# The Free High School Science Texts: Textbooks for High School Students Mathematics

FHSST Authors The Free High School Science Texts: Textbooks for High School Students Studying the Sciences Mathematics Grades 10 - 12 Version 0 September 17, 2008 ii iii Copyright 2007 “Free High School Science Texts” Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no FrontCover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”. STOP!!!! Did you notice the FREEDOMS we’ve granted you? Our copyright license is different! 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Thousands of hours went into making them and they are a gift to everyone in the education community. iv FHSST Core Team Mark Horner ; Samuel Halliday ; Sarah Blyth ; Rory Adams ; Spencer Wheaton FHSST Editors Jaynie Padayachee ; Joanne Boulle ; Diana Mulcahy ; Annette Nell ; René Toerien ; Donovan Whitfield FHSST Contributors Rory Adams ; Prashant Arora ; Richard Baxter ; Dr. Sarah Blyth ; Sebastian Bodenstein ; Graeme Broster ; Richard Case ; Brett Cocks ; Tim Crombie ; Dr. Anne Dabrowski ; Laura Daniels ; Sean Dobbs ; Fernando Durrell ; Dr. Dan Dwyer ; Frans van Eeden ; Giovanni Franzoni ; Ingrid von Glehn ; Tamara von Glehn ; Lindsay Glesener ; Dr. Vanessa Godfrey ; Dr. Johan Gonzalez ; Hemant Gopal ; Umeshree Govender ; Heather Gray ; Lynn Greeff ; Dr. Tom Gutierrez ; Brooke Haag ; Kate Hadley ; Dr. Sam Halliday ; Asheena Hanuman ; Neil Hart ; Nicholas Hatcher ; Dr. Mark Horner ; Mfandaidza Hove ; Robert Hovden ; Jennifer Hsieh ; Clare Johnson ; Luke Jordan ; Tana Joseph ; Dr. Jennifer Klay ; Lara Kruger ; Sihle Kubheka ; Andrew Kubik ; Dr. Marco van Leeuwen ; Dr. Anton Machacek ; Dr. Komal Maheshwari ; Kosma von Maltitz ; Nicole Masureik ; John Mathew ; JoEllen McBride ; Nikolai Meures ; Riana Meyer ; Jenny Miller ; Abdul Mirza ; Asogan Moodaly ; Jothi Moodley ; Nolene Naidu ; Tyrone Negus ; Thomas O’Donnell ; Dr. Markus Oldenburg ; Dr. Jaynie Padayachee ; Nicolette Pekeur ; Sirika Pillay ; Jacques Plaut ; Andrea Prinsloo ; Joseph Raimondo ; Sanya Rajani ; Prof. Sergey Rakityansky ; Alastair Ramlakan ; Razvan Remsing ; Max Richter ; Sean Riddle ; Evan Robinson ; Dr. Andrew Rose ; Bianca Ruddy ; Katie Russell ; Duncan Scott ; Helen Seals ; Ian Sherratt ; Roger Sieloff ; Bradley Smith ; Greg Solomon ; Mike Stringer ; Shen Tian ; Robert Torregrosa ; Jimmy Tseng ; Helen Waugh ; Dr. Dawn Webber ; Michelle Wen ; Dr. Alexander Wetzler ; Dr. Spencer Wheaton ; Vivian White ; Dr. Gerald Wigger ; Harry Wiggins ; Wendy Williams ; Julie Wilson ; Andrew Wood ; Emma Wormauld ; Sahal Yacoob ; Jean Youssef Contributors and editors have made a sincere effort to produce an accurate and useful resource. Should you have suggestions, find mistakes or be prepared to donate material for inclusion, please don’t hesitate to contact us. We intend to work with all who are willing to help make this a continuously evolving resource! www.fhsst.org v vi Contents I Basics 1 1 Introduction to Book 1.1 II 3 The Language of Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . Grade 10 3 5 2 Review of Past Work 7 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 What is a number? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.3 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.4 Letters and Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.5 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.6 Multiplication and Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.7 Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.8 Negative Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.9 2.8.1 What is a negative number? . . . . . . . . . . . . . . . . . . . . . . . . 10 2.8.2 Working with Negative Numbers . . . . . . . . . . . . . . . . . . . . . . 11 2.8.3 Living Without the Number Line . . . . . . . . . . . . . . . . . . . . . . 12 Rearranging Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.10 Fractions and Decimal Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.11 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.12 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.12.1 Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.2 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.3 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.12.4 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.13 Mathematical Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.14 Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.15 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3 Rational Numbers - Grade 10 23 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.2 The Big Picture of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.3 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 vii CONTENTS CONTENTS 3.4 Forms of Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.5 Converting Terminating Decimals into Rational Numbers . . . . . . . . . . . . . 25 3.6 Converting Repeating Decimals into Rational Numbers . . . . . . . . . . . . . . 25 3.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.8 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 4 Exponentials - Grade 10 29 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 4.3 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.3.1 Exponential Law 1: a0 = 1 . . . . . . . . . . . . . . . . . . . . . . . . . 30 4.3.2 Exponential Law 2: am × an = am+n . . . . . . . . . . . . . . . . . . . 30 4.3.3 Exponential Law 3: a−n = 4.3.4 4.4 m 1 an , a n 6= 0 . . . . . . . . . . . . . . . . . . . . 31 Exponential Law 4: a ÷ a = am−n . . . . . . . . . . . . . . . . . . . 32 4.3.5 Exponential Law 5: (ab)n = an bn . . . . . . . . . . . . . . . . . . . . . 32 4.3.6 Exponential Law 6: (am )n = amn . . . . . . . . . . . . . . . . . . . . . 33 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 5 Estimating Surds - Grade 10 37 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 5.2 Drawing Surds on the Number Line (Optional) . . . . . . . . . . . . . . . . . . 38 5.3 End of Chapter Excercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 6 Irrational Numbers and Rounding Off - Grade 10 41 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 6.2 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 6.3 Rounding Off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 6.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 7 Number Patterns - Grade 10 7.1 45 Common Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 7.1.1 Special Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 7.2 Make your own Number Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . 46 7.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 7.3.1 7.4 Patterns and Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 8 Finance - Grade 10 53 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 8.2 Foreign Exchange Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 8.3 8.2.1 How much is R1 really worth? . . . . . . . . . . . . . . . . . . . . . . . 53 8.2.2 Cross Currency Exchange Rates 8.2.3 Enrichment: Fluctuating exchange rates . . . . . . . . . . . . . . . . . . 57 . . . . . . . . . . . . . . . . . . . . . . 56 Being Interested in Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 viii CONTENTS 8.4 Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 8.4.1 8.5 8.6 8.7 CONTENTS Other Applications of the Simple Interest Formula . . . . . . . . . . . . . 61 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 8.5.1 Fractions add up to the Whole . . . . . . . . . . . . . . . . . . . . . . . 65 8.5.2 The Power of Compound Interest . . . . . . . . . . . . . . . . . . . . . . 65 8.5.3 Other Applications of Compound Growth . . . . . . . . . . . . . . . . . 67 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 8.6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 8.6.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 9 Products and Factors - Grade 10 71 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2 Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.1 Parts of an Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.2 Product of Two Binomials . . . . . . . . . . . . . . . . . . . . . . . . . 71 9.2.3 Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 9.3 More Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 9.4 Factorising a Quadratic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 9.5 Factorisation by Grouping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 9.6 Simplification of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 9.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 10 Equations and Inequalities - Grade 10 83 10.1 Strategy for Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 10.2 Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 10.3 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 10.4 Exponential Equations of the form ka(x+p) = m . . . . . . . . . . . . . . . . . . 93 10.4.1 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 10.5 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 10.6 Linear Simultaneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.1 Finding solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.2 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 10.6.3 Solution by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . 101 10.7 Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 10.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 10.7.2 Problem Solving Strategy . . . . . . . . . . . . . . . . . . . . . . . . . . 104 10.7.3 Application of Mathematical Modelling . . . . . . . . . . . . . . . . . . 104 10.7.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 106 10.8 Introduction to Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . 107 10.9 Functions and Graphs in the Real-World . . . . . . . . . . . . . . . . . . . . . . 107 10.10Recap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 ix CONTENTS CONTENTS 10.10.1 Variables and Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 10.10.2 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 10.10.3 The Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 10.10.4 Drawing Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 10.10.5 Notation used for Functions . . . . . . . . . . . . . . . . . . . . . . . . 