Maths Handbook and Study Guide Grade 11 Kevin Smith PDF
                     MATHS MADE EASY 
A comprehensive guide to 
Grade 11 Mathematics 
Simple, logical and colour-coded notes 
with step-by-step examples 
I NOW INCLUDES EXERCISES WITH WORKED THROUGH SOLUTIONS I 
COVERS THE FUNDAMENTALS OF GRADE 11 MATHEMATICS 
AND COMPLEMENTS THE MATERIAL IN ANY CLASS TEXT 
GRADE 11 
KEVIN SMITH 
SUITABLE FOR IEB AND NATIONAL CURRICULUM STUDENTS 
Dr. Belinda Huntley B.Sc(Hons)(WITS.), H.Dip. Ed.(PG)(WITS), M.Ed.(WITS) , PhD (University of Pretoria)  
HANDBOOK and 
5 T U D yTM G U IDE 
MATHS MADE EASY 
A comprehensive guide to 
Grade 11 Mathematics 
Simple, logical and colour-coded notes 
with step-by-step examples 
COVERS THE FUNDAMENTALS OF GRADE 11 MATHEMATICS 
AND COMPLEMENTS ANY CLASS TEXT 
GRADE 11 
KEVIN SMITH 
BSe Eng {Electrieal)w'Ts 
Now includes exercises with worked through solutions 
CAPS COMPLIANT 
SUITABLE FOR IEB AND NATIONAL CURRICULUM STUDENTS 
© Copyright Kevin Smith I Berlut Books CC 2012 
ABOUT THE AUTHOR 
-:1 £41:110" i: I 
Kevin Smith is a young Eledrical Engineer who is passionate about Education, 
specifically Maths and Science. He has been giving extra lessons in Maths and Science for 
the past 8 years and is the founder of The Lab Maths and Science Academy which is 
attended by students across Johannesburg. 
As an extra lesson teacher, Kevin has established the critical areas where students 
struggle and has written the Maths Handbook and Study Guides with these areas in 
mind. 
The books are written in a clear, simple, visual and logical manner. The colour coding 
facilitates explanations, definitions, formulas, recaps of previous work, hints and ideas. 
They are easy to read, easy to understand and easy to apply what has been learnt. They 
work in conjunction with all other Maths Books. They are welcome additions to the 
Handbook and Study Guide series. 
Kevin's objective is for the Maths Handbook and Study Guides to demystify Maths and 
help students to reach their potential in this challenging subject. The subtitles of the 
books are 'Maths made Easy' and this is what he aims to do. Kevin ensures that his 
work is up to date at all times and that it is suitable for IEB and National Curriculum 
students. 
Beryl Lutrin - aeriut Books 
Author of the Afrikaans Handbook and Study Guide 
Co-author of the English Handbook and Study GuIde 
December 2011 
Acknowledg ments 
A very special thanks to my past Grade 11 and 12 teachers, Mrs Nadia Wilmans and Mrs 
Christine Geere, for making Mathematics come alive in their outstanding Maths classes 
and for Their contribution towards the information contained within this textbook. 
I feel that Professor Rex van Olst also needs a special thank you for his advice, feedback 
and encouragement whilst the idea of this textbook was still in the developmental stages. 
I would also like to extend my thanks and appreciation to Beryl Lutrin for her invaluable 
advice which has helped to make this book what it is today. 
I would like to thank my fomily for their continual support and encouragement. My 
mother Vera Smith deserves a special mention for her assistance in the graphic design of 
this textbook. I could not have done it without her. 
Thank you to my proof readers and advisors Mr Ari Levin and Dodor Belinda Huntley for 
helping to ensure that the information contained in this book is accurate and CAPS 
compliant 
Lastly, I would like to dedicate this book to all of my past and present students, 
wishing them the best of luck in Mathematics and in life. 
Ii:> Copyright Kevin Smith I Berlot Book. cc 
TABLE OF CONTENTS 
Tips on how to study with The Maths Handbook and Study Guide ........................ 1 
CHAPTER 1 - THE NUMBER SYSTEM, EXPONENTS AND SURDS 
A. The Number System ................................................................................................ 3 
• Natural Numbers ................................................................................................... 3 
• Whole Numbers .................................................................................................... 3 
• Integers ................................................................................................................. 3 
• Rational Numbers ..................................................................................................3  
• Irrational Numbers ................................................................................................. 3 
• Non Real I InIaginary Numbers .............................................................................. 3 
B. Definitions and Laws of Exponents and Surds ..................................................... 4 
• Definitions and I..aws Revision ...................................................................................... 4 
• Grade 10 Revision ......................................................................................................... 5 
• Addition and Subtraction of Fractions with Variable Exponents .................................. 6 
• Simplification of Surds .................................................................................................. 7 
• Simple Equations involving Surds ................................................................................. 10 
• Steps to solving Surd Equations .................................................................................... 10 
• Equations with Rational Exponents ............................................................................... 11 
CHAPTER 2 - EQUATIONS AND INEQUALITIES 
A. Grade 10 Revision ................................................................................................... 15 
• Factorisation ......................................................................................................... 15 
• Factorisation by Grouping ...................................................................................... 17 
• Factorisation of the Sum and Difference of CUbes .................................................... 18 
• Rules of Fractions .................................................................................................. 19 
B. Quadratic Equations ............................................................................................... 20 
• Steps to Solving Quadratic Equations ............................................................................ 20 
• Equations with Fractions ................................................................................................ 21 
• Equations with substitution (K-method) ........................................................................ 23 
• Surd Equations which lead to Quadratic Equations ....................................................... 25 
• Steps to Solving Surd Equations which lead to Quadratic Equations ............................ 25 
• Exponential Equations ................................................................................................... 28 
C. Simultaneous Equations ......................................................................................... 30 
D. Completing the Square ........................................................................................... 32 
• Completing the square on a quadratic expression .......................................................... 32 
• Completing the square on a quadratic equation ............................................................. 34 
• Applications of completing the square. .......................................................................... 36 
E. The Quadratic Formula .......................................................................................... 38 
• Derivation of the Quadratic Formula ............................................................................. 38 
• Nature of Roots .............................................................................................................. 39 
• The Discriminant ........................................................................................................... 40 
• Determining a Quadratic Equation from the Roots of an Equation ............................... 41 
F. Inequalities ............................................................................................................... 42 
• Revision of Number Lines and Notation ....................................................................... 42 
• Revision of Linear In.equalities ...................................................................................... 42 
• Quadratic Inequalities .................................................................................................... 43 
• In.equalities with Repeated Factors ................................................................................ 45 
• Unusual In.equalities ....................................................................................................... 46 
• In.equalities which do not Factorise ............................................................................... 47 
• Mixed Exatllple .............................................................................................................. 48 
© Copyright Kevin Smith I Berlot Books CC i 
G. Mathematical modelling ......................................................................................... 49 
• Steps to Solving Word Problems ................................................................................... 49 
• Number and Age Problems (Revision of Grade 10) ...................................................... 49 
• Shopping problems ........................................................................................................ 50 
• Speed, Distance an.d Time Problems .............................................................................. 51 
• Area Problems ................................................................................................................ 52 
• Rate of Flow Problems ................................................................................................... 53 
CHAPTER 3 - NUMBER PATTERNS 
A. Arithmetic (linear) Sequences ................................................................................ 55 
B. Quadratic Sequences ............................................................................................... 58 
CHAPTER 4 - ANALYTICAL GEOMETRY 
A. Summary of Analytical Formulas ......................................................................... 64 
B. Properties of Quadrilaterals ................................................................................... 65 
• Quadrilaterals ................................................................................................................. 65 
C. Grade 10 Revision of Midpoint, Distance and Gradient Formulas .................... 67 
D. Finding The Equation of a Straight Line .............................................................. 69 
E. Angle of Inclination of a Straight Line .................................................................. 70 
• Angle of Inclination when the gradient of a line is positive (m > 0) ............................. 70 
• Angle of Inclination when the gradient of a line is negative (m < 0) ............................. 70 
• Using the Angle of Inclination to determine the angle between two lines .................... 71 
CHAPTER 5 - FUNCTIONS 
A. Grade 10 Revision ................................................................................................... 76 
• Axes of Sym.metry ................................................................................................. 76 
• Asymptotes ........................................................................................................... 76 
• Domain and Range ................................................................................................7 6 
• Function an.d Mapping Notation ............................................................................. 76 
• Grade 10 Revision of The Parabola. .. o •• o •• o •• o •• o •• o •• o •• o •••••••••••••••••••••••••••••••••••••••••••••••• 77 
• Grade 10 Revision of the Hyperbola .......................................................................7 9 
• Grade 10 Revision of the Exponential Function ....................................................... 80 
B. The Parabola (Quadratic Function) ...................................................................... 81 
• The Parabola in the form of y=a(x_p)2+Q ..............................................................8 1 
2 
• The Parabola in the form. of y=ax + bx +c .................................................................. 84 
• Determining the Equation of a Parabola .................................................................. 87 
• Case 1: Given the turning point and one other point. ................................................ 87 
• Case 2: Given the x-intercepts and one other point ....................................................... 88 
• Case 3: Given the y-intercept and two other points ....................................................... 89 
C. The Hyperbola ......................................................................................................... 90 
• Axes of Symmetry an.d Points of Intersection ............................................................... 90 
• Determining the Equation of a Hyperbola ..................................................................... 93 
D. The Exponential Function ...................................................................................... 95 
• Steps to Sketching the Exponential Function ................................................................ 96 
• Determining the Equation of an Exponential Graph ...................................................... 99 
E. Translations and Reflections of Graphs ................................................................ 100 
F. Graph Interpretation .............................................................................................. 102 
• Lengths of vertical and horizontal lines ................................................................... 102 
• Interpretation and Deductions from Graphs ................................................................... 103 
• Real World Applications ................................................................................................ 107 
• Average Gradient (Average Rate of Change) ................................................................ 108 
© Copyright Kevin Smith I Berlot Books CC ii 
G. Trigonometric Functions ........................................................................................ 109 
• Grade 10 Revision of the Sin, Cos and Tan Graphs .................................................. 109 
• The effects of a (Change in Amplitude) .................................................................. 110 
• The effects of q (Vertical Shifts) ............................................................................. 110 
• The effects of p (Horizontal Shifts) ......................................................................... 111 
• The effects of k (Change in Period) ......................................................................... 114 
H. Interpretation of Trigonometric Functions .......................................................... 116 
CHAPTER 6 - TRIGONOMETRY 
A. Grade 10 Revision ................................................................................................... 119 
• Trigonometric Ratios ............................................................................................. 119 
• Special Angles ...................................................................................................... 119 
• Trigonometric Ratios in the Cartesian Plane ............................................................ 120 
B. Identities ................................................................................................................... 123 
• Proofs ................................................................................................................... 123 
• Application of Identities ......................................................................................... 124 
C. Reduction Formulas ................................................................................................ 127 
• Negative Angles .................................................................................................... 128 
• Co-Functions ......................................................................................................... 129 
• Using Reduction Formulas, Co-functions and Negative Angles with Numbers ........... 130 
D. Trigonometric equations ........................................................................................ 133 
• General Solutions .................................................................................................. 133 
• Types of General Solutions .................................................................................... 135 
• Equations in a Specified Domain ............................................................................ 139 
E. Restrictions in Trigonometric Equations .................................................................... 140 
CHAPTER 7 - MEASUREMENT 
A. Surface Area and Volume Formulas .................................................................... 142 
• Right Prisms ......................................................................................................... 142 
• Scaling Factors of Prisms ....................................................................................... 142 
• Cylinders .............................................................................................................. 143 
• Pyramids ............................................................................................................... 143 
• Spheres ................................................................................................................. 144 
• Hemispheres ......................................................................................................... 144 
• Cones ................................................................................................................... 145 
• Mixed Examples .................................................................................................... 146 
CHAPTER 8 - EUCLIDEAN GEOMETRY 
A. Revision of Lines and Triangles ............................................................................. 150 
• Straight Lines ........................................................................................................ 150 
• Para1lellines ........................................................................................................ 150 
• Triangles ............................................................................................................... 150 
• Congruency ........................................................................................................... 151 
B. Circle Terminology ................................................................................................. 152 
© Copyright Kevin Smith I Berlot Books CC iii 
C. Circle Theorems ...................................................................................................... 153 
• Theorem 1: line segment through centre and midpoint ............................................ 153 
• Theorem 1: (Converse) .......................................................................................... 153 
• Theorem 2: Perpendicular bisector of a chord ... o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o •• o ••• 154 
• Theorem 3 Angle at centre is twice the angle at the circumference ............................ 156 
• Theorem 4: Angles in same segment are equal ......................................................... 158 
• Theorem 4: (Converse) .......................................................................................... 159 
• Theorem 3 and 4: (Corollaries) ............................................................................... 159 
• Theorem 5: Opposite angles of a cyclic quadrilateral ............................................... 161 
• Theorem 5: (Converse) .......................................................................................... 161 
• Theorem 5: (Corollary) .......................................................................................... 162 
• Proving that a Quadrilateral is Cyclic ...................................................................... 162 
• Theorem6: Tangents from the same point are equal ................................................ 164 
• Theorem7: Tan-Chord Theorem ............................................................................ 166 
• Theorem7: (Converse) .......................................................................................... 166 
• Detennining which is the angle in the alternate segment ........................................... 167 
D. Mixed Examples ...................................................................................................... 169 
• Tips for Solving Riders .......................................................................................... 173 
• Summary of Theorems, Converses and Corollaries .0 •• 0 •• 0 •• 0 •• 0 •• 0 •• 0 •• 0 •• 0 •• 0 •• 0 .................. 173 
CHAPTER 9 - SINE, COSINE AND AREA RULES 
A. Revision .................................................................................................................... 177 
• Sides, angles and areas of triangles ......................................................................... 177 
• Angles of elevation and depression ......................................................................... 177 
B. The Area Rule .......................................................................................................... 178 
• Proof .................................................................................................................... 178 
C. The Sine Rule ........................................................................................................... 179 
• The Sine Rwe Proof .............................................................................................. 180 
• The Ambiguous Case ............................................................................................. 181 
D. The Cosine Rule ...................................................................................................... 182 
• The Cosine Rwe Proof. .......................................................................................... 182 
• Directions on a l>ea.ring .......................................................................................... 183 
• Mixed Ex.arnples .................................................................................................... 184 
E. Summary .................................................................................................................. 187 
• The Area Rule ....................................................................................................... 187 
• The Sine Rule ........................................................................................................ 187 
• The Cosine Rwe .................................................................................................... 187 
CHAPTER 10- FINANCIAL MATHEMATICS 
A. Simple and Compound Growth and Decay .......................................................... 189 
• Simple Growth an.d Decay ...................................................................................... 189 
• Compound Growth and Decay ................................................................................ 190 
B. The Effects of Different Compounding Periods ................................................... 192 
C. Time Lines ................................................................................................................ 195 
• Interest Rate Changes ............................................................................................ 195 
• Additional With.drawals or Deposits ....................................................................... 196 
• Mixed Ex.arnples .................................................................................................... 197 
• Nominal and Effective Interest Rates ...................................................................... 198 
© Copyright Kevin Smith I Berlot Books CC iv 
CHAPTER 11 - PROBABILITY 
A. Grade 10 Revision ................................................................................................... 201 
• Probability of an Event Occurring and Sample Space ............................................... 201 
• Theoretical and Experimental Probability ................................................................ 201 
• Notation ................................................................................................................ 201 
• Mutually Exclusive Events ..................................................................................... 201 
• The Addition Rule ................................................................................................. 202 
• Complementary Events .......................................................................................... 202 
B. Dependent and Independent Events ...................................................................... 204 
• The Product Rule ................................................................................................... 204 
C. Venn Diagrams ........................................................................................................ 206 
D. Tree Diagrams ......................................................................................................... 210 
E. Contingency Tables ................................................................................................. 212 
CHAPTER 12 - STATISTICS 
A. Revision of Terminology ......................................................................................... 215 
• Measures of Central Tendency ............................................................................. 215 
• Five Number Summary .......................................................................................... 215 
• Measures of Dispersion .......................................................................................... 215 
• Calculation of the Mean ......................................................................................... 215 
• Calculation of quartiles and percentiles ................................................................. 215 
B. Histograms and Frequency Polygons .................................................................... 216 
• Frequency Polygons .............................................................................................. 217 
C. Ogives ....................................................................................................................... 220 
D. Variance and Standard Deviation ......................................................................... 224 
• Standard Deviation with Grouped Data ................................................................... 226 
E. Symmetrical and Skewed Data .............................................................................. 227 
• Identification of Outliers ........................................................................................ 228 
CHAPTER 13 - CALCULATOR TIPS 
A. Number Patterns ..................................................................................................... 231 
B. Sketching Graphs .................................................................................................... 232 
C. Trigonometric Equations ....................................................................................... 233 
D. Determining Roots of an Equation ........................................................................ 234 
CHAPTER 14 - MAKING THE MOST OF MATHS 
A. Maths in the Classroom .......................................................................................... 236 
B. Exam Study Techniques ......................................................................................... 236 
C. The Morning of the Exam ...................................................................................... 236 
D. Exam Writing Techniques ...................................................................................... 237 
CHAPTER 15-
GLOSSARY, EXERCISES, CAPS DOCUMENTATION AND COPYRIGHT 
Glossary of Important Terms ...................................................................................... 238 
Glossary of Important Symbols ................................................................................... 244 
Exercises Table of Contents ......................................................................................... 246 
Curriculum Assessment Policy Statement - CAPS ..................................................... 394 
Term Planner ................................................................................................................ 396 
Term Planner Summary. .............................................................................................. 397 
© Copyright Kevin Smith I Berlot Books CC v 
TIPS ON HOW TO STUDY WITH 
THIS STUDY GUIDE 
1. Read slowly. A mathematics book does not flow like a novel - even missing a 
single word could cause a misunderstanding in a fundamental concept. 
2. Take note of the colours used. If something is in colour, it is there for a reason. 
3. The environment in which you study is extremely important. A quiet room 
with no distractions is conducive to learning. You will get very little out of a 
mathematics book if you cannot concentrate. 
4. Ensure that each sentence or line of an equation is understood before 
proceeding. 
5. Do not leave out anything. Each example illustrates a specific concept and 
these are rarely repeated. There is very little chance of understanding missed 
information by reading further in the textbook. 
6. All symbols in mathematics have a very specific meaning. If there is any 
uncertainty about the meaning of a term or symbol, do not set it aside, look it up 
or ask a teacher to explain it to you. The glossary at the back of this book will 
explain most of the important terms you need to know. 
7. Always study with a pencil and paper. Work out proofs and derivations 
yourself, using the textbook as a guide. After reading an example, cover it up 
and attempt tbe problem yourself, taking note of any errors you made. 
8. Matbematics is a hands-on subject. A sportsman doesn't become a 
professional by watching on the sidelines. He/she practises for hours every day. 
In the same way, you cannot become a mathematician by watching others do 
maths. Practice, practice and more practice is the only way to truly master the 
subject. 
9. Your brain also needs to rest. Do not study for hours at a time without a break. 
Effective studying requires concentration. If you are tired take a nap. 
10. When preparing for an exam or test, make sure that you do as many past 
papers as you can. If you are unsure how to answer a question, try to find the 
method in this book before looking at the answer. This will prove very satisfying 
l and build your confidence. Before your exam, practise at least two papers under exam conditions and within the allocated time. 
© Copyright Kevin Smith I Berlot Books CC I 




CHAPTER 1 
THE NUMBER SYSTEM, 
EXPONENTS AND SURDS 
WHAT YOU WILL LEARN IN THIS CHAPTER 
III ASSESSMENTS CONTENT YOU WILL BE ABLE TO: 
3 A. The Number System Take note that there exist 
3 • Natural Numbers numbers other than those on the 
real number line, the so-called 
3 • Whole Numbers 
non-real numbers. It is possible 
3 • Integers to square certain non-real 
3 • Rational Numbers numbers and obtain negative 
real numbers as answers. 
3 • Irrational Numbers 
3 • Non RealI Imaginary • Apply the laws of exponents to 
Numbers expressions involving rational 
4 B. Definitions and Laws of exponents. 
Exponents and Surds • Add, subtract, multiply and 
4 • Definitions and Laws divide simple surds. 
Revision 
5 • Grade 10 Revision 
6 • Addition and Subtraction 
of Fractions witb 
Variable Exponents 
7 • Simplification of Surds 
10 • Simple Equations 
involving Surds 
10 • Steps to solving Surd 
Equations 
11 • Equations witb Rational 
Exponents 
© copyright Kevin Smith I Berlot Books CC 2 
A. THE NUMBER SYSTEM 
The real number system consists of many different classifications. 
This section recaps on the types of numbers that you have learnt so far and introduces 
the concept of non-reall imaginary numbers. 
NATURAL NUMBERS 
{1;2;3;4; ... } 
Natural numbers start at 1 and increase in ones up to infinity (00). 
The symbol for a set of natural numbers is: N 
WHOLE NUMBERS 
{0;1;2;3;4; ... } 
Whole numbers start at 0 and increase in ones up to infinity (00). 
The symbol for a set of whole numbers is: No 
INTEGERS 
{. .. ; -4; - 3; - 2; - 1; 0; 1; 2; 3; 4; ... } 
Integers extend whole numbers to include negative numbers. 
Integers start at - 00 and increase in ones up to +00. 
The symbol for a set of integers is: Z 
RATIONAL NUMBERS 
3 • 
Examples of rational numbers are: - ; 0,3 ; 0,75 
4 
Rational numbers are numbers which can be written as a fraction (where the numerator 
and denominator are integers) or as a recurring decimal or as a decimal that 
terminates. In order for a fraction to be a rational number, the numerator and 
denominator have to be integers and the denominator cannot be equal to zero. 
The symbol for a set of rational numbers is: IQ 
IRRATIONAL NUMBERS 
Examples of irrational numbers are: 1r ;..[i; W; 2,35408 ... 
Irrational numbers cannot be written as a fraction or with terminating or recurring 
decimals. In other words irrational numbers are non-recurring and non-terminating. 
Examples of irrational numbers are pi and the square root of numbers that are not 
perfect squares or cube roots of numbers that are not perfect cubes. 
The symbol for a set of irrational numbers is: IQ' 
NON REAL / IMAGINARY NUMBERS 
Examples of non real numbers are: -/-2; v=9; 1'?J-25 
A non real number occurs when you take the even root of a negative number. Non-real 
numbers are known as complex numbers. 
The odd root of a negative number has a real solution. For example: ~ = -2 'il 
I ~ Complete Exercise 1 on page 247. 
© Copyright Kevin Smith I Berlot Books CC 3 
B. DEFINITIONS AND LAWS OF EXPONENTS AND 
SURDS 
Notation: 
The table below summarises the laws and definitions of exponents and surds that you 
have learnt so far. 
DEFINITIONS AND LAWS REVISION @... 
Definition/Law I Example Explanation 
Any base raised to the power 
of zero is equal to 1. 
( x*-o as 0° is undefined) 
A base raised to a negative 
exponent is equal to its 
2. x- m =_1_ reciprocal raised to the same 
? positive exponent. 
(once again x*-O) 
When multiplying like bases 
keep the base and add the 
exponents. 
When dividing like bases 
? 7 m-n 8x = 2x7-3 = 2x4 keep the base and subtract 4. -= x 
xn 4x3 the exponents. 
(numbers divide normally) 
When raising exponents to a 
(? r 3 3 3x3 3 5. = XnIXn (2a b)3 = 2 a b1x3 = 8a9b power, keep the base and 
multiply the exponents. 
When a fraction is raised to 
6 
6. (;r ? x an exponent, the numerator = -ym (X:J=  i and denominator must be 
raised to that exponent. 
IT the exponent is a fraction, 
m 2 the base is raised to the 
7. x-;; = ~xm 3 = ~272 = ~729=9 numerator and rooted to the 27
denominator. 
(x > 0 and n> 0) 
When surds are multiplied 
they can be split apart and 
rooted individually. 
When taking the root of a 
root, it is the same as taking 
the single root to the product 
of both roots. 
AU of the above laws and definitions also apply in reverse. 
© Copyright Kevin Smith I Berlot Books CC 4 
GRADE 10 REVISION @... 
1 
-- 0 
Example 1: Simplify 16 4 + (-10) 
1 
164 + (_10)° ~ split base into its prime factors and apply law 1 
1 
=(24)4 +1 ~applylaw 5 
= Z-1 +1 ~ apply law 2 
=1.+1 
2 
3 
= 
2 
102 2% 3 % 
Example 2: Simplify '2 1 
25%.6 % 
2x 2x 
--1-0--;::2.3---,-1 ~ sp li t b ases m. to the·l r pn.m e "la ctors 
25x.6 x-
2x 
(5.2 )2X .3
= r ~ apply law 5 (52 .(3.2)2x-l 
2x 2x 2x 
5 2 3
= 2 ; 1- 2 1 -7 apply laws 3 and 4 
5 x.3 x- .2 x-
=5 2x-2x. 32x-( 2x-l) .22x-( 2x-l) 
= 5° ,32x-2x+l.2 2x-2x+l ~ apply law 1 
= 1 1 1.3.2 
=6 
• • 92%+1.22%+2 
Example 3: Sunplify 3 2 1 
3 % .12%+ 
92x+1 22x+2 
· 1 ~ split bases into their prime factors 
3 3 x-2.12x+ 
tX 1 (32 + 22x+2 
= r+1 ~  apply law 5 33x 2- .(3 .22 
34x+2 22x+2 
= 3 2 . 1 2 2 -7 apply laws 3 and 4 
3 x- .3x+ .2 x+ 
=3 4x+2-(3x-2+x+l).22x+2-{2x+2) 
=3 4x+2-3x+2-x-l.22x+2-2x-2 
© Copyright Kevin Smith I BerlDt Books CC 5 
ADDITION AND SUBTRACTION OF FRACTIONS WITH 
VARIABLE EXPONENTS 
During addition and subtraction of fractions with variable exponents, law 3 on 
pg. 4 is generally used in reverse to split apart exponents. In most cases 
factorisation can then be used to simplify the expression. 
