -

=-

+ -

]

] ] ] ]

]

] ]

g

g g g g

g

g g

(v) x

x x

x

x

x1

1 2 1

3 1

2 1

3 53+

+ +

+

=+

+-3

] g

(w) x

x

x

x

x x

x

2 3

2 1

2 3

1

2 1 2 3

2 12 2-

-

-

-=

- -

- +2-

] ]g g

(x) x

x

x x

x x

x x

x x

9

1

9

9 1

1 9

9 24

2

2

2

2 3

2

-

+

-

- +=

+ -

- - -2-

]

]

]

]g

g

g

g

2. 81

3. 195

- 4. 0, 1x = 5. 9, 3x = -

6. 18 8 0x y- + = 7. 17 25 19 0x y- - =

x

x x x

x

x x

5 1

6 5 1 2 9 5 2 9

5 1

2 9 20 512

2 3

2

2

+

+ - - -=

+

- +

]

] ] ]

]

] ]

g

g g g

g

g g

Answer S7-S8.indd 590 7/31/09 12:08:52 PM

591ANSWERS

Test yourself 8

1. (a)

(b)

2. 10 3dx

dyx= - 3. (a) 42 9 2 8

dx

dyx x x5 2= - + -

(b) 2 1

11dx

dy

x 2=

+] g (c) 8( ) ( )

dx

dyx x x9 2 4 4 22= + + -

(d) 40 5 5 (10 1)dx

dyx x x x x2 1 2 1 2 13 4 3= - + - = - -] ] ]g g g

(e) 2

5dx

dy x3

= (f) 10

dx

dy

x3= -

4. dtdv

t4 3= - 5. (a) 1 (b) 20 6. 10 7. 42

8. (a) 2x = - (b) 1x = (c) 2x =

9. (a) 32 4 9f x x 3= +l] ]g g (b) 3

5dx

dy

x 2= -

-] g

(c) dx

dyx x9 1 3 1= - -] ]g g (d)

4dx

dy

x2= -

(e) f xx5

145

=l] g

10. y

11. 9 7 0x y- - = 12. (2, 3) 13. drdS

r8r=

14. ( 2- , 71), (5, 272- ) 15. 4 6 0x y- - = 16. 3525

17. 9 18. x y12 4 0+ - = 19. ,51

dtds

u at t= + =

20. 107


1. ,f f1 3 1 36= - = -l] ]g g 2. 1813

-

3. ; , .dtdx

t t t8 300 0 37 53 2= + = -

4. , ,x y x y x y2 0 3 3 0 6 12 0+ = - - = - + =

5. , , , , 12 26 0, 12 170 0x y x y2 2 2 14- - + - = + + =^ ^h h

6. 43

7. 5 5 1 9 15 9 5 110 5 1 9 (4 13)x x x x

x x x

3 4 5 2

2 4+ - + - +

= + - -

] ] ] ]

] ]

g g g g

g g

8. x

x x x

x

x4 9

2 4 9 16 2 1 4 9

4 9

2 12 17

8

4 3

5

-

- - + -

=-

- +

]

] ] ]

]

]

g

g g g

g

g

9. x12

6 2046

3 51! !=

-=

- 10. 2 25 0x y+ - =

11. 271

a = - 12. ,P 241

6161

= -d n 13. ,x31

31 13!

=

14. 21

15. , , ,x y Q PQ3 5 0 0 5 10- + = = =^ h

16. 8n = 17. , , x y11211

23 3

12 3 012 31- + =e o

18. , ,x21

121

153

= - 19. (a) ,x 90 270c c=

(b) y

x1

90c 180c 270c 360c

20. ,4 73- -^ h 21. 3 9 14 0x y- - = 22. x x

x

4 3 2

4 534 -

-] g

23. (a) ,x y x y16 32 1 0 4 2 1 0+ + = - - =

(b) 2m m21

1

1 2$ #= -

= -

So perpendicular

Answer S7-S8.indd 591 7/31/09 12:08:53 PM


24. 0, 2, 6x = 25. ,a b14 7= - = 26. 22

5 22

27. 121

p = 28. drdV

38 3r

= 29. 4k = 30. 4 0x y- - =

31. 4 13 0x y- - = 32. 481

- 33. , ,a b c1 2 4= - = =

34. 8 28S r rhr r r= +-

35. (a) 6 5 3 1 3 5x x x2 3- - -] ]g g (b) x x

x

3 2 1

5 64- +

+-

]

]

g

g

36. x6

4 13!=

37. (a) 7 80 0x y+ - =

(b) ,Q 471

12491

= -d n


1. 0.77- 2. 1 3. 5 2 1 0x y+ - = 4. ,2 2-^ h

5. 0.309- 6. (a) 3 cm2 (b) , 1AC BD13 cm cm= =

7. 1; ,m m A43

68

1 121

1 2 #= - = - = -d n 8. x 15c=

9. 127

10.

11.

12. ’45 49c 13. Domain: all real ;x21

! range: all

real y 0!

14.

15.

16. sin4 i 17. 2 units 18. 8 15 0x y- + =

19. ,120 240c ci = 20. 132

- 21. 2 22. 11 565ca = l

23. .y 16 5= 24. 3 5 0x y+ - = 25. x132

31 1

26. 7 27. 3x = 28. 3-

29. Show perpendicular distance from ,0 0^ h to the line is 2 units, or solving simultaneous equations gives only one solution.