110 10.11Characteristics of Functions - All Grades . . . . . . . . . . . . . . . . . . . . . . 112 10.11.1 Dependent and Independent Variables . . . . . . . . . . . . . . . . . . . 112 10.11.2 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.11.3 Intercepts with the Axes . . . . . . . . . . . . . . . . . . . . . . . . . . 113 10.11.4 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.5 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.6 Lines of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 10.11.7 Intervals on which the Function Increases/Decreases . . . . . . . . . . . 114 10.11.8 Discrete or Continuous Nature of the Graph . . . . . . . . . . . . . . . . 114 10.12Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 10.12.1 Functions of the form y = ax + q . . . . . . . . . . . . . . . . . . . . . 116 10.12.2 Functions of the Form y = ax2 + q . . . . . . . . . . . . . . . . . . . . . 120 10.12.3 Functions of the Form y = a x + q . . . . . . . . . . . . . . . . . . . . . . 125 10.12.4 Functions of the Form y = ab(x) + q . . . . . . . . . . . . . . . . . . . . 129 10.13End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 11 Average Gradient - Grade 10 Extension 135 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11.2 Straight-Line Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 11.3 Parabolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 11.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 12 Geometry Basics 139 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 12.2 Points and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 12.3 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 12.3.1 Measuring angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 12.3.2 Special Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 12.3.3 Special Angle Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 12.3.4 Parallel Lines intersected by Transversal Lines . . . . . . . . . . . . . . . 143 12.4 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.4.1 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 12.4.2 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 12.4.3 Other polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 12.4.4 Extra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 12.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 12.5.1 Challenge Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 x CONTENTS 13 Geometry - Grade 10 CONTENTS 161 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 13.2 Right Prisms and Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 13.2.1 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 13.2.2 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 13.3 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.3.1 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 13.4 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 13.4.2 Distance between Two Points . . . . . . . . . . . . . . . . . . . . . . . . 172 13.4.3 Calculation of the Gradient of a Line . . . . . . . . . . . . . . . . . . . . 173 13.4.4 Midpoint of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 13.5 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13.5.1 Translation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 13.5.2 Reflection of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 13.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 14 Trigonometry - Grade 10 189 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 14.2 Where Trigonometry is Used . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 14.3 Similarity of Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 14.4 Definition of the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . 191 14.5 Simple Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . 195 14.5.1 Height and Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 14.5.2 Maps and Plans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 14.6 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 199 14.6.1 Graph of sin θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 14.6.2 Functions of the form y = a sin(x) + q . . . . . . . . . . . . . . . . . . . 200 14.6.3 Graph of cos θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 14.6.4 Functions of the form y = a cos(x) + q . . . . . . . . . . . . . . . . . . 202 14.6.5 Comparison of Graphs of sin θ and cos θ . . . . . . . . . . . . . . . . . . 204 14.6.6 Graph of tan θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 14.6.7 Functions of the form y = a tan(x) + q . . . . . . . . . . . . . . . . . . 205 14.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 15 Statistics - Grade 10 211 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2 Recap of Earlier Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2.1 Data and Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . . 211 15.2.2 Methods of Data Collection . . . . . . . . . . . . . . . . . . . . . . . . . 212 15.2.3 Samples and Populations . . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3 Example Data Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 xi CONTENTS CONTENTS 15.3.1 Data Set 1: Tossing a Coin . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3.2 Data Set 2: Casting a die . . . . . . . . . . . . . . . . . . . . . . . . . . 213 15.3.3 Data Set 3: Mass of a Loaf of Bread . . . . . . . . . . . . . . . . . . . . 214 15.3.4 Data Set 4: Global Temperature . . . . . . . . . . . . . . . . . . . . . . 214 15.3.5 Data Set 5: Price of Petrol . . . . . . . . . . . . . . . . . . . . . . . . . 215 15.4 Grouping Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 15.4.1 Exercises - Grouping Data . . . . . . . . . . . . . . . . . . . . . . . . . 216 15.5 Graphical Representation of Data . . . . . . . . . . . . . . . . . . . . . . . . . . 217 15.5.1 Bar and Compound Bar Graphs . . . . . . . . . . . . . . . . . . . . . . . 217 15.5.2 Histograms and Frequency Polygons . . . . . . . . . . . . . . . . . . . . 217 15.5.3 Pie Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 15.5.4 Line and Broken Line Graphs . . . . . . . . . . . . . . . . . . . . . . . . 220 15.5.5 Exercises - Graphical Representation of Data . . . . . . . . . . . . . . . 221 15.6 Summarising Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 15.6.1 Measures of Central Tendency . . . . . . . . . . . . . . . . . . . . . . . 222 15.6.2 Measures of Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 15.6.3 Exercises - Summarising Data . . . . . . . . . . . . . . . . . . . . . . . 228 15.7 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 15.7.1 Exercises - Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . 230 15.8 Summary of Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 15.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 16 Probability - Grade 10 235 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 16.2 Random Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 16.2.1 Sample Space of a Random Experiment . . . . . . . . . . . . . . . . . . 235 16.3 Probability Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 16.3.1 Classical Theory of Probability . . . . . . . . . . . . . . . . . . . . . . . 239 16.4 Relative Frequency vs. Probability . . . . . . . . . . . . . . . . . . . . . . . . . 240 16.5 Project Idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 16.6 Probability Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 16.7 Mutually Exclusive Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 16.8 Complementary Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 16.9 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 III Grade 11 17 Exponents - Grade 11 249 251 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 17.2 Laws of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 √ m 17.2.1 Exponential Law 7: a n = n am . . . . . . . . . . . . . . . . . . . . . . 251 17.3 Exponentials in the Real-World . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 17.4 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 xii CONTENTS CONTENTS 18 Surds - Grade 11 18.1 Surd Calculations . . . . . . . . . . √ √ √ 18.1.1 Surd Law 1: n a n b = n ab √ p n a 18.1.2 Surd Law 2: n ab = √ . . n b √ m 18.1.3 Surd Law 3: n am = a n . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 255 . . . . . . . . . . . . . . . . . . . . . . . . 256 18.1.4 Like and Unlike Surds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 18.1.5 Simplest Surd form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 18.1.6 Rationalising Denominators . . . . . . . . . . . . . . . . . . . . . . . . . 258 18.2 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 19 Error Margins - Grade 11 261 20 Quadratic Sequences - Grade 11 265 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 20.2 What is a quadratic sequence? . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 20.3 End of chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 21 Finance - Grade 11 271 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 21.2 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 21.3 Simple Depreciation (it really is simple!) . . . . . . . . . . . . . . . . . . . . . . 271 21.4 Compound Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 21.5 Present Values or Future Values of an Investment or Loan . . . . . . . . . . . . 276 21.5.1 Now or Later . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 21.6 Finding i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 21.7 Finding n - Trial and Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 21.8 Nominal and Effective Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . 280 21.8.1 The General Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 21.8.2 De-coding the Terminology . . . . . . . . . . . . . . . . . . . . . . . . . 282 21.9 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 21.9.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 21.9.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 21.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 22 Solving Quadratic Equations - Grade 11 287 22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 22.2 Solution by Factorisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 22.3 Solution by Completing the Square . . . . . . . . . . . . . . . . . . . . . . . . . 