Sl+2x -3 S2x 
Example 1: Simplify • 
2Sx.4 
SI+2x -3 S2x 
----.- -7 split bases apart using law 3 in reverse 
2Sx.4 
2x 
= SI.S2x - 3.S -7 °th WI a ddi ti°o n and  sub  tracti°o n alw ays 1 k fi HCF 00 or a 
(S2 t .4 
~(S-3) 
=~~2x-.4-'-
2 
=-
4 
1 
= 
2 
• • 32x+2 + 32x+l 
Example 2: Simplify 2 2 2 1 
3 x+ -3 x+ 
32x+2 + 32x+1 
2 2 2 1 -7 do not cancel over a '±' sign, split bases apart using law 3 in reverse 
3 x+ -3 x+ 
2x 2 2x 1 
= 3 3 + 3 32 • 2 2 · 1 -7 with addition and subtraction always look for a HCF 
3 x.3 - 3 x.3 
_ ~(32+3) 
-~(32-3) 
12 
= 
6 
=2 
I ~ Complete Exercise 2 on page 247. 
© Copyright Kevin Smith I Berlot Books CC 6 
SIMPLIFICATION OF SURDS 
A surd is the root of a number that would result in an irrational number. 
For example, ..fi is a surd as the answer is irrational. However, J4 would not be 
referred to as a surd as it has an answer which is rational (2). 
All the laws on page 4 for rational exponents also apply to surds. ~ 
Example 1: Simplify the following, writing aU answers in simplest surd form: 
a) .,fii c) .Jii + J48 -.fi5 
b) .J98 - Jj2 -.Ji8 d) 6+.Ji8 
3 
a) JTi -7 break apart into a perfect square and another number 
= .J36x2 -7 perfect square, apply law 8 
=,J36x..fi 
=6..fi 
b) .J98 -m - .J18 -7 break apart into a perfect square and another number 
=.J49x2-.J16x2 -.J9x2 -7 perfect square, apply law 8 
=7 ..fi - 4..fi - 3..fi 
=o..fi 
=0 
c) .J12 +../48 -m -7 break apart into a perfect square and another number 
= .J4x3 +.J16x3 -.J25x3 -7 perfect square, apply law 8 
=2 ./3 +4./3 -5./3 
=./3 
6+.J18 . 
d) -7 break apart mto a perfect square and another number 
3 
6+.J9x2 
_...:c..._ -7 perfect square, apply law 8 
3 
6+3..fi 
----7 take out a HCF of 3 
3 
3, (2+..fi) 
3, 
=2+..fi 
It is important to remember the following properties of surds: 
1) Fx ~O; x ~O 
2) FxxFx =x; x~O 
© Copyright Kevin Smith I Berlot Books CC 7 
Example 2: Simplify (J7 + 2) (J7 - 2) without the use of a calculator. 
2 2 
(J7 + 2)( J7 - 2) ~difference of squares (a+b)(a-b) = a _b
= (J7)2 _(2)2 
=7-4 
=3 
Example 3: Simplify (M +.J3) ( M -.J3) without the use of a calculator. 
2 2 
(2v'2 +.J3)( 2v'2 - .J3) ~ difference of squares (a+b)(a-b) = a _b
= (2v'2)2 _ (.J3)2 
=8-3 
=5 
Example 4: Simplify (3.J3 - .J8)2 without the use of a calculator. 
(3.J3 -,J8t ~ perfect square, don't forget the middle tenn: (a - b)2 a2 = - 2ab 2 + b
= 27 - 6.J24 +8 ~ 2x3.J3x,J8 = 6.J3x8 = 6.J24 
= 35- 6.J4x6 ~6x.J4x6 =6x2-/6 =12-/6 
=35-12-/6 
Example 5: Simplify (2.J2 -6)(.J8 +6) without the use ofa calculator. 
(2v'2 - 6) (,J8 + 6) ~ multiply out 
= 2M +12v'2 -6,,[8 -36 ~-6,J8 = -6.J4x2 =-12v'2 
= 8 + 12v'2 -12v'2 - 36 
=-28 
Example 6: Simplify ~64x6 -#Slx8 +~25x4 without the use ofa calculator. 
~64x6 - # 81x8 +~25x4 ~ rewrite using law 7 
1 1 1 
=( 64x6)3 -(81x8 )"4 + 4 (25x )2 ~split numbers into their prime factors 
1 1 1 
3 6
= (4 x )3 4 8 2 4-(3 x )4 + (5 x )2 ~uselaw 5 
=4x2 -3x2 +5x2 
2 
=6x
© Copyright Kevin Smith I Berlot Books CC 8 
E xampI e  7 : SI· mpl ify  6.[2;8C -2fJi 8;8 W'It ho ut  the  use 0 f a calculato r. 
8 
v8x
6W -2.J8j . . 
CO -7 split numbers mto a perfect square and another number 
8 
vSx
6W -2~4X2xx8 
= I -7 perfect squares use law S 
v 4x2x x8 
6x4.J2-4x4 .J2 
= 
2x4 .J2 
-~ 
-
=1 
Recap: Rationalising the denominator 
It is often useful to write fractions with rational numbers in the denominator. 
When a surd is present in a denominator it is possible to write the denominator 
as a rational number by multiplying the numerator and denominator by the 
surd. 
x 
2,Ja 2b Ua(l+b} . 
Example 8: Prove that --+ r = Without the use of a calculator. 
a va a 
2~ 2b , 
LHS=--+ , -7LCDof ava 
a va 
2~(~)+2ab 
= 
a~ 
2a +2ab . 
= , -7 take out a highest common factor of 2a 
ava 
2,.a' (I+b} 
= 
,.a'~ 
= 2 (~b) -7 rationalise the denominator by multiplying by I in the form of ~ 
2(I+b} ~ 
--'--;=--"-x-
~ ~ 
2~(I+b} 
= =RHS 
a 
Complete Exercise 3 on pages 247 and 248. 
© Copyright Kevin Smith I Berlot Books CC 9 
SIMPLE EQUATIONS INVOLVING SURDS 
In Grade 11 you are expected to be able to solve equations with surds. 
Simple surd equations can be solved by inspection or by squaring both sides of the 
equation as shown in the examples below. 
Whenever dealing with surd equations you have to check your answers by substituting 
your answer into the original equation. ~ 
STEPS TO SOLVING SURD EQUATIONS 
1. Isolate the surd. 
2. Square both sides. 
3. Solve for x. 
4. Check your answer. 
Example 1: Determine the value of x if ../ x + 2 - 3 =O .  
../x+2 -3 = 0 ~isolate the surd 
: . ../ x + 2 = 3 ~ square both sides and note that ( ..r af = a 
(../x+2 )2 = (3)2 
:.x+2=9 
:.x=7 
Check: ../7 + 2 - 3 = 0 ~ answer is valid 
Example 2: Determine the value of x if ../ x - 5 + 4 =8  . 
../x-5 +4 =8 ~isolate the surd 
: . ../ x - 5 = 4 ~ square both sides 
:. (../x_5 )2 =(4)2 
:.x-5=16 
:.x=21 
Check: ../21-5 + 4 = 8 ~ answer is valid 
Example 3: Determine the value ob if 3../40-x -5 =1 3. 
3../40 - x - 5 = 13 ~ isolate the surd 
.'. 3../40 - x = 18 ~ square both sides 
:. (3../40_x )2 = (18 )2 
:. 9(40-x) =324 ~don't forget that the 3 also had to be squared 
:. 360-9x = 324 
:.-9x=-36 
:.x=4 
Check: 3../40 - 4 - 5 = 13 ~ answer is valid 
I ~ Complete Exercise 4 on page 248. 
© Copyright Kevin Smith I Berlot Books CC 10 
EQUATIONS WITH RATIONAL EXPONENTS 
In Grade 11 you will deal with equations which have rational exponents 
(exponents with fractions). Simple exponential equations are also shown in this 
section. These will be explored in more detail further on in the year. 
I'.-
If an equation is in the form of: x q =a  
• There will be a positive and negative solution if p is even and q is odd. 
• There will be 1 solution if p is odd. 
Dlustrative Example: Solve for x if x% - 22 =O . 
There are two methods that can be used to solve the above equation: 
Method 1: 
x% - 22 = 0 ~ difference of squares 
:. xX' = 2 or xX' = -2 ~ raise both sides to the power of 3 
:.(xX'J =(2)3 or (xX' J = {_2)3 
:.x=8 or x=-8 
Method 2: 
x%=  22 ~ P is even and q is odd, :. there will be a positive and negative solution 
:. (x tl/tJ %= ±(2t )~ t  ~bothsideswereraisedtothepowerof % 
:.x= ±23 =±8 
Example 1: Solve for x if x% = 24. 
x%=  24 ~ raise both sides to the power of % 
:. (x  %t J% =± (2 4)Y3 z ~ don't forget the '±' as p is even and q is odd 
Example 2: Solve for x if xYz = 125 
xYz =1 25 ~ convert 125 to a prime base 
:. xYz =5 3 ~ raise both sides to the power of % 
:.(x Yz )% = (53 )% ~ only 1 solution (no '±') as p is odd 
:.x=52 =25 
© Copyright Kevin Smith I Berlot Books CC 11 
Example 3: Solve fou if (x - 3)% = 9. 
( x - 3)% = 9 ---) convert 9 to a prime base 
:. (x-3)% 2 =3 ---)raise both sides to the power of % 
:.[(X-3)% t 'i =±(3t )Yt ---) don't forget the '±' as p is even and q is odd 
:.x-3=27 or x-3=-27 
:.x=30 or x=-24 
Example 4: Solve for x if 3Yz = 9. 
3~ = 9 ---) convert 9 to a prime base 
:. 3~ 2 = 3 ---) if the bases are the same the exponents have to be equal 
:.~=2 
:.x=4 
8 
Example 5: Solve for x Ys 2if x 5 = II 
3 
41 28 
xiS =I f ---) raise both sides to the power of % 
3 
:. (x Ys f~ = ± ( !: t ---) don'tforget the '± ' 
210 5 
:. x = 310 ---) 8X"4 =1 0 
Example 6: Solve for x if 4.3Yz = 9.2Yz 
4.3~ = 9.2~ ---)isolate the bases with variables (divide both sides by 2~ and 4) 
~ 
:. 3 x2l  = -9 ---) convert 9 and 4·m to the·r r pn.m e b ases 
272 4 
3~ 32 
:. 2~ = 22 
:. (%)~ = (% r-- -) as the bases are the same the exponents are equal 
... ~ = 2 ---) multiply both sides by 2 
2 
:.x=4 
~ Complete Exercise 5 and the mixed exercise on pages 248 and 249. 
II) Copyright Kevin Smith I Berlut Books CC 12 




CHAPTER 2-
EQUATIONS AND INEQUALITIES 
WHAT YOU WILL LEARN IN THIS CHAPTER 
1:1 CONTENT III ASSESSMENTS YOU WILL BE ABLE TO: 
15 A. Grade 10 Revision 4 I. Factorise expressions. 
15 • Factorisation 
• Solve quadratic equations 
17 • Factorisation by Grouping 
18 • Factorisation of the Sum • Solve quadratic inequalities in 
and Difference of Cubes one variable and interpret the 
solution graphically 
19 • Rules of Fractions 
20 B. Quadratic Equations • Solve equations in two 
20 • Steps to Solving Quadratic unknowns, one of which is 
linear the other quadratic, 
Equations algebraically or graphically. 
21 • Equations with Fractions 
23 • Equations with substitution • Complete the square on 
quadratic expressions and 
(K-method) 
equations 
25 • Surd Equations which lead 
to Quadratic Equations • Solve equations using the 
25 • Steps to Solving Surd quadratic formula 
Equations which lead to • Determine the nature of roots 
Quadratic Equations 
28 • Exponential Equations 
30 C. Simultaneous Equations 
32 D. Completing the Square 
32 • Completing the square on 
a quadratic expression 
34 • Completing the square on 
a quadratic equation 
36 • Applications of completing 
the square 
38 E. The Quadratic Formula 
38 • Derivation of the 
Quadratic Formula 
39 • Nature of Roots 
40 • The Discriminant 
41 • Determining a Quadratic 
Equation from the Roots of 
an Equation 
© Copyright Kevin Smith I Berlot Books CC 13 
CHAPTER 2-
EQUATIONS AND INEQUALITIES 
WHAT YOU WILL LEARN IN THIS CHAPTER 
III III ASSESSMENTS CONTENT YOU WILL BE ABLE TO: 
42 F. Inequalities 
42 • Revision of Number lines I 
and Notation 
42 • Revision of linear 
Inequalities 
43 • Quadratic Inequalities 
4S • Inequalities with Repeated 
Factors 
46 • Unusual Inequalities 
47 • Inequalities which do not 
Factorise 
48 • Mixed Example 
49 G. Mathematical modelling 
49 • Steps to Solving Word 
Problems 
49 • Number and Age Problems 
(Revision of Grade 10) 
50 • Shopping problems 
51 • Speed, Distance and Time 
Problems 
52 • Area Problems 
53 • Rate of Flow Problems 
© copyright Kevin Smith I Berlot Books CC 14 
A. GRADE 10 REVISION 
FACTORISATION 
In this section you will revise the factorisation that you learnt in Grade 10. 
Factorisation by taking out a Highest Common Factor 
Example 1: Factorise x 2 -2x. 
x2 - 2x ~ x is common to both terms 
=x(x-2) 
Example2: Factorise 2x(x-2)-3(x-2). 
2x(x-2) -3(x-2) ~ (x-2) is common to both terms 
=(x-2)(2x-3) 
Factorisation of the DiHerence of Squares 
When factorising the difference of two squares, rewrite the two squares as the 
products of their roots. 
2 2 
x _a =(N -~)(N +~)=(x-a)(x+a) 
Only use the difference of two squares when the squares are separated by a minus sign. ~ 
Example 1: Factorise x 2 -4. 
x2 -4 
=(x-2)(x+2) 
Example 2: Factorise 27 x 2 -12. 
27x2-12 
2 
= 3 (9x - 4 ) ~ always look for a HCF fIrst 
=3(3x-2)(3x+2) 
Example 3: Factorise 8lx4 -16. 
81x4 -16 
= (9x2 + 2 4) (9x - 4 ) ~ this can be factorised further 
(9x2 = +4)(3x-2)(3x+2) 
© Copyright Kevin Smith I Berlot Books CC 15 
Factorisation of Trinomials 
2 
The general form a of a quadratic trinomial is ax ± bx ± c. 
There are three types of trinomials that you will be expected to factorise: 
1. Trinomials of the form x 2 ±bx +c -7 (a =1). 
2. Trinomials of the form x 2 ±bx - c -7 (a = 1). 
2 
3. Trinomials of the form ax ±b x ±c  -7 ( a *- 1) • 
1. Factorising Trinomials in the form of x 2 ±b x + c. 
x2 ± bx t c ... if the sign before the constant is a '+' then you 
L...J know both signs in the brackets will be the same 
Lb oth signs in the brackets will be the same as the 
sign of the coefficient of x 
Dlustrative Example 1: Factorise x 2 + 8x + 15. 
x2 +8x+15 
Both signs will be the same because the coefficient of 15 (the constant) is a '+'. 
Both signs will be '+' as the coefficient of x is a '+'. 
(x+O)(x+O) 
In order to find the missing numbers you look for the factors of 15 (the constant) that 
will add up to 8, the coefficient of x. 
Factors of 15 are: 15 ~ -7 5 + 3 = 8 
1 3 
... x2 +8x+15 = (x +5)(x+3) 
2. Factorising Trinomials in the form of x 2 ±b x - c. 
x2 +-Lbx +c .... if the sign before the constant is a '-' then you L...J know there will be opposite signs in the brackets 
the larger of the two factors will get the 
sign of the coefficient of x 
Dlustrative Example 2: Factorise x 2 + 2x -15. 
x2 +2x - 15 
The signs will be opposite because the coefficient of 15 (the constant) is a '- '. 
The larger factor will get a '+' as the coefficient of x is a '+'. 
(x+O)(x-O) 
In order to find the missing numbers you look for the factors of 15 (the constant) where 
the difference between the factors will be equal to 2, the coefficient of x. 
Factors of 15 are: 15 [51-75 - 3 = 2 
l W 
:. x2 + 2x - 15 = (x - 3)(x+5) -75 gets a '+' sign as it is the larger factor 
© Copyright Kevin Smith I Berlot Books CC 16 
3. Factorising Trinomials in the form ofax2 ± bx - c. 
When the coefficient (a) of ~ is not equal to one, the cross method is used to 
factorise the expression. Ibis is illustrated in the example below. 
Dlustrative Example 1: Factorise 2x2 +7x +3. 
In this method you draw a cross with the factors of 2 (a) on the left side and the factors 
of 3 (the constant) on the right side. You then multiply diagonally across the cross. The 
factors which add or subtract to equal 7 (the coefficient of x) will be the correct factors. 
(2' 3) (2. X 1) 
(1. X 1) (1. 3) 
2xI = 2 and 3xl=3 2x3 = 6 and Ixl=I 
Incorrect as you cannot correct: 6 + 1 = 7 
make 7 with a 2 and a 3 
:. 2x2 +7x+3 = (2x+I)(x+3) 
Dlustrative Example 2: Factorise 2 6x - x - 2. 
2 (6. 2) 2 2
(l'X ) (2'X ) (3'X - ) 
(6. I) (I. X I) (3, I) (2, I) 
Ixl = I and 6x2= 12 6xl= 6 and Ix2= 2 2xl = 2 and 3x2= 6 3xl= 3 and 2x2= 4 
Incorrect as you ClIJl1lot Incorrect as you ClIJl1lot Incorrect as you cannot Correct: you can get -I 
make -I with a I and 12 make -I with a 6 and 2 make -I with a 2 and 6 with a 3 and 4. However. 
the 4 must be negative. 
2 
:. 6x -x-2 = (3x-2)(2x+l) 
~ Always look for a HCF before using the cross method. ~ 
FACTORISATION BY GROUPING 
The idea behind grouping is to group terms together and then, if possible, take 
out an HCF in order to obtain a common bracket. 
2 
Example 1: Factorise 6x - 2ax - 3xy +ay. 
2 
6x - 2ax - 3xy + ay 
= 3x (2x - y) - a (2x - y) -7 group tenns to obtain a common bracket 
=(2x- y)(3x-a) 
~ Grouping is generally used if there are more than 3 terms in an expression. 
© Copyright Kevin Smith I Berlot Books CC 17 
FACTORISATION OF THE SUM AND DIFFERENCE OF CUBES 
The sum and difference of perfect cubes always factorise into the form shown 
below: 
1" bracket: cube root each term and keep the same sign 
2"" bracket: square the first term 
change the sign and multiply both terms together 
square the last term (the sign of the last term is always positive) 
The sum and difference of cubes always Cactorises into a binomial and a trinomial 'if 
Example 1: Factorise 27x3 + y3 
3
27x +l 
= (3x+ y}(9x2 -3xy + y2) 
Example 2: Factorise 54x3 -16 
54x3 - 16 ~ always look for a RCF when factorising (2 is common to both tenns) 
= 2(27x3-8) 
= 2(3x-2}(9x2 +6x +4) ~ you cannot drop the 2 
Example 3: Factorise (x - y f - 3 64x
(x - y)3 _ 64x3 ~ difference of two perfect cubes 
2 
= [(x- y)-4xJ [(x- y)2 +4x(x- Y)+16x ] 
= (x - y _4X)(X2 -2xy + y' +4x2 -4xy + 16x2) ~ gather like terms 
= (-3x - y}(21x2 -6xy + y2) 
a;) In factorisation of the sum and difference of two cubes the trinomial (in the 
l{ second bracket) will never factorise further in the real number system. 
Complete Exercise 1 number 1 on page 258. 
© Copyright Kevin Smith I Berlot Books CC 18 
RULES OF FRACTIONS 
Rule Explanation 
a e ac When multiplying fractions, multiply the 
-x-=- numerators and the denominators. 
1 b d bd 
Factorise numeraton and denominaton if 
Restrictions: b *- 0 , d*-O possible and try to cancellirst. Ci( 
a e a d ad When dividing fractions, Invert whatever 
-+-=-x-=- Is after the divide sign and change the 
2 b d b e be operation from division to multiplication. 
Restrictions: b *- 0 , c*-O , d*-O (This is known as the tip and times rule.) 
a+e _ate Fractions can only be added (or 
- ---- subtracted) if they have the same 
3 b b b denominators. If the denominators are the 
same, keep the denominator and add the 
Restrictions: b *- 0 numerators together. 
. x 2lif +5x-6 x2+8x+12 
Example 1: SImpY2 :2 • 
x +2x-15 x +7x+l0 
~+~-6 ~+b+12 . . 
2 : 2 -7 factonse numerators and denommators 
x +2x-IS x +7x+1O 
_ (x+6}(x-l) . (x+6)~ 
- (x+S}(x-3) . (x+S) ~ 
(~ (x-l) (~~ 
= ~uJ x~-'J -7 'tip and times' 
~(x-3) ~ 
x-I 
= 
x-3 
Example 2: SI· mpl ify  x+ 5 +-9- +-3x- . 
x-3 3-x 
9 3x 
x+S+--+-- -7take out a '-'to make a LCD of x-3 
x-3 3-x 
9 3x 
=x+S+--- -- -7LCD=x-3 
x-3 x-3 
=- ('x-+--S-}'-("x'--3--)-+-9:'--3_x- 
x-3 
x2 +2x-lS+9-3x 
x-3 
x2 -x-6 
x-3 
_~(x+2) 
- 0 
=x+2 
I ~ Complete Exercise 1 number 2 on page 258. 
© Copyright Kevin Smith I Berlot Books CC 19 
B. QUADRATIC EQUATIONS 
A quadratic equation is an equation of the form: ax' + bx + c =0  
In most cases a quadratic equation will have two solutions. 'ir 
STEPS TO SOLVING QUADRATIC EQUATIONS 
1. Make the right hand side or left hand side equal to zero. 
2. Factorise (always look for a common factor first). 
3. Solve for x. 
U) The x-values of the solutions to a quadratic equation are known as roots or 
l' zeros. These represent the x-intercepts of a function. 
Example 1: Solve for x if x 2 -16 = O. 
x2 -16=0 
:. (x-4)(x+4) = 0 
.'. x - 4 = 0 or x + 4 = 0 -7 it is not necessary to show this step 
:.x=4 or x=-4 
Example 2: Solve for x 2 if x -16x =O . 
x2 -16x = 0 -7 take out aHCF of x 
:. x {x-16) = 0 -7 do not divide by x as x could be equal to 0 
:. x = 0 or x-16 = 0 -7 it is not necessary to show this step 
:.x=O or x=16 
Example 3: Solve for the roots of x ( x - 9) = -18. 
x{x-9) = -18 -7LHS orRHS must be equal to 0 
:. x2 -9x+18 =0 -7factorise 
:. (x-6)(x-3) = 0 
... x - 6 = 0 or x - 3 = 0 -7 it is not necessary to show this step 
:.x=6 or x=3 
Example 4: Solve for x if -5x2 - 5x + 30 = O. 
-5x2 -5x+30=0 -7 take outaHCF of -5 
:. -5 (x2 +x-6) =0 -7 divide both sides by -5 
:.x2 +x-6=0 
:. (x+3)(x-2) =0 
:.x=-3 or x=2 
© Copyright Kevin Smith I Berlot Books CC 20 
Example 5: Solve for x if 3 X - 2X2 + X = o. 
X3_ 2x2+X=O 
:. X(X2 -2x+l) = 0 -Hlways look for aRCF first 
:. x(x-l)(x-l) = 0 ~ do not divide by x asx could be equal to 0 
:. x=O or x-l=O 
:. x=O or x=l 
I ~ Complete Exercise 2 on page 258. 
EQUATIONS WITH FRACTIONS 
Example 1: Solve for x if _1_ = 1-~. 
x+l x-I 
_1_=1_ 2x 
x+l x-I 
LCD: (x-l)(x+l) x *- 1 and x*- -1 ~ you have to put in restrictions 
:.l(x-l) = l (x-l)(x+l) -2x(x+l) ~ multiply both sides by the LCD 
:. x-I = x2 -1-2x2 -2x 
:. x-I = 2 _x -2x-l ~ quadratic equation, make one side equal 0 
:. x2 +3x = 0 ~ take out a RCF ofx 
:.x(x+3)=O 
:.x=O or x=-3 
x 2 -1 
Example 2: Solve for x if 2. 
x+l 
LCD: (x+l) x *- -1 ~ you have to put in restrictions 
:. x2 -1 = 2(x+ 1) ~ multiply both sides by the LCD 
:. x2 -1 = 2x + 2 ~ quadratic equation, make one side equal 0 
:.x2-2x-3=O 
:.(x-3)(x+l)=O 
:.x=3 or x=-1 
but from the restriction above x *- -1 
:.x=3only 
© Copyright Kevin Smith I Berlot Books CC 21 
• 2x x 3 
Example 3: Solve for x if -----= --. 
x+3 x-3 3-x 
2x x 3 
-----=--
x+3 x-3 3-x 
h x 3 . 