30. (a) ,g g2 1 3 6= - = -] ]g g

(b)

31. 3 4x x2 - 32. 2

1- 33. 17.5 m

34. ,x y2 17= - = - 35. (a) 7.0AB m= (b) 27.8 m 2

36. cos3 i 37. (a) 2 4 0x y- + = (b) , ,,P Q2 0 0 4-^ ^h h

(c) 4 units 2

38. 127 m 39. 15 units 2 40. ( )

( )

f x x xx xf x

33

6 2

6 2

- = - - - -

= - -=

] ]g g

41. 16x x x x x1 1 18 1 2 12 22 2 2 2 2 33 4+ + + ++ =^ ^ ^ ^h h h h

Answer S7-S8.indd 592 7/31/09 12:08:54 PM

593ANSWERS

42. y431

9# #- 43. 3

x2-

44. (a) 3 4 0x y- - = (b) 2 0x y- - =

(c) 3 10 0x y+ + = (d) ,R 10 0= -^ h

45. 138

units 46. Domain: all ;x 4!- range: all y 0!

47. 2 7

1

x - 48. 4.9 km 49. 8 7 10x x 3- - -

50. 1

5

x 2+] g 51. 2 3x - 52.

x x

x

x x

x

5

17 2

5

17 22 2+

- -=

+

+- ] g

53. 6 56 0x y+ - = 54. ,f f2 45 2 48- = - - =l] ]g g

55. ,a b2 9= = - 56. 7 5 9 0x y- + =

57. 47 109 0x y- + = 58. 0.25x = -

59. (a) domain: x21

$ range: y 0$

(b) domain: all real x 7!- range: all real y 0!

(c) domain: x2 2# #- range: y2 0# #-

60. (a) (1, 1) (b) 2 13 units (c) 121

-

(d) 3 2 5 0x y+ - =

61. (a) 62. (b), (d) 63. (a) 64. (c) 65. (c)

Chapter 9: The quadratic function

Exercises 9.1

1. Axis of symmetry 1,x = - minimum value 1-

2. Axis of symmetry 1.5,x = - minimum value 7.5-

3. Axis of symmetry 1.5,x = - minimum value 0.25-

4. Axis of symmetry 0,x = minimum value 4-

5. Axis of symmetry 83

,x = minimum point ,83

167

d n

6. Axis of symmetry 1,x = maximum value 6-

7. Axis of symmetry 1,x = - maximum point ,1 7-^ h

8. Minimum value ,1- 2 solutions

9. Minimum value 3.75, no solutions

10. Minimum value 0, 1 solution

11. (a) ;x 3= - (-3, -12) (b) ;x 4= - (-4, 17)

(c) ; ,x 141

141

381

= d n (d) ; ,x 141

141

1341

= - - -d n

(e) ; ,x 3 3 23= - - -^ h

12. (a) (i) x 1= - (ii) -3 (iii) (-1, -3)

(b) (i) 1x = (ii) 1 (iii) (1, 1)

13. (a) Minimum (-1, 0) (b) Minimum (4, -23) (c) Minimum (-2, -7) (d) Minimum (1, -1) (e) Minimum (2, -11)

(f) Minimum ,41

381

- -d n

(g) Maximum (-1, 6)

(h) Maximum (2, 11)

(i) Maximum , 721

43

d n

(j) Maximum (1, -3)

14. (a) (i) -2 (ii) Minimum 0 (iii) y

x-4 -3 -2 -1 2

3

2

1

4

5

-2

-3

-11

(b) (i) -1, 3 (ii) Minimum -4

(iii) y

x-4 -3 -2 -1 2 3 4

2

1

3

4

5

-3

-2

-11

-4

-5

Answer S7-S8.indd 593 7/31/09 12:08:55 PM


(c) (i) 5.83, 0.17 (ii) Minimum -8

(iii) y

x-4 -3 -2 -1 2 3 4 5 6

4

2

6

8

10

-6

-4

-21

-8

-10

(d) (i) -2, 0 (ii) Minimum -1

(iii) y

x-4 -3 -2 -1 2

2

1

3

4

5

-3

-2

-11

(e) (i) 3! (ii) Minimum -18

(iii) y

x-2-3-4 -1 1 2 5

1

2

-6

-8

-10

-12

-14

-16

-18

-4

-243

(f) (i) -1, 32

(ii) Minimum 21

12-

(iii) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-4

-5

-6

-2

-1

-2

1

112

23

(g) (i) 1.65, -3.65 (ii) Maximum 7

(iii) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

7

6

-3

-2

-11

(h) (i) 1.3, -2.3 (ii) Maximum 341

(iii) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-2

-11

3 14

Answer S9-S10.indd 594 8/1/09 8:43:34 PM

595ANSWERS

(i) (i) 0.56, -3.56 (ii) Minimum 441

(iii) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

-3

-2

-11

4 14

(j) (i) 2.87, -0.87 (ii) Maximum 7

(iii) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

6

7

-3

-2

-11

15. (a) 4 (b) None

(c) y

x-4 -3 -2 -1 2 3 4 5

2

1

3

4

5

6

7

-3

-2

-11

16. (a) None (b) 643

(c) y

x-4 -3 -2 -1 2 3 4 5

4

2

6

8

10

12

14

-3

-2

-11

17. (a) 387

- (b) None

(c) y

x-4 -3 -2 -1 2 3 4 5

1

2

-18

-16

-14

1

-12

-10

-8

-6

-4

-2

18. (a) y

x-4 -3 -2 -1 2 3 4 5

4

2

6

8

-3

-2

-11

(b) ,x x2 31 2 (c) x2 3# #

Answer S9-S10.indd 595 8/1/09 8:43:44 PM


19.