290 22.4 Solution by the Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . 293 22.5 Finding an equation when you know its roots . . . . . . . . . . . . . . . . . . . 296 22.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 xiii CONTENTS CONTENTS 23 Solving Quadratic Inequalities - Grade 11 301 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 23.2 Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 23.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 24 Solving Simultaneous Equations - Grade 11 307 24.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 24.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 25 Mathematical Models - Grade 11 313 25.1 Real-World Applications: Mathematical Models . . . . . . . . . . . . . . . . . . 313 25.2 End of Chatpter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 26 Quadratic Functions and Graphs - Grade 11 321 26.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 26.2 Functions of the Form y = a(x + p)2 + q . . . . . . . . . . . . . . . . . . . . . 321 26.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 26.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 26.2.3 Turning Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 26.2.4 Axes of Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 26.2.5 Sketching Graphs of the Form f (x) = a(x + p)2 + q . . . . . . . . . . . 325 26.2.6 Writing an equation of a shifted parabola . . . . . . . . . . . . . . . . . 327 26.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 27 Hyperbolic Functions and Graphs - Grade 11 329 27.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 27.2 Functions of the Form y = a x+p +q . . . . . . . . . . . . . . . . . . . . . . . . 329 27.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 27.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 27.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332 27.2.4 Sketching Graphs of the Form f (x) = a x+p + q . . . . . . . . . . . . . . 333 27.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 28 Exponential Functions and Graphs - Grade 11 335 28.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 28.2 Functions of the Form y = ab(x+p) + q . . . . . . . . . . . . . . . . . . . . . . . 335 28.2.1 Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 28.2.2 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 28.2.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 28.2.4 Sketching Graphs of the Form f (x) = ab(x+p) + q . . . . . . . . . . . . . 338 28.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 29 Gradient at a Point - Grade 11 341 29.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 29.2 Average Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 29.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 xiv CONTENTS 30 Linear Programming - Grade 11 CONTENTS 345 30.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.1 Decision Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.2 Objective Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 30.2.3 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.2.4 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.2.5 The Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 30.3 Example of a Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.4 Method of Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5 Skills you will need . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5.1 Writing Constraint Equations . . . . . . . . . . . . . . . . . . . . . . . . 347 30.5.2 Writing the Objective Function . . . . . . . . . . . . . . . . . . . . . . . 348 30.5.3 Solving the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 30.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 31 Geometry - Grade 11 357 31.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 31.2 Right Pyramids, Right Cones and Spheres . . . . . . . . . . . . . . . . . . . . . 357 31.3 Similarity of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 31.4 Triangle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 31.4.1 Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361 31.5 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368 31.5.1 Equation of a Line between Two Points . . . . . . . . . . . . . . . . . . 368 31.5.2 Equation of a Line through One Point and Parallel or Perpendicular to Another Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 31.5.3 Inclination of a Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371 31.6 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 31.6.1 Rotation of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 31.6.2 Enlargement of a Polygon 1 . . . . . . . . . . . . . . . . . . . . . . . . . 376 32 Trigonometry - Grade 11 381 32.1 History of Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 32.2 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . 381 32.2.1 Functions of the form y = sin(kθ) . . . . . . . . . . . . . . . . . . . . . 381 32.2.2 Functions of the form y = cos(kθ) . . . . . . . . . . . . . . . . . . . . . 383 32.2.3 Functions of the form y = tan(kθ) . . . . . . . . . . . . . . . . . . . . . 384 32.2.4 Functions of the form y = sin(θ + p) . . . . . . . . . . . . . . . . . . . . 385 32.2.5 Functions of the form y = cos(θ + p) . . . . . . . . . . . . . . . . . . . 386 32.2.6 Functions of the form y = tan(θ + p) . . . . . . . . . . . . . . . . . . . 387 32.3 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 32.3.1 Deriving Values of Trigonometric Functions for 30◦ , 45◦ and 60◦ . . . . . 389 32.3.2 Alternate Definition for tan θ . . . . . . . . . . . . . . . . . . . . . . . . 391 xv CONTENTS CONTENTS 32.3.3 A Trigonometric Identity . . . . . . . . . . . . . . . . . . . . . . . . . . 392 32.3.4 Reduction Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 32.4 Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 399 32.4.1 Graphical Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 32.4.2 Algebraic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 32.4.3 Solution using CAST diagrams . . . . . . . . . . . . . . . . . . . . . . . 403 32.4.4 General Solution Using Periodicity . . . . . . . . . . . . . . . . . . . . . 405 32.4.5 Linear Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . 406 32.4.6 Quadratic and Higher Order Trigonometric Equations . . . . . . . . . . . 406 32.4.7 More Complex Trigonometric Equations . . . . . . . . . . . . . . . . . . 407 32.5 Sine and Cosine Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 32.5.1 The Sine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 32.5.2 The Cosine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 32.5.3 The Area Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414 32.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416 33 Statistics - Grade 11 419 33.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2 Standard Deviation and Variance . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2.1 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 33.2.2 Standard Deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 33.2.3 Interpretation and Application . . . . . . . . . . . . . . . . . . . . . . . 423 33.2.4 Relationship between Standard Deviation and the Mean . . . . . . . . . . 424 33.3 Graphical Representation of Measures of Central Tendency and Dispersion . . . . 424 33.3.1 Five Number Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 33.3.2 Box and Whisker Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . 425 33.3.3 Cumulative Histograms . . . . . . . . . . . . . . . . . . . . . . . . . . . 426 33.4 Distribution of Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 33.4.1 Symmetric and Skewed Data . . . . . . . . . . . . . . . . . . . . . . . . 428 33.4.2 Relationship of the Mean, Median, and Mode . . . . . . . . . . . . . . . 428 33.5 Scatter Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 33.6 Misuse of Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 33.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435 34 Independent and Dependent Events - Grade 11 437 34.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 34.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437 34.2.1 Identification of Independent and Dependent Events . . . . . . . . . . . 438 34.3 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441 IV Grade 12 35 Logarithms - Grade 12 443 445 35.1 Definition of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445 xvi CONTENTS CONTENTS 35.2 Logarithm Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446 35.3 Laws of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 35.4 Logarithm Law 1: loga 1 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 35.5 Logarithm Law 2: loga (a) = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 35.6 Logarithm Law 3: loga (x · y) = loga (x) + loga (y) . . . . . . . . . . . . . . . . . 448 35.7 Logarithm Law 4: loga xy = loga (x) − loga (y) . . . . . . . . . . . . . . . . . 449 35.8 Logarithm Law 5: loga (xb ) = b loga (x) . . . . . . . . . . . . . . . . . . . . . . . 450 √ 35.9 Logarithm Law 6: loga ( b x) = logab(x) . . . . . . . . . . . . . . . . . . . . . . . 450 35.10Solving simple log equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 35.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 35.11Logarithmic applications in the Real World . . . . . . . . . . . . . . . . . . . . . 454 35.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 35.12End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455 36 Sequences and Series - Grade 12 457 36.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 36.2 Arithmetic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457 36.2.1 General Equation for the nth -term of an Arithmetic Sequence . . . . . . 458 36.3 Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 36.3.1 Example - A Flu Epidemic . . . . . . . . . . . . . . . . . . . . . . . . . 459 36.3.2 General Equation for the nth -term of a Geometric Sequence . . . . . . . 461 36.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 461 36.4 Recursive Formulae for Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 462 36.5 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.5.1 Some Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.5.2 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463 36.6 Finite Arithmetic Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 36.6.1 General Formula for a Finite Arithmetic Series . . . . . . . . . . . . . . . 