:. -----= - --~ take out a '-' to make a common denommator of x - 3 
x+3 x-3 x-3 
LCD: (x - 3)( x + 3) x""' ±3  ~ you have to put in restrictions 
:. 2x{x-3)-x{x+3) =-3{x+3) ~ both sides were multiplied by the LCD 
:. 2x2 -6x-x2 - 3x=-3x- 9 ~ don't forget to distribute the '-' 
:.x2-6x+9=0 
:. (x-3)(x-3) =0 ~ this could also be written as {x_3)2 
:.x=3 
but x ""' 3 from the above restrictions 
:. No solution 
. 3x+6 4 -4 
Example 4: Solve for x if 2 +-- = 2 • 
x -x-2 2-x x +4x+3 
3x+6 +-4 - -4 ". lin de . 2 = 2 ~ lactonse non- ear normnators 
x -x-2 2-x x +4x+3 
3x+6 4 -4 ~ tak " did' I 3 b :. ( )() (){) e out a - as you III examp e a ove 
x-2 x+1 x-2 x+3 x+1 
LCD: (x+3)(x+1)(x-2), x,,", -3 and x ""' -1 and x ""' 2 ~ put in restrictions 
:. (3x+6 )(x+3) -4{x+3)(x+ 1) = -4{x-2) ~ both sides were multiplied by the LCD 
2 
:. 3x + 15x+ 18-4(x2 +4x+3) =-4x+8 
:. 3x2 + 15x+ 18 - 4x2 - 16x- 12 =-4x+8 ~ don't forget to distribute the '-' 
:.-x2+3x-2=0 
2
:.x -3x+2=0 
:.{x-1)(x-2)=0 
:.x=1 or x=2 
but x ""' 2 from the above restrictions 
:.x=10nly 
You can only multiply both sides by the LCD (commonly known as dropping 
~ the LCD) if you are dealing with an equation. Always put your restrictions in 
and check that none of your answers are equal to your restrictions. 
I ~ Complete Exercise 3 on page 258. 
© Copyright Kevin Smith I Berlot Books CC 22 
EQUATIONS WITH SUBSTITUTION (K-METHOD) 
When an equation has an expression which is repeated, it is often simpler to 
replace the repeated expression with a single variable. 
The variable that is normally used is Ie, hence the name k-method. 
2 
Example 1: Solve for x if (x -3xt  -S(x2 -3x )-20= O. 
(x2 _3x)2 -8(x 2 -3x) -20 =0 ~repeated expression, use the k-method 
2 
Letk =x -3x 
:. k 2 -8k -20=0 
:.(k -lO){k +2)=0 
:. k =lO or k =-2 
2 
Now substitute k = x - 3x back into your solutions for k 
~ ~-h=IO oc ~-h=4 
2 
or x -3x+2=0 
:.(x-5){x+2)=0 or (x-2){x-I)=0 
:.x=5 or x=-2 or x=2 or x=1 
2 
Example 2: Solve for x if ;0 + 5 = x + 1. 
x +1 
+ = 2 ; 0 5 x + I ~ repeated expression, use the k-method 
x +1 
2
Letk=x +1 
:, 50 +5= k 
k 
2 
LCD =k , k "" 0,:. x + I,," 0 ~ you have to put in restrictions 
:.50+5k =k2 
:. k 2 -5k-50 =0 
:. (k -lO){k +5) =0 
:. k =lO or k =-5 
2 
Now substitute k = x + 1 back into your solutions for k 
:. x2+1 =1O or x2+1 =-5 
.' . x2 = 2 9 or x = -6 ~ no real solutions as you cannot square root - 6 
.' . x = ±3 ~ don't forget the '±' and remember that this is two answers, + 3 and - 3 
£1) If you obtain an X4 when simplifying an equation, the k-method is the most 
1'" likely method to use provided that you cannot take out a ReF. 
© Copyright Kevin Smith I Berlot Books CC 23 
2 
Example 3: Solve for x if X - 4x = 60 17 
x 2-4x' 
x2 
60 
-4x =17 2 -Hepeated expression, use the k-method 
x -4x 
Letk=x2-4x 
:. k =17- 60 
k 
2 
LCD =k, k "- 0,:. x -4x "- 0 :. x "-0 andx"- 4 -7 you have to put in restrictions 
:. k2 =17k -60 
:. e -17k +60=0 
:. (k -5)(k -12) = 0 
:. k =5 or k =12 
Now substitute k = x2 - 4x back into your solutions for k 
:.x2-4x=5 or x2-4x=12 
:.x2-4x-5=0 or x2 -4x-12=0 
:.(x-5)(x+l)=0 or (x-6)(x+2)=0 
:.x=5 or x=-1 or x=6 or x=-2 
Example 4: Solve for x if 2x2 -4x =3 6 ;70 
3x -6x 
2x2 -4x = 36 ; 70 -7 take out a HCF to obtain the repeated expression 
3x -6x 
2
:.2(x -2x)=36 ( ~70 ) -7 repeated expression, use the k-method 
3 x -2x 
Letk=X2_2x 
:. 2k = 36- 270 
3k 
LCD =3k, k "- 2 0,:. x -2x "- 0 :. x"- 0 andx"- 2 -7 you have to put in restrictions 
:.6k2 =lOSk -270 
:. 6k2 - lOSk + 270 = 0 -7 take out a HCF of 6 
:. 6(k2 -ISk +45) =0 -7 divide both sides by 6 and factorise 
:. (k -15)(k -3) = 0 
:. k =15 or k =3 
2 
Now substitute k = x - 2x back into your solutions for k 
:. x2 -2x=15 x2 or -2x = 3 
:. x2 -2x-15 = 0 x2 or -2x-3 =0 
:.(x-5)(x+3)=0 or (x-3)(x+l)=0 
:.x=5 or x=-3 or x=3 or x=-1 
I ~ Complete Exercise 4 on page 259. 
© Copyright Kevin Smith I Berlot Books CC 24 
SURD EQUATIONS WHICH LEAD TO QUADRATIC EQUATIONS 
(SOLVING EQUATIONS BY SQUARING BOTH SIDES) 
In Chapter 1 you dealt with simple linear surd equations (pg. 10). You will now 
deal with surd equations which lead to a quadratic equation. It is important to 
note that in this case you may introduce an answer which is not a solution to the 
original surd equation. 
For example consider the case if you solve for x in the equation ,jx  + 2 = x : 
,jx  + 2 = x ~ square both sides (line 1: original surd equation) 
:.x+2=x2 (line 2: new quadratic equation) 
:.x2 -x-2=0 
:.(x-2){x+l}=0 
:.x=2 or x=-I~N/A 
Two solutions have been obtained for x. Both solutions are valid for line 2 of the 
above equation. However, if the solutions are substituted back into the original 
equation (line 1) it can be seen that x = -1 is not a solution as it will cause the square 
root of a number to be negative which is not possible. 
For this reason it is essential that you always check your answers by substituting 
your solutions back into the original equation when dealing with surd equations. 
STEPS TO SOLVING SURD EQUATIONS WHICH LEAD TO 
QUADRATIC EQUATIONS 
1. Isolate the surd. 
2. Square both sides. 
3. Solve for x as you would in a normal equation. 
4. Check your answers 
Example 1: Solve for x if ,j2x + 8 - x =O . 
,j2x + 8 - x = 0 ~ isolate the surd 
.'. ,j2x + 8 = x ~ square both sides 
2 
:.2x+8=x
:.~-2x-8=0 
:.(x-4){x+2}=0 
:.x=4 or x=-2~N/Aseecheckbelow 
:.x=4 only 
Check 
x=4 x=-2 
LHS=~2(4}+8-4 LHS=~2(-2)+8-(-2} 
=0 =4 
=RHS i'RHS 
:. x = 4 is a solution :.xi'-2 
~ It is not necessary to show your check, only which answer is not applicable. 
© Copyright Kevin Smith I Berlot Books CC 25 
Example 2: Solve for x if 3../ x + 4 - x =o . 
3../ x + 4 - x = 0 ~ isolate the surd 
:. 3../ x + 4 = x ~ square both sides, don't forget to square the 3 as well 
2 
:. 9(x+4) = x
2 
:.9x+36=x
:. x2 -9x-36 = 0 
:. (x-12)(x+3) = 0 
:.x=12 or x=-3~N/Aseecheckbelow 
:.x=12 only 
Check 
x=12 x=-3 
UIS=3~(12)+4-12 LHS =3~( -3+4) -(-3) 
=0 =6 
=RHS *-RHS 
:. x = 12 is a solution :.x*--3 
Example 3: Solve for x if 2../ x-I + 6 =2 x. 
2../ x-I + 6 = 2x ~ isolate the surd 
:. 2../x-l = 2x-6 ~ square both sides, don't forget to square the 2 as well 
:. 4(x-l) = (2x-6)2 
Recap: 
:. 4x-4 =4x2 -24x +36 ~don't forget the middle term 
:. 4x2 - 28x + 40 = 0 ~ take out a HCF of 4 and factorise 
:. 4(x2 -7x+l0) =0 ~divide both sides by4 and factorise 
:. (x-5)(x-2) = 0 
:. x = 5 or x = 2 ~ N/A see check below 
:.x=5only 
Check 
x=5 x=2 
UIS =2~(5)-1+6 LHS =2~(2)-1+6 
=10 =8 
RHS=2(5) RHS=2(2) 
=10 =4 
=UIS *-UIS 
:. x = 5 is a solution :.x*- 2 
© Copyright Kevin Smith I Berlot Books CC 26 
Example 4: Solve for x if 2..) x + 1 + 1 = ~. 
x+l 
2..) x + 1 + 1 = ~ ~ repeated expression, use k-method 
2 x+l 
Letk=2..)x+l 
:. k +l= 20 
k 
LCD = k, k '" 0 ~ you have to put in restrictions 
:. k2 +k =20 
:. k2 +k -20=0 
:.(k -4)(k +5)=0 
:. k =4 or k =-5 
Now substitute k = 2..) x+ I back into your solutions fur k 
:.2..) x+ 1= 4 or 2..) x+ 1= -5 ~ N/A a square root cannot be a negative number 
:.4(x+l)=16 
:.4x+4=16 
:.4x=12 
:.x=3 
Check 
x=3 
LHS =2~(3)+1+1 RHS = 20 
2~(3)+1 
=5 =5 
=LHS 
:. x = 3 is a solution 
I ~ Complete Exercise 5 on page 259. 
© Copyright Kevin Smith I Berlot Books CC 27 
EXPONENTIAL EQUATIONS 
An exponential equation is an equation in which the variable you are solving for 
is an exponent. When solving exponential equations we use the property that: 
if aX =a ' , then x = y 
It is also possible for exponential equations to lead to quadratic equations. Cf 
Example 1: Solve for x if 27x =!.. 
9 
27X =!. 
9 
:. (33 r r 2 = ~ ensure that you have the same base on both sides of the equation 
3x 
:.3 =r2 
:.3x=-2 
2 
:.X=--
3 
2 
Example 2: Solve for x if 125x .25x- =1 . 
125x.25x 2 - =1  
2 
:. (53 r( 52 r- = 5° ~ write numbers as their prime bases, 5° = 1 (law/definition 1) 
:. 53x.52x-4 = 50 ~ don't forget to distribute the 2 
.·.53x+2x-4 = 50 
.·.3x+2x-4=O 
.·.5x=4 
4 
.·.X=-
5 
Example 3: Solve for x if 3x+2 + 3x+3 - 3x =3 15. 
3x+2 + 3x+3 x _ 3 = 315 ~ split bases apart using law 3 in reverse 
:. 3x.32 +3x.33 _ 3x =315 ~ take out a HCF of 3x 
x 2 + 3 :. 3 (3 3 -1) = 315 
x 
:. 3 (35) = 315 ~ divide both sides by 35 
:.3x =9 
:.3x 
2 
=3
:.x=2 
© copyright Kevin Smith I Berlot Books CC 28 
Example 4: Solve for x if 22.< - 2'< - 2 =O . 
22.< _ 2x _ 2 = 0 -722x = (2X)2 , use the k-method 
Letk=2x 
:. k2 - k - 2 = 0 -7 factorise 
:.(k -2}(k +l}=0 
:. k =2 or k =-l 
Now substitute k = 2x back into your solutions for k 
:.2X =21 x or 2 = -1 -7 NtA  an exponent cannot cause a number to be negative 
:.x=l 
Example 5: Solve for x if 32.< -10.3'< + 9 =O . 
32x -1O.3x 2x x+9 = 0 -7 3 = (3 )2 , use the k-method 
x 
Letk=3
2 
:. k -10k + 9 = 0 -7 factorise 
:.(k -9}(k -l}=0 
:. k =9 or k =l 
Now substitute k = 3x back into your solutions for k 
x x 
:.3 = 9 or 3 = 1-7 get the same bases on both sides of each equation 
x 2 x 
:.3 = 3 or 3 = 3° -7 any number raised to the power of 0 is equal to I 
:.x=2 or x=O 
Example 6: Solve for x if 22. 0 
:. the inequality is always greater than or equal to 7. 
:. There is no solution 
Example 3: Solve for x if x 2 -6x+14> 0 
x2-6x+14>0 
CV: x = 6 ± J=W ---) no solution to critical values, therefore complete the square 
2 
:. [x2 -6x+(-3)2] _ ( _3)2 +14> 0 ---) GX4 J = (2)2 was added and subtracted 
:. (x -3)2 +5 > 0 ---) this will always be 1. the coefficient of x 
2 
AIl you have seen previously, this inequality has a minimum value of 5 ---) a > 0 and q > 0 
:. the inequality is always greater than or equal to 5. 
:.XE lR. 
© Copyright Kevin Smith I Berlot Books CC 47 
MIXED EXAMPLE 
~x2-x-6 
Example: For what values of x are the roots of the expression : 
x+2 
a) Real. 
b) Non-real. 
c) Undef'med. 
a) For real roots x2 -x-6;::: 0 and x;,,-2 
:. (x-3)(x+2);::: 0 
:. Critical values: x = 3 or x=-2 
E 0 . , 
+ I 
E I + .. 
-2 3 
:. x < -2 or x;::: 3 ~ x;" 2 as it would cause a 0 in the denominator 
This can also be written as XE (-oo;-2) or x=[3;oo) 
b) For non-real roots x2 - x - 6 < 0 
:.(x-3)(x+2) 0) 
Example: Determine the angle of inclination of the line y = x - 2 • 
y=x-2 Recap: 
if tan8 = m then 8 =arctan(m) 
x 
arctan is displayed on a calculator as tan-I 
and is accessed by pressing Is hift II tan I. 
The gradient of the line y = lx - 2 is equal to 1: 
.o. tan (} = 1-+ in order to solve for (} use Is hift I I tan 1111 on your calculator 
.o. (} = 45° 
ANGLE OF INCLINATION WHEN THE GRADIENT OF A LINE IS NEGATIVE (m < 0) 
It is important to note that a calculator will only provide answers for an angle 
between ±9 0° • A negative angle means that the angle is measured clockwise. 
This means that if a line has a negative gradient you will have to add 180° to the 
answer on your calculator in order to obtain the angle which is measured anti-
clockwise from the horizontal as demonstrated in the example below. 
Example: Determine the angle of inclination of the line y = -x + 2. 
Y 
x 
y =-x+2 
The gradient of the line y = - lx + 2 is equal to -1: 
.o. tan (} = -1 -+ in order to solve for (} use Is hift I rtla n--"l_lr-1:11 on your calculator 
As the gradient is negative the calculator will give an answer of -45°. This is the angle 
measured clockwise from the horizontal as shown in the above diagram. In order to 
obtain (} you have to add 180° , as adjacent angles on a straight line add up to 180° . 
.o. (} = -45° + 180° 
.o. (} =1 35° 
Complete Exercise 3 numbers 1 and 2 on page 293. 
© Copyright Kevin Smith I Berlot Books CC 70 
USING THE ANGLE OF INCLINATION TO DETERMINE THE 
ANGLE BETWEEN TWO LINES 
An important application of the angle of inclination is to determine the angle 
formed between two straight lines. 
Example 1: Determine the value of p , the angle formed by lines AB and BC in 
the diagram below if the gradient of line AB is !. and the gradient of 
2 
line BC is -!. . 
3 
A 
Method 1: Remember that the exterior angle of a triangle is equal to the sum of the 
opposite interior angles. 'iP 
Bae = P+  (JA Jj ~ exterior angle of a triangle 
I 1 
tan (JAJj = - tan Bae = --
2 3 
••• (JAB = 26,57" :. Bae =- 18,43° + 180° ~ gradient is negative 
Bae = P+  (JAJj :. Bae = 161,57° 
:. P = BaC- (JAJj 
:. P = 161,57° - 26,57° 
:·P=135° 
Method 2: Extend lines AD and BC and draw in a horizontal line through B. 
A 
By extending the lines it can easily be seen that P = {/De - {/A D 
1 1 
tan (JAJj = - tan Bae = --
2 3 
:. (JAB = 26,57° :. Bae = -18,43° + 180° ~ gradient is negative 
P= Bae - (JAJj :. Bae = 161,57° 
:. P = 161,57° - 26,57° 
:·P=135° 
© Copyright Kevin Smith I Berlot Books CC 71 
Example2: Triangle PQR has coordinates P(-3;3);Q(4;-1);R(-2;-3) as 
shown below. Determine: 
P(-3;3) 
Q(4;-1) x 
a) The gradient of PQ, QR and PRo 
b) IT triangle PQR is a right-angled triangle. 
c) The size of P correct to one decimal place. 
- Yq -Yp Yr-Yq Yr-Yp 
a) mPQ- mQR = mpR= 
Xq-Xp xr-xq xr-xp 
-1-3 -3+1 -3-3 
= = -- = 
4+3 -2-4 -2+3 
4 1 
= -- = - =--6 
7 3 
b) As none of the products of the gradients are equal -1 Ll.PQR is not a right-angled 
triangle. 
c) In order to determine P draw in a horizontal line at R as shown below: 
Q 
From the diagram it can be seen that P = BpR - 8QR 
LI _ 1 tan BpR = -6 
tan uQR - -
3 :. BpR = -80,54° + 180° ~ gradient is negative 
:. 8QR = 18, 43° :.~ =99,46° 
P= BpR - 8QR 
:. P = 99,46° - 18,43° 
:. P = 81, 0° ~ correct to 1 decimal place 
© Copyright Kevin Smith I Berlot Books CC 72 
Example 3: The graphs of f (x) =- 3x + 6 and g (x) =2 x - 4 are sketched below. 
Determine the size of angle p. 
y 
g(x) 
x 
From the diagram below it should be clear that p = 0g(x) + O/(x) 
y 
g(x) 
x 
tan 0g(x) = 2 
:. Og(x) =6 3,43° 
tan O/(x) = -3 
:. O/(x) = -71,57° 
It is important to understand that O/(x) is not the angle of inclination of f(x) but the 
size of the angle measured clockwise from the x-axis. In this case it is not necessary to 
calculate the angle of inclination of f(x) by adding 180°, as p= 0g(x) + O/(x) 
:. p =6 3,43° + 71,57° 
:·P=135° 
Complete Exercise 3 number 3 on page 293 and the 
Mixed Exercise on pages 293-295. 
© Copyright Kevin Smith I Berlot Books CC 73 




CHAPTER 5 - FUNCTIONS 
WHAT YOU WILL LEARN IN THIS CHAPTER 
ASSESSMENTS 
CONTENT 
YOU WILL BE ABLE TO: 
76 A. Grade 10 Revision Extend Grade 10 work on the 
76 • Axes of Symmetry relationships between variables 
76 in terms of numerical, graphical, • Asymptotes verbal and symbolic 
76 • Domain and Range representations of functions and 
76 • Function and Mapping convert flexibly between these 
Notation representations (tables, graphs, 
words and formulae). Including 
77 • Grade 10 Revision of The linear and quadratic polynomial 
Parabola functions, exponential functions, 
79 • Grade 10 Revision of the some rational functions and 
Hyperbola trigonometric functions. 
80 • Grade 10 Revision of the • Generate as many graphs as 
Exponential Function necessary, initially by means of 
81 B. The Parabola (Quadratic point-by-point plotting, 
supported by available 
Function) 
tecbnology, to make and test 
81 • The Parabola in the fonn conjectures and bence generalise 
of y=a(x_p)2 + q the effects of the parameter 
84 which results in a horizontal • The Parabola in the fonn 
shift and that which results in a 
of y=~+bx+c 
horizontal stretch andlor 
87 • Determining the Equation reflection about the y- axis. 
of a Parabola 
87 • Problem solving and graph work • Given the turning point involving the prescribed 
and one other point functions. 
88 • Given the x-intercepts and 
• Determine the average gradient one other point 
between points and develop an 
• Given the y-intercept and 
89 intnitive understanding of 
two other points gradient at a point. 
90 C. The Hyperbola 
90 • Axes of Symmetry and 
Points of Intersection 
93 • Determining the Equation 
of a Hyperbola 
95 D. The Exponential Function 
96 • Steps to Sketching the 
Exponential Function 
99 • Determining the Equation 
of an Exponential Graph 
100 E. Translations and Reflections 
of Graphs 
© Copyright Kevin Smith I Berlot Books CC 74 
CHAPTER 5 - FUNCTIONS 
WHAT YOU WILL LEARN IN THIS CHAPTER 
1:1 CONTENT III ASSESSMENTS YOU WILL BE ABLE TO: 
102 F. Graph Interpretation 
102 • lengths of vertical and I 
horizontal lines 
103 • Interpretation and 
Deductions from Graphs 
107 • Real World Applications 
108 • Average Gradient 
(Average Rate of Change) 
109 G. Trigonometric Functions 
109 • Grade 10 Revision of the 
Sin, Cos and Tan Graphs 
110 • The effects of a (Change in 
Amplitude) 
110 • The effects of q (Vertical 
Shifts) 
111 • The effects of p 
(Horizontal Shifts) 
114 • The effects of k (Change in 
Period) 
116 H. Interpretation of 
Trigonometric Functions 
© copyright Kevin Smith I Berlot Books CC 75 
A. GRADE 10 REVISION 
AXES OF SYMMETRY 
If a function has a line of symmetry, it means that the function is a mirror image 
of itself about the line of symmetry. In other words, if the function were folded 
along the line of symmetry it would duplicate the graph on the other side of the line. 
ASYMPTOTES 
Asymptotes are imaginary lines that a graph approaches, but never touches or 
cuts. 
DOMAIN AND RANGE 
Domain: The domain refers to the possible x-values for which a graph is defined. 
Range: The range refers to the possible y-values for which a graph is defined. 
Example: The graphs of f (x) and g (x) are sketched below. Using the graphs 
clearly state the domain and range of each graph. It is given that A is 
the turning point of the parabola f (x) . 
f(x) 
x 
A(2;-3) 
f(x}: g(x} : 
Domain: xe lR Domain: x e lR 
This can also be written as x e ( --00 ; 00 ). This can also be written as x e ( --00 ; 00 ). 
Range: ye [-3 ;00} Range: ye (1;00) 
This can also be written as y ;::>: -3. This can also written as y > 1. 
Notice that the turning point of a parabola and the asymptote of an exponential function 
determine the range of the functions. if( 
FUNCTION AND MAPPING NOTATION 
In Grade 10 you were introduced to different ways of representing functions. 
The different notations are summarised below: 
y = ... ~ equation notation 
f (x) = ... ~ function notation 
f : x ~ ... ~ mapping notation 
AU of the above representations mean the same thing. if( 
© Copyright Kevin Smith I Berlot Books CC 76 
GRADE 10 REVISION OF THE PARABOLA 
In Grade 10 2 you were introduced to the parabola in the form of ,= ax + q . 
It was shown that a determines the shape of a parabola whilst q resulted in a 
vertical shift (up or down) and was the ,-intercept. The properties of a parabola 
that you learnt in Grade 10 are summarised below: 
Concave up (happy face) .. ~ Concave down (sad face) (OJ 
As the value of a increases from 0 As the value of a decreases from 0 
the graph is compressed and gets the graph is compressed and gets 
narrower. This is demonstrated in narrower. This is demonstrated in 
the graphs below where a is the graphs below where a is 
increased from 0,1 to O,l and then decreased from -0,1 to -0,2 and 
to 1. then to -1. 
y 
x 
x 
The value of q results in a vertical shift. If the value of q is greater than 0 
the graph will be shifted up. If q is less than 0 the graph will be shifted 
down. 
The domain of a parabola is: 
XE lR ~ also written as XE (-00; co) 
The range of a parabola is: 
• y~ q if a >O ~alsowrittenasYE[q ;oo} 
• y~ q if a O '" /(+,fa";-rJ,;" +q) ~, / 
q>o " , , 
..............; ; ,. ............. . , ,,  x .. ................ ................  
a 
In Grade 10 it was shown that the hyperbola was symmetrical about 
the lines y = x + q and y = -x + q . In Grade 11 the hyperbola will be 
shifted left and right and the axes of symmetry will be shifted in the 
same manner. 
If a > ° the axis of symmetry If a < ° the axis of symmetry 
y = x + q intersects the hyperbola y = -x+ q intersects the 
at the points : hyperbola at the points : 
• [+ta ;+Ia+qJ  • [-ta ;+Ia+qJ 
• [- ta ;-Ia+qJ • [+ta ;-Ia+qJ 
As the function is shifted left and right in Grade II, the points of 
intersection will also be shifted left and right. 
The sign of a is not taken into account when calculating the points of 
intersection. 'i( 
The domain of a hyperbola in Grade 10 was: 
xe JR, x,,",O -7also written asxe (--00;00) x,,",O 
This will change in Grade 11 as the graph is shifted left and right. 'i? 
The range of a hyperbola is: 
ye JR, x,,", q -7 also written as ye (--00;00), y ""' q 
© Copyright Kevin Smith I Berlot Books CC 79 
GRADE 10 REVISION OF THE EXPONENTIAL FUNCTION 
In Grade 10 you were introduced to the exponential function in the form of: 
y =ab" +q where b > Oand b ",0 
It was shown that a determines the shape of the graph as well as on which side of 
the asymptote the exponential function lies. Together with a the value of b 
determines whether the function is increasing or decreasing. 
The value of q resulted in a vertical shift (up or down) and the line y =q  is the 
horizontal asymptote. The properties of the exponential function that you learnt 
in Grade 10 are summarised below: 
graph lies above the The graph lies below the 
horizontal asymptote and is an horizontal asymptote and is a 
increasing function. decreasing function. 
y y 
x 
y=q 
---------- --------
x 
graph above the The graph lies below the 
horizontal asymptote and is a horizontal asymptote and is an 
decreasing function. increasing function. 
y y 
" 
---~ y=q 
-------- ------------_. 
x 
The domain of an exponential function is: 
x e 1R ~ also written as x e (--00; 00 ) 
The range of an exponential function is: 
y> q if a > 0 ~ also written as ye (q ;oo) 
y  0 -7 happy face parabola (concave up) ~ 
1. Test for shape: 
if a < 0 -7sad face parabola (concave down) C:i) 
2. Determine the x and y coordinates of the turning point (TP) from the equation. 
3. Determine the y-intercept by letting x equal to O. 
4. Determine thex-interceptls by lettingy equal to O. 
a > 0: Happy face parabola a < 0: Sad face parabola 
(Concave up) (Concave down) 
If a > 0 the y value of the TP will be the minimum value of the graph. 