Graph is always above the x-axis so y 02 for all xx x3 2 4 02

` 2- + for all x

20. y

x-4 -3 -2 -1 2 3 4 5

4

2

6

8

-6

-4

-21

Graph is always above the x -axis so y 02 for all x x x 2 02` 2+ + for all x

21. y

x-4 -3 -2 -1 2 3 4 5

2

4

-18

-10

-12

-14

-16

-8

-6

-4

-21

Graph is always below the x -axis so y 01 for all x x x2 7 02` 1- + - for all x

22. y

x-4 -3 -2 -1 2 3 4 5

1

2

-5

-6

-7

-4

-3

-2

-11

Graph is always below the x -axis so y 01 for all x x x5 4 1 02` 1- + - for all x

Exercises 9.2

1. ,x x3 31 2- 2. 1 0n ##- 3. 0, 2a a# $

4. ,x x2 21 2- 5. y0 6# # 6. 0 2t 11

7. 4, 2x x1 2- 8. 3, 1p p# $- - 9. ,m m2 41 2

10. 3, 2x x# $- 11. h121

21 1 12. 4 5x ##-

13. 2 7k21# #- 14. ,q q 631 2 15. All real x

16. ,n n4 3# $- 17. x3 51 1- 18. t6 2# #-

19. ,y y31

51 2- 20. ,x x2 4# $-

Exercises 9.3

1. (a) 20 (b) -47 (c) -12 (d) 49 (e) 9 (f) -16 (g) 0 (h) 64 (i) 17 (j) 0

2. (a) 17 unequal real irrational roots (b) -39 no real roots (c) 1 unequal real rational roots (d) 0 equal real rational roots (e) 33 unequal real irrational roots (f) -16 no real roots (g) 49 unequal real rational roots (h) -116 no real roots (i) 1 unequal real rational roots (j) 48 unequal real irrational roots

3. 1p = 4. k 2!= 5. b87

# - 6. p 22 7. k 2121

2 -

8. a 3 02=

b ac4 1 4 3 7

830

2 2

1

- = - -

= -

] ] ]g g g

So x x3 7 02 2- + for all x

9. ,k k5 3$# - 10. k0 41 1 11. ,m m3 31 2-

12. ,k k1 1# $- 13. 3

p1

1 - 14. b0 221

# #

y

x-4 -3 -2 -1 2 3 4 5

4

2

6

8

-6

-4

-21

Answer S9-S10.indd 596 8/1/09 8:41:30 PM

597ANSWERS

15. ,p p2 6# $-

16. Solving simultaneously: 2 6y x= + (1)

3y x2= + (2)

Substitute (2) in (1):

x xx xb ac

3 2 62 3 04 2 4 1 3

160

2

2

2 2

2

+ = +

- - =

- = - - -

=

] ] ]g g g

So there are 2 points of intersection

17. 3 4 0x y+ - = (1) 5 3y x x2= + + (2) From (1): 3 4y x= - + (3) Substitute (2) in (3):

5 3 3 48 1 0

4 8 4680

x x xx x

b ac 1 1

2

2

2 2

2

+ + = - +

+ - =

- = - -

=

] ]g g

So there are 2 points of intersection

18. 4y x= - - (1) y x2= (2) Substitute (2) in (1):

44 0

4 1 415

0

x xx x

b ac 1 4

2

2

2 2

1

= - -

+ + =

- = -

= -

] ]g g

So there are no points of intersection

19. 5 2y x= - (1) 3 1y x x2= + - (2) Substitute (2) in (1):

x x xx x

b ac

3 1 5 22 1 0

4 2 4 1 10

2

2

2 2

+ - = -

- + =

- = - -

=

] ] ]g g g

So there is 1 point of intersection the line is a tangent to the parabola

20. 341

p =

21. (c) and (d)

Exercises 9.4

1. (a) , ,a b c1 2 6= = = - (b) , ,a b c2 11 15= = - = (c) , ,a b c1 1 2= = = - (d) , ,a b c1 7 18= = = (e) , ,a b c3 11 16= = - = - (f) , ,a b c4 17 11= = = (g) , ,a b c2 12 9= = - = - (h) , ,a b c3 8 2= = - = (i) , ,a b c1 10 24= - = = - (j) , ,a b c2 0 1= - = = -

2. , ,m p q2 5 2= = - =

3. 4 5 2 2 1 3 4x x x x x2 - + = - - + + +] ]g g

4. a x x b x cx x x

x x x xx x

2 3 21 2 3 1 2 17

3 2 6 2 172 9

RHS

RHS

2

2

= - + + - +

= - + + - +

= + - - + - +

= + +

=

] ] ]

] ] ]

g g g

g g g

true

5. , ,A B C1 5 6= = = - 6. , ,a b c2 1 1= = = -

7. , , .K L M1 6 7 5= = = 8. 12 5 2 3 65 2x x 2+ + - - -] ]g g

9. , ,a b c0 4 21= = - = -

10. (a) 5y x x2= - - (b) 3y x x2= -

(c) 2 3 7y x x2= - + (d) 4 9y x x2= + -

(e) 2 1y x x2= - - +

Exercises 9.5

1. (a) ,2 1a b ab+ = - = (b) . ,1 5 3a b ab+ = = - (c) . , .0 2 1 8a b ab+ = = - (d) ,7 1a b ab+ = - =

(e) ,232

1a b ab+ = =

2. (a) 3 (b) 6- (c) 0.5- (d) 21

3. (a) 3 10 0x x2 + - = (b) 4 21 0x x2 - - = (c) 5 4 0x x2 + + = (d) x x 08 112 - + = (e) 2 27 0x x2 - - =