466 36.6.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 36.7 Finite Squared Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468 36.8 Finite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469 36.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470 36.9 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471 36.9.1 Infinite Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . 471 36.9.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 36.10End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 37 Finance - Grade 12 477 37.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 37.2 Finding the Length of the Investment or Loan . . . . . . . . . . . . . . . . . . . 477 37.3 A Series of Payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478 37.3.1 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479 xvii CONTENTS CONTENTS 37.3.2 Present Values of a series of Payments . . . . . . . . . . . . . . . . . . . 479 37.3.3 Future Value of a series of Payments . . . . . . . . . . . . . . . . . . . . 484 37.3.4 Exercises - Present and Future Values . . . . . . . . . . . . . . . . . . . 485 37.4 Investments and Loans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 37.4.1 Loan Schedules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 37.4.2 Exercises - Investments and Loans . . . . . . . . . . . . . . . . . . . . . 489 37.4.3 Calculating Capital Outstanding . . . . . . . . . . . . . . . . . . . . . . 489 37.5 Formulae Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 37.5.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 37.5.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 37.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490 38 Factorising Cubic Polynomials - Grade 12 493 38.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 38.2 The Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493 38.3 Factorisation of Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 494 38.4 Exercises - Using Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 496 38.5 Solving Cubic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 38.5.1 Exercises - Solving of Cubic Equations . . . . . . . . . . . . . . . . . . . 498 38.6 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 39 Functions and Graphs - Grade 12 501 39.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.2 Definition of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 39.3 Notation used for Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 39.4 Graphs of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502 39.4.1 Inverse Function of y = ax + q . . . . . . . . . . . . . . . . . . . . . . . 503 39.4.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.3 Inverse Function of y = ax2 . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504 39.4.5 Inverse Function of y = ax . . . . . . . . . . . . . . . . . . . . . . . . . 506 39.4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 39.5 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 40 Differential Calculus - Grade 12 509 40.1 Why do I have to learn this stuff? . . . . . . . . . . . . . . . . . . . . . . . . . 509 40.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 40.2.1 A Tale of Achilles and the Tortoise . . . . . . . . . . . . . . . . . . . . . 510 40.2.2 Sequences, Series and Functions . . . . . . . . . . . . . . . . . . . . . . 511 40.2.3 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512 40.2.4 Average Gradient and Gradient at a Point . . . . . . . . . . . . . . . . . 516 40.3 Differentiation from First Principles . . . . . . . . . . . . . . . . . . . . . . . . . 519 xviii CONTENTS CONTENTS 40.4 Rules of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 40.4.1 Summary of Differentiation Rules . . . . . . . . . . . . . . . . . . . . . . 522 40.5 Applying Differentiation to Draw Graphs . . . . . . . . . . . . . . . . . . . . . . 523 40.5.1 Finding Equations of Tangents to Curves . . . . . . . . . . . . . . . . . 523 40.5.2 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 40.5.3 Local minimum, Local maximum and Point of Inflextion . . . . . . . . . 529 40.6 Using Differential Calculus to Solve Problems . . . . . . . . . . . . . . . . . . . 530 40.6.1 Rate of Change problems . . . . . . . . . . . . . . . . . . . . . . . . . . 534 40.7 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535 41 Linear Programming - Grade 12 539 41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.2.1 Feasible Region and Points . . . . . . . . . . . . . . . . . . . . . . . . . 539 41.3 Linear Programming and the Feasible Region . . . . . . . . . . . . . . . . . . . 540 41.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 42 Geometry - Grade 12 549 42.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2 Circle Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549 42.2.2 Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550 42.2.3 Theorems of the Geometry of Circles . . . . . . . . . . . . . . . . . . . . 550 42.3 Co-ordinate Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 42.3.1 Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566 42.3.2 Equation of a Tangent to a Circle at a Point on the Circle . . . . . . . . 569 42.4 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571 42.4.1 Rotation of a Point about an angle θ . . . . . . . . . . . . . . . . . . . . 571 42.4.2 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573 42.4.3 Characteristics of Transformations . . . . . . . . . . . . . . . . . . . . . 573 42.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574 43 Trigonometry - Grade 12 577 43.1 Compound Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577 43.1.1 Derivation of sin(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 577 43.1.2 Derivation of sin(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578 43.1.3 Derivation of cos(α + β) . . . . . . . . . . . . . . . . . . . . . . . . . . 578 43.1.4 Derivation of cos(α − β) . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.5 Derivation of sin 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.6 Derivation of cos 2α . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579 43.1.7 Problem-solving Strategy for Identities . . . . . . . . . . . . . . . . . . . 580 43.2 Applications of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . 582 43.2.1 Problems in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 582 xix CONTENTS CONTENTS 43.2.2 Problems in 3 dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 584 43.3 Other Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.1 Taxicab Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.2 Manhattan distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.3 Spherical Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 43.3.4 Fractal Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588 43.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 44 Statistics - Grade 12 591 44.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.2 A Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.3 Extracting a Sample Population . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 44.4 Function Fitting and Regression Analysis . . . . . . . . . . . . . . . . . . . . . . 594 44.4.1 The Method of Least Squares . . . . . . . . . . . . . . . . . . . . . . . 596 44.4.2 Using a calculator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 44.4.3 Correlation coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 44.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600 45 Combinations and Permutations - Grade 12 603 45.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.1 Making a List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.2 Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3.1 The Factorial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.4 The Fundamental Counting Principle . . . . . . . . . . . . . . . . . . . . . . . . 604 45.5 Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.1 Counting Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.2 Combinatorics and Probability . . . . . . . . . . . . . . . . . . . . . . . 606 45.6 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 45.6.1 Counting Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 607 45.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 45.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 V Exercises 613 46 General Exercises 615 47 Exercises - Not covered in Syllabus 617 A GNU Free Documentation License 619 xx CONTENTS CONTENTS 43.2.2 Problems in 3 dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 584 43.3 Other Geometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.1 Taxicab Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.2 Manhattan distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 43.3.3 Spherical Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587 43.3.4 Fractal Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588 43.4 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 44 Statistics - Grade 12 591 44.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.2 A Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591 44.3 Extracting a Sample Population . . . . . . . . . . . . . . . . . . . . . . . . . . . 593 44.4 Function Fitting and Regression Analysis . . . . . . . . . . . . . . . . . . . . . . 594 44.4.1 The Method of Least Squares . . . . . . . . . . . . . . . . . . . . . . . 596 44.4.2 Using a calculator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 44.4.3 Correlation coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 44.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600 45 Combinations and Permutations - Grade 12 603 45.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.1 Making a List . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603 45.2.2 Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.3.1 The Factorial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 604 45.4 The Fundamental Counting Principle . . . . . . . . . . . . . . . . . . . . . . . . 