If a < 0 the y value of the TP will be the maximum value of the graph. l 
It is important to note that: 
• If the graph is 'happy' and the y-value of the turning point is positive, 
the graph will not have any x-intercepts. 
'1r 
• If the graph is 'sad' and the y-value of the turning point is negative, the 
graph will not have any x-intercepts. 
y 
a>O 
x 
q>O 
a 0 ---) happy face Shape: a < 0 ---) sad face 
2) Turning point: (3; - 4) Turning point: (-3 ; 2 ) 
3) y-int: letx=O y-int: letx=O 
y=(0-3)2_4 =5 y =-2(0+3)2 +2=-16 
:. y-int:(0;5) :. y-int:(0;-16) 
4) x-int: lety=O x-int: lety=O 
0=(x-3)2_ 4 0=-2(x+3)2+ 2 
:.0=x2-6x+9-4 :. 0=-2(x2 +6x+9)+2 
:.0=x2-6x+5 :. 0=-2x2 -12x-18+2 
:.0=(x-l}(x-5) 
:.0= -2(x2+6x+8) ---)divideby -2 
:.x=1 or x=5 
2
:.x-int's:(I;O) and (5;0) :.0=x +6x+8 
:.0=(x+2}(x+4) 
:.x=-2 or x=-4 
:.x-int's:(-2;0) and (-4;0) 
J(x) 
x 
(0;-16) 
h(x) 
I ~ Complete Exercise 1 number 1 on page 303. 
© Copyright Kevin Smith I Berlot Books CC 83 
2 
THE PARABOLA IN THE FORM OF Y =ax + bx +c 
When the parabola is in the form 2 y = ax + bx + c the value of c is the y-intercept 
and as before a tells you about the shape of the parabola. 
In order to determine the turning point of the parabola you can complete the 
square on the general equation as shown below. 
y=ax2+bx+c 
2
:.y=a(x +! x+ :) -7add and subtract Gx!J =(:J2 
:.y=a[(x+~)2 _ b22 +~] -7LCD=4a2 
2a 4a a 
:. y = a[( x+ :a J+ 4a:~b2 ] -7 multiply a back in 
2 
:. y =a(x+~)2 + 4ac-b -7 x-coordinate of the TP and y-coordinate of the TP 
2a 4a 
From this it can be seen that x = -~ is the x-coordinate of the turning point. 
2a 
This line is also the axis of symmetry. The parabola is always symmetrical about 
its turning point. This means that it is a mirror image of itself over a vertical line 
passing through the turning point. 
STEPS TO SKETCHING THE PARABOLA: 
The only difference in sketching the parabola in the form =a x2 y + bx + c is that 
you have to determine the turning point by using the axis of symmetry (AS) or by 
completing the square. 
if a > 0 -7 happy face parabola (concave up) .. 
1. Test for shape: '" 
if a < 0 -7sad face parabola (concave down) (oJ 
2. Determine the x and y coordinates of the turning point. 
a) Determine the Axis of Symmetry (AS) using the equation x = -b • 
2a 
b) Substitute the AS into the equation of the parabola to fmd the y-coordinate 
of the turning point. 
3. Determine the y-intercept by letting x equal to O. 
4. Determine thex-intercept/s by lettingy equal to O. 
The parabola only has 1 axis of symmetry (the line of x = -b ) and has no 
~ 2a 
asymptotes. 
© Copyright Kevin Smith I Berlot Books CC 84 
Example 1: On the same set of axes, sketch the graphs of g ( x ) = _2x2 -12x - 10 
= 2 and f(x) x -6x+ S. 
g(x) : f(x): 
1) Shape: a < 0 ~ sad face Shape: a > 0 ~ happy face 
-b -(-12) b -(-6) 
2) AS:x=-= -3 AS:x=-= 3 
2a 2(-2) 2a 2(1) 
g (-3) = -2( _3)2 -12( -3) -10 = 8 1(3) = (3)2 -6(3)+5 = -4 
:. TP :( -3 ;8) :. TP:(3;-4) 
3) y-int: (0;- 10) y-int: (0; 5) 
4) x-int: lety=O x-int: lety=O 
0=-2x2-12x-1O 0=x2-6x+5 
:.0= -2(x2+6x+5) ~divideby -2 :.0=(x-l)(x-5) 
:.x=1 or x=5 
:.0=x2+6x+5 
:.x-int's:(I;O) and (5;0) 
:.0=(x+5)(x+l) 
:.x=-5 or x=-1 
:. x-int 's:( -5;0) and (-1;0) 
y 
f(x) 
x 
(3;-4) 
g(x) 
© Copyright Kevin Smith I Berlot Books CC 85 
2 
Example 2: On the same set of axes, sketch the graphs of f (x) = x - 2x - 3 and 
g(x) = _!.x2 -3x-8. 
4 
f(x): g(x): 
1) Shape: a > 0 ---) happy face Shape: a < 0 ---) sad face 
b -(-2) 
2) AS:x=- =1 AS:x=-..!!..= -(-3) =-6 
2a 2(1) 2a 2(-~} 
f (1) = (1)2 -2(1) -3 = -4 g (--6) = -!.(- -6)2 -3(--6) -8 = 1 
4 
:. TP: (1;-4) 
:. TP: (-6 ;1) 
3) y-int: (0; -3) y-int: (0; -8) 
4) x-int: lety=O x-int: lety = 0 
0=x2 -2x-3 0= -!.x2-3x-8 ---)multiplyby -4 
:.0= (x+ 1)(x-3) 4 
:.0 = x2 + l2x+32 
x=-l or x=3 
:.x-int's:(-l;O) and (3;0) :.0=(x+8)(x+4) 
:.x=-8 or x=-4 
:. x-int's:( -8;0) and (-4;0) 
y 
f(x) 
x 
g(x) 
~ Complete Exercise 1 number 2 on page 303. 
© Copyright Kevin Smith I Berlot Books CC 86 
DETERMINING THE EQUATION OF A PARABOLA 
There are usually three ways in which you will be expected to find the equation 
of a parabola (quadratic function): 
1. You will be given the turning point and one other point on the curve. 
2. You will be given the x-intercepts and one other point on the curve. 
3. You will be given the y-intercept and two other point on the curve. 
1. GIVEN THE TURNING POINT AND ONE OTHER POINT 
In this case, use the equation of the parabola in turning point form: y = a (x _ P )2 + q 
Example: Determine the equation of f (x) shown below, if the point 
(-4;4) is its turning point and (-2;2) is a point on the graph. Write 
2 
your answer in the form f (x) = ax + bx + c . 
x 
Turning Point (TP) : ( -4; 4) 
:. y = a[ x- ( -4)J2 + 4 
:. y = a(x+ 4)2 +4 
Now substitute the other known point (-2; 2) into the above equation to solve for a. 
:.2= a(-2+4)2+ 4 
:.2= 4a +4 
:.-2=4a 
1 
.'. a =-2- 
1 2 
:.y= --(x+ 4) +4 
2 
2 
:. y = -~ ( x +8x +16)+4 ~ don't forget the middle term 
= 1 2 :. y --x -4x-8+4 
2 
:. f(x) = _.l x2 -4x-4 ~a1ways write the equation in standard form 
2 
Complete Exercise 2 number 1 on page 303. 
© Copyright Kevin Smith I Berlot Books CC 87 
2. GIVEN THE X -INTERCEPTS AND ONE OTHER POINT 
Another form of the equation of the parabola is y =a  (x - R1 ) (X - R2 ) • 
In this form Rl and R2 are the roots (x-intercepts) of the equation. 
When given the x-intercepts and another point, use the equation in this form in 
order to determine the equation of the parabola. 
Example: Determine the equation of the parabola shown below. Points 
A( -2;0) and B(5;0) are the x-intercepts and C(2;-24) is a point on 
the graph. 
y 
A(-2;O) B(5;O) 
x 
:. y = a [ x-( -2)][x-5] 
:. y = a (x+ 2}(x-5) 
Now substitute the other known point (2; -24) into the above equation to solve for a 
:. -24 =a (2+2}(2-5) 
:.-24=-12a 
:. a=2 
:. y = 2(x+ 2}(x-5) 
:. y = 2( 2 x -3x-1O) 
:. y = 2x2 - 6x - 20 ~ always write the equation in standard form 
I ~ Complete Exercise 2 number 2 on page 303. 
© Copyright Kevin Smith I Berlot Books CC 88 
3. GIVEN THE Y -INTERCEPT AND TWO OTHER POINTS 
2 
In this case use the equation of the parabola in the form of y =a x + bx + c 
and use simultaneous equations to solve for the values of a and b. 
Example: A parabola's y-intercept is at pointA(0;-2). H points B(2;8) and 
C( -6;-8) are points which lie on the graph, determine the equation 
of the parabola. 
:. y = ax2 +bx +c -7 c is the y-intercept (-2) 
y=ax2+bx-2 
Substitute the point B (2; 8) into the equation: 
:.8 = a(2)2 +b(2)-2 
:.8 = 4a+2b-2 
:.10=4a+2b -7 ill 
Substitute the point C ( -6; - 8) into the equation: 
:. -8 = a( --6)2 +b( -6)-2 
:.-8=36a-6b-2 
:.6b=36a+6 
:.b=6a+I -7~ 
Substitute ~ into ill: 
:.10=4a+2(6a+l) 
:.1O=4a+12a+2 
:.8=16a 
:.a=.!. -7@j 
2 
Substitute @j into ~: 
:.b=6(Yz )+1 
:.b=4 
:. y =.!.x2 +4x-2 
2 
I ~ Complete Exercise 2 number 3 on page 304. 
© Copyright Kevin Smith I Berlot Books CC 89 
c. THE HYPERBOLA 
a 
In Grade 11 the hyperbola will be in the form of y = --+ q • 
x- p 
The hyperbola will either lie in the lot and 3n1 quadrants formed by the 
asymptotes, or in the 2nd and 4th quadrants formed by the asymptotes. This is 
summarised below: 
a -7 a is a constant which detennines the quadrants that the hyperbola will lie in 
If a > 0 the hyperbola will lie in the 1s t and 3rd quadrants 
If a < 0 the hyperbola will lie in the 2nd and 4th quadrants 
p -7 P is the vertical asymptote (the line x = p) 
q- 7 q is the horizontal asymptote (the line y = q) 
It is important to note that the value of p shifts the graph horizontally while q shifts 
the graph vertically. 
AXES OF SYMMETRY AND POINTS OF INTERSECTION 
The hyperbola has 2 axes of symmetry: 
y=x- p +q and y= - (x- p )+ q =- x + p +q 
The axes of symmetry will cut the hyperbola at the following points: 
If a > 0 the line y = x - p + q will cut the hyperbola at: 
(+..Ja + p;+..Ja +q) and (-..Ja + p;-.Ja +q) 
If a < 0 the line y = - x + p + q will cut the hyperbola at: 
(-..Ja + p;+..Ja +q) and (+..Ja + p;-.Ja +q) 
The sign of a is not taken into account when calculating the points of intersection. 'if 
Hyperbolas with p and q equal to 0 are shown below: 
y y 
,J . ~ 
' , '''-x 
' , '''' x 
,,~  , , , 
, 
, , , , x x 
(-.Ia+p;- -./a+q,) , ' 
, , , , ,( +-./a+p;--./a+q) 
, , , , 
, , , , , , 
, , 
, , , , , 
Figure 1: a> 0 Figure 2: a < 0 
~ The sign of p is always opposite to the sign ofp  in the denominator. 
© Copyright Kevin Smith I Berlot Books CC 90 
STEPS TO SKETCHING THE HYPERBOLA: 
1. Determine the vertical and horizontal asymptotes and which quadrants the 
hyperbola lies in. 
2. Determine the x-intercept by letting y =0  and the y-intercept by letting x =O . 
3. If asked, determine the equations of the axes of symmetry. 
4. If asked, determine the points of intersection of the axes of symmetry with the 
hyperbola. 
Example 1: Sketch the graph of y =_4_+ 4 showing both axes of symmetry 
x- 2 
and their point(s) of intersection with the graph. 
1) Determine the asymptotes and quadrants: 
a =+4 ~a>0 Asymptotes: x = 2 ~ vertical 
:. The hyperbola lies in the 1st and 3rd quadrants y = 4 ~ horizontal 
2) Determine the x and y-intercepts: 
x-int: let y = 0 y-int: let x = 0 
4 4 
0=--+4 y=--+4 
x-2 0-2 
:. y = 2 
:. -4 = _4_ ~ multiply both sides by x - 2 
x-2 :. y-int: (0;2) 
:.-4(x-2) = 4 
:.-4x+8=4 
:.x=1 
:. x-int: (1;0) 
3) Determine the equations of the axes of symmetry: 
y=x- p +q y= - (x- p )+ q 
:. y =x- 2+4 :. y = - (x- 2)+ 4 
:.y=x+2 
:.y=-x+6 
4) Determine the points of intersection: 
(+.Ja + p ;+.Ja +q) and (-.Ja + p ;-fa +q) ~a >0 
= (.J4+2;.J4+4) and (-.J4+2;- .J4+4) 
= (4;6) and (0;2) 
• • y • : ",'y=x+2 •. . .  I .' 
• • I .' • •. . 
• 
.  I .' I .' 
I • (4;6) 
.... I # 4' 4' 
.... I, # 
- --- - - --- - - --,'f. . - - --- - - --- - - - -
# # I .... 
# I ...... 
I '. 
(1 ,0) ' • 
.. • • .I .  x • 
• • I 
• •
• I •
  
.. • • Ix~2 ........ y=-x+6 
© copyright Kevin Smith I BerlDt Books CC 91 
Example 2: Sketch the graph of ,= ~ + 4 showing both axes of symmetry 
x +2 
and their point(s) of intersection with the graph. 
1) Determine the asymptotes and quadrants: 
a =-4 -7a 0 then the graph will lie above the asymptote. 
If a < 0 then the graph will lie below the asymptote. 
p ~ p shifts the graph left or right. 
q ~ The line y = q is the horizontal asymptote (the vertical shift). 
In order to demonstrate the shape and y-intercept of the exponential function we will let 
a =1 , P = 0 and q = O. This results in the standard graph of y = X b which is sketched 
below: 
a =1 and b >1 a =1 and O< b x ----~-->x 
By letting x = 0 the y-intercept of the graph is: 
y =b O ~ any number raised to the power of 0 is equal to 1 
:. y =1 
:. The y-intercept is at the point (0;1). 
So, for the standard exponential graph where a = 1 the x-axis is the horizontal 
asymptote and the y-intercept will be at (0;1). If a = -1 the graph is reflected about the 
x-axis and the y-intercept will be at (0; -1) and the graph will lie below the horizontal 
asymptote, as sketched below. 
a =-1 and b >1 a =-1 and O< b , 
The domain of an exponential function is: 
x E lR ~ also written as x E (--00; 00 ) 
The range of an exponential function is: 
• y> q if a > 0 ~ also written as yE (q;oo) 
• y  0 ~ graph is above the asymptote 
if a < 0 ~ graph is below the asymptote 
2. Check if the function is increasing or decreasing: 
if b > I and a > 0 ~ graph is increasing (as x increases so does the value of y) 
if 0 < b < I and a > 0 ~ graph is decreasing (as x increases the value of y decreases) 
if b > I and a < 0 ~ graph is decreasing (as x increases the value of y decreases) 
if 0 < b < I and a < 0 ~ graph is increasing (as x increases so does the value of y) 
Note: if b is raised to a negative lit is the same as writing i. ~ b -n = ;n 'f 
3. Find the y intercept by setting x equal to O. 
4. Find the horizontal asymptote (y =q ). 
5. Find the x-intercept by settingy equal to O. 
6. Substitute a value for x in order to obtain another point on the graph and 
obtain the correct shape. 
~ The exponential function will only have an x-intercept if: 
a > 0 and q < 0 ~ the horizontal asymptote lies below the x-axis 
y a >O 
p  0 ~ the horizontal asymptote lies above the x-axis 
y 
------------- ---------
© Copyright Kevin Smith I Berlot Books CC 96 
2 
Example 1: Sketch the graph of f (x) =4 (2t- +1. 
1) a > 0 .... graph is above the horizontal asymptote 
2) b > 1 and a > 0 :. graph is increasing 
3) Find the y-intercept: 
y-int:x=O 
2
:.y=4(2t +1 
2
:.y=4(2r + 1 
:. y = 4(,7;;) + 1 
:.y=2 
:. y-int: (0;2) 
4) y = 1 is the horizontal asymptote 
5) x-intercept: none ~ a > 0 and q > 0 
6) Substitute a value for x in order to obtain another point on the graph: 
substitute x = 2 ~ x = p is a good choice 
:. y =4(2)2-2 + 1 
:.y=4(1)+1 
:.y=5 
:.Additionalpoint: (2;5) 
y 
y=! 
x 
© Copyright Kevin Smith I Berlot Books CC 97 
1 ( )-x+2 
Example 2: Sketch the graph of y =- 2" 4 + 1. 
Note the minus sign in front of the x. Always rewrite the equation so that x is positive by 
using the rule that b -x =G J. 'if' 
:.y= --1 ( 4 )- (x-2) + 1 
2 
1 ( I J"'-2 
:·y= -2 4 + 1 
1) a < 0 -+ graph is below the horizontal 2) 0 < b < 1 and a < 0 :. graph is 
asymptote increasing 
3) Find the y-intercept: 4) y = 1 is the horizontal asymptote 
y-int:x=O 
2 
:.y=-M:f +1=-7 
:. y-int: (0;-7) 
5) Find the x-intercept by setting y = 0: 6) Substitute a value for x in order to 
obtain another point on the graph: 
:.0 = -.!. ( 4 )-{x-2) + 1 
2 substitute x = 2 -7 P is a good choice 
2 2 
:.-I=-~Gr-2 :. y=_.!.(4r + +1 -7 multiply by -2 2 
t 2 :. y=-.!.(I)+1 2 2 :.2=(2- -7,X = 2- 2 
1 
:. 21 = 2-2x+4 ... y=2- 
:.1=-2x+4 
:. Additional point: (2;~) 
:.2x=3 
...  x= 132/  
:. x-int: (Yz;O) 
y y~l 
------------ ------(z~~-----------------. 
x 
I ~ Complete Exercise 5 on page 304. 
© Copyright Kevin Smith I Berlot Books CC 98 
DETERMINING THE EQUATION OF AN EXPONENTIAL GRAPH 
In most cases, when fmding the equation of an exponential graph you will be 
given the horizontal asymptote or enough information to fmd it. You will also be 
given a point or points which lie on the graph. Once you have the horizontal 
asymptote it can be substituted into q and the remaining unknowns can then be 
found by substituting the given point or points into the equation. 
Example: The exponential function below has the equation f (x) = 2x- P +q. 
The line of y =2  is the horizontal asymptote and A (3;3) is a point 
on the graph. Determine the equation of the graph. 
y 
_____ ~_~_=_=_=_=_~_~_~_=_ ___________ l=2 
x 
f{x) = 2x- P +q 
:. f (x) = 2x- P + 2 ~ Y = q is the horizontal asymptote 
Sub A{3;3) 
:.3=23 p- +2 
:. 1 = 23 p - ~ you need to get the same bases on both sides 
:. 2° =2 3-p ~ any number raised to the power 0[0 is equal to 1 (aO =1) 
:.0=3-p 
:.p=3 
:. f{x) = 2x 3 - +2 
I ~ Complete Exercise 6 on page 305. 
© Copyright Kevin Smith I Berlot Books CC 99 
E. TRANSLATIONS AND REFLECTIONS OF GRAPHS 
As you have seen in the graphs that you have dealt with, graphs can be 
translated (shifted) horizontally and vertically. 
It has been shown that the value of p is responsible for horizontal shifts and the 
value of q is responsible for vertical shifts. 
A reflection in the x-axis causes all the y-values to change sign and leaves the 
x-values the same. 
A reflection in the y-axis causes all the x-values to change sign and leaves the 
y-values the same. 
Illustrative Example 1: Sketch the graph of f (x) = x2 • Then, on the same set of 
axes sketch the graphs of: 
a) f(x-2) b) f(x+2) 
a) f(x-2) = (x_2)2 b) f(x+2)=(x+2)2 
Ibis is a parabola with a turning Ibis is a parabola with a turning 
point at (2; 0) and y-intercept of point at (-2; 0) and y-intercept of 
(0;4). Ibis is the graph of f(x) (0;4) .Ibis is the graph of f(x) 
shifted 2 units to the right. shifted 2 units to the left. 
f(x+2) y f(x) f(x-2) 
(-2;0) (0;0) (2;0) x 
From the above graphs it should be clear that: 
• The graph of f (x - p) shifts the graph of f (x) P units to the right. 
• The graph of f ( x + p ) shifts the graph of f (x ) p units to the left. 
Illustrative Example 2: Given the function f (x) = (x - 3)2 + 4 : 
a) Determine the turning point of f. b) Determine the turning point of 
g(x) if g(x)=f(x-l) 
a) From the equation of f (x) it can be seen that the turning point is (3; 4) 
b) f (x  -I) shifts the graph of f (x ) I unitto the right. 
:.TPofg(x) is (4 ;4) 
© Copyright Kevin Smith I Berlot Books CC 100 
IDustrative Example 3: The graph of f (x) = x2 is sketched below. 
Thegraphsof g(x)=f(x)+5 and h(x)=f(x)-5 are 
also sketched on the same set of axes. 
g(x} 
f(x} 
h(x} 
x 
From the graphs it can be seen that g (x) shifts f (x) up 5 units. 
The graph of h (x) shifts f (x) down 5 units. 
In summary: 
• The graph of f (x) + q shifts the graph of f (x) q units up. 
• The graph of f (x) - q shifts the graph of f (x) q units down. 
IDustrative Example 4: The graph of f (x) = (x+2)2 +3 is sketched below. 
The graphs of g(x)=f(-x) and h(x)=-f(x) are 
also sketched on the same set of axes. 
y 
(-2;3) (2;3) 
(-2;-3) x 
h(x} 
From the graphs it can be seen that g (x) is a reflection of f (x) about the y-axis and 
the graph of h (x) is a reflection of f (x) about the x-axis. 
In summary: 
• The graph of f (-x) is a reflection of the graph of f (x) about the y-axis. 
• The graph of - f (x) is a reflection of the graph of f (x) about the x-axis. 
I ~ Complete Exercise 7 on page 305. 
© Copyright Kevin Smith I Berlot Books CC 101 
F. GRAPH INTERPRETATION 
LENGTHS OF VERTICAL AND HORIZONTAL LINES 
The length of any vertical line AB is: AB = Ytop - Ybottom 
The length of any horizontal line CD is: CD = xright - x/eft 
Example 1: Determine the length of line AB and CD below: 
y 
C(-5;3) n(4;3) 
x 
AB = Ytop - Ybot CD = xright - Yleft 
= Ya -Yb 
=8-2 =4-(-5) 
= 6 units = 9 units 
Example 2: Line AB below can move anywhere between points C and D. It is 
given that f(x) =2x2 -8x+6 and g(x) =-2x+6. Write down an 
expression for the length of line AB in terms of x. Hence, determine 
the length of line AB when x = 2. 
x 
AB = g (x) - f(x) -7 Ytop - Ybot 
:.AB = -2x+6 - ( 2x2 -8x+6) 
:.AB = -2x+6-2x2 +8x-6 
:. AB = _2x2 + 6x -7 this is the length of line AB anywhere between points C and D 
Now substitute x = 2 into the equation for AB above: 
:.AB = _2(2)2 +6(2) 
= 4 units atx= 2 
© Copyright Kevin Smith I Berlot Books CC lO2 
INTERPRETATION AND DEDUCTIONS FROM GRAPHS 
2 
Example 1: In the diagram below, f (x) = x +3x -4 and g(x) = x+4. 
L x 
g(x) 
D 
Determine: 
a) The length of CD if OL = 2 units • b) The length of OT if EF = 7 units • 
c) The coordinates of A and B. d) The coordinates of P. 
e) The values ofx for which f (x) ~ O. 
f) The values ofx for which f (x) S g(x). 
g) The values ofx for which f (x ).g(x) > O. 
a) CD= g(x) - f(x) 
= x+4 -(~ +3x-4) 
=x+4-x2-3x+4 
=-x2-2x+8 
Sub x = - 2 ~ OL is 2 units but L is on the negative x-axis :. the x-value at L is - 2 
:. CD =-( _2)2 -2(- 2)+8 
:. CD = 8 units 
b) EF= f(x) - g(x) 
:. 7 = x2+3x-4-(x+4) ~ EF=7units 
:.7=x2+2x-8 
:.x2+2x-15=O 
:. (x+5)(x-3) = 0 
:. x = -5 or x = 3 ~ N/A as T is on the negative x-axis 
:. OT = 5 units ~ positive as a length cannot be negative 
© Copyright Kevin Smith I Berlot Books CC 103 
L x 
g(x} 
c) x-int: let y = 0 D 
:. x2 +3x-4 =0 
:. (x+4}(x-l) = 0 
:.x=-4 or x=l 
:. A is the point (-4;0) and B is the point (1;0) 
d) P is a point of intersection of the graphs 
:. f(x} = g(x} 
2
:. x +3x-4 =x+4 
:.x2 +2x-8=0 
:.(x-2}(x+4}=0 
:. x = 2 or x = -4 ~ this is the x-value at point A (the other point of intersection) 
:. The x-value at point P is 2 
Sub x = 2 into g (x) ~ f (x) could also be used (use the easier equation) 
:.g(2}=2+4=6 
:. P is the point (2;6) 
e) f (x) ~ 0 ~ this occurs where the graph of f is above the x-axis 
:.x~-4 or x~l ~alsowrittenasxE (-00;-4] or [1;00) 
f) f (x) ~ g (x) ~ this occurs where the graph of f is below or intersects g 
:. f(x} ~ g(x} from points A to P 
:. -4~ x~ 2 ~ also written as XE [-4;2] 
g) f(x}.g(x}>O 
There are 2 possible ways this can occur: 
l)f(x} andg(x} arebothpositive ~ '+' X '+'='+' 
This occurs from point B onwards. 