4. 0.5m = 5. 32k = - 6. 4b = 7. 1k = 8. 13p =

9. 5k = - 10. m 3!= 11. 1k = - 12. ,n 1 3= -

13. ,p r2 7= = - 14. ,b c6 8= - = 15. ,a b0 1= = -

16. 11

`ab ba

= =

17. (a) 1k = - (b) 1, 0k = - (c) 1.8k = - (d) 3k =

(e) ,k k1 0# $-

18. (a) p 2 3!= (b) ,p p2 3 2 3# $-

(c) p2

3 3!=

19. (a) k 2= (b) 3k = - (c) 2k =

20. (a) 1m = (b) ,m m2

3 102

3 101 2

- +

(c) 3m = -

Exercises 9.6

1. (a) ,x 1 4= - - (b) 2, 5y = (c) 4, 2x = - (d) 1, 4n = - (e) 3, 5a = - (f) 3, 4p = (g) ,x 2 4= - (h) 5, 12k = (i) ,t 6 4= - (j) ,b 12 4= - -

Answer S9-S10.indd 597 8/1/09 6:52:33 PM


2. (a) 2, 3x = - (b) 2, 3x = (c) 4, 5x = (d) 3, 5x =

(e) 121

x = , 4

3. (a) x 3!= (b) ,y 2 2! != (c) x2

1 5!=

(d) . , . , . , .x 1 37 4 37 0 79 3 79= - - (e) ,a 2 2 6!= - -

4. (a) 0, 3x = (b) 1p = (c) 1x = (d) 1x = (e) 1, 3x =

5. 2,x 1! != 6. 1x = -

7. . , . , . , .x 2 19 0 46 1 93 0 52! ! ! !=

8. (a) , , ,x 0 90 180 360c c c c= (b) , ,x 90 180 270c c c= (c) , ,x 90 210 330c c c= (d) , , ,x 60 90 270 300c c c c= (e) , , ,x 0 180 270 360c c c c=

9. (a) , , , ,x 0 45 180 225 360c c c c c= (b) , ,x 0 180 360c c c= (c) , , , ,x 0 30 150 180 360c c c c c= (d) 45 , 60 ,135 , 120 , 225 , 240 , 315 , 300x c c c cc c c c= (e) 30 , 60 , 120 , 150 , 210 , 240 , 300 , 330x c c c c c c c c=

10.

( ) ( ) ( ) ( )

xx

x xx

x x

x xx x

33

25

3 33

23 5 3

3 2 5 33 5 3 2 0

2

2

# # #

+ ++

=

+ + ++

+ = +

+ + = +

+ - + + =

]

] ]

] ]

g

g g

g g

Let 3u x= +

u ub ac

5 2 04 5 4 1 2

170

2

2 2

2

- + =

- = - -

=

] ] ]g g g

So u has 2 real irrational roots. x 3` + and so x has 2 real irrational roots

Test yourself 9

1. (a) x0 3# # (b) ,n n3 31 2- (c) 2 2y ##-

2. , ,a b c1 9 14= = - = 3. (a) 2x = (b) 3-

4. ab ac1 0

42 4 1 724

0positive definite

2

# #

`

2

1

D

=

= -

= - -

= -

2] g

5. (a) 6 (b) 3 (c) 2 (d) 18 (e) 30 6. ,x 132

31

=

7. (a) iv (b) ii (c) iii (d) ii (e) i

8.

( ) ( )

ab ac

1 04

3 4 1 47

0

2

2# #

1

1

D

= -

= -

= - - -

= -

x x4 3 02` 1- + - for all x

9. (a) 41

x = - (b) 681

10. 3 2 12 3 41x x2- + + -] ]g g 11. , ,x 30 150 270c c c=

12. (a) 341

k = (b) 1k = (c) 3k = (d) 3k = (e) 2k =

13. ,x21

3= - 14. m169

1 - 15. ,x 0 2=

16. (a) i (b) i (c) iii (d) i (e) ii

17. (a) iii (b) i (c) i (d) ii

18.

ac

kk

1

1

1

For reciprocal roots

LHS RHS

ba

ab

aa

=

=

=

= =

∴ roots are reciprocals for all x .