604 45.5 Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.1 Counting Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 45.5.2 Combinatorics and Probability . . . . . . . . . . . . . . . . . . . . . . . 606 45.6 Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606 45.6.1 Counting Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 607 45.7 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608 45.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 V Exercises 613 46 General Exercises 615 47 Exercises - Not covered in Syllabus 617 A GNU Free Documentation License 619 xx Part I Basics 1 Chapter 1 Introduction to Book 1.1 The Language of Mathematics The purpose of any language, like English or Zulu, is to make it possible for people to communicate. All languages have an alphabet, which is a group of letters that are used to make up words. There are also rules of grammar which explain how words are supposed to be used to build up sentences. This is needed because when a sentence is written, the person reading the sentence understands exactly what the writer is trying to explain. Punctuation marks (like a full stop or a comma) are used to further clarify what is written. Mathematics is a language, specifically it is the language of Science. Like any language, mathematics has letters (known as numbers) that are used to make up words (known as expressions), and sentences (known as equations). The punctuation marks of mathematics are the different signs and symbols that are used, for example, the plus sign (+), the minus sign (-), the multiplication sign (×), the equals sign (=) and so on. There are also rules that explain how the numbers should be used together with the signs to make up equations that express some meaning. 3 1.1 CHAPTER 1. INTRODUCTION TO BOOK 4 Part II Grade 10 5 Chapter 2 Review of Past Work 2.1 Introduction This chapter describes some basic concepts which you have seen in earlier grades, and lays the foundation for the remainder of this book. You should feel confident with the content in this chapter, before moving on with the rest of the book. So try out your skills on the exercises throughout this chapter and ask your teacher for more questions just like them. You can also try making up your own questions, solve them and try them out on your classmates to see if you get the same answers. Practice is the only way to get good at maths! 2.2 What is a number? A number is a way to represent quantity. Numbers are not something that you can touch or hold, because they are not physical. But you can touch three apples, three pencils, three books. You can never just touch three, you can only touch three of something. However, you do not need to see three apples in front of you to know that if you take one apple away, that there will be two apples left. You can just think about it. That is your brain representing the apples in numbers and then performing arithmetic on them. A number represents quantity because we can look at the world around us and quantify it using numbers. How many minutes? How many kilometers? How many apples? How much money? How much medicine? These are all questions which can only be answered using numbers to tell us “how much” of something we want to measure. A number can be written many different ways and it is always best to choose the most appropriate way of writing the number. For example, “a half” may be spoken aloud or written in words, but that makes mathematics very difficult and also means that only people who speak the same language as you can understand what you mean. A better way of writing “a half” is as a fraction 1 2 or as a decimal number 0,5. It is still the same number, no matter which way you write it. In high school, all the numbers which you will see are called real numbers and mathematicians use the symbol R to stand for the set of all real numbers, which simply means all of the real numbers. Some of these real numbers can be written in a particular way and some cannot. Different types of numbers are described in detail in Section 1.12. 2.3 Sets A set is a group of objects with a well-defined criterion for membership. For example, the criterion for belonging to a set of apples, is that it must be an apple. The set of apples can then be divided into red apples and green apples, but they are all still apples. All the red apples form another set which is a sub-set of the set of apples. A sub-set is part of a set. All the green apples form another sub-set. 7 2.4 CHAPTER 2. REVIEW OF PAST WORK Now we come to the idea of a union, which is used to combine things. The symbol for union is ∪. Here we use it to combine two or more intervals. For example, if x is a real number such that 1 < x ≤ 3 or 6 ≤ x < 10, then the set of all the possible x values is (1,3] ∪ [6,10) (2.1) where the ∪ sign means the union (or combination) of the two intervals. We use the set and interval notation and the symbols described because it is easier than having to write everything out in words. 2.4 Letters and Arithmetic The simplest things that can be done with numbers is to add, subtract, multiply or divide them. When two numbers are added, subtracted, multiplied or divided, you are performing arithmetic 1 . These four basic operations can be performed on any two real numbers. Mathematics as a language uses special notation to write things down. So instead of: one plus one is equal to two mathematicians write 1+1=2 In earlier grades, place holders were used to indicate missing numbers in an equation. 1+=2 4−=2 + 3 − 2 = 2 However, place holders only work well for simple equations. For more advanced mathematical workings, letters are usually used to represent numbers. 1+x=2 4−y =2 z + 3 − 2z = 2 These letters are referred to as variables, since they can take on any value depending on what is required. For example, x = 1 in Equation 2.2, but x = 26 in 2 + x = 28. A constant has a fixed value. The number 1 is a constant. The speed of light in a vacuum is also a constant which has been defined to be exactly 299 792 458 m·s−1 (read metres per second). The speed of light is a big number and it takes up space to always write down the entire number. Therefore, letters are also used to represent some constants. In the case of the speed of light, it is accepted that the letter c represents the speed of light. Such constants represented by letters occur most often in physics and chemistry. Additionally, letters can be used to describe a situation, mathematically. For example, the following equation x+y =z (2.2) can be used to describe the situation of finding how much change can be expected for buying an item. In this equation, y represents the price of the item you are buying, x represents the amount of change you should get back and z is the amount of money given to the cashier. So, if the price is R10 and you gave the cashier R15, then write R15 instead of z and R10 instead of y and the change is then x. x + 10 = 15 (2.3) We will learn how to “solve” this equation towards the end of this chapter. 1 Arithmetic is derived from the Greek word arithmos meaning number. 8 CHAPTER 2. REVIEW OF PAST WORK 2.5 2.5 Addition and Subtraction Addition (+) and subtraction (-) are the most basic operations between numbers but they are very closely related to each other. You can think of subtracting as being the opposite of adding since adding a number and then subtracting the same number will not change what you started with. For example, if we start with a and add b, then subtract b, we will just get back to a again a+b−b=a 5+2−2=5 (2.4) If we look at a number line, then addition means that we move to the right and subtraction means that we move to the left. The order in which numbers are added does not matter, but the order in which numbers are subtracted does matter. This means that: a+b = b+a a−b 6= b − a if a 6= b (2.5) The sign 6= means “is not equal to”. For example, 2 + 3 = 5 and 3 + 2 = 5, but 5 − 3 = 2 and 3 − 5 = −2. −2 is a negative number, which is explained in detail in Section 2.8. Extension: Commutativity for Addition The fact that a + b = b + a, is known as the commutative property for addition. 2.6 Multiplication and Division Just like addition and subtraction, multiplication (×, ·) and division (÷, /) are opposites of each other. Multiplying by a number and then dividing by the same number gets us back to the start again: a×b÷b=a 5×4÷4=5 (2.6) Sometimes you will see a multiplication of letters as a dot or without any symbol. Don’t worry, its exactly the same thing. Mathematicians are lazy and like to write things in the shortest, neatest way possible. abc = a·b·c = a×b×c a×b×c (2.7) It is usually neater to write known numbers to the left, and letters to the right. So although 4x and x4 are the same thing, it looks better to write 4x. In this case, the “4” is a constant that is referred to as the coefficient of x. Extension: Commutativity for Multiplication The fact that ab = ba is known as the commutative property of multiplication. Therefore, both addition and multiplication are described as commutative operations. 2.7 Brackets Brackets2 in mathematics are used to show the order in which you must do things. This is important as you can get different answers depending on the order in which you do things. For 2 Sometimes people say “parenthesis” instead of “brackets”. 