2)f(x} and g(x} are both negative ~ '-' x '-' = '+' 
This does not occur on this graph . 
... x > 1 ~ also written as x E (1; 00 ) ~ x '" 1 as the question said greater than 0 
A graph is positive where it lies above tbe x-axis and negative where it 
lies below tbe x-axis. 
© Copyright Kevin Smith I Berlot Books CC 104 
2 
Example 2: The graph below represents y = _x +8x with turning point A(4;16). 
y A(4;J6) 
x 
2 
For what values of t will the equation _x + 8x + t = 0 have two roots 
which are positive and greater than O? 
In this question it is important to understand that t will shift the graph vertically up or 
down. 
In order to have two positive roots the graph will have to lie between the two dotted 
graphs shown below. This means that the graph has to be shifted down, but by no more 
than 16 units. 
A(4;J6) 
x 
:.-16 < t < 0 ~ also written as tE (-16;0) 
© Copyright Kevin Smith I Berlot Books CC 105 
Example3: It is given that f(x}=-x2 +6x 2and g(x}=x -4x+3. The line PQ 
can lie anywhere between points A and B. 
x 
f(x} 
a) Write down an expression for the length of line PQ. 
b) Hence, determine the maximum length of PQ. 
a) PQ = f(x) - g(x) 
:. PQ = _x2 +6x-( x2 -4x+3) 
:.PQ =_x2 +6x-x2 +4x-3 
... PQ =- 2x2 + 1O x - 3 ---) this expression represents the length of line PQ 
The above equation is a parabola, where the ,-value represents the length of line 
PQ for a particular x-value between A and B. It was shown in graph sketching that 
the ,-coordinate of the turning point is the maximum or minimum value of the 
graph. Therefore, the ,-coordinate of the turning point of PQ will be the maximum 
length of PQ. 
b) PQ=-2x2+lOx-3 
AS:x= -b 
2a 
:. x = (10) = ~ ---) this is the x-value where the length of PQ will be a maximum 
2 -2 2 
Sub x = ~ into the equation of PQ to obtain the maximum value: 
2 
:.PQmax =_2(,%)2 +10(,%)-3 
19 . 
=- umts 
2 
:. The maximum length of PQ is 9,5 units. 
Complete Exercise 8 on pages 305 and 306. 
© Copyright Kevin Smith I Berlot Books CC 106 
REAL WORLD APPLICATIONS 
Example: A Grade 11 mathematics student builds a water rocket and launches it 
from the roof of his school. He determines that the height above the 
ground from the moment it is launched can be modelled by the 
function H (t) =- 5t2 + 20t + 20 , which is sketched below: 
H 
Time (s) t 
a) Determine the height of the school roof. 
(Hint: This is the height the rocket was launched from) 
b) Determine the rocket's maximum height above the ground. 
c) Determine the time the rocket was in the air for correct to the nearest second. 
a) The y-intercept of the graph represents its initial starting height: 
:. H(O) =_5(0)2 +20(0)+20 
= 20 metres 
b) The y-coordinate of the turning point represents the rocket's maximum height: 
H(t) = -5t2 +20t+20 
-b 
AS:x=-
2a 
x = -(20) = 2 ~ this is the time when the rocket will reach its maximum height 
2 -5 
:. H(2) =_5(2)2 +20(2)+20 
=40 metres 
:. The maximum height the rocket reaches is 40 metres. 
c) The x-intercepts of the graph represent when the rocket is on the ground: 
0= -5t2 + 20t + 20 ~ divide both sides by - 5 
... 0 2 = t - 4t - 4 ~ use the quadratic formula 
-( -4)±~( -4)2 -4(1)( -4) -b±~b2 -4ac 
:.t 2(1) --)X 2a 
4±4.J2 
:.t 
2 
:. t = 4,83 or t = -0,83 
but t *- -0,83 ~ time cannot be negative 
:. t = 5 seconds ~ the question said the answer must be correct to the nearest second 
:. The rocket was in the air for 5 seconds. 
© Copyright Kevin Smith I Berlot Books CC 107 
AVERAGE GRADIENT (AVERAGE RATE OF CHANGE) 
The average gradient (m) is the average rate of change between two points. 
The average gradient between the points (~; YI) and (Xz; Y2) is given by: 
average gradient =m  = Y2 - Yl 
x2 -xl 
The average gradient between two points on a curve is the gradient of a straight 
line which joins the two points together. 
IDustrative Example: 
Given the function f (x) = x 2 - 4, determine the average gradient of the function 
between x =1  and x =3 . 
In order to determine the average gradient you need the y values where x = 1 and x = 3. 
y • 
5 ••••• 
j(1)=(1)2 -4= -3 
j(3)=(3)2_4= 5 
j(3)- j(l } 5-(-3) ~= 4 3 x 
3- 1 2 
Nowi et  us exaJDlne the  2l'IUd l'e n t b et w een the  f1o IOI WlD2Po'mt s on the fu1 nc ti on: 
Point A PointB Gradient 
5-(-3} 
(1;-3) (3;5) m= 4 
3-1 
0-(-3} 
(1;-3) (2;0) m= =3 
2-1 
-1,75-( -3} 
(1;-3) (1,5;-1,75) m= 2,5 
1,5-1 
-2,56-(-3} 
(1;-3) (1,2;-2,56) m= 2,2 
1,2-1 
-2,79-(-3} 
(1;-3) (1,1;-2,79) m= =21 
1,1 -1 ' 
-2,8975-( -3} 
(1;-3) (1,05;-2,8975) m= 2,05 
1,05-1 
(1,005;-2, 989975} -2,989975-( -3} 
(1;-3) m= 2,005 
1,005-1 
From the table it can be seen that as you calculate the average gradient using points 
closer to A, the gradient starts to approach 2. 
As the points get closer together, the gradient of line starts to 
approximate the gradient of the graph at a point. From this it can be seen 
that the gradient at point A is equal to 2. 
I ~ Complete Exercise 9 on pages 306 and 307, 
© Copyright Kevin Smith I BerlDt Books CC 108 
G. TRIGONOMETRIC FUNCTIONS 
GRADE 10 REVISION OF THE SIN, COS AND TAN GRAPHS 
y =s in x where x e [ -360· ; 360· ] 
x 
360· 
The of the standard sin is 360·. 
The domain of the standard unrestricted sin graph is x e JR. 
In the above graph the domain has been restricted to xe [ -360·; 360· ] 
The range of a standard sin graph is: ye [-1; 1] 
y =c os x where x e [ _360· ; 360· ] 
The of the standard cos is 360·. 
The domain of the standard unrestricted cos graph is x e JR. 
In the above graph the domain has been restricted to x e [ - 360· ; 360· J. 
The range of a standard cos graph is: ye [-1; 1] 
y = tan x where xe [-360· ;360· ] 
I !: .: -. 
; -225' 
-360· 
T T r 
The of the standard tan is 80·. 
The domain of the standard unrestricted tan graph is: 
xe JR, x'" ±90· + 180· k where ke Z 
In the above graph the domain has been restricted to xe [ -360·; 360· ] 
but x '" ±90· and x '" ±270· 
The range of a standard tan graph is: ye JR. -7 also written as y E ( --00; 00 ) 
© Copyright Kevin Smith I Berlot Books CC 109 
THE EFFECTS OF a (CHANGE IN AMPLITUDE) 
If there is a change in amplitude the y-values of the standard graphs will be 
multiplied by a. This causes a change in the amplitude by a factor of a. A change 
in amplitude stretches or compresses the graphs vertically. 
The graphs will be in the forms of: 
y =a sin (x} y =a cos (x} y =a tan (x} 
Example: On separate axes sketch the graphs of: 
y = -3sin x, y = ! cos x and y = 2tan x where XE [-360· ;360· ] 
y 
-1 
y=-3sin x 
r The standard tan graph (where 
a is equal to 1) has ay-value of 
1 at 45 degrees. In this case, as 
a is equal to 2, the y-value at 45 
~ degrees is equal to 2. 
The domain is restricted in the 
question. Therefore, the graph 
L is drawn from -360 to 360 degrees. y= 2 tan x 
THE EFFECTS OF q (VERTICAL SHIFTS) 
If q is added to the graph it will shift up by q units. 
If q is subtracted from the graph it will shift down by q units. 
The graphs will be in the forms of: 
y= sin(x} ±q y= cos (x} ±q y= tan (x} ±q 
Example 1: Sketch the graph of y = sin x + 2 where x E [-360·; 360· ] 
y 
4 
···························3· 
x 
-360· -270 _180· _90· 90· 180· 270· 360· 
-1 
y= sin x +2 
You can think of the line y = q as the new x-axis. 
© Copyright Kevin Smith I Berlot Books CC 110 
Example 2: On separate axes sketch the graphs of: 
y= sin x +1, y= cos x -2 and y= tan x +1 wherexE [0·;360·J 
y 
x 
90· 180· 270· 360· 
x 
90· 180· 270· 360· 
-1 
y= sin x +l y= cos x -2 
y 
3 r The standard tan graph (where 
2 q is equal to 0) bas a y-value of 
- I- --x- 1 at 45 degrees. In this case, as 
~ q is equal to 1, the y-value at 45 
-1 degrees is equal to 2. This is 
-2 because one is added to all of the 
L y-values. y= tan x +l 
THE EFFECTS OF P (HORIZONTAL SHIFTS) 
In the case of a left shift, po will be added to the angle. 
The graphs will be in the forms of: 
y=sin(x+ p' ) y=cos (x+ p' ) y=tan (x+ p' ) 
Example 1: Sketch the graph of y = sin ( x+ 30' ) where XE [-30';330· ] 
This is a shift of 30· left. To sketch the graph, take every key value (intercepts and 
turning points) and subtract 30· from the x-coordinates. 
Original Point Shift New point 
(0· ;0) (0· - 30· ;0) (-30· ;0) 
(90·;1) (90· _30· ;1) (60· ;1) 
(180· ;0) (180· - 30· ;0) (150· ;0) 
(270· ;-1) (270· - 30· ;-1) (240· ;-1) 
(360· ;0) (360· _30· ;0) (330· ;0) 
y 
1 
(-30' ;0) 0,5 
~-r---+----~~-----r--~~~ 
-30.:.0 5 60' 240' 330' x , 
-1 
© copyright Kevin Smith I Berlot Books CC 111 
Example 2: On separate axes sketch the graphs of: 
y =cos (X + 60' ) wherexE [-30';330° ] andy = tan (X+ 4S' ) where XE [ -45°;315° ] 
As you did in the previous example draw up a table and subtract 60' and 45' from the 
x-coordinates of the key values of the cos and tan graph respectively. 
y = cos ( x+ 60') y=tan (x+45') 
Original Original New 
Shift New point Shift 
Point Point point 
(0';1 ) (0° _ 60' ;1) (-{jO' ;I) (0' ; 0) (0° _ 45' ;0) (-45° ;0) 
(90';0) (90' - 60' ;0) (30' ;0) (45';l ) (45' - 45' ;1) (0' ;1 ) 
asymptote asymptote asymptote 
(180' ;-1) (180' _ 60' ;-1) (120' ;-1) 
x=90' x=90' _ 45' x= 45' 
(270' ;0) (270' - 60' ;0) (210' ;0) (180';0) (180' _ 45' ;0) (135' ;0) 
asymptote asymptote asymptote 
(360';1) (360' _ 60' ;1) (300' ;1) 
x=270' x=270' _ 45' x= 225' 
(360' ;0) (360' - 45' ;0) (315' ;0) 
-{;O' 
y=cos(x+ 60' ) 
-45' 
--{),5 r T' 
.............. -1 
········1··· ········································1··· 
y=tan (x+ 45' ) 
© copyright Kevin Smith I Berlot Books CC 112 
In the case of a right shift, pO will be subtracted from the angle. 
The graphs will be in the forms of: 
y=sin(x- p' ) y=cos(x- p' ) y=tan(x- p' ) 
Example: Sketch the graph of y=cos(x- 90· ) where xe[-360·;360·] 
This is a shift of 90· right. To sketch the graph, take every key value (intercepts and 
turning points) and add 90· to the x-coordinates. 
y=cos(x-90·) 
Original Original New 
Shift New point Shift 
Point Point point 
(-450·;0) (-450· +90· ;0) (-360· ;0) (0·;1) (0· +90· ;1) (90· ;1) 
(-360·;1) (-360· +90· ;1) (-270· ;1) (90·;0) (90·+90· ;0) (180· ;0) 
(-270·;0 ) (-270· +90· ;0) (-180· ;0) (180·;-1) (180· +90· ;-1) (270· ;-1) 
(-180· ;-1) (-180·+90· ;-1) (-90· ;-1) (270· ;0) (270· +90· ;0) (360· ;0) 
(-90·;0) (-90· +90· ;0) (0· ;0) (360· ;1) (360· +90· ;1) (450· ;1) 
You do not sketch the point (450°;1) as it is not within the domain that was given in the 
question. ~ 
y 
'i' Notice that a graph of cosO shifted 90' to the right is a graph of sinO. 
© Copyright Kevin Smith I Berlot Books CC 113 
THE EFFECTS OF k (CHANGE IN PERIOD) 
A change in period stretches or compresses the graph horizontally. 
In the case of a change in period the angle will be multiplied by a factor of k. 
y = sin (kx ) y = cos (kx ) y = tan (kx ) 
H 0  1 the graph is compressed and the period decreases. 
Period of a graph: 
In order to calculate the period of a graph use the formula: 
.o d original period 
newpen = k 
Example 1: Sketch the graph of y =sin (2x) where XE [-360';360' ] 
new period = 360' = 180' ~ one cycle is completed in 180' instead of 360' 
2 
In order to sketch the graph, take every key value (intercepts and turning points) and 
divide the x-coordinates by 2. 
Oriltinal Point New point Oriltinal Point New point 
(±720' ;0) (±360' ;0)  (±360';0 ) (±180' ;0 ) 
(±630' ;+1) (±315' ;+1) (±270' ;+1) (±135' ;+1) 
(±540' ;0) (±270' ;0)  (±180';0 ) (±90' ;0) 
(±450';±1) (±225' ;±1) (±90' ;±1) (±45' ;±1) 
Starting point stays at: ( 0' ; 0 ) 
The starting point of a graph does not change with a change in period. 'if 
y 
© Copyright Kevin Smith I Berlot Books CC 114 
Example 2: Sketch the graph of y =c os (~x ) where xE [ -360·; 360· ] 
new period = 3rz· = 720· -7 one cycle is completed in 720· instead of 360· 
In order to sketch the graph, take every key value (intercepts and turning points) and 
divide the x-coordinates by a ~ (this is the same as multiplying by 2). 
Orieinal Point New point Orieinal Point New point 
0 0 0 0 
(-180 ;-1) (-360 ;-1) (90 ;0 ) (180 ;0) 
0 0 0 0 
(-90 ;0) (-180 ;0) (180 ;-1) (360 ;-1) 
Starting point stays: ( 00 ; 1)  
y 
90· 360. x 
0 
Example 3: Sketch the graph of y =t an (2x) where x E [ 0·; 360 ] 
new period = 180· = 90· -7 one cycle is completed in 90· instead of 180· 
2 
In order to sketch the graph, take every key value (intercepts and turning points) and 
divide the x-coordinates by 2 . 
Original Point New point Original Point New point 
0 0 
(00;0) (00 ;0) (360 ;0) (180 ;0) 
o asymptote asymptote 
(4s ;1 ) (22,so ;1 ) o 
x=4S0° x= 22S
asymptote asymptote 0
o (S40° ;0) (270 ;0 ) 
x=90° x= 4S
0 0 asymptote asymptote 
(180 ;0) (90 ;0) o 
x=630° x= 31S
asymptote asymptote 0 0 
o (720 ;0) (360 ;0) 
x=270° x= 13S
y 
1 
x 
22,S" 145. 60· 
-1 
© Copyright Kevin Smith I Berlot Books CC 115 
H. INTERPRETATION OF TRIGONOMETRIC 
FUNCTIONS 
Example 1: The graphs of I(x)=asin(x) and g(x)=cos(kx) are sketched 
below for x E [ 0° ;180° J: 
y 
A 
2 
1 r-::::-:::::: D 
45' 90' 135' 158,53' 180' x 
-1 
B 
a) Determine the values of a and k. 
b) Determine the length of line AB. 
c) Determine the coordinates of point D correct to 2 decimal places. 
d) For which values ofx is 1 (x) ~g(x). 
e) For which values ofx is ~~:~ SO. 
a) The standard sin graph starts at (0; 0) and has an amplitude of 1. The above graph 
starts at (0; 0) and has an amplitude of 2. 
:.a=2 
The standard cos graph has a period of 360' . The above graph has a period of 180'. 
:. 360' = 180' 
k 
:.k=2 
b) AB= /(x) - g(x) 
:.AB=2-(-1) 
:. AB = 3 units 
c) In order to solve for the y-value of D substitute the given x-value into the equation 
of either I(x) org(x). 
1(158,53') = 2sin(158,53') ~ g(x) could also be used 
= 2sin(158,53') 
=0,73 
:. D is the point (158,53' ;0,73) 
d) 1 lies above or on g from points C to D. 
:.21,47° :'>x:'>158,53° 
e) ~ ~ :~ :'> 0 when one graph is postive and the other is negative. 
:. 45' < x < 135' ~ x "* 45' or 135' as g (x) = 0 at these points (you can't divide by 0) 
© Copyright Kevin Smith I Berlot Books CC 116 
Example 2: The graphs of f (x) = tan (x - p) + q and g ( x ) =acos x are sketched 
below for x E [0° ;315° J: 
1 
-1 
-2 
a) Determine the equation of g (x) . 
b) Determine the equation of f (x) • 
c) For which values ofx is f (x) = g(x)? 
d) For which values ofx is g( x) S f (x)? 
e) Write down the equation of h (x) if h (x) is the graph of g (x) 
shifted 30' to the left and 2 units up. 
a) The standard cos graph starts at (0;1). It can be seen on the graph that this is a 
standard cos graph which has been reflected about the x-axis. 
:.a =-1 
:.g(x)=-cosx 
b) The standard tan graph has an asymptote at 90'. In the above graph the asymptote is 
at 135' . This means the graph has been shifted right by 45' . 
:. p =45' 
It can also be seen that the graph has been shifted one unit down. 
:. q =-1 
:. f (x) = tan ( x-45')-1 
c) This is where the graphs intersect each other. 
:.x=90' or x=270' 
d) This occurs where the graph of g is below or equal to the graph of f . 
:.90' S x -· 
Figure 2: Angle of depression 
Note that the angle of elevation is always equal to the angle of depression 
due to alternate angles between parallel lines. 
© Copyright Kevin Smith I Berlot Books CC 177 
B. THE AREA RULE 
The area rule is used to determine the area of triangles which are not right 
angled triangles or triangles where the perpendicular height is not known. The 
area rule is summarised in the diagrams below: 
Angle A known Angle B known Angle C known 
1 A 1 A 1 A 
A"ABC = 2 bc sin A A"ABC = 2ac sin B A"ABC = 2 absin C 
Use the Area Rule when given two sides and an included angle. 
An included angle is the angle between the two given sides. 
PROOF 
The proof of the area rule is shown below for an acute triangle and an obtuse 
triangle. 
Acute Angled Triangle Obtuse Angled Triangle 
B 
o 
o 
~ o o o h: o o 
A D C o 
o 
•L .•  ____I_S_o_o_-A__ ¥ -A'-_ _- ;-_.....:::::... 
b DAb C 
Area of .6.ABC = .!.bxh ~ ill Area of .6.ABC = .!.bxh ~ ill 
2 2 
In .6.ADB: In .6.ADB: In 
A h 
sinA=- sin(180-A) = h ~sin(180-A)=sinA 
c c 
.". h = csinA ~1ll .". h = csinA ~1ll 
Sub 2 into ill : Sub 2 into ill : 
1 A 1 A 
.". Area of .6.ABC=-bcsinA .". Area of .6.ABC=-bcsinA 
2 2 
Similarly it can be proved that: Similarly it can be proved that: 
1 A 1 A 
Area of .6.ABC = -acsinB Area of .6.ABC = - a c sin B 
2 2 
1 A 1 A 
Area of .6.ABC=-absinC Area of .6.ABC =-absinC 
2 2 
L 
© Copyright Kevin Smith I Berlot Books CC 178 
Example 1: Determine the area of the triangles shown below correct to 1 decimal 
place where necessary: 
1) 2) 3) 
C B 
A~ C 140'/ X-
8 A 30mm C 
1) 2) 3) 
A=!(S)(6)sin 30' A =~(21)(25)Sin A A = ~ (30)(19)sinC 
2 
= 12units2 A =IS0' -( 20' +23') =137' C =140' -7S' = 62' 
.'. A =!(21)(25)sin I37' .'. A = !(30)(19)sin 62' 
2 2 
2 
.'.A = 179,Ocm2 .'. A = 251,6 mm
Example 2: Determine the sizels of A B 
if .6.ABC alongside has 
2
an area of 50m • 
1 A 
Area = 2 (AB)(AC)sin A 
.'.50 =! (10)(20)sin A 
2 A 
A 1 
.'.sinA =-
2 
.'. A = 30' ~ this is an acute angle (less than 90' ) C 
It is important to note that it is possible for A to be an obtuse angle. You have to 
give both possible solutions to A . 
.'. A = 30' or A = ISO' - 30' = ISO' ~ obtuse angle 
I ~ Complete Exercise 1 on page 358. 
c. THE SINE RULE 
The Sine Rule is used to determine an unknown length or angle of a triangle 
which is not right angled. 
a b c sin A sinB sinC 
--=--=-- or --=--=--
sin A sinB sinC a b c 
Use the Sine Rule when given two sides and an angle that is not between 
the given sides or if two angles and a side are given. 
© Copyright Kevin Smith I Berlot Books CC 179 
THE SINE RULE PROOF 
The proof of the Sine rule is shown below for an acute triangle and an obtuse 
triangle: 
Acute Angled Triangle Obtuse Angled Triangle 
B 
,
,, 
 
 
Lh ,,  h:, ,  
A D C ,,  
L,.  ____1_8_0o_-_A_ _~ A --_:_.L.........;::• • 
b D A b C 
In .6.ADB: In .6.ADB: 
A h 
sin A =- sin (180· -A) = h ~ sin (180· -A) = sin A 
c c 
:. h = csin A ~ ill :. h = c sin A ~ ill 
In .6.CDB: In .6.CDB: 
A h 
sin C = h sin C =-
a a 
:.h= a sin C ~1ll :.h= a sin C ~1ll 
:. ill = III :. ill = III 
:. c sin A =a sinC :. csin A = a sin C 
sin A sin C sin A sinC 
.. --=-- .. --=--
a c a c 
Similarly it can be proved that: Similarly it can be proved that: 
sin A sin C sinB sin A sin C sinB 
--=--=--
a c b a c b 
The Sine Rule can also be proved using the Area Rule as shown below: 
Acute Angled Triangle Obtuse Angled Triangle 
~ 
A b e 
C 
1 A 1 A 1 A 
From the area rule: -bc sin A =-absinC =-acsinB 
222 
1 sin A sin C sinB 
Now divide all sides by - abc : .. --= --=--
2 a c b 
© Copyright Kevin Smith I Berlot Books CC 180 
B 
Example: Solve lI.ABC in the diagram alongside: 
When a question asks you to solve a triangle you have to solve for 
ail unknown angles and sides. 'if 
AB BC 
= 
sin e sin A 
:. AB = 26 sin 67° 26, 41 units 
sin 65° 
A median 
Left skew (negatively skewed): 
Left Skew (negatively skewed) Left skewed 
I· I: ·  ..  .• · . .  
l 
I I 
I I 
Min Q, Q, Q,Max AI I l
mean < median 
H a data set is normally distributed: 
• Approximately 68% of the data falls within 1 standard deviation of the 
~ mean. 
• Approximately 95% of the data falls within 2 standard deviations of 
the mean. 
© Copyright Kevin Smith I BerlDt Books CC 227 
IDENTIFICATION OF OUTLIERS 
An outlier is a data value which is much larger or smaller than the rest of the 
values in the data set. This value will not follow the general trend of the data. 
A value is defined to be an outlier if it fulf'Ils one of the conditions below: 
• The value is less than Ql -(1,5xIQR) 
• The value is greater than Q3 +(1,5xIQR) 
Outliers can also be seen from scatter plots of bivariate data (data which involves two 
variables). In a scatter plot a general trend can be seen from the plotted points. Points 
which do not follow the general trend are most likely outliers in the data set. The scatter 
plot below shows the amount of time that a Grade 11 class spent studying for a test and 
the marks they obtained out of 50. 
50 
45 
"ll 40 .. 
. ~ 35 
.. 30 
.g 25 
1l 20 .. 
~ 15 • 
10 
5 
20 40 60 80 100 120 140 
Time spent studying (min) 
In the scatter plot it can be seen that the marks follow a linear pattern except for the 
point (140; 15) . This point clearly lies outside the trend of the general data and is an 
outlier. In Grade 12 you will deal with scatter plots in more detail. In Grade 11 you 
have to be able to identify outliers from scatter plots. 
Example 1: The table below shows the Prime Interest rate in South Africa from 
2006 up until 2012. 
a) Determine the five number summary for the data in the table. 
b) Draw a box and whisker plot for the data. 
c) Comment on whether or not the data is skewed. 
d) Are there outliers in the data? 
a) You always have to write the data in order when dealing with a 5 number summary. 