19. (a) 3 28 0x x2 + - = (b) 10 18 0x x2 - + =

20. 1, 3x =


1. k 4 02$D= -] g and a perfect square ∴ real, rational roots

2. y x x5 42= - + 3. , ,a b c4 3 7= = - = 4. x 2!=

5. 11 6. 2.3375n = - 7. .p 0 752 8. Show 0D =

9. x 1!=

10. 2, 19, 67 2, 13, 61A B C A B Cor= = - = = - = = -

11. 2

4 12

31

1

x x

xx x2 - -

+=

-+

+

12. ,k k2

1 212

1 21# $

- +

13. , ,x 30 90 150c c c= 14. ,x 12

3 5!=

15. , , ,x 60 90 270 300c c c c= 16. 23-

Chapter 10: Locus and the parabola

Exercises 10.1

1. A circle 2. A straight line parallel to the ladder.

3. An arc 4. A (parabolic) arc 5. A spiral

6. The straight line 2 2 | | 2x xor1 1 1-

7. A circle, centre the origin, radius 2 (equation 4x y2 2+ = i

8. lines y 1!= 9. lines x 5!= 10. line 2y =

11. Circle 1x y2 2+ = (centre origin, radius 1)

12. Circle, centre , ,1 2-^ h radius 4 13. 5y = -

Answer S9-S10.indd 598 8/1/09 6:52:45 PM

599ANSWERS

14. Circle, centre (1, 1), radius 3 15. x 7= - 16. 3x =

17. y 8!= 18. x 4!=

19. Circle, centre , ,2 4-^ h radius 6

20. Circle, centre , ,4 5-^ h radius 1

Exercises 10.2

1. x y 12 2+ = 2. 2 2 79 0x x y y2 2+ + + - =

3. 10 4 25 0x x y y2 2- + + + = 4. 8 6 13 0x y- + =

5. 12 26 1 0x y- - = 6. y x!=

7. 3 32 3 50 251 0x x y y2 2- + - + =

8. 5 102 5 58 154 0x x y y2 2- + + - =

9. 4 20 36 0x x y2 - + - = 10. 20 0x y2 - =

11. 8 32 0y x2 + - = 12. 2 8 7 0x x y2 - + - =

13. 12 0x y2 + = 14. 5 2 11 0x x y y2 2- + - - =

15. 3 4 0x x y y2 2+ + - - =

16. 2 17 0x x y y2 2+ + - - =

17. 2 4 2 6 47 0x x y y2 2+ + - + =

18. 2 2 2 4 27 0x x y y2 2+ + + + =

19. 3 4 25 0, 3 4 15 0x y x y+ + = + - =

20. ,x y x y12 5 14 0 12 5 12 0- - = - + =

21. x y2 3 5 5 0!- - =

22. 7 9 0, 7 5 0x y x y- + = + - =

23. 7 4 30 0, 32 56 35 0x y x y- - = + - =

24. 16 7 40 0xy x y- - + =

25. 6 3 12 9 0x x y y2 2- - - + =

Problem

,x y x y12 5 40 0 12 5 38 0+ - = + + =

Exercises 10.3

1. (a) Radius 10, centre (0, 0) (b) Radius 5 , centre (0, 0) (c) Radius 4, centre (4, 5) (d) Radius 7, centre (5, −6) (e) Radius 9, centre (0, 3)

2. (a) 16x y2 2+ = (b) 6 4 12 0x x y y2 2- + - - = (c) 2 10 17 0x x y y2 2+ + - + = (d) 4 6 23 0x x y y2 2- + - - = (e) 8 4 5 0x x y y2 2+ + - - = (f) 4 3 0x y y2 2+ + + = (g) 8 4 29 0x x y y2 2- + - - = (h) 6 8 56 0x x y y2 2+ + + - = (i) 4 1 0x x y2 2+ + - = (j) 8 14 62 0x x y y2 2+ + + + =

3. 18 8 96 0x x y y2 2- + + + =

4. 4 4 8 0x x y y2 2+ + + - = 5. 2 48 0x x y2 2- + - =

6. 6 16 69 0x x y y2 2+ + - + =

7. 10 4 27 0x x y y2 2- + + + = 8. 9 0x y2 2+ - =

9. 2 10 25 0x x y y2 2- + - + =

10. 12 2 1 0x x y y2 2+ + - + =

11. 8 6 22 0x x y y2 2- + - + = 12. 6 1 0x y y2 2+ + + =

13. (a) Radius 3, centre (2, 1) (b) Radius 5, centre (−4, 2) (c) Radius 1, centre (0, 1) (d) Radius 6, centre (5, −3) (e) Radius 1, centre (−1, 1) (f) Radius 6, centre (6, 0) (g) Radius 5, centre (−3, 4) (h) Radius 8, centre (−10, 2) (i) Radius 5, centre (7, −1) (j) Radius 10 , centre (−1, −2)

14. Centre ,3 1-^ h , radius 4 15. Centre ,2 5^ h , radius 5

16. Centre ,1 6- -^ h , radius 7 17. Centre (4, 7), radius 8

18. Centre ,121

1-d n , radius 221

19.

20. Show perpendicular distance from the line to ,4 2-^ h is 5 units, or solve simultaneous equations.

21. (a) Both circles have centre ,1 2-^ h (b) 1 unit

22. 2 2 23 0x x y y2 2+ + + - = 23. 34 units

24. (a) 5 units (b) 3 units and 2 units (c) XY is the sum of the radii. The circles touch each other at a single point, ,0 1^ h .

25. Perpendicular distance from centre ,0 0^ h to the line is equal to the radius 2 units; perpendicular distance from centre ,1 2-^ h to the line is equal to the radius 3 units.

26. (a) 2 6 15 0x x y y2 2+ + - - = (b) , ,,2 7 1 2- -^ ^h h (c) ,Z 1 8= -^ h

(d) m m31

3

1

zx yx# #= -

= -

ZXY 90` c+ =

27. (a) 4 units (b) 4 10 13 0x x y y2 2- + + + =

Answer S9-S10.indd 599 8/1/09 6:52:57 PM


Exercises 10.4

1. (a) 20x y2 = (b) 36x y2 = (c) 4x y2 = (d) 16x y2 =

(e) 40x y2 = (f) 12x y2 = (g) 24x y2 = (h) 44x y2 =

(i) 8x y2 = (j) 48x y2 =

2. (a) x y42 = - (b) 12x y2 = - (c) 16x y2 = -

(d) 28x y2 = - (e) 24x y2 = - (f) 36x y2 = -

(g) 32x y2 = - (h) 8x y2 = - (i) 60x y2 = -

(j) 52x y2 = -

3. (a) (i) (0, 1) (ii) y 1= - (b) (i) (0, 7) (ii) 7y = - (c) (i) (0, 4) (ii) 4y = - (d) (i) (0, 9) (ii) 9y = - (e) (i) (0, 10) (ii) 10y = - (f) (i) (0, 11) (ii) 11y = -

(g) (i) (0, 3) (ii) 3y = - (h) (i) (0, 121

c m (ii) 1y21

= -

(i) (i) 0, 221

c m (ii) 2y21

= - (j) (i) 0, 343

c m

(ii) 3y43

= -

4. (a) (i) (0, −1) (ii) 1y = (b) (i) (0, −6) (ii) 6y = (c) (i) (0, −2) (ii) 2y = (d) (i) (0, −12) (ii) 12y = (e) (i) (0, −5) (ii) 5y = (f) (i) (0, −4) (ii) 4y = (g) (i) (0, −8) (ii) 8y = (h) (i) (0, −10) (ii) 10y =