9 2.8 CHAPTER 2. REVIEW OF PAST WORK example (5 × 5) + 20 = 45 (2.8) 5 × (5 + 20) = 125 (2.9) whereas If there are no brackets, you should always do multiplications and divisions first and then additions and subtractions3 . You can always put your own brackets into equations using this rule to make things easier for yourself, for example: a×b+c÷d = 5 × 5 + 20 ÷ 4 = (a × b) + (c ÷ d) (5 × 5) + (20 ÷ 4) (2.10) If you see a multiplication outside a bracket like this a(b + c) (2.11) 3(4 − 3) then it means you have to multiply each part inside the bracket by the number outside a(b + c) = ab + ac (2.12) 3(4 − 3) = 3 × 4 − 3 × 3 = 12 − 9 = 3 unless you can simplify everything inside the bracket into a single term. In fact, in the above example, it would have been smarter to have done this 3(4 − 3) = 3 × (1) = 3 (2.13) 3(4a − 3a) = 3 × (a) = 3a (2.14) It can happen with letters too Extension: Distributivity The fact that a(b + c) = ab + ac is known as the distributive property. If there are two brackets multiplied by each other, then you can do it one step at a time (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd (a + 3)(4 + d) = a(4 + d) + 3(4 + d) (2.15) = 4a + ad + 12 + 3d 2.8 2.8.1 Negative Numbers What is a negative number? Negative numbers can be very confusing to begin with, but there is nothing to be afraid of. The numbers that are used most often are greater than zero. These numbers are known as positive numbers. A negative number is simply a number that is less than zero. So, if we were to take a positive number a and subtract it from zero, the answer would be the negative of a. 0 − a = −a 3 Multiplying and dividing can be performed in any order as it doesn’t matter. Likewise it doesn’t matter which order you do addition and subtraction. Just as long as you do any ×÷ before any +−. 10 CHAPTER 2. REVIEW OF PAST WORK 2.8 On a number line, a negative number appears to the left of zero and a positive number appears to the right of zero. negative numbers -3 -2 positive numbers -1 0 1 2 3 Figure 2.1: On the number line, numbers increase towards the right and decrease towards the left. Positive numbers appear to the right of zero and negative numbers appear to the left of zero. 2.8.2 Working with Negative Numbers When you are adding a negative number, it is the same as subtracting that number if it were positive. Likewise, if you subtract a negative number, it is the same as adding the number if it were positive. Numbers are either positive or negative, and we call this their sign. A positive number has positive sign (+), and a negative number has a negative sign (-). Subtraction is actually the same as adding a negative number. In this example, a and b are positive numbers, but −b is a negative number a − b = a + (−b) (2.16) 5 − 3 = 5 + (−3) So, this means that subtraction is simply a short-cut for adding a negative number, and instead of writing a + (−b), we write a − b. This also means that −b + a is the same as a − b. Now, which do you find easier to work out? Most people find that the first way is a bit more difficult to work out than the second way. For example, most people find 12 − 3 a lot easier to work out than −3 + 12, even though they are the same thing. So, a − b, which looks neater and requires less writing, is the accepted way of writing subtractions. Table 2.1 shows how to calculate the sign of the answer when you multiply two numbers together. The first column shows the sign of the first number, the second column gives the sign of the second number, and the third column shows what sign the answer will be. So multiplying or a + + - b + + - a × b or a ÷ b + + Table 2.1: Table of signs for multiplying or dividing two numbers. dividing a negative number by a positive number always gives you a negative number, whereas multiplying or dividing numbers which have the same sign always gives a positive number. For example, 2 × 3 = 6 and −2 × −3 = 6, but −2 × 3 = −6 and 2 × −3 = −6. Adding numbers works slightly differently, have a look at Table 2.2. The first column shows the sign of the first number, the second column gives the sign of the second number, and the third column shows what sign the answer will be. a + + - b + + - a+b + ? ? - Table 2.2: Table of signs for adding two numbers. 11 2.8 CHAPTER 2. REVIEW OF PAST WORK If you add two positive numbers you will always get a positive number, but if you add two negative numbers you will always get a negative number. If the numbers have different sign, then the sign of the answer depends on which one is bigger. 2.8.3 Living Without the Number Line The number line in Figure 2.1 is a good way to visualise what negative numbers are, but it can get very inefficient to use it every time you want to add or subtract negative numbers. To keep things simple, we will write down three tips that you can use to make working with negative numbers a little bit easier. These tips will let you work out what the answer is when you add or subtract numbers which may be negative and will also help you keep your work tidy and easier to understand. Negative Numbers Tip 1 If you are given an equation like −a + b, then it is easier to move the numbers around so that the equation looks easier. For this case, we have seen that adding a negative number to a positive number is the same as subtracting the number from the positive number. So, −a + b = −5 + 10 = b−a (2.17) 10 − 5 = 5 This makes equations easier to understand. For example, a question like “What is −7 + 11?” looks a lot more complicated than “What is 11 − 7?”, even though they are exactly the same question. Negative Numbers Tip 2 When you have two negative numbers like −3 − 7, you can calculate the answer by simply adding together the numbers as if they were positive and then putting a negative sign in front. −c − d = −7 − 2 = −(c + d) −(7 + 2) = −9 (2.18) Negative Numbers Tip 3 In Table 2.2 we saw that the sign of two numbers added together depends on which one is bigger. This tip tells us that all we need to do is take the smaller number away from the larger one, and remember to put a negative sign before the answer if the bigger number was subtracted to begin with. In this equation, F is bigger than e. e−F = 2 − 11 = −(F − e) (2.19) −(11 − 2) = −9 You can even combine these tips together, so for example you can use Tip 1 on −10 + 3 to get 3 − 10, and then use Tip 3 to get −(10 − 3) = −7. Exercise: Negative Numbers 1. Calculate: (a) (−5) − (−3) (d) 11 − (−9) (g) (−1) × 24 ÷ 8 × (−3) (j) 3 − 64 + 1 (m) −9 + 8 − 7 + 6 − 5 + 4 − 3 + 2 − 1 12 (b) (−4) + 2 (e) −16 − (6) (h) (−2) + (−7) (k) −5 − 5 − 5 (c) (−10) ÷ (−2) (f) −9 ÷ 3 × 2 (i) 1 − 12 (l) −6 + 25 CHAPTER 2. REVIEW OF PAST WORK 2.9 2. Say whether the sign of the answer is + or (a) −5 + 6 (d) −5 ÷ 5 (g) −5 × −5 (j) 5 × 5 2.9 (b) −5 + 1 (e) 5 ÷ −5 (h) −5 × 5 (c) −5 ÷ −5 (f) 5 ÷ 5 (i) 5 × −5 Rearranging Equations Now that we have described the basic rules of negative and positive numbers and what to do when you add, subtract, multiply and divide them, we are ready to tackle some real mathematics problems! Earlier in this chapter, we wrote a general equation for calculating how much change (x) we can expect if we know how much an item costs (y) and how much we have given the cashier (z). The equation is: x+y =z (2.20) So, if the price is R10 and you gave the cashier R15, then write R15 instead of z and R10 instead of y. x + 10 = 15 (2.21) Now, that we have written this equation down, how exactly do we go about finding what the change is? In mathematical terms, this is known as solving an equation for an unknown (x in this case). We want to re-arrange the terms in the equation, so that only x is on the left hand side of the = sign and everything else is on the right. The most important thing to remember is that an equation is like a set of weighing scales. In order to keep the scales balanced, whatever, is done to one side, must be done to the other. Method: Rearranging Equations You can add, subtract, multiply or divide both sides of an equation by any number you want, as long as you always do it to both sides. So for our example we could subtract y from both sides x+y x+y−y = = x = x = = z z−y (2.22) z−y 15 − 10 5 so now we can find the change is the price subtracted from the amount handed over to the cashier. In the example, the change should be R5. In real life we can do this in our head, the human brain is very smart and can do arithmetic without even knowing it. When you subtract a number from both sides of an equation, it looks just like you moved a positive number from one side and it became a negative on the other, which is exactly what happened. Likewise if you move a multiplied number from one side to the other, it looks like it changed to a divide. This is because you really just divided both sides by that number, and a 13 2.9 CHAPTER 2. REVIEW OF PAST WORK x+y z x+y−y z−y divide the other side too. Figure 2.2: An equation is like a set of weighing scales. In order to keep the scales balanced, you must do the same thing to both sides. So, if you add, subtract, multiply or divide the one side, you must add, subtract, multiply or divide the other side too. number divided by itself is just 1 a(5 + c) a(5 + c) ÷ a a × (5 + c) a 1 × (5 + c) 5+c c = 3a = 3a ÷ a a = 3× a = 3×1 = 3 (2.23) = 3 − 5 = −2 However you must be careful when doing this, as it is easy to make mistakes. The following is the wrong thing to do 5a + c 5+c = 6= 4 3a (2.24) 3a ÷ a Can you see why it is wrong? It is wrong because we did not divide the c term by a as well. The correct thing to do is 5a + c = 5+c÷a = c÷a = Exercise: Rearranging Equations 14 3a 3 3 − 5 = −2 (2.25) CHAPTER 2. REVIEW OF PAST WORK 2.10 1. If 3(2r − 5) = 27, then 2r − 5 = ..... 2. Find the value for x if 0,5(x − 8) = 0,2x + 11 3. Solve 9 − 2n = 3(n + 2) 4. Change the formula P = A + Akt to A = 5. Solve for x: 2.10 1 ax + 1 bx =1 Fractions and Decimal Numbers A fraction is one number divided by another number. There are several ways to write a number divided by another one, such as a ÷ b, a/b and ab . The first way of writing a fraction is very hard to work with, so we will use only the other two. We call the number on the top, the numerator and the number on the bottom the denominator. For example, 1 5 numerator = 1 denominator = 5 (2.26) Extension: Definition - Fraction The word fraction means part of a whole. The reciprocal of a fraction is the fraction turned upside down, in other words the numerator becomes the denominator and the denominator becomes the numerator. So, the reciprocal of 32 is 32 . A fraction multiplied by its reciprocal is always equal to 1 and can be written b a × =1 b a (2.27) This is because dividing by a number is the same as multiplying by its reciprocal. Extension: Definition - Multiplicative Inverse The reciprocal of a number is also known as the multiplicative inverse. A decimal number is a number which has an integer part and a fractional part. The integer and the fractional parts are separated by a decimal point, which is written as a comma in South 14 Africa. For example the number 3 100 can be written much more cleanly as 3,14. All real numbers can be written as a decimal number. However, some numbers would take a huge amount of paper (and ink) to write out in full! Some decimal numbers will have a number which will repeat itself, such as 0,33333 . . . where there are an infinite number of 3’s. We can write this decimal value by using a dot above the repeating number, so 0,3̇ = 0,33333 . . .. If there are two repeating numbers such as 0,121212 . . . then you can place dots5 on each of the repeated numbers 0,1̇2̇ = 0,121212 . . .. These kinds of repeating decimals are called recurring decimals. Table 2.3 lists some common fractions and their decimal forms. 