I 8,5 9 10 11 13 14 15 
Minimum Value: 8,5 
Medi an: POS·I·ti on = -n+-l = -7+-1 = 4 :. 11 I'S  the me di an 
2 2 
Low er Quart'il e : P OS.I.ti on = -n+-l = -3+-1 = 2 :. 9 I'S  the  1o wer qu artil e 
2 2 
Upper Quartile: Position = median positions + 2 = 6 :. 14 is the upper quartile 
Maximum Value: 15 
© Copyright Kevin Smith I Berlot Books CC 228 
b) ~r: -----r, ------------~I--~I 
IIIII i IIII I III1111111111111111 111111111111111111111111111111IIIIIIII1111111 1111111111111111 
8 8,5 9 10 11 12 13 14 15 
Q, Q, Q, 
e) It can be seen from the box and whisker plot that the data is skewed to the right. 
d) IQR= 14 - 9 =5 
(2J -(1,5xIQR) = 9 -(1,5x5} = 1,5 and Q3 +(1,5xIQR) = 14 +(1,5x5} = 21,5 
There are no values less than 1,5 or greater than 21,5. Therefore, there are no 
outliers in the data set. Although the data is slightly skewed to the right there are no 
values that are high enough or low enough to be considered outliers. 
Example 2: The table below shows the percentage distribution of the South 
African population by province in 2010. 
a) Determine the five number summary for the data in the table. 
b) Draw a box and whisker plot for the data. 
e) Are there any outliers in the data? 
a) You always have to write the data in order when dealing with a 5 number summary. 
I 2 6 6 7 11 11 14 21 22 
Minimum Value: 2 
Medi an: PoS· .l tionn+=l --9=+-1- =5: . 11 i·s  th eme di an 
2 2 
Lo wer Q u artil e: P oS..l tionn+=l- -=4+-1- = 2,5  :. -6+-6=  6 i·s  th e 1o werqua rtil e 
22 2 
Upper Quartile: Position = median position +2,5 = 5+2,5 = 7,5 
.. 14+21 --- 17,5 is the upper quartile 
2 
Maximum Value: 22 
b) I , , ! I 
I I I I I I I I I I I I I 
2 4 6 8 10 11 12 14 16 17,518 19 20 21 22 
Q, Q, Q, 
e) IQR = 17,5 - 6 = 11,5 
(2J -(1,5xIQR) = 6 -(1,5x11,5) =-11,25 
Q3 +(1,5xIQR) = 17,5 +(1,5x11,5} = 34,75 
There are no values less than -11,5 or greater than 34,75. Therefore, there are no 
outliers in the data set. 
Complete Exercise 4 on page 385 and 386 
the Mixed Exercise on pages 386 and 387. 
© Copyright Kevin Smith I Berlot Books CC 229 



CHAPTER 13 - CALCULATOR TIPS 
WHAT YOU WILL LEARN IN THIS CHAPTER 
III CONTENT III ASSESSMENTS YOU WILL BE ABLE TO: 
231 A. Number Patterns I-Use your calculator effectively I 
232 B. Sketching Graphs to check solutions and save 
233 C. Trigonometric Equations time in tests and exams. 
234 D. Determining Roots of an 
Equation 
© copyright Kevin Smith I Berlot Books CC 230 
A. NUMBER PATTERNS 
In grade 12 you will learn that your calculator can determine a line of best fit for 
bivariate data. For now it is sufficient for you to understand tbat if you enter 
points on a line for a linear, exponential or quadratic function, your calculator 
can determine tbe equation of tbe function that best fits tbe trend of tbe points. 
Example 1: Determine the nth term of tbe linear sequence 5 ; 8 ; 11 ; ... 
As the sequence is linear, the equation has the form Tn = bn + a ---) y = mx + c 
This is the form of a straight line graph, where the term is a y-value and the term 
number (n) is the x-value. 
In order to obtain the equation press: 
Mode ---) STAT ---) A + BX. 
In this form X is the same as n in the equation and A is the same as a and B is the same 
as b. 
Now enter the points of the sequence into the calculator. Remember that when 
n=l ; Tn =8---)x=l; y=8 
x---)n y ---) Tn 
1 5 
2 8 
3 11 
Now press: AC ---) Shift ---) STAT ---) REG ---) B ---)= 
:. B = 3 
Now press: Shift ---) STAT ---) REG ---) A ---)= 
:. A = 2 
:.Tn =3n+ 2 
Example 2: Determine tbe nth term of tbe quadratic sequence 6 ; 13 ; 24 ; ••• 
2 
As the sequence is quadratic Tn =an +bn+c 
In order to obtain the equation press: 
Mode ---)STAT ---) _ + CX2. 
In the calculator the equation is in the form y = cx2 + bx + a . This means that the 
2 
answer for a in the calculator is the answer for c in the equation Tn = an + bn + c and 
the answer for c in the calculator is the answer for a in the above formula 
x---)n y ---) Tn 
1 6 
2 13 
3 24 
Now press: AC ---) Shift ---) STAT ---) REG ---) C ---) = 
:.a = 2 ---) 2 this is a in the equation Tn =an +bn+ c 
Now press: Shift ---) STAT ---) REG ---) B ---)= 
:. b = 1 
Now press: Shift ---) STAT ---) REG ---) A ---)= 
:. c 2 = 3 ---) this is c in the equation Tn =an +bn +c 
:.Tn =2n2 +1n+ 3 
© Copyright Kevin Smith I Berlot Books CC 231 
B. SKETCHING GRAPHS 
It is possible to sketch any graph using the table method. Your calculator has a 
table mode which can draw up a table of points in order to sketch a graph. This 
is especially useful when drawing trigonometric functions. 
Example 1: Use a calculator to draw a table of points that can be used to sketch 
the graph of y =sin ( x + 30· ) where x E [ -30';330' ] 
In order to obtain a table press: Mode ~ TABLE. 
Now type in the equation. You can obtain an x by pressing: ALPHA ~ x. 
0
• The calculator will now ask you where on the x-axis to start. Choose _30 • 
0
• The calculator will now ask you where on the x-axis to end. Choose 330 • 
• You now have to choose the steps to increase in. Provided that there is no change in 
0 
period choose steps of 90 for graphs of sinx and cosx. For graphs of tanx choose 
steps of 450 • 
From the calculator: 
Example 2: Use a calculator to draw a table of points that can be used to sketch 
the graph of y =sin (2x) where XE [-360';225' ] 
In order to obtain a table press: Mode ~ TABLE. 
Now type in the equation. You can obtain an x by pressing: ALPHA ~ x. 
0
• The calculator will now ask you where on the x-axis to start. Choose -360 • 
0
• The calculator will now ask you where on the x-axis to end. Choose 225 • 
• You now have to choose the steps to increase in. In this case k = 2 , this means that 
the period of the graph is 360· = 180· . You would usually increase in steps of 900 • 
2 
As k = 2 , increase in steps of 90· = 90· = 45· . 
k 2 
From the calculator: 
If the re I.S  a cha nge m.  pen·o d,  m. crease m.  s tep s 0 f -90· ".otr he  graph  s 0 f 
k 
45' 
sinx and cosx. For the graph of tanx increase in steps of 
k 
© Copyright Kevin Smith I Berlot Books CC 232 
C. TRIGONOMETRIC EQUATIONS 
When solving trigonometric equations in a specified domain it was seen that you 
had to substitute values for k in order to solve for the specific solution. Your 
calculator can draw up a table for various values of k. 
Example: Solve for fJ if sinfJ = sin 40· and fJE [-360"; 7201 
sin 8 = sin 40· 
RA=4O· 
:.8= 40· + 360·k or 8=180· _ 40· +360·k; kE Z 
= 140· + 360· k 
You can now use your calculator to draw up a table of the values of fJ for 
different values of k as shown below: 
In order to obtain a table press: Mode ~ TABLE. 
Now type in the equation f (x) = 40· + 360· x ~ use an x instead of k. 
• The calculator will now ask you where to start. A value of -3 is usually sufficient. 
However, if the value of 8 lies within the specified domain when k = ±3 you have 
to test k=±4. 
• The calculator will now ask you where to end. A value of 3 is usually sufficient. 
• You now have to choose the steps to increase in. Always choose 1. 
Now repeat the process with the equation: f (x) = 140· + 360· x 
-3 -1040· x -940· x 
-2 --680· x -580· x 
-1 -320· ~ -220· ~ 
0 40· ~ 140· ~ 
1 400· ~ 500· ~ 
2 760· x 860· x 
3 1120· x 1220· x 
Now select the values of 8 which lie in the given interval of [-360"; 720"] . Remember 
that in the calculator the value of 8 is f (x). 
:.8=-320· or 8=40· or 8=400· or 8=-220· or 8=140· or 8=500· 
this is also written as 8 E { -320· ; - 220· ; 40· ; 140·; 400·; 500· } 
© Copyright Kevin Smith I Berlot Books CC 233 
D. DETERMINING ROOTS OF AN EQUATION 
Your calculator can also determine the roots of a quadratic equation. This 
functionality is only available on the Casio 991.fx-ESPlus. 
Example 1: Solve for x if 2x2 +7x+ 3 = o. 
Press: Mode -tEQN -t aX2 + bX+ c =0. 
Now enter the values of a, b and c pressing the =b utton after each entry. Once all 
coefficients have been entered press the = button to obtain the first root and then 
press the = button again to obtain the second root. 
1 
:.X=-- or x=-3 
2 
Note that you cannot simply write down the answer. You must work backwards 
from the answer as you did on page 41 to determine the factors showing all steps. 
This is shown below: 
2x2+ 7x+ 3=O 
:. (2x+l)(x+3) = 0 
1 
...  x=-2- or x=-3 
Example 2: Solve for x if 2x2 - 2x - 4 = 0 . 
If you can, always take out a HCF before using a calculator. 
2
:.2(lx - lx - 2)=O 
:. lx2 - 1x- 2 =O 
Now Press: Mode -tEQN -t aX2 + bX+ c =0. 
Now enter the values of a, b and c pressing the =b utton after each entry. Once all 
coefficients have been entered press the = button to obtain the first root and then 
press the = button again to obtain the second root. 
:.x=-l or x=2 
Now write down the full solution: 
2
:.2(lx - 1x- 2)=O 
:. lx2 - 1x- 2 =O 
:.(x+l)(x-2)=O 
:.x=-l or x=2 
The functions of your calculator are not limited to what has been 
demonstrated in this section. Your calculator manual will show you all the 
functions that your calculator is capable of. Read through the manual and 
make sure you are familiar with your calculator. 
© Copyright Kevin Smith I Berlot Books CC 234 



CHAPTER 14 
MAKING THE MOST OF MATHS 
WHAT YOU WILL LEARN IN THIS CHAPTER 
III III ASSESSMENTS CONTENT YOU WILL BE ABLE TO: 
236 A. Maths in the Classroom I • Get the most out of I 
236 B. Exam Study Techniques mathematics in your classroom. 
236 C. The Morning of the Exam 
237 D. Exam Writing Techniques • Study with confidence and use 
your time effectively. 
• Relax and walk into your exam 
with confidence. 
• Write your exam neatly and use 
your time efficiently. 
© Copyright Kevin Smith I Berlot Books CC 235 
A. MATHS IN THE CLASSROOM 
• Up to 80% of learning should be achieved in the classroom. Try to sit as close 
as possible to the front of the classroom where you can see and hear properly. 
• Concentrate and avoid the temptation to talk to your friends. Missing an 
explanation could result in a misunderstanding of an entire section. 
• Do not be shy! If you don't understand an explanation that your teacher has 
given, put your hand up and say so. Ask your teacher to explain it again until 
you do understand. If you did not understand a concept, it is likely that the 
majority of your class is in the same position. 
B. EXAM STUDY TECHNIQUES 
• Practising past exam papers is the most helpful and practical form of revision. 
Past papers expose you to the required level of work, exam format and the 
necessary time limitations imposed on you. 
• Make sure you start studying early. You should start with past papers at least 
two weeks before the exam. Make sure that as you do each question you check 
your answer. If you check your answers after you have completed the whole 
paper, you cannot remember what each question was about. This means that you 
do not learn from your mistakes. It is important that you revise each section 
before starting past papers. 
• In order to achieve the correct pace, while timing yourself, do an entire past 
paper and only check your answers at the end of the paper. This should be done 
two to three days before your exam. 
• Ensure the room that you are studying in is brightly lit and properly ventilated. 
• Cellular phones can be your worst enemy whilst studying. Turn your phone otT 
and check your messages only when you have finished studying. 
C. THE MORNING OF THE EXAM 
• Make sure that everything you need for your exam is packed the night before 
and ready to take in the morning. Ensure your calculator is in the correct mode 
with charged batteries. It is a good idea to take a spare calculator into your exam. 
• Sleep is very important. A void going to bed too late or waking up very early to 
study. A maths test or exam requires a fresh brain. 
• Eat Breakfast! Hunger affects concentration. 
• Avoid discussing work before the exam. This is likely to cause confusion. 
• You've worked hard and are prepared - now relax and enjoy the exam. 
© Copyright Kevin Smith I Berlot Books CC 236 
D. EXAM WRITING TECHNIQUES 
• Read through the entire question paper before you start. You can answer 
the questions in any order you like. Answer the questions that you are most 
confident in first. However, you must answer an entire question in the same 
section, don't answer parts of a question in different places in your answer book. 
If you are unsure of a certain part of the question, leave it out and come back to 
it. 
• H you leave out part of a question, make sure you leave enough space to 
answer it. Always leave at least half a page open if the question is worth more 
than two marks. If you are not sure how much space will be needed, be safe and 
leave a full page. 
• Mark questions that you have left out on the question paper, so that you 
remember to come back to them. 
• Neatness is crucial. Keeping your solutions neat and legible during an exam 
may result in up to a 5% increase in your results compared to an untidy and 
illegible answer book. Neat workings also assist you, as it is easier to check 
answers and find mistakes. In addition, neatness helps to avoid silly mistakes. 
Do not start a new question at the end of a page. 
• Do not attempt to save space in your exams! Cramped work is messy and 
difficult to mark. Leave at least one line between answers. Similarly, do not 
write complicated fractions on a single line. 
• Never cross out an answer, unless you have already written another answer to 
replace it. Once you have what you feel is the correct answer, only then cross 
out the incorrect workings. If you leave two solutions without crossing one out 
only the first solution will be marked. 
• Do not use any type of correction fluid (eg. tippex) in a test or exam. 
Rather neatly cross out your answer and carry on. Erasable pens should not be 
be used. 
• Follow the layout shown in this handbook. Keep to the top to bottom layout and 
the exam will be a pleasure to write and mark. 
• An answer book that has the fmal answer of each question underlined with an 
arrow is very pleasant for a teacher to mark. However, this can be very costly in 
terms of time. H you are a slow or untidy worker, use a ruler and neatly 
underline your answers with arrows, only once you are finished your test or 
exam and have checked through your solutions. Make sure that you don't cross 
out your answer when doing this. 
• Finally, relax and be positive. It is important that you don't panic at any stage 
during the exam. 
Good luck - You have earned it! 
© Copyright Kevin Smith I Berlot Books CC 237 
CHAPTER 15 
GLOSSARY OF IMPORTANT TERMS 
AND SYMBOLS 
Term Explanation ~'" 
A perpendicular line 
drawn from one side of a 
altitude 
figure to the vertex of the 
other side. 
3 
The length of maximum 
deviation from the zero 
2 
amplitude line. 
Amplitude = Ymax-Ymin 1 
2 ~~~~~~+ 
90 180 270 360 x 
The angle between the 
horizontal and the line of 
angle of depression 
sight when looking down 
at an obiect. 
I Object I 
The angle between the 
horizontal and the line of 
angle of elevation 
sight when looking up at 
Angle of 
an obiect. 
elevation 
A ..L  __ .l!.wi?!lntal 
y 
The angle measured 
anticlockwise from the 
angle of inclination positive side of the x axis. 
It is always between 0' 
x 
and 180'. 
A sequence where the In the sequence: 
diHerence between any 2,4,6. .. 
arithmetic sequence 
two sequential terms is The constant diHerence d 
constant. is Tn-T. .... l=2. 
© Copyright Kevin Smith I BerlDt Books CC 238 
Term Explanation ~ 
y 
An Imaginary line that a 
asymptote 
graph can never touch. 1_ _____ _ 
--.l _~ __- -,x,+ 
A line drawn so that the Y Axis of symmetry 
graphs or parts of the 
axis of symmetry graph on either side of 
the line are mirror 
images of each other. 
x 
Data that is given in The time a student 
ordered pairs, an x and y spends studying and the 
bivariate data coordinate. It is usually mark that he or she 
represented using a obtains is an example of 
scalier plot. bivariate data. 
A visual representation of 
the median (QJ, lower 
quartlle(Q,), upper whisker box whisker ~[===JI~~~ 
box and quartile (Q.), as _II as 
I I 
whisker plots the highest and lowest :-low~'e s-t 25 50 75 highest 
values In a set of data, to value Ql Q, Q3 value 
show the spread around 
the median. 
Y 
Quadrant Quadrant 
Four quadrants that are n I 
Cartesian Plane separated by an x and y 
x 
axis 
Quadrant Quadrant 
III IV 
I ---< 
I I 
The physical distance 
circumference around a circle. Formula: 
C=2trr or C=1rd 
I 
----I 
I 
In the expression 
The number in front of a 
coefficient 
variable. 2x2 + 4 the coefficient of 
~is 2. 
1 L 
© Copyright Kevin Smith I Berlot Books CC 239 
Term Explanation ~ 
When 3 points lie on the 
collinear points same line they are 
collinear. 
I 
T T 
The term in an expression In the expression 
constant term that does not contain any 2x2 + bx + 4 the constant 
variables. term is 4. 
I , 
T 
The vertices of a cyclic 
cyclic quadrilateral quadrilateral lie on the 
circumference of the same 
circle 
I I 
3 
The number at the bottom In the fraction - , x is the 
denominator 
of a fraction. x 
denominator. 
I , 
A line throug h the centre 
of the circle that touches diameter 
diameter 
the circumference of the 
eircle at two points. 
t 1 
y 
The domain of a function 
domain is the possible x values -1 x 
that the function can have. 
In the above graph 
the domain is x ~ -1 
The equivalent annual 
If the Interest rate 
Interest rate that would 
received is 10% p.a. 
be received due to the 
compounded monthly, the 
effective Interest rate effects of compounding 
effective rate Is: 
Interest. 
ieff = (1+ 0X2)12 -1 
l+ ieff = (1+ ~om r 
© Copyright Kevin Smith I Berlot Books CC 240 
Term Explanation ~ 
In the term xa, a is the 
exponent The power that a number 
(and base) or variable is raised tao exponent and x is the 
base. 
y 
The area within a linear o Feasible Region 
feasible region programming graph 
where all the constraints 
are satisfied. 
~ x t 
______ ____ __ __ f.l~~~~ 
An equation or graph 
function where there are multiple 
(many-to-one) x values for a single y 
value. x 
y 
An equation or graph 
function where there Is a single y 
(one-to-one) value for a single x value. 
A sequence where the ratio In the sequence: 
of any two sequential tenns 
geometric sequence 4, a, 16, ••• 
is conslant. 
the constant ratio is 2 
1 
The average gradient 
The role of change of Y with 
between the points (5;2) 
respect x. 
gradient (m) and (10;5) is: 
m= y, -y, 5-2 3 
Xl -Xl m=--=-
10-5 5 
y 
to 2. 
A graph without gaps that ~ 15 
histogram uses reclangles to represent .r..  1. 
freq~ • 
0-1 SJ 20-38 40-59 x 
Data 
© Copyright Kevin Smith I Berlnt Books CC 241 
Term Explanation ~ 
y 
The point where a line 
Intercept 
cuts the X-axis (x-Intercept) 
(r or y) x-intercept 
and y-axis (y-Intercept). 
I I 
~ 
I I 
A line that Is perpendicular 
normal to a tangent at the point 
of contact. 
t ~ 1 x+S 
--
2 
The number or expression 
numerator x+5 is the numerator 
at the top of a fraction. and 2 is the 
denominator 
.... 
y 
A graph where the data 
range (x) Is plotted ]t-
ogive against the cumulative 
.;~ 
s'" 
frequency (y). aE 
..l....- Data Range • 
A value In a set of data 
that is much higher or 2,100,125,126,130 
outlier 
lower than any of the In the above set of data 
(of a data set) 
other values in the data points 2, is an outlier. 
set. 
1 
A value in a set of data 
parallel lines that Is much higher or 
lower than any of the 
(II) 
other values in the data 
set. 
The number of degrees It Graphs of sine and 
takes for a trigonometric cosine have a period of 
period 
function to complete one 360· whilst tan has a 
cycle. period of 180· 
© Copyright Kevin Smith I Berlot Books CC 242 
Term Explanation _ 
In a 90· triangle the 
Pythagoras (hypotenuse)'= sum of the 
(Theorem of Pythagoras) square of the opposite two 
sides. a 
I 
I I 
The possible y-values that 
range a graph or function can x 
have. 
In the above graph 
I I the range Is y ~-1 
The reciprocal of a number 
The recipracal of 
is one divided by the 
reciprocal number. A number 1 
multiplied by lis reciprocal x 15-x 
Is 1. 
root The value of a variable 
(of an equation) The raot of x-5 =0  is 5 that satisfies an equation. 
An irrational number that 
is written with a root ,J5ii written in simplest 
surd form 
symbol (usually in its surd form is 5../2 
simplest form). 
A straight line that tauches 
a graph at a single point. 
tangent The gradient f:I the tangent 
is equal to the gradient of 
the graph at the point of 
CIOnIact. 
y 
trend line A line that shows the 
(line of best fit) general shape of a scatter 
1 
plot, thus showing the 
(regression line) 1 
overall trend of the data. 
.3't...,-,-----:------:--:-:--7> 
Independent variable x 
© Copyright Kevin Smith I Berlot Books CC 243 
Term Explanation ~ 
tuming point 
A point on a curve where 
the gradient at the point 
turning point is equal to zero. At the 
TP the derivative of the 
function equals zero. 