(i) (i) 0,21

-c m (ii) 21

y = (j) (i) 0, 521

-c m (ii) 521

y =

5. (a) 28x y2 = (b) 44x y2 = (c) 24x y2 = - (d) 8x y2 = (e) 12x y2

!= (f) 32x y2!=

(g) 32x y2 = (h) 71

x y2 =

6. (a) Focus , ,0 2^ h directrix 2,y = - focal length 2 (b) Focus , ,0 6^ h directrix 6,y = - focal length 6

(c) Focus , ,0 3-^ h directrix 3,y = focal length 3

(d) Focus , ,021

d n directrix ,y21

= - focal length 21

(e) Focus , ,0 143

-d n directrix 143

,y = focal length 143

(f) Focus , ,081

d n directrix ,y81

= - focal length 81

7. 2y = 8. ,4 4^ h 9. ,X 121

83

= - -d n

10. 4, 2-^ h and 4, 2- -^ h ; 8 units

11. (a) 12x y2 = - (b) 3y = (c) 3331

units

12. (a) Substitute the point into the equation.

(b) 3 4 3 0x y+ - = (c) ,243

-d n

13. (a) 4 2 0x y- + = (b) 0, 1^ h does not lie on the line (c) 4 2 1 0x x y y2 2- + - + = (d) Substitute ,0 1^ h into the equation of the circle.

14. (a) Substitute Q into the equation of the parabola. (b) 1 2 2 0q x qy aq2 - - + =_ i (c) Equation of latus rectum is .y a= Solving with 4x ay2 = gives two endpoints , , ,A a a B a a2 2-^ ^h h . Length of 4AB a= .

Exercises 10.5

1. (a) 8y x2 = (b) 20y x2 = (c) 56y x2 = (d) 36y x2 = (e) 32y x2 = (f) 24y x2 = (g) 28y x2 = (h) 12y x2 = (i) 16y x2 = (j) 4y x2 =

2. (a) y x362 = - (b) 16y x2 = - (c) 40y x2 = - (d) y x242 = - (e) 8y x2 = - (f) 48y x2 = - (g) 44y x2 = - (h) y x202 = - (i) 12y x2 = - (j) 28y x2 = -

3. (a) (i) (2, 0) (ii) x 2= - (b) (i) (3, 0) (ii) 3x = - (c) (i) (4, 0) (ii) 4x = - (d) (i) (1, 0) (ii) 1x = - (e) (i) (7, 0) (ii) 7x = - (f) (i) (8, 0) (ii) 8x = - (g) (i) (6, 0) (ii) 6x = - (h) (i) (9, 0) (ii) x 9= -

(i) (i) 41

, 0c m (ii) 41

x = - (j) (i) 421

, 0c m (ii) 4x21

= -

4. (a) (i) (−2, 0) (ii) 2x = (b) (i) (−3, 0) (ii) 3x = (c) (i) (−7, 0) (ii) 7x = (d) (i) (−1, 0) (ii) 1x = (e) (i) (−6, 0) (ii) 6x = (f) (i) (−13, 0) (ii) 13x =