5 or a bar, like 0,12 15 2.11 CHAPTER 2. REVIEW OF PAST WORK Fraction 1 20 Decimal Form 0,05 1 16 0,0625 1 10 0,1 1 8 0,125 1 6 0,166̇ 1 5 0,2 1 2 0,5 3 4 0,75 Table 2.3: Some common fractions and their equivalent decimal forms. 2.11 Scientific Notation In science one often needs to work with very large or very small numbers. These can be written more easily in scientific notation, which has the general form a × 10m (2.28) where a is a decimal number between 0 and 10 that is rounded off to a few decimal places. The m is an integer and if it is positive it represents how many zeros should appear to the right of a. If m is negative then it represents how many times the decimal place in a should be moved to the left. For example 3,2 × 103 represents 32000 and 3,2 × 10−3 represents 0,0032. If a number must be converted into scientific notation, we need to work out how many times the number must be multiplied or divided by 10 to make it into a number between 1 and 10 (i.e. we need to work out the value of the exponent m) and what this number is (the value of a). We do this by counting the number of decimal places the decimal point must move. For example, write the speed of light which is 299 792 458 ms−1 in scientific notation, to two decimal places. First, determine where the decimal point must go for two decimal places (to find a) and then count how many places there are after the decimal point to determine m. In this example, the decimal point must go after the first 2, but since the number after the 9 is a 7, a = 3,00. So the number is 3,00 × 10m , where m = 8, because there are 8 digits left after the decimal point. So the speed of light in scientific notation, to two decimal places is 3,00 × 108 ms−1 . As another example, the size of the HI virus is around 120 × 10−9 m. This is equal to 120 × 0,000000001 m which is 0,00000012 m. 2.12 Real Numbers Now that we have learnt about the basics of mathematics, we can look at what real numbers are in a little more detail. The following are examples of real numbers and it is seen that each number is written in a different way. √ 3, 1,2557878, 56 , 10, 2,1, 34 − 5, − 6,35, − 1 90 (2.29) Depending on how the real number is written, it can be further labelled as either rational, irrational, integer or natural. A set diagram of the different number types is shown in Figure 2.3. 16 CHAPTER 2. REVIEW OF PAST WORK 2.12 N Z Q R Figure 2.3: Set diagram of all the real numbers R, the rational numbers Q, the integers Z and the natural numbers N. The irrational numbers are the numbers not inside the set of rational numbers. All of the integers are also rational numbers, but not all rational numbers are integers. Extension: Non-Real Numbers All numbers that are not real numbers have imaginary components. We√will not see imaginary numbers in this book but you will see that they come from −1. Since we won’t be looking at numbers which are not real, if you see a number you can be sure it is a real one. 2.12.1 Natural Numbers The first type of numbers that are learnt about are the numbers that were used for counting. These numbers are called natural numbers and are the simplest numbers in mathematics. 0, 1, 2, 3, 4 . . . (2.30) Mathematicians use the symbol N to mean the set of all natural numbers. The natural numbers are a subset of the real numbers since every natural number is also a real number. 2.12.2 Integers The integers are all of the natural numbers and their negatives . . . − 4, −3, −2, −1, 0, 1, 2, 3, 4 . . . (2.31) Mathematicians use the symbol Z to mean the set of all integers. The integers are a subset of the real numbers, since every integer is a real number. 2.12.3 Rational Numbers The natural numbers and the integers are only able to describe quantities that are whole or complete. For example you can have 4 apples, but what happens when you divide one apple into 4 equal pieces and share it among your friends? Then it is not a whole apple anymore and a different type of number is needed to describe the apples. This type of number is known as a rational number. A rational number is any number which can be written as: a b (2.32) where a and b are integers and b 6= 0. The following are examples of rational numbers: 20 , 9 −1 20 , , 2 10 17 3 15 (2.33) 2.12 CHAPTER 2. REVIEW OF PAST WORK Extension: Notation Tip Rational numbers are any number that can be expressed in the form ab ; a, b ∈ Z; b 6= 0 which means “the set of numbers ab when a and b are integers”. Mathematicians use the symbol Q to mean the set of all rational numbers. The set of rational numbers contains all numbers which can be written as terminating or repeating decimals. Extension: Rational Numbers All integers are rational numbers with denominator 1. You can add and multiply rational numbers and still get a rational number at the end, which is very useful. If we have 4 integers, a, b, c and d, then the rules for adding and multiplying rational numbers are c a + b d a c × b d = = ad + bc bd ac bd (2.34) (2.35) Extension: Notation Tip The statement ”4 integers a, b, c and d” can be written formally as {a, b, c, d} ∈ Z because the ∈ symbol means in and we say that a, b, c and d are in the set of integers. Two rational numbers ( ab and dc ) represent the same number if ad = bc. It is always best to simplify any rational number so that the denominator is as small as possible. This can be achieved by dividing both the numerator and the denominator by the same integer. For example, the rational number 1000/10000 can be divided by 1000 on the top and the bottom, which gives 8 (Figure 2.4). 1/10. 23 of a pizza is the same as 12 8 12 Figure 2.4: 8 12 2 3 of the pizza is the same as 2 3 of the pizza. You can also add rational numbers together by finding a lowest common denominator and then adding the numerators. Finding a lowest common denominator means finding the lowest number that both denominators are a factor 6 of. A factor of a number is an integer which evenly divides that number without leaving a remainder. The following numbers all have a factor of 3 3, 6, 9, 12, 15, 18, 21, 24 . . . and the following all have factors of 4 4, 8, 12, 16, 20, 24, 28 . . . 6 Some people say divisor instead of factor. 18 CHAPTER 2. REVIEW OF PAST WORK 2.12 The common denominators between 3 and 4 are all the numbers that appear in both of these lists, like 12 and 24. The lowest common denominator of 3 and 4 is the number that has both 3 and 4 as factors, which is 12. For example, if we wish to add 43 + 23 , we first need to write both fractions so that their denominators are the same by finding the lowest common denominator, which we know is 12. We can do this by multiplying 43 by 33 and 32 by 44 . 33 and 44 are really just complicated ways of writing 1. Multiplying a number by 1 doesn’t change the number. 3 2 + 4 3 = = = = = 3 3 2 4 × + × 4 3 3 4 3×3 2×4 + 4×3 3×4 9 8 + 12 12 9+8 12 17 12 (2.36) Dividing by a rational number is the same as multiplying by its reciprocal, as long as neither the numerator nor the denominator is zero: a c a d ad ÷ = . = b d b c bc (2.37) A rational number may be a proper or improper fraction. Proper fractions have a numerator that is smaller than the denominator. For example, −1 3 −5 , , 2 15 −20 are proper fractions. Improper fractions have a numerator that is larger than the denominator. For example, −10 13 −53 , , 2 15 −20 are improper fractions. Improper fractions can always be written as the sum of an integer and a proper fraction. Converting Rationals into Decimal Numbers Converting rationals into decimal numbers is very easy. If you use a calculator, you can simply divide the numerator by the denominator. If you do not have a calculator, then you unfortunately have to use long division. Since long division, was first taught in primary school, it will not be discussed here. If you have trouble with long division, then please ask your friends or your teacher to explain it to you. 2.12.4 Irrational Numbers An irrational number is any real number that is not a rational number. When expressed as decimals these numbers can never be fully written out as they have √ an infinite number of decimal places which never fall into a repeating pattern, for example 2 = 1,41421356 . . ., π = 3,14159265 . . .. π is a Greek letter and is pronounced “pie”. Exercise: Real Numbers 19 2.13 CHAPTER 2. REVIEW OF PAST WORK 1. Identify the number type (rational, irrational, real, integer) of each of the following numbers: (a) dc if c is an integer and if d is irrational. (b) 23 (c) -25 (d) 1,525 √ (e) 10 2. √ Is the following pair of numbers real and rational or real and irrational? Explain. 4; 18 2.13 Mathematical Symbols The following is a table of the meanings of some mathematical signs and symbols that you should have come across in earlier grades. Sign or Symbol > < ≥ ≤ Meaning greater than less than greater than or equal to less than or equal to So if we write x > 5, we say that x is greater than 5 and if we write x ≥ y, we mean that x can be greater than or equal to y. Similarly, < means ‘is less than’ and ≤ means ‘is less than or equal to’. Instead of saying that x is between 6 and 10, we often write 6 < 10. This directly means ‘six is less than x which in turn is less than ten’. Exercise: Mathematical Symbols 1. Write the following in symbols: (a) (b) (c) (d) 2.14 x is greater than 1 y is less than or equal to z a is greater than or equal to 21 p is greater than or equal to 21 and p is less than or equal to 25 Infinity Infinity (symbol ∞) is usually thought of as something like “the largest possible number” or “the furthest possible distance”. In mathematics, infinity is often treated as if it were a number, but it is clearly a very different type of “number” than the integers or reals. When talking about recurring decimals and irrational numbers, the term infinite was used to describe never-ending digits. 