The roots of 
a~?+bx+c = 0 are: 
Used to solve for the roots 
quadratic formula 
of any quadratic equation. -b±~b2-4ac 
x = ----'-:---
2a 
1 
GLOSSARY OF IMPORTANT SYMBOLS 
Symbol Meaning Symbol Meaning 
element 
F square root E (of a set) 
rq n'h root 01- not equal to 
... approximately equal N natural numbers 
to 
No whole numbers A>B A Is greater than B 
Z Integers A Copyrisbt Kevin Smith I Berlut Boob CC 253 
Simple Surd Equations 
Ex'n:t. .. 4: Exemjse 4 is on page 248 
1) 
8) "/x+3=10 
e) 2~~+3+4=12 
... x+3=100 
... x=97 
... 2~~+3=8 
b) ../x+2 =../fi ."~~+3=4 
... x+2=17 
... x=15 ... ~+3=16 
2 
... ~=13 
c) "/8-x =-2 2 
... No Solution as ../8 - x ~ 0 ... x=26 
d) "/3x-5+8=9 I) 'J./..Jx+3=9 
... ../3x-5 =1 ... 9(..Jx+3) =81 
... 3x-5 = 1 
... ..Jx+3=9 
... 3x=6 
... ..Jx=6 
... x=2 
... x=36 
2) 
8) 2../2x+3 +2(5) = 35 2 b) Ao =7rr
... 4(2x+3) = 625 
... 357r = 7r( ../x+5)2 
625 
... 2x+3=- ... 357r= 7r(x+5) 
4 
... 35=x+5 
... 2x= 613 
4 ... x=30 
613 
:.x=-
8 
... x=76,63 
Equations with Rational Exponents 
Exerd"S: El£ernjse 5 is on page 248 
8) (Xl' b) 42x+1 = Ys c) 2x 1 - =64 =4 
... 2x-1 6 ... 24x+2 =2-3 =2
... 2-x =22 
... x-l=6 
... x=-2 ... 4x+2=-3 
... x=7 
...  x=-%4  
d) 5,-2 = Is x e) 9x81 = 27 I) 7x9x =7 
5' 
x 2 ... 3
4x =31 2x 
... 5 - =S-x ... 3 =1 
.. .4x=1 2x 
... x-2=-x ... 3 =30 
... x=1 ... x=~ ... x=O 
4> Copyrisbt Kevin Smith I Berlut Boob CC 254 
g) 52,+3 = 125,+5 h) 22>+2 = (0,5)2,--4 i) 25.2' = 4.5' 
:.52>+3 = 53,+15 2
:.22,+2 = (2-1 )2>--4 :.5 .2' = 22.5' 
:. 2x+3=3x+15 
:.22,+2 = 2-2>+4 :.(Ys) x (2)2 :.x=-12 = 5" 
:.2x+2=-2. Copyrisbt Kevin Smith I Berlut Boob CC 255 
8X .4X+1 4+.Ji2 ~X.p 
e) g) 
5x 3 i) 2 +
(1+.J3f x-%yl~ 
23X X (22r+1 4+2.J3 xX),~ 
25x+3 1+2.J3+3 
x-%y~ 
23x.22x+2 4+2.fj 
25x+3 4+2.fj =xX+%.y.%-~ 
= 23x+2x+2-5x-3 =1 =xy 
-2-1 - 1/ 
- -72 .J5 j) 
b) (2.J3- ~)(.fj+ ~) 
(6+.J5)(6-.J5) 
1 
5+2../50 =6+2-1--
f) 
5 .J5 3 =--
36-5 20 
5+10../2 = 
5 .J5 
3 
--
31 2 ab2
t(I+2.J2) k) (a b+ t 
t =1 
=1+2.J2 
3) 
a) .Jx+2 =4 b) ~7+..Ix =3 c) 2.J8-x = 2 
:.x+2=16 :. 7+..Ix =9 : . .J8-x =1 
:. x = 14 
: . ..Ix =2 :.8-x=1 
Check:.J14 + 2 = 4 :.x=4 :.x=7 
~7+.J4 Check: 2.J8 -7 = 2 Check: = 3 
4) (p+ ~r =l 
2 1 2 
P +2+-=q 
p2 
5) 
b) x+ Y+z.JxY = 7 +.J4xI2 
a) (..Ix +.JY)2 = (~7 +.J4st 
:.x+y+2,JxY =7+2.Ji2 
:. x+z.JxY + y =7+.J4s :. xy = 12 
:.x+y+2,JxY =7+.J4s 
:.x+y=7 
6) 
a) Ao = 1I:r2 b) Circumference = 21I:r 
:.100 = 1I:r2 
= 211:( 1r) 
2 100 
:.r =- 2011: 
11: =-cm 
10 lo-Jii .JK 
... r = JH em --7 ----;- = =2o-Jiicm 
The Mixed Exercise is on page 249 
© Copyright Kevin Smith I Berlot Books CC 256 
7) 
2 
a) x +3x-6 = 0 
. _ -3 + ~r-3( ) -::-2- -4-(1) --(- 6) -3-J(3f -4(1)(-6) 
.. x- 2(1) or x 
2(1) 
-3+.J33 -3-.J33 
:.X= or x 
2 2 
b) _2x2-3x+6=0 
_( -3) + ~r--(- -'3)2:-_4-(--2-)(-6) -(-3)-J(-3)2 -4(-2)(6) 
:.x= 2(-2) or x 2(-2) 
-3+.J5'i -3-.J5'i 
:.x= or x 
4 4 
9) 
22020 + 22019 120 121 3 +3
a) b) 
22019 _22017 3122 _3121 
2017 120
2 (23 +22) 3 (1+3) 
2017 120 2 1
2 (22 -1) 3 (3 _3 ) 
=N =~ 
=2 2 2 .J3 2.J3 =-=-x-=-
.J3 .J3.J3 3 
1 
10) A .. =-b.  .l h 
2 
:. 20 =~(if2 -2.J3)(h) 
:.40 = (4'-'2-2.J3)(h) 
40 
:. h = '-'2 .J3 -+ rationalise the denonrinalor 
4 2-2 3 
. h= 40 x4'-'2+2.J3 
.. 4'-'2-2.J3 4'-'2+2.J3 
40( 4'-'2 +2.J3) 
:.h= 20 
:.h=8'-'2+Wem 
The Mixed Exercise is on page 249 
© Copyright Kevin Smith I BerlDt Books CC 257 
Chapter 2- Equations and Inequalities 
Grade 10 Revision 
Exercise 1: Exercise 1 solutions are on pages 264 and 265 
1) Factorise the fonowing expressions: 
a) x2-2x-IS e) 4x2 -16x+16 i) 6x2+Sx-4 
b) x2 +x-12 f) x2+lOx+2S j) a3 _a2b+b3 _ab2 
x2 -Sx+12 2 2 2 c) g) 2x2 -Sx-12 k) x _a -2ab-b
d) x2 -6x+9 b) 6x2+13x-S I) 2~-3~-Sx+12 
2) Simplify the fonowing expressions: 
4 S 2 2 
a) -+-- d) x +Sx+6 x -4 
S 2 3-Sx 
g) ---+--
x+l x-2 2x2 -9 . x2 -Sx+6 x+3 3-x x -9 
b) _1_+ 2 1 123 
e) --+----
x-2 (x_l)2 x-I x-3 x+l 3-x 
x2 +x-6 . ~ +Sx+6 xiS f) ---+--
c) 2 • ---=-2- -- x-2 x+2 4_x2 
x -6x+S x -3x-IO 
Quadratic Equations 
Exerdse2: Exercise 2 solutions are on page 265 
1) Solve for x in the fonowing equations: 
a) x2 -Sx+6=O c) (x-S}(x+12)=0 e) 4x2-12x+1O=1 
b) x2 +3x = 0 d) (x-S}(x+12) = -70 f) -~ -Sx2 -ISx = 0 
2) Solve for the unknown variable in the fonowing equations: 
a) (y+3)2=l+6y+9 b) ~+16=0 
Equations with Fractions 
Exerdse3: Exercise 3 solutions are on pages 266 and 267 
1) Solve for x in the fonowing equations: 
IS IS 
a) x+-=ll f) x-l+--=S 
x x+2 
6 
b) 2x+4=-- 3x-24 +_9_=0 g) 
2x-l 9 3x-6 
2 6 2x+4 Sx+lO 
c) ----=-- b) _3_+~=1 
x+l 2-x x+l x+3 2-x x2+x-6 
4x-14 2x-4 S 2y-6 10 
d) -----=0 i) --+ 
x+4 x+2 y+l l+3y+2 4-y
2  
e) 2x +_1_=1 
x-I x+l 
2) Solve for the unknown variable in each of the fonowing equations: 
3a-l 3a 2 -0 6p2 -S3p+1O 
a) +-+--- 3p Sp c) o 
a2 -2a-3 a+l 3-a p2 -3p-1O 5-p p+2 
5y+4 2y+3 2l-17 
b) --+ --= -==------='---
y+2 l-y l+y-2 
© copyright Kevin Smith I Berlot Books CC 258 
Equations w1th Substitution 
Exercise 4: Solve for the unknown variable In each of the foUowlng equations: 
d) (q2+2qt -2(q2+ 2q)-3=0 
e) p2 +5p +5 = 12 
2 p2+5p 
2 4 
c) x -3x+4=---- f) (x2-xt -8x2+8x+12=0 
x2 -3x 
Exercise 4 solutions are on pages 267 and 268 
Quadratic Surd Equations 
Exercise 5: Solve for the x in each of the following equations: Exercise 5 solutions are on pages 268 and 269 
a) -Fx+x=O f) x-7=·hx+l 3 j) .J2x-l -2 
b) .J2x-l x=.Jx+6 g) .J22-7x-x=-4 
c) 2x-.J32-8x =0 k) .J12+x =-Fx+3 
b) 4.Jx-3 -3= 2x-9 
d) x+ 2 = .J9x+ 18 I) 2.J4-x=2-Fx-4 
i) .Jx-5-1=_6-
e) x-4=.J56-5x .Jx-5 
Exponential Equations 
Exercise 6: Solve for the unknown variable in each of the foUowlng equations: 
813y+2 
d) __ =34 
24rY 
b) 23m.Tm=T2m b) 9'_3' =6 2
e) 5 ,-4 =1 
c) S-3'.5' =.!:. f) 32,-4 = U 
5 
Exercise 6 solutions are on page 269 
Simultaneous Equations 
Exercise 7: Exercise 7 solutions are on pages 270 and 271 
1) Solve for x and y in each of the following equations: 
a) y+6x = 4 and f) y-x=-6 and 
2 2
2y = 5x -22x+8 2x +5xy=-2i 
b) 2x-y=8 and g) y-x=5 and 
y =x2 +4x-23 xy-3y = 8x-20 
c) y+x=6 and b) 2'+Y = 256 and 
x2+i=36 xy+3x+2y = 36 
d) 2x-3y = 2 and i) 5,-2y-1 = 1 and 
x2-2i=8 ~+2xy-2x-4y=0 
e) y-2=2(x-l)2 and j) (x+5)2 +(y_3)2 =49 and 
y-x=2 y+x=-9 
2) Given the diagram below solve the simultaneous equations of y = - 8x - 24 and y = 2x2 - 2x - 24. 
y =-8x-24 
© copyright Kevin Smith I BerlDt Books CC 259 
Completing the Square 
Exercise 8: Exercise 8 solutions are on pages 272 aod 273 
1) Complete the square on the foUowing quadratic expressions: 
y =x2 a) +6x-S d) y=x2-Sx g) y=-2x2 -3x+S 
b) y =x2 -4x+9 e) y=2x2_4x+6 2h) y=3x +10x-2 
2
c) y=x -12x+2 f) y = -3x
2 -6x+ 12 
2) Solve for x by completing the square on the foUowing quadratic equations: 
a) x2-6x-S=0 d) -Sx2+1Sx+l=0 g) ~+2x+3=0 
2 2
b) x +2x-7=0 e) x2+2hx+3c=O h) ax +3bx-1S=O 
2 
c) 2~+3x-S=O f) x -2x-3c = 0 
Applications of Completing the Square 
Exercise 9: Complete the square to determine the maximum or mioimum value of: 
2
a) x -2x-1O c) ~-12x+40 
b) _x2 -Sx+3 
Exercise 9 solutions are on page 274 
The Quadratic Formula and Nature of Roots 
Exercise 10: Exercise 10 solutions are on page 274 
1) Solve for x in the foUowing equations correct to 1 decimal place: 
4~+Sx-3=O -3x2a) c) -4x+S=O 
b) 7~-3x-2=O d) 2~-6x-12=O 
2) Given that (x -2,5)( x -vCl)(x +2)( x -Ji)( x +.,/3) = 0, solve for x if: 
x Z c) X E Q' a) E 
b) x E R d) is non-real 
3) Given that 2x2 (x-2)+5x(x-2)-5(x-2) = 0, solve for x if: 
a) x E /Ii b) x E Q' 
4) Determine the nature of roots of the equations below without solving the equations: 
a) 14x2-6x+S=O c) SO~+40x+S=O 
b) 9x2 +1Sx+3=0 d) -2Sx2 -lOx-l=0 
Inequalities 
Exercise 11: Exercise 11 solutions are on pages 274. 275 and 276 
1) Solve for x in the foUowing inequalities: 
a) x2-4~O g) (x-3)+IS0 (x-l)2(X-S) 
k) >0 
b) x2
x+2 
-x-6>O x+S 
2 h) (x+l)(x-3) >1 
c) x -3x-1O:S;O I) x2+Sx+ll +1~O x-S x-2 
2
d) x +9x+1S:S;0 i) (x-2)(x+l) no 
e) _x2 +2x+1S < 0 x-4 
(x-2)2(X+S) (x-3)(x+2) :s; 0 
f) j) 0 d) (q+3)2+50) :. !..=...Ix 
2 
:. k2 -k-6=0 1 
:. X=-
:. (k-3)(k+2) =0 4 
:. k = 3 or k =-2 
:."x-5 =3 or "x-5 =-2--7 no solution I) 2"4-x = 2...1x-4 
:. x-5 =9 :.4(4-x) = 4x-I6.../x+16 
:. x=14 
:.16-4x-4x-16 =-I6.../x 
: . -8x = -16.../x 
j) "2x - l ~ -2 
2x-l :.64~ =256x 
Letk ="2x-l :.64~-256x=0 
2
:. k-i= - 2, (k>0) :.x -4x=0 
2 :. x(x-4)=0 
:. k +2k-3=0 
:. x=4 (x .. O) 
:. (k-l)(k+3)=0 
:. k=1 or k=-3 
:."2x-l=1 or "2x-l=-3--7N/A 
: . 2x-I=1 
:.x=1 
Exercise:; is OD page 259 
Exponential Equations 
Eurdse6: Ex~i~ f! i~ Q!! RI&!i 252 
2
a) T20. = 7 - 3-4 = 1 b) 9"'-3'=6 
:. -2a=2 - 3a :. 52, -4 =SJ :.32, _3' -6=0 
:. a=2 :. 2x-4=0 Let k =3' 
23m.2-m = 2- 2m :. 2.< = 4 b) :.k2 - k-6=0 
3m m 2m : . .-4 = X :.k = 3 or k=-2 
:. m=O 1 :.32, -4 = 3-2 :.3' =3 or 3' =-2 
3 :. 2x-4= - 2 
c) r '.5' =!.. :.x=1 
(3, .. -2) 
5 :. 2x=2 
: . r 2, =r1 : . .: 0 
5±.fi9 
:.x=4 or x=1 or x 
2 :.(x-l)2+6 ;;>:0 
:. XE IR 
The Mixed Exercise is on pages 262 and 263 
© Copyright Kevin Smith I Berlot Books CC 281 
x2
2 
( +1)(3-2x) j) 2x +3x-2,;; 0 
i) .",, z b) a = 3 --+ (2a = 6) 
:.T 2 n =3n +bn+c 
6 6 6 6 6 6 3=b+c--+ill 
x=16-10 and y=54+28 and z=156+46 4=2b+c --+~ 
=6 =82 =202 ~-ill : 
l=b--+1] 
I] into ill: 
:.c=2 
:.T =3n2 n +n+2 
Ta Tg TIO 
8) ... ; -15 '>."" -8 '>."" 1 ; ... 
7'>. "" 9 
2 
a=I--+(2a=2) 
:.Tn =n2+bn+c 
-79=8b+c--+ill 
-89=9b+c--+~ 
~-ill: 
-lO=b --+1] 
I] into ill: 
:.c=1 
:.Tn =n2-10n+l 
The Mixed Exercise is on pages 285 and 286 
© Copyright Kevin Smith I Berlot Books CC 290 
Honeycombs: 1;3; 6; 10; 15; ... ---+ Quadratic (second difference = 1) 
Walls: 6 ;15; 27; 42; 60; ... ---+ Quadratic (second difference = 3) 
Honeycombs:l;3; 6; 10; 15; ... Walls: 6 ;15; 27; 42; 60; ... 
Sequence is quadratic (second difference = 1) Sequence is quadratic (second difference = 3) 
a= ~ ---+(2a=2) a=.% ---+(2a=3) 
:.Tn = ~n2+bn+c 
2 
:.Tn = .%n +bn+c 
li=b+c---+ill %= b+c---+ill 
l=2b+c---+~ 9=2b+c---+~ 
~-ill: ~-ill: 
li =b ---+ I] %= b---+I] 
I] into ill: I] into ill: 
:.c=O :.c=O 
2 
:.Tn = ~n2+ lin---+numberofhoneycombs :. Tn = .% n + %n ---+ number of walls 
but each wall contains 10 ice-cream sticks 
:. the nth tenn for the number of ice-cream sticks is: 
2 
Tn = 15n + 45n 
10) Determine the general equation for the nth term of the sequence: 
2'\,,/7'\,,/15'\,,/2~"..-4~,/57 a) x = T5 = 40 cards 
5'\,,/8 '\,,/11'\,,/14'\,,/17 y: 
3 3 3 3 2155= .%n + ~n 
a=.%---+(2a=3) :.3n2 +n-31O=0 
2
:.Tn = %n +bn+c :. (n-1O)(3n+31) = 0 
li=b+c---+ill :. n = 10 or n = -3K ---+ N/A 
:. y = 10 stories 
1 = 2b+c ---+ ~ 
~-ill: 
b) From the above pattern 5 stories is the most 
li =b ---+ I] 
that can be built from a single deck. (T6 = 57) 
I] into ill: 
:.c=O c) 
2
:.Tn=%n +lin 
E 
22.5cm 22.5cm 
H2 =(45)2_(22,5)2 
:.H=38,97cm 
The Mixed Exercise is on pages 285 and 286 
© Copyright Kevin Smith I Berlot Books CC 291 
Chapter 4- Analytical Geometry 
Grade 10 Revision 
Exercise 1: Exercise 1 solutions are on pages 296 and 297 
1) Determine the lengths of the line segments joining the following points: 
a) A(2;-5) and B(-I;-8) c) S(1O;-6) and R(-3;2) 
b) P(5;5) and Q( -2;3) d) M(-I;-5) and N(I;-10) 
2) Determine the midpoints of the line segments joining the following points: 
a) A(2;-5) and B(-I;-8) c) S(1O;-6) and R(-3;2) 
b) P(5;5) and Q(-2;3) d) M(-I;-5) and N(I;-1O) 
3) Determine the gradients of the line segments joining the following points: 
a) A(2;-5) and B(-I;-8) b) P(5;5) and Q(-2;3) 
4) lntriangleABC,withA(4;6), B(x;y) and C(-4;-6), P(5;1) isthemidpointofAB.Determine: 
a) The coordinates of b) The perimeter of the c) If the triangle is right-
point B. triangle. angled. 
5) Determine two possible values of x if AB is .J185 units and A is the point (4; 6) and B is the point 
(x;-5). 
6) Determine if AB and CD are parallel, perpendicular or neither given the points: 
a) A(-8;-4) and B(2;4) and C(-4;O) and 0(-20;20) 
b) A(5;6) and B(4;2) and C(8;24) and 0(4;8) 
7) Determine whether or not the points A(-5;-7) and B(4;11) and C(28;59) are collinear. 
8) If A is the point A (x; 3) and B is the point B (-2; y) determine the valueJs of : 
a) x if AD is perpendicular to the line y=-3x+6 and 0 is the point 0(4;2). 
b) x andy if the midpoint ofABis (1;6). 
Finding the equation of a straight line 
Exercise 2: Exercise 2 solutions are on pages 297 and 298 
1) Determine the equation of the line which has the gradient given below and passes through the given 
point. 
a) m = -2 and point (3;8) b) m=4 and point (-5;3) c) m=-.!. and point (3;1) 
3 
2) Determine the equation of the line which: 
a) Passes through the points (-2;5) and (-8;12). 
b) Is parallel to 3y -12x = -9 and passes through the point (2;2). 
c) Is perpendicular to 4y-x-8 = 0 and passes through the point (-2;6). 
d) Is a vertical line through the point (3;5). 
e) Is a horizontal line through the point (-8;-12). 
© Copyright Kevin Smith I Berlot Books CC 292 
Angle of Inclination 
Exercise 3: Exercise 3 solutions are on page 299 
1) Correct to 1 decimal place, determine the angle of inclination of the straight lines whicb pass through 
the points: 
a) (1;1) and (4;7) c) (6;4) and (-8;-3) e) (-2;3) and (-3;6) 
b) (5;4) and (1;8) d) (-1;5) and (1;-3) f) (7;-7) and (8;7) 
2) Determine the gradient of a straight line with an angle of inclination of: 
a) 30· e) 60· e) o· 
b) 45· d) 150· f) 180· 
3) Determine the magnitude (size) of angle a in each of the following diagrams. Give your answer 
correct to 1 decimal place. You may assume that all dotted lines are parallel to the x·axis. 
y b) Y c) 
y=0,5%+1 
B 
x A x 
A 
B Y Y Y 
d) e) A f) 
x x 
B 
1 
y=--x+3 
2 
b) i) Y 
C x 
y=2%+1 
y=3% 
y =-2%+4 
Mhed ExercIse: The Mixed Exercise solutions are on pages 299. 300. 301 and 302 
1) 1f a quadrilateral bas vertices A(-2;2), B(-3;-2), C(3;-1) and 0(4;3), prove that: 
a) ABCD is parallelogram. 
b) ABCD is not a rhombus or a square. 
© Copyright Kevin Smith I BerlDt Books CC 293 
2) The diagram below shows triangle PQR with vertices P(-2;8),Q(4;4) andR(l;--{j). Determine 
the foUowing, leaving your solutions in simplest surd form where necessary. 
P(-2;8) y 
Q(4;4) 
x 
R(1;-6) 
a) The perimeter of triangle PQR correct to I decimal place. 
b) The gradients ofPR and QR. 
c) The angle of inclination of PR and QR 
d) Hence determine angle PRQ. 
3) Points P(a;10), Q(2;b) and R(c;S)of triangle PQR are shown below. PQ is produced its own 
length to point S (-3; 0) • It is given that PQ = PRo The diagram is not drawn to scale. 
Q(2;b)I;L-__~ R(c;5) 
S(-3;0) 
a) Determine the values of a and b. 
b) Hence, determine the value/s of c. 
c) Determine the equation of PS. 
4) In the diagram below, M is the midpoint of line segment OP. The diagram is not drawn to scale. 
y B 
o A C x 
a) Determine the coordinates of M. 
b) Determine the gradient of OP. 
c) Determine the value of P. the angle of inclination ofOP. 
d) Determine the gradient of AB correct to one decimal place. 
e) What is the area of triangle POC? 
5) Given that P(-2;6), Q(14;2) and R(x;4) are points in the Cartesian plane, calculate the valuests 
of x, if: 
a) QR is perpendicular to the x-axis. 
b) R is the ntidpoint ofPQ. 
c) The gradient of QR is equal to 2. 
d) The length of QR is 2.J2 units. 
The Mixed Exercise solutions are on pages 299 300 301 and 302 
© Copyright Kevin Smith I Berlot Books CC 294 
6) In the diagram below QS is a straigbt line with equation 2y -41' =4  and SQR =2 0'. Determine, 
correct to one decimal place where applicable: 
Y S 
Q 
x 
a) The angle of inclination of QS. 
b) The gradient ofline PQ. 
7) P(-4;-2), Q(8;3) and R(4;10) are the vertices of triangle PQR. QN is perpendicular to PR and 
M is the midpnint of PRo The diagram is not drawn to scale. 
Y R(4;IO) 
Q(8;3) 
x 
P(-4;-2) 
a) Determine the length of PRo 
b) Determine the coordinates of M the midpoint of PRo 
c) Calculate the gradient of PRo 
d) Hence, determine the equation of QN. 
e) Calculate the area of triangle PQR ifN is the point (2;7). 
f) Determine the size of angle p correct to one decimal place. 
8) A(-3;-I), B(-1;9) and C(3;7) are the vertices of triangle ABC. AB.lPQ and BR is 
perpendicular to the x-axis. 
y 
C(3;7) 
x 
A(-3;-I) 
a) Determine the coordinates of Q, the midpoint of AC. 
b) Determine the equation of AB. 
c) Hence, deterroine the equation of PQ. 
d) Determine 8AB , the angle of inclination of AB. 
e) Hence, deterroine the value of P if P = ABR. 
9) The equation ofa line can be written as (S-2p)x+(p+2)y =10. 
a) Rewrite the above equation in the form of y = mx + C. 
b) Determine the equation of the line if the line is parallel to y = 3x + 2. 
c) Determine the value ofp if the line passes through the point (-4;8). 
d) Determine the value of P if the line is parallel to the y-axis. 
e) Determine the value of p if the line is parallel to the x-axis. 
The Mixed Exercise solutions are on pageS 299. 300. 301 and 302 
© Copyright Kevin Smith I Berlut Books CC 295 
Chapter 4- Analytical Geometry (Memo) 
Grade 10 Revision 
Exercise 1: Exercise 1 is on page 292 
1) 
a) AB =~(Xz -xd +(Y2 - Yi)2 c) SR =~(Xz -xd +(Y2 -yd 
:. AB = ~( _1_2)2 +(_8+5)2 :.SR = ~(_3_1O)2 +(2+6)2 
:. AB = .Jl8 = 3,fi units :. SR = .fij3 units 
b) PQ=~(x2-xd2+(Y2-Yi)2 d) MN=~(Xz-xd+(Y2-yd 
:. PQ = ~( _2_5)2 +(3_5)2 :.MN =~(1+1)2 +(_10+5)2 
:. PQ = .J53 units :. MN =..fi9 units 
2) 
a) M(Xz+Xl.Y2+YlJ b) M(Xz+Xl.Y2+Yi J c) M( Xz +xl. Y2 +Yl J d) M( Xz +xl. Y2 + Yl J 
2 ' 2 2 ' 2 2 ' 2 2 ' 2 
=M (--1_+2. -_8--5J =M(-2+5.3+5J =M(-3+10.2-6J =Me-1 --·1O---5J 
2 ' 2 2 ' 2 2 ' 2 2' 2 
=M(L_13J 
2' 2 =M(%;4J =MG;-2J 
=M(0 ; -~5J 
3) 
Y2 -Yi 
a) mAS =-- b) T1TQ = Y2 -Yl 
Xz -Xl Xz -Xl 
-8+5 3-5 
:.mAS= -1-2 :.T1TQ = -2-5 
:.mAS =1 2 
:.T1TQ =7 
4) 
a) P is the midpoint of AB 
:. xA +xB 5 and YA +YB 1 
2 2 
:.4+xB =10 and 6+YB =2 
:.xB =6 and YB =-4 
:.Bisthepoint (6;-4) 
b) Perimeter = AB + BC + AC 
AB =~(Xz 2 2 2 2 -xd +(Y2 - Yi)2 Be =~(Xz -xd +(Y2 - Yi)2 AC = ~(Xz -xd +(Y2 - Yd
:. AB = ~(6_4)2 + {-4-6)2 :. BC =~(6+4)2 +(-4+6)2 :.AC=~{-4-4)2 +(_6_6)2 
:.AB =2,fi6 :.BC=2,fi6 :.AC=4.Jf3 
:. Perimeter = 2,fi6 +2,fi6 +4.Jf3 =34,82 units 
c) For the triangle to be right angled: AC2 = AB2 + BC2 -+ Pythagoras 
2 
AC = 208 units 
AB2 +BC2 =104+104= 208 units 
:.Ll.ABC is right angled at B 
The properties of perpendicular lines could also have been used to prove that B is a right angle. 