(g) (i) (−15, 0) (ii) 15x = (h) (i) 21

, 0-c m (ii) 21

x =

(i) (i) 621

, 0-c m (ii) 621

x = (j) (i) 141

, 0-c m (ii) 141

x =

5. (a) 20y x2 = (b) 4y x2 = (c) 16y x2 = - (d) 12y x2 =

(e) 36y x2!= (f) 8y x2

!= (g) 12y x2 = (h) 21

y x2 =

6. (a) Focus , ,2 0^ h directrix 2,x = - focal length 2

(b) Focus , ,1 0^ h directrix 1,x = - focal length 1

(c) Focus , ,3 0-^ h directrix 3,x = focal length 3

(d) Focus , ,121

0d n directrix 121

,x = - focal length 121

(e) Focus , ,141

0-d n directrix 141

,x = focal length 141

(f) Focus , ,121

0d n directrix ,x121

= - focal length 121

7. 4x = (latus rectum) 8. , , ,,12 3 6 3 6-^ ^h h

9. , ,,9 6 81 18-^ ^h h

10. (a) 5 12 25 0x y- - = (b) ,5 461

- -d n (c) 10125

units 2

(d) 4132

units (e) 11.7 units 2

Exercises 10.6

1. (a) yx 3 8 32- = +] ^g h (b) 5 4 6x y2- = +] ^g h

(c) x y1 4 32 = +-] ^g h (d) 12x y4 32- = - -] ^g h

(e) 6 8 7x y2- = +] ^g h (f) 16x y7 32+ = - -] ^g h

(g) 4x y2 52- = - -] ^g h (h) 9 12 6x y2+ = +] ^g h

(i) 1 8 1x y2+ = +] ^g h (j) x y3 4 22- = - -] ^g h

2. (a) 4 4 4y x2- = +^ ]h g (b) 1 8 2y x2- = +^ ]h g

(c) y x2 12 12+ = +^ ]h g (d) 10 4 29y x2- = - -^ ]h g

(e) 3 16 1y x2+ = - -^ ]h g (f) 6 8 4y x2- = +^ ]h g

(g) 5 24 2y x2+ = - -^ ]h g (h) 12 4 36y x2+ = +^ ]h g

(i) y x2 20 12- = - -^ ]h g (j) 4 8 2y x2+ = - -^ ]h g

AnswerS10.indd 600 7/31/09 2:55:56 PM

601ANSWERS

3. (a) 2 8 9 0x x y2 + - + = (b) x x y8 4 16 02 + - + =

(c) 4 8 12 0x x y2 - - - = (d) 6 8 41 0x x y2 - - + =

(e) 4 16 20 0x x y2 + - + = (f) 2 16 1 0x x y2 + + + =

(g) x x y8 20 4 022 - + - = (h) 10 8 1 0x x y2 + + + =

(i) 6 12 45 0x x y2 + + + = (j) x y4 24 02 + + =

(k) 6 12 3 0y y x2 - - - = (l) 8 4 8 0y y x2 - - + =

(m) 8 32 0y x2 - + = (n) y y x4 16 0122 + - =-

(o) 2 8 7 0y y x2 + - - = (p) y y x8 12 042 + + + =

(q) 2 4 11 0y y x2 - + - = (r) 6 16 25 0y y x2 - + + =

(s) 4 2 5 0y y x2 - + + = (t) y y x2 2 062 - + =-

4. (a) (i) (3, −2) (ii) 4y = - (b) (i) (1, 1) (ii) y 3= -

(c) (i) (−2, 0) (ii) 2y = - (d) (i) (4, 2) (ii) 4y = -

(e) (i) (−5, −1) (ii) 5y = - (f) (i) (3, 1) (ii) 3y =

(g) (i) (−1, 0) (ii) 4y = (h) (i) (2, 0) (ii) 2y =

(i) (i) (4, −2) (ii) 4y = (j) (i) (−2, −3) (ii) 5y =

5. (a) (i) (0, −1) (ii) 2x = - (b) (i) (2, 4) (ii) 4x = - (c) (i) (0, 3) (ii) 4x = - (d) (i) (3, −2) (ii) x 5= - (e) (i) (7, 1) (ii) 5x = - (f) (i) (1, −5) (ii) 5x = (g) (i) (11, −7) (ii) 13x = (h) (i) (−3, 6) (ii) 7x =

(i) (i) (−7, 2) (ii) 9x = (j) (i) 1021

, 3- -c m (ii) 921

x =

6. 12 36 0x y2 - + =

7. ,x x y x x y4 8 4 0 4 8 12 02 2+ - - = + + + =

8. 2 4 19 0x x y2 - - - = 9. 12 12 12 0y y x2 - + + =

10. 2 16 1 0x x y2 - - + = 11. 2 28 29 0x x y2 - - + =

12. 4 24 44 0y y x2 + + - = 13. 6 32 9 0y y x2 - - + =

14. 6 8 15 0x x y2 - + - = 15. 2 16 49 0y y x2 + - + =

16. 6 4 7 0x x y2 + + - = 17. 4 12 8 0x x y2 - - - =

18. 2 16 95 0y y x2 + + - =

19. (a) Vertex ,2 1-^ h , focus ,2 3-^ h , directrix 1y = -

(b) Vertex ,3 2^ h , focus ,3 5^ h , directrix 1y = -

(c) Vertex ,1 1-^ h , focus ,1 2-^ h , directrix 0y =

(d) Vertex ,3 4^ h , focus ,7 4^ h , directrix x 1= -

(e) Vertex ,0 2-^ h , focus ,6 2-^ h , directrix 6x = -

(f) Vertex ,5 0-^ h , focus ,7 0-^ h , directrix x 3= -

20. Vertex ,1 4-^ h , focus 1, 3- -^ h , directrix 11,y = axis 1,x = - maximum value 4

21. 4 8 12 0x x y2 - - + = or 4 8 36 0x x y2 - + - =

22. (a) 8 9 72 0x y2 + - = (b) , , y0 73223

8329

=d n

23. (a)

(b) , , y1 843

941

- - = -d n

24. 4 8 20 0x x y2 + + - = 25. 0.3 m

Exercises 10.7

1. 31

m = 2. m 4= - 3. m 1= - 4. 21

m =

5. dx

dyx= 6. 2 0x y- - = 7. 2 12 0x y- + =

8. 6 0, 18 0x y x y+ - = - - =

9. 2 2 0, 2 9 0x y x y- - = + - =

10. ,,x y M 187

21

4 8 0+ - = = d n

11. , ,x y P9 0 18 27+ - = = -^ h

12. 33, 60.5Q = ^ h

13. , ,x y x y4 144 0 4 2 9 0+ + = + + = , .18 40 5-^ h ; show the point lies on the parabola by substituting it into the equation of the parabola

14. , ,x y R4 0 4 0- - = = ^ h

15. (a) Substitute P into the equation of the parabola (b) 2 0x py p p3+ - - = (c) Substitute 0, 1^ h into the equation of the normal.

( )Since 0, 1 0

p p pp pp p

p p

0 2 00

1

3

3

2

2!

+ - - =

= +

= +

+ =

Test yourself 10

1. 8 6 29 0x y+ - = 2. 4 8 4 0x x y2 - - - =

3. Centre , ,3 1^ h radius 4 4. (a) ,1 3-^ h (b) 4, 3-^ h

5. 25x y2 2+ = 6. (a) 2y = (b) ,0 2-^ h

7. 3 10 0x x y y2 2+ + - - = 8. 8 16 16 0x x y2 - + - =

9. (a) (i) ,1 1^ h (ii) ,1 2^ h (b) 0y =

10. 2 3 6 0x y+ + = 11. 14 units

12. 24y x2 = - 13. 8 16 0x y2 - + =

14. ,x y x y4 3 16 0 4 3 14 0- - = - + =

15. ,y x y x= = - 16. 20y x2 = 17. (a) 21

- (b) 2

18. (a) 4 72 0x y- + = (b) ,9 2041

d n

19. Sub ,0 4^ h: 7 0 3 4 12 0LHS RHS# #= - + = =

20. , 792

-d n

21. (a) 3 0x y- - = (b) ,R 0 3= -^ h (c) ,F FP FR0 3 6= = =^ h

AnswerS10.indd 601 7/31/09 2:55:57 PM



1. (a) 8 6 29 0x y+ - = (b) Midpoint of AB lies on line; m m 11 2 = -

2. (a) 2 6 15 0x x y y2 2- + - - = (b) Put 0y = into equation

3. ,221

3-d n

4. (a) ;x y x y4 2 9 0 2 24 0- + = + - = (b) 1m m1 2 = - (c) , .X 3 10 5= ^ h (d) 3 4 8 0;x y- + = focus ,0 2^ h lies on the line

5. ,0 0^ h

6. (a) ;x y x y2 4 1 0 2 4 0- - = + + = (b) Point lies on line 1y = -

7. 2 4 2y x x2= - + - 8. 3 2 0x y+ + =

9.