© Copyright Kevin Smith I BorlDt Books CC 296 
5) AB = ~(X2 -xtl2 +(Y2 - Yl)2 
:. AB2 = (x_4)2 +(_5_6)2 
:.185 x2 = -8x+16+121 
:.x2 -8x-48=0 
:. (x-12)(x+4) = 0 
:.x=12 or x=-4 
6) 
Y2 -Yl b) m = Y2 - YI Y2-Yl a) m =--
AB "'co=--AB 
"2 - xI "2 - xI "2 -xI 
4+4 2-6 8-24 
:.m =-- :.m . "'co=--AB AB 2+8 = 4-5 .. 4-8 
4 :.m
-5 AB 
=4 :. "'co = 4 
.• mAB 
:. mAB = "'co 
:. mAB X "'co =-1 :.ABIICD 
:.AB.LCD 
7) 
Y2 -Yl Y2-YI 
a) m =-- b) 
AB ","C =--
"2 -xI "2 - xI 
11+7 59-11 
:.mAB =--
4+5 :. ","C = 28-4 
:.mAB =2 :. ","C = 2 
:. the points are collinear as mAD = ","C 
8) 
b) x-2=1 and 3+y =6 
2 2 
:.x-2=2 and 3+y=12 
.. -=--
3 x-4 :.x=4 and y=9 
:.x-4=3 
:.x=7 
Finding the equation of a straight line 
Exerdse2: Exercise 2 is on page 292 
1) or 
a) y=mx+c a) Y-YI =m(x-xtl 
:.y=-2x+c :.Y-YI =-2(x-xtl 
sub (3;8) sub (3;8) 
:.8 = -2(3) +c :.y-8=-2(x-3) 
:.c =14 :.y-8=-2x+6 
:.y=-2x+14 :. Y =-2x+14 
b) y=mx+c b) Y-Yl =m(x-xtl 
:.y=4x+c :. Y- Yl =4(x-xJl 
sub (-5;3) sub (-5;3) 
:.3=4(-5)+c :. y-3 = 4(x+5) 
:.c =23 :. y-3 = 4x+20 
:.y=4x+23 :.y=4x+23 
© copyright Kevin Smith I BerlDt Books CC 297 
or 
c) y=mx+c c) Y-YI=m(x-xd 
1 
:.y=--x+c 1 :'Y-YI =--(x-xd 
3 3 
sub (3;1) sub (3;1) 
1 1 
:.1=-"3(3)+c :.y-1=-"3(x-3) 
:.c=2 1 
:.y-1=--x+1 
1 
:.y=--x+2 3 
3 1 
:.y=--x+2 
3 
~ oc 
a) y=mx+c a) Y- YI =m(x-xd 
12-5 7 
m=--=-- 12-5 7 m=--=--
-8+2 6 -8+2 6 
7 
:.y=--x+c 7 
6 :'Y-YI =-"6(x-xd 
sub (-2;5) --7 (-8;12) could also be used sub (-2;5)--7(-8;12) could also be used 
:.5=-~(-2)+C :.y-5=_2(x+2) 
6 
8 
:.c=- 7 7 
3 :.y-5=-"6-"3 
7 8 
:.y=--x+- 7 8 :.y=--x+-
6 3 6 3 
b) 3y-12x=-9 b) 3y-12x=--9 
:.y=4x-3 :.y=4x-3 
:.mline=4 :. mline= 4 
:.y=4x+c :. Y- YI = 4(x-xd 
sub (2;2) sub (2;2) 
:.2=4(2)+c :. y-2=4(x-2) 
:.c=--{) :. y-2=4x-8 
:.y=4x-6 :. Y =4x-6 
c) 4y-x-8=O c) 4y-x-8=O 
1 1 
:.y=-x+2 :.y=-x+2 
4 4 
:.mline =-4 :. mline =-4 
:.y=-4x+c :. Y- YI = -4(x-xd 
sub (-2;6) sub (-2;6) 
:.6=-4(-2)+c :.y-6=-4(x+2) 
:.c=-2 :.y-6=-4x-8 
:.y=-4x-2 :.y=-4x-2 
d) A vertical line has the equation of x = constant 
:.x=3 
e) A horizontal line has the equation of y = constant 
:. y =-12 
Exercise 2 is on page 292 
© Copyright Kevin Smith I Berlot Books CC 298 
Angle of Inclination 
Exercise 3: Exercise 3 is on page 293 
1) 
a) m= 7-1=2 8-4 _ -3-4 _ 1/ b) m=-=-I 
4-1 1-5 c) m- -8-6- 72 
:.tan8=2 :. tan 8 =-1 :.tanli= Yz 
0 0 0 0 
:.8 = 63,4 :. li = -45 +180 = 135,0 0 
:.8= 26,6
-3-5 6-3 
d) m=--=-4 e) m=--=-3 f) m= 7+7 =14 
1+1 -3+2 8-7 
:. tanli =-4 :. tanli=-3 :. tanli = 14 
0 0 0 0 0 0 0 
:.8=-75,96 +180 =104,0 :.8= -71,57 +180 = 108,4 :.8=85,9
2) 
0 
a) tan 30 = J3 b) tan4S'=1 c) tan60° = J3 
3 
O 
e) tan180° = 0 
d) tan150° = _ J3 e) tanO = 0 
3 
3) 
3-2 1 5+4 
0) :.m"=8_6=2: b) tana=0,5 c) :.m =--=3 
AD 2+1 
tana=mAB :. a=26,6° tana=3 
1 
:. tana=- :.a=71,6° 2 
:.a=26,6° 
1 -4-1 
d) :.tana=-2 e) :. tan a =-- f) :.m. . =--=-1 
2 3+2 
0 0 0 
:. a = -63,43 :. a = -26,57 + 180 :.tana=-I 
0 0 
0 0 
:. a= 63,4 :. a = 153,4 :. a = -45 + 180
0 
:. a= 135
g) a = liAB - liCD h) a = liCD -liAB i) a = liAB + liCD 
tanliAB = 2 tanliCD =-2 tanliAB = 3 
0 0 0 0 
:. liAB = 63,43 :. liCD = -63,43 + 180 :. liAB = 71,57
1 0 tanli =-3 
tanli :. liCD = 16,57 CD CD =-
2 0 tanliAB = 2 :. liCD = -71,57
0 
:. liCD = 26,57 0 0 0 
:. liAB = 63,43 :. a = 71,57 +71,57
0 0 
:. a = 63,43 -26,57 0 0 
:. a =116,57 -63,43 :.a=143,t 
0 
:. a =36,9
:. a =53,t 
Mixed Exercise: The Mixed Exercise is on pages 293, 294 and 295 
1) 
a) MAC=(3;2;-1;2) (-3+4 -2+3) 
~B;.~ -2-;-2- A(-2;2D D(4;3) 
=(~;~) (2'2) 
C(3;-I) 
:. Midpoint of AC = Midpoint of BD B(-3;-2) 
:. ABCD is a parm (diags bisect each other) 
© Copyright Kevin Smith I Berlot Books CC 299 
b) Diagonals of a rhombus and square intersect at 90" 
-1-2 3 3+2 5 
mAC =--=-- and mBO =--=-
3+2 5 4+3 7 
353 
mAcxfflao =--x-=--
577 
:.mACXmBO ,,-1 
:. ABeD is not a rhombus or a square as diagonals do not intersect at 90". 
2) 
a) Perimeter = PQ +QR + PR 
PQ = J{4-8)2 +{4+2)2 QR = JH-4)2 +{1_4)2 PR =JH-8)2 +(1+2)2 
:. PQ = 2.J13 units :. QR =.JW9 units :. PR = ../205 units 
:. Perimeter = 2.J13 +../109 +../205 = 32,0 units 
-6-8 14 Yr -Yq --6-4 10 
"'QR 
1+2 3 xr -Xq 1-4 3 
14 10 
c) tani9pR =-- tan8QR =-
3 3 
:.~ =-77,91" +180" =102,09" :.~ =73,3" 
d) PRQ=~ -8QR 
= 102,09" - 73, 3" 
=28,79" 
3) In tbis question it is important to note tbat it is given tbat PQ = PR = QS. 
a) a-3 =2 and 10-0 =b 
2 2 
:.a=7 :.b=5 
or 
b) PQ2 ={7_2)2 +{1O_5)2 b) QR is a vertical line as the y-values 
=50 of Q and R are the same. 
:.PR2 =50 :. The x-value of the midpoint of QR is 7 
:.50 = {c_7)2 +{5 _10)2 . 2+c =7 .. 2 
2 
:.50=c -14c+49+25 :.c=12 
:.c2 -14c+24=0 
:. (c-12){c- 2) = 0 
:.c=12 or c=2~N/A 
10-0 10-0 
c) mps =1 c) fflps =1 
7+3 7+3 
:.y=lx+c :.Y-YI =l{x-xd 
sub {-3;0) ~points P or Q could be used sub {- 3;0) ~ points P or Q could be used 
:.0=-3+c :. y-0=x+3 
:.c=3 :.y=x+3 
:.y=x+3 
The Mixed EXercise is on pages 293 294 and 295 
© Copyright Kevin Smith I Berlot Books CC 300 
4) 
a) M=C2;0; 8;0)=(6;4) d) BAB = P+30° -7 ext. augle ofa triangle 
:. BAB = 33,69° +30° = 63,69° 
Yp -Yo 8-0 2 
mop mAB =tanBb) AB = 
xp -xo 12-0 3 :. mAB = tau 63,69° = 2,0 
c) tanp=% e) AI. = ~ X base>< -L height 
:·P=33,69° :.AI. = 'x(12)(8) 
2 
:. AI. = 48 units
5) 
a) x=14 d) QR
2 =(x_14)2+(4_2)2 
2 
:.8 = x -28x+196+4 
14-2 2
b) x=-- :.x -28x+192=0 
2 :. (x-16)(x-12) = 0 
:.x=6 
:.x=16 or x=12 
Yr-Yq 
c) "'QR = 
xr-xq 
4-2 
:.2=--
x-14 
:.x-14=1 
:. x = 15 
6) 
a) 2y=4x+4 b) 200=~-~ 
:.y=2x+2 :.~ =63,4° _20° 
tan~s =2 
:.~ =43,4° 
:. BQs = 63,4° 
:.fflpQ =tan/lpQ = tan 43,4° =0,9 
7) 
a) PR =J(10+2f +(4+4)2 e) QN=J(8-2)2+(3-7)2 =2v'i3units 
:. AB = 4v'i3 units APQR = ,XXPRXQN 
:. APQR = ~ ( 4v'i3)( 2v'i3) 
2 
:. APQR = 52 units
3+2 5 
I) fflPQ = 8+4 = 12 
2 5 3 
d) "'QN = -'3 -7 QN -L PR tan~ = 12 aud tan/ipR =-2 
2 
:.y=--x+c :.~ =22,62° aud ~ =56,31° 
3 
Sub Q(8;3) :.P=/ipR -~ 
:. P = 56,3!" -22,62° = 33,7° 
:.3 = -~(8)+C 
25 
.'. c=3- 
2 25 
:.y=--x+-
3 3 
The Mixed Exercise is on pages 293 294 and 295 
© Copyright Kevin Smith I Berlot Books CC 301 
8) 
d) tan0AB =5 
a) Q=(3;3;7;1)=(0;3) 
:.OAB = 78,69' 
9+1 
b) mAB=--=5 e) P = 90' -OAB --7 angels in a triangle 
-1+3 
:.y=5x+c :. p = 90' -78,69' 
Sub A( -3;-1) --7 B( -1;9) can also be used :·P=11,31' 
:.-1=5(-3)+c 
:. c = 14 
:.y=5x+14 
1 
c) "'PQ = -5 --7 PQ 1- AB 
1 
:.y=--x+c 
5 
Sub Q(0;3) 
:.3 = -~(O)+c 
:.c=3 
9) 
a) (p+2)y=-(5-2p)x+1O 
c) y=[2P-5Jx+~ 
:.y=[2P-5Jx+~ p+2 p+2 
p+2 p+2 Sub (-4;8) 
:.8=[2P-5J(-4)+~ 
b) 2p-5=3 p+2 p+2 
p+2 :.8(p+2) = -8p+20+10 
:.2p-5=3(p+2) :.8p+16=-8p+30 
:.2p-5=3p+6 :.16p =14 
:.p=-l1 14 7 
:.P=16=g 
:.y=[2(-11)-5]x+~ 
-11+2 -11+2 
d) gradient is undefined 
10 
:.y=3x--
9 :.p+2=0 
:.p=-2 
e) gradient = 0 
:.2p-5=0 
...  p=5712  
The Mixed Exercise is on pages 293. 294 and 295 
© Copyright Kevin Smith I Berlot Books CC 302 
Chapter 5- Functions 
Sketching the Parabola 
Exercise 1: Exercise I solutions are on pages 310 and 311 
1) Sketch the graphs of the following parabolas showing the turning point and intercepts with the mres: 
a) y = 2(x-3)2 -2 c) h(x) = _(x_3)2 +5 e) m(x) = -(x+2)2_1 
b) J(x)=-(x+2)2+ 4 d) p(x)= Yz(x+2)2+ 2 I) J:x--7(x+3)2_4 
2) Sketch the graphs of the following parabolas showing the turning point and intercepts with the mres: 
a) y=x2+4x+3 c) J(x)=-2~+6x e) y=-9- x2  +27x+27- 
2 2 
2
b) J(x)=-x +2x+3 I 2 
I) y=--x +2x+2 
2 
Determining the Equation of a Parabola 
Exercise 2: Exercise 2 solutions are on page 312 
1) Determine the equations of the parabolas below. In each of the graphs, A is the turning point and B 
is a point on the curve. 
a) b) c) 
y y y 
B(7;1) 
x 
x 
B(4;-2) 
A(-2;-2) 
d) e) I) 
y y y 
A(-I;-3) B(7;-I) 
A(6;-5) 
2) Determine the equations of the parabolas below. In each of the graphs, A and B are the x-intercepts 
and C is a point on the curve. 
b) c) 
y 
x 
© copyright Kevin Smith I Berlot Books CC 303 
3) Determine the equations of the parabolas below. In each of the graphs A and B are points on the 
curve and C is the y·intercept. 
a) b) c) 
y y 
AH;6) B(2;6) B(3;9) 
A(-5; 13) B(2;13) 
x C(O;3) 
x 
C(O;-6) A(-I;-7) C(O;3) 
x 
Exercise 2 solutions are on page 312 
Sketching and Finding the Equation of the Hyperbola 
Exercise 3: Exercise 3 solutions are on pages 313 aod 314 
1) Draw a neat sketcb of the grapbs below sbowing all intercepts with the axes, asymptotes, and both 
axes of symmetry. 
-4 --{i 
a) g(x)=--2 c) f(x)= X~3-1 e) y=-+3 
x+2 x-I 
I 6 -4 
b) f:x~---2 d) y=--3 f) y=---I 
x+1 x-I x-2 
2) For each of the graphs in question 1, determine the coordinates of the points of intersections of the 
axes of symmetry and the graphs. 
Exercise 4: Determine the equations of the hyperbolas shown below: Exercise 4 solutions are on page 314 
c) Y e) 
a).y~~ ••• _ ~ •. j~ . .,  .....(-.3.;5.) •:. ........ , 
y=-l- -----_&• . 
x 
(-2;0) x 
x 
:x=2 
b) Y f) Y 
x 
, 
(5;-4) ·y-;:"i··· _. -,- _. . 
, 
~_-_- - - e - - - - --
: (4;-6) 
Sketching and Finding the Equation of the Exponential Function 
Exercise 5: Sketcb tbe graphs below sbowing all Intercepts with the axes and asymptotes. 
I
a) y=2x+I +2 c) f:x~Z-x+I+2 e) f(x)=GT+ -2 
2 
d) f(x)=-Gf +4 
Exercise 5 solutions are on page 315 
© Copyright Kevin Smith I BerlDt Books CC 304 
Exercise 6: Exercise 6 solutions are on page 316 
In each of the graphs below, solve for the values of b, p or q and write down the equation of the graph. 
a) y b) y c) y 
.. j ... _----_. 
(0;3 
x x 
0;-2) 
y=-2 
- --- - --- - --- - --- - ---- -~~~~-------------
,=3X P - +q x-2 h:x-+-2 (b  ) +q 
Translations and Reflections of Graphs 
Exercise 7: Exercise 7 solutions are on page 316 
1) The graph of / (x) = x 2 + 2x + 5 is given. Write down the new equation of h (x) and describe the 
translation of / (x) if: 
a) h(x)=f(x-2) c) h(x)=-f(x) 
b) h(x) = f(x)-2 d) h(x)=f(-x) 
2) The graph of /(x) =~+1 is given. Write down the new equation of the graph if: 
x 
a) The graph is shifted 2 units down. e) The graph is shifted 2 units right and I 
b) The graph is shifted 2 units left. unit down. 
c) The graph is reflected about the y-axis. f) The graph is shifted 2 units left and I unit 
d) The graph is reflected about the x-axis. up. 
3) The parabola f{x) has turning point A{-3;2) and point B{-I;-6) Jies on the graph. 
a) Determine the equation of the parabola in the form / (x) = a ( x _ p)2 + q . 
b) If the parabola is shifted 2 units to the right and I unit down, determine the new turning point. 
c) If g(x)=f(x-l),whatisthey-interceptof g(x)? 
d) If g(x)=-/(x)+l ,whatis the turning point of g(x)? 
Graph Interpretation 
Exercise 8: Exercise 8 solutions are on pages 317 and 318 
1) Draw a rough sketeh of the graph of / (x) =ax2 +bx+c if: 
a) a < O;b>O and c >0 b) aO;b > 0 and c  0 --+ happy face 
a > 0 --+ happy face 
TP: (-2;-1) TP: (-3;-4) 
TP: (-2;2) 
y-int: y = _(0+2)2 -1=-5 y-int: Y = (0+3)2 -4 = 5 
y-int:y= Yz(0+2)2 +2=4 
:. y-int:(0;-5) :. y-int:(0;5) 
:. y-int:(0;4) 
x-int: none (a < 0 andq < 0) x-int: 0=(x+3)2_4 
x-int: none (a> 0 and q > 0) 
:.~+6x+5=0 
y 
:.(x+l)(x+5)=0 
x :.x=-1 or x=-5 
:.x-int:(-I;O) or (-5;0) 
x 
x 
© Copyright Kevin Smith I Berlot Books CC 310 
2) 
a) y=x2+4x+3 2b) f(x)=-x +2x+3 c) f(x)=-2~+6x 
a > 0 --+ happy face a < 0 --+ sad face a < 0 --+ sad face 
4 
TP:x=--=-2 2 TP:x=---=l 
2 2(-1) TP:x=- 2(~2) = % 
:. y = (_2)2 +4(-2)+3 =-1 f (1) = _(1)2 + 2(1) +3 = 4 
f(%)=-2(%t +6(%)=% 
:. TP: (-2;-1) 
:. TP: (1;4) 
y-int: (0;3) :.TP: (%;%) 
y-int: (0;3) 
x-int:O = x2 +4x+3 y-int: (0;0) x-int:O= _x2 +2x+3 
:.(x+1)(x+3)=0 
:.x2
x-into 0 = _2x2 + 6x 
-2x-3=0 
:.x=-l or x=-3 :. -2x(x-3) 
:. (x+1)(x-3) = 0 
:. x-int: (-1;0) or (-3;0) :.x=O or x=3 
:.x=-l or x=3 
:.x-int:(O;O) or (3;0) 
:. x-int: (-1;0) or (3;0) 
y (~.2.) 
2·2 
(-1;0) (3;0) 
x x 
x 
Exercise 1 is on page 303 
2 9 1 27 1 2 
d) h(x)=-2x +8x+24 e) y=--r+27x+- f) y=--x +2x+2 
2 2 2 
a < 0 --+ sad face a < 0 --+ sad face a < 0 --+ sad face 
8 27 
TP:x=---=2 TP:x 2 2 
2(-2) TP:x= 2(-%) 3 2(-Ji) 
:. y = _2(2)2 +8(2)+24 = 32 
:.y=_%(3)2+ 27 (3)+2; =54 :. y = -Ji (2)2 +2(2)+2 =4 
:. TP: (2;32) 
:. TP: (2;4) 
:. TP: (3;54) 
y-int: (0;24) 
y-int: (0;2) 
x-int:O = -2~ +8x+24 y-int: (0;2Yz) 2
x-int:0=-Jix +2x+2 
:. 0=-2(x2 -4x-12) x-int:O=-%~ +27x+ 2Yz 
:.0=~-4x-4 
:.(x+2)(x-6)=0 :.0=-9x
2+54x+27 
-(-4)±~(-4 -4(1)( -4) 
:.x=-2 or x=6 -54 ± ~r-(5-)42 =-_-4-(- 9) -2(7 -) 
:.X= 2(1) 
:. x-into (-2;0) or (6;0) :.x 2(-9) 
:.x=2-2.fi or x=2+2.fi 
:.x=3-2.J3 or x=3+2.J3 
y (2;32) :. x-int: (2- 2.fi ;O)or( 2+2.fi;0) 
:. x-int:( 3-2,[3;0) or(3+2.J3 ;0) 
(O;24) 
Y (3;S4) 
(-2;0) (6;0) 
(O;~) 
x 
(3-2-'3;0) (3+2-'3;0) 
x 
Exercise 1 is on page 303 
© Copyright Kevin Smith I BerlDt Books CC 311 
Determining the Equation of a Parabola 
Exercise 2: Exercise 2 is on pages 303 and 304 
1) 
a) y=a(x+2)2_2 b) y =a(x-3)2 +1 c) y = a(x-6)2 +S 
SubB(-4;6) Sub B( 4;-2) Sub B(7;1) 
:.6=a(-4+2)2_ 2 :.-2=a(4-3)2 +1 :.I=a(7-6)2+S 
:.8=4a :.-3=a :.-4=a 
:.a=2 :. y =-3(x-3)2 +1 :. y =-4(x-6)2 +S 
:. y = 2(x+2)2_2 
:. y =-3(~ -6x+9)+1 :. y =-4(~ -12x+36)+S 
:. y = 2(x2 +4x+4)-2 2 
:. y =-3x +18x-26 :. y =-4x2 +48x-139 
:. y = 2x2 +8x+6 
d) y=a(X+l)2-3 e) y =a(x+3)2 +2 f) y=a(x-6)2_ S 
Sub B(O;-2) Sub B( -2;4) Sub B(7;-I) 
:.-2=a(O+I)2 -3 :.4=a(-2+3)2+ 2 :.-I=a(7-6)2 -S 
:.a =1 :.2=a :.4=a 
:.y=l(x+l)2_3 :.y=2(x+3)2+ 2 :. y =4(x-6)2_S 
:.y=x2+2x-2 :. y =2(x2 +6x+9)+2 :. y =4(~ -12x+36)-S 
2x2 +12x+20 2 :. y = :. y = 4x -48x+ 139 
2) 
a) y=a(x+2)(x-S) b) Y =a(x+3)(x-2) c) y=a(x+S)(x-6) 
Sub C(4;-3) Sub C(I;-4) Sub (4;9) 
:. -3 = a(4+2)(4-S) :. -4 = a(1 +3)(1-2) :.9=a(4+S)(4-6) 
:. -3 = a(-6) :. -4 = a (-4) :.9 = a( -18) 
:.a= Yz :.a=1 :.a=-Yz 
:. y =1(x+3)(x-2) 
:. y = Yz(x+2)(x-S) :. y = - Yz(x+S)(x-6) 
2
2 :.y=x +x-6 2 
:.y= Yzx -%x-s :. y = - Yz x + Yz x+ IS 
3) 
a) y=ax2+bx-6 b) y=ax2 +bx+3 2c) y=ax +bx+3 
SubA(-{i;6) SubA(-I;-7) Sub A(-S;13) 
:.6=a(-{i)2 +b(-6)-6 :. -7 =a(_1)2 +b(-I)+3 :.13 = a(_S)2 +b( -S)+3 
:.b=6a-2~m :.b=a+lO~m :.b=Sa-2~m 
Sub B(2;6) Sub B(3;9) Sub B(2;13) 
:.6 = a(2)2 +b(2)-6 :.9 = a(3)2 +b(3)+3 :.13 = a(2)2 +b(2) +3 
:.12=4a+2b~~ :.6=9a+3b~~ :.1O=4a+2b~~ 
Sub minto~: Sub minto~: Sub minto~: 
12=4a+2(6a-2) :.6=9a+3(a+1O) :.1O=4a+2(Sa-2) 
:.16=16a :.6=12a+30 :.14=14a 
a=l~rn a=-2~rn a=l~rn 
Sub rn into ill: Sub rn into ill: Sub rn into ill: 
b = 6(1)-3 = 4 b=-2+10=8 b=S(I)-2=3 
x2 :. y = +4x-6 :.y=-2~+8x+3 :.y=x2 +3x+3 
© copyright Kevin Smith I Berlot Books CC 312 
Sketching and Finding the Equation of the Hyperbola 
Exercise 3: Exercise 3 is on page 304 
1) 
-4 I 9 
a) g(x)=--2 b) f:x--7---2 c) f(x)=--1 
x+2 x+l x+3 
a < 0 --7 2'" and 4" quadrants a < 0 --7 2'" and 4" quadrants a> 0 --71" and 3" quadrants 
asymptotes: x = -2 and y =-2 asymptotes: x = -1 and y = -2 asymptotes: x = -3 and y =-1 
. -4 . -I 
x-mt: 0=---2 x-mt: 0=---2 x-int: 0=_9_ _ 1 
x+2 x+1 x+3 
. 2=-=i.. :.2=2 '1=_9_ 
.. x+2 x+1 .. x+3 
:.2x+4=-4 :.2x+2=-1 :.x+3=9 
:. x = -4 --7 (-4;0) :.x=-Yz --7(-'%;0) :.x=6--7(6;0) 
-4 
y-int:y=---2=-4 -I y-int:y=_9_-1=2 
0+2 y-int:y=--2=-3 0+3 
0+1 
:. y-int: (0;-4) :. y-int: (0;2) 
:. y-int: (0;-3) 
AS: y=xandy=-x-4 AS: y=x+2andy=-x-4 
AS: y=x-Iandy=-x-3 
x =,-2 y y 
. 
x=-l' 
• 
6 -6 -4 
d) y=--3 e) y=-+3 f) y=---1 
x-I x-I x-2 
a > 0 --71" and 3" quadrants a < 0 --7 2'" and 4" quadrants a < 0 --7 2'" and 4" quadrants 
asymptotes: x = 1 and y = - 3 asymptotes: x = 1 and y = 3 asymptotes: x = 2 and y = -1 
x-int: 0=~-3 x-int: 0= -6 +3 x-int: 0= -4 -I 
x-I x-I x-2 
:.3=_6_ :.-3=~ :.1= -4 
x-I x-I x-2 
:.3x-3=6 :.-3x+3=-6 :.x-2=-4 
:. x = 3 --7 (3;0) :.x=3--7(3;0) :.x=-2--7(-2;0) 
y-int:y=~-3=-9 -4 y-int:y= -6 +3=9 y-int:y=---I=1 
0-1 0-1 0-2 
:. y-int: (0;-9) :. y-int: (0;9) :. y-int: (0;1) 
AS: y=x-4andy=-x-2 AS: y=x+2andy=-x+4 AS: y=x-3andy=-x+l 
x 
x 
y =-x-2 
y=-x+4 
© Copyright Kevin Smith I Berlot Books CC 313 
2) 
a) aO; p =1 andq=-3 
Points of intersection: Points of intersection: 
=(-J4-2;v'4-2) and (v'4-2;-v'4-2) (-16+1;-16-3) and (-.16+1;--16-3) 
= (-4;0) and (0;-4) 
b) a