10. (a) 4 10 21 0x x y y2 2+ + - + =

(b) 2 5 8;x y2 2+ + - =] ^g h centre , ;2 5-^ h

28 2radius = =

11. 3

2 3-

12. (a) 4 16 52 0y y x2 + - + = (b) 2 6 0x y- - =

13. 4 2 units 14. 2 2 0x y y2 2+ - - =

15. 696 mm from the vertex

16. ;x y x y141 127 32 0 219 23 58 0+ + = + + =


1. ≤ , ≥m m2 3 2. 4 3 16 0x y+ - =

3. Centre , ,3 5-^ h radius 7

4. (a) 32

(b) 31

- (c) 191

5. Focus , ,0 2-^ h directrix 2y =

6. 5x = - or 6- 7. 1k = -

8. ,x y x y3 4 14 0 3 4 16 0- - = - + =

9. Vertex ,4 17- -^ h , focus , .4 16 75- -^ h

10. ,x 0 3= 11. 2 2 0x y+ + = 12. b 2$ -

13. 16,x y2 2+ = circle centre ,0 0^ h and radius 4

14. 4 6 12 0x x y y2 2+ + + - =

15. x x y y3 6 17 02 2- + - - =

16. 0.75- 17. 5 54 5 20 79 0x x y y2 2- + + - =

18. , ,a b c2 1 0= = =

19. 9 x

x2

--

20. 4 16 20 0x x y2 - - + =

21. and (given)


AC BC CD CE

CDAC

CEBC

ACB ECD

`

+ +

= =

=

=

since two sides are in proportion and their included angles are equal, Δ ABC is similar to Δ CDE 5.3 cmy =

22. 4 0x y- - =

23. 2 16 15 0x x y2 + - - = 24. ,x 0 2=

25. 04

1 4( 1)( 9)35

0

ab ac2

2

1

1

D = -

= - - -

= -

Since a 01 and 0, 9 0x x21 1D - + - for all x

26. ( )( ) ( )x x x8 3 2 5 3 2 51 3 4+ + +- ( )x x30 7 2 5 3= + +] g

27. sec cosecx x

28. Centre , ,5 3-^ h radius 2

29.

( )( )

ab ac0

41 4 1 3

110

2

2

2

1

D = -

= - -

= -

] g

Since 0a 2 and ,01D x x 3 02 2- + for all x

30. 1k = 31. 3 2 9 0x y+ - =

32. (a) 217 km (b) 153c

33. , ,a b c3 18 34= = - = - 34. ,x x4 32 1

35. ’95 44ci =

36. 361 0 and a perfect squareT 2= ^ h

37. 2 9 0x y+ + = 38. k 3#

AnswerS10.indd 602 7/31/09 2:55:58 PM

603ANSWERS

39. 5 4 41 0x y- - = 40. 22

3 6 10 3 3 5- + -

41. 4.9 , 11.1x ycm cm= = 42. 1x = 43. 8.25 units

44. 4.5 m 45. 2187128

46. °, °, °, °x 60 120 240 300=

47. 2 3 3 0x y+ - = 48. ,y 131

21

= - 49. 162c

50. °, °, °, °x 45 135 225 315=

51. 1, 2 or , 4x y x y41

41

= - = = - =

52. a b a ab b2 2 42 2+- +] ^g h 53. 43x = 54. 311

-

55. 1.8 units 56. tan i

57. 8 2 5 ( 1) 2( 1)x x x x2 3 2 4+ - + -] g ( ) ( )x x x2 1 9 20 12 3 2= - + -

58. 41

59. 2 3 25 0x x y y2 2+ + - - =

60. Focus (2, 1), directrix 5y =

61. 2 36 0x y- - =

62. Distance from centre ,0 0^ h to line is

| |d

a b

ax by c

1040

4radius

line is tangent

2 2

1 1

`

=+

+ +

=

=

=

63. k 221

= -

64. ,x x2 21 2-

65. Radius 3; 9x y2 2+ =

66. , ,a b c3 14 9= = - =

67. Domain: all real x ; range: y 3$ -

68. )|ED(

,

ACB ECDABC CED

AC CDABC CDEby AAS

vertically opposite anglesalternate angles

given

AB|

`

+ ++ +

/D D

=

=

=

^

^

h

h

69. 46 m 2 70. 3 0x y+ - =

71. x x x12 36 62 2- + = -] g

72. . , .y y2 5 6 5$ # -

73. (a) x y9 16 0- + = (b) x y9 20 0+ + =

(c) ,Q 20 0= -^ h

74. (a) 8 129 0x y- + = (b) ,R 781

17641

= d n

75. , ,a b c1 3 1= = - = -

76. (c) 77. (d) 78. (b) 79. (a) 80. (c) 81. (c)

AnswerS10.indd 603 8/1/09 6:27:02 PM

AnswerS10.indd 604 7/31/09 2:56:00 PM

complete textbook

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