20 CHAPTER 2. REVIEW OF PAST WORK 2.15 2.15 End of Chapter Exercises 1. Calculate (a) 18 − 6 × 2 (b) 10 + 3(2 + 6) (c) 50 − 10(4 − 2) + 6 (d) 2 × 9 − 3(6 − 1) + 1 (e) 8 + 24 ÷ 4 × 2 (f) 30 − 3 × 4 + 2 (g) 36 ÷ 4(5 − 2) + 6 (h) 20 − 4 × 2 + 3 (i) 4 + 6(8 + 2) − 3 (j) 100 − 10(2 + 3) + 4 2. If p = q + 4r, then r = ..... 3. Solve x−2 3 =x−3 21 2.15 CHAPTER 2. REVIEW OF PAST WORK 22 Chapter 3 Rational Numbers - Grade 10 3.1 Introduction As described in Chapter 2, a number is a way of representing quantity. The numbers that will be used in high school are all real numbers, but there are many different ways of writing any single real number. This chapter describes rational numbers. 3.2 The Big Picture of Numbers Real Numbers Rationals Integers All numbers inside the grey oval are rational numbers. 3.3 Whole Natural Irrationals Definition The following numbers are all rational numbers. 10 , 1 21 , 7 −1 , −3 10 , 20 −3 6 (3.1) You can see that all the denominators and all the numerators are integers1 . Definition: Rational Number A rational number is any number which can be written as: a b where a and b are integers and b 6= 0. 1 Integers are the counting numbers (1, 2, 3, ...), their opposites (-1, -2, -3, ...), and 0. 23 (3.2) 3.4 CHAPTER 3. RATIONAL NUMBERS - GRADE 10 Important: Only fractions which have a numerator and a denominator that are integers are rational numbers. This means that all integers are rational numbers, because they can be written with a denominator of 1. Therefore, while √ 2 , 7 −1,33 , −3 π , 20 −3 6,39 (3.3) are not examples of rational numbers, because in each case, either the numerator or the denominator is not an integer. Exercise: Rational Numbers 1. If a is an integer, b is an integer and c is not an integer, which of the following are rational numbers: (a) 2. If 3.4 a 1 5 6 (b) a 3 (c) b 2 (d) 1 c is a rational number, which of the following are valid values for a? √ (d) 2,1 (a) 1 (b) −10 (c) 2 Forms of Rational Numbers All integers and fractions with integer numerators and denominators are rational numbers. There are two more forms of rational numbers. Activity :: Investigation : Decimal Numbers You can write the rational number 12 as the decimal number 0,5. Write the following numbers as decimals: 1. 2. 3. 4. 5. 1 4 1 10 2 5 1 100 2 3 Do the numbers after the decimal comma end or do they continue? If they continue, is there a repeating pattern to the numbers? You can write a rational number as a decimal number. Therefore, you should be able to write a decimal number as a rational number. Two types of decimal numbers can be written as rational numbers: 1. decimal numbers that end or terminate, for example the fraction 24 4 10 can be written as 0,4. CHAPTER 3. RATIONAL NUMBERS - GRADE 10 3.5 2. decimal numbers that have a repeating pattern of numbers, for example the fraction can be written as 0,333333. 1 3 For example, the rational number 65 can be written in decimal notation as 0,83333, and similarly, the decimal number 0,25 can be written as a rational number as 41 . Important: Notation for Repeating Decimals You can use a bar over the repeated numbers to indicate that the decimal is a repeating decimal. 3.5 Converting Terminating Decimals into Rational Numbers A decimal number has an integer part and a fractional part. For example, 10,589 has an integer part of 10 and a fractional part of 0,589 because 10 + 0,589 = 10,589. The fractional part can be written as a rational number, i.e. with a numerator and a denominator that are integers. Each digit after the decimal point is a fraction with denominator in increasing powers of ten. For example: • • 1 10 is 0,1 1 100 is 0,01 This means that: 0 3 1 + + 10 100 1000 103 = 2 1000 2103 = 1000 2,103 = 2 + Exercise: Fractions 1. Write the following as fractions: (a) 0,1 3.6 (b) 0,12 (c) 0,58 (d) 0,2589 Converting Repeating Decimals into Rational Numbers When the decimal is a repeating decimal, a bit more work is needed to write the fractional part of the decimal number as a fraction. We will explain by means of an example. If we wish to write 0,3 in the form follows a b (where a and b are integers) then we would proceed as x = 0,33333 . . . 10x = 9x = 3,33333 . . . 3 3 1 = 9 3 x = (3.4) multiply by 10 on both sides subtracting (3.4) from (3.5) 25 (3.5) 3.7 CHAPTER 3. RATIONAL NUMBERS - GRADE 10 And another example would be to write 5,432 as a rational fraction x = 1000x = 999x = x = 5,432432432 . . . (3.6) 5432,432432432 . . . 5427 5427 201 = 999 37 multiply by 1000 on both sides (3.7) subtracting (3.6) from (3.7) For the first example, the decimal number was multiplied by 10 and for the second example, the decimal number was multiplied by 1000. This is because for the first example there was only one number (i.e. 3) that recurred, while for the second example there were three numbers (i.e. 432) that recurred. In general, if you have one number recurring, then multiply by 10, if you have two numbers recurring, then multiply by 100, if you have three numbers recurring, then multiply by 1000. Can you spot the pattern yet? The number of zeros after the 1 is the same as the number of recurring numbers. But √ not all decimal numbers can be written as rational numbers, because some decimal numbers like 2 = 1,4142135... is an irrational number and cannot be written with an integer numerator and an integer denominator. However, when possible, you should always use rational numbers or fractions instead of decimals. Exercise: Repeated Decimal Notation 1. Write the following using the repeated decimal notation: (a) 0,11111111 . . . (b) 0,1212121212 . . . (c) 0,123123123123 . . . (d) 0,11414541454145 . . . 2. Write the following in decimal form, using the repeated decimal notation: (a) 32 3 (b) 1 11 (c) 4 65 (d) 2 91 . . . 3. Write the following decimals in fractional form: (a) 0,6333 . . . (b) 5,313131 (c) 11,570571 . . . (d) 0,999999 . . . 3.7 Summary The following are rational numbers: • Fractions with both denominator and numerator as integers. • Integers. • Decimal numbers that end. • Decimal numbers that repeat. 26 CHAPTER 3. RATIONAL NUMBERS - GRADE 10 3.8 3.8 End of Chapter Exercises 1. If a is an integer, b is an integer and c is not an integer, which of the following are rational numbers: (a) (b) (c) (d) 5 6 a 3 b 2 1 c 2. Write each decimal as a simple fraction: (a) 0,5 (b) 0,12 (c) 0,6 (d) 1,59 (e) 12,277 3. Show that the decimal 3,21̇8̇ is a rational number. 4. Showing all working, express 0,78̇ as a fraction 27 a b where a, b ∈ Z. 3.8 CHAPTER 3. RATIONAL NUMBERS - GRADE 10 28 Chapter 4 Exponentials - Grade 10 4.1 Introduction In this chapter, you will learn about the short-cuts to writing 2 × 2 × 2 × 2. This is known as writing a number in exponential notation. 4.2 Definition Exponential notation is a short way of writing the same number multiplied by itself many times. For example, instead of 5 × 5 × 5, we write 53 to show that the number 5 is multiplied by itself 3 times and we say “5 to the power of 3”. Likewise 52 is 5 × 5 and 35 is 3 × 3 × 3 × 3 × 3. We will now have a closer look at writing numbers using exponential notation. Definition: Exponential Notation Exponential notation means a number written like an when n is an integer and a can be any real number. a is called the base and n is called the exponent. The nth power of a is defined as: an = 1 × a × a × . . . × a (n times) (4.1) with a multiplied by itself n times. We can also define what it means if we have a negative index, −n. Then, a−n = 1 ÷ a ÷ a ÷ . . . ÷ a (n times) (4.2) Important: Exponentials If n is an even integer, then an will always be positive for any non-zero real number a. For example, although −2 is negative, (−2)2 = 1 × −2 × −2 = 4 is positive and so is (−2)−2 = 1 ÷ −2 ÷ −2 = 14 . 29 4.3 4.3 CHAPTER 4. EXPONENTIALS - GRADE 10 Laws of Exponents There are several laws we can use to make working with exponential numbers easier. Some of these laws might have been seen in earlier grades, but we will list all the laws here for easy reference, but we will explain each law in detail, so that you can understand them, and not only remember them. a0 m a × an a−n am ÷ an n (ab) (am )n 4.3.1 = 1 (4.3) = am+n 1 = an = am−n (4.4) (4.5) (4.6) n n (4.7) (4.8) 1, (a 6= 0) (4.9) = a b = amn Exponential Law 1: a0 = 1 Our definition of exponential notation shows that a0 = For example, x0 = 1 and (1 000 000)0 = 1. Exercise: Application using Exponential Law 1: a0 = 1, (a 6= 0) 1. 160 = 1 2. 16a0 = 16 3. (16 + a)0 = 1 4. (−16)0 = 1 5. −160 = −1 4.3.2 Exponential Law 2: am × an = am+n Our definition of exponential notation shows that am × an = 1 × a× ...× a ×1 × a × . . . × a = 1 × a× ...× a = (m times) (n times) (m + n times) am+n For example, 27 × 23 = (2 × 2 × 2 × 2 × 2 × 2 × 2) × (2 × 2 × 2) = 210 = 27+3 30 (4.10) CHAPTER 4. EXPONENTIALS - GRADE 10 4.3 teresting This simple law is the reason why exponentials were originally invented. In the Interesting Fact Fact days before calculators, all multiplication had to be done by hand with a pencil and a pad of paper. Multiplication takes a very long time to do and is very tedious. Adding numbers however, is very easy and quick to do. If you look at what this law is saying you will realise that it means that adding the exponents of two exponential numbers (of the same base) is the same as multiplying the two numbers together. This meant that for certain numbers, there was no need to actually multiply the numbers together in order to find out what their multiple was. This saved mathematicians a lot of time, which they could use to do something more productive. Exercise: Application using Exponential Law 2: am × an = am+n 1. x2 .x5 = x7 2. 2x3 y × 5x2 y 7 = 10x5 y 8 3. 23 .24 = 27 [Take note that the base (2) stays the same.] 4. 3 × 32a × 32 = 32a+3 4.3.3 Exponential Law 3: a−n = 1 ,a an 6= 0 Our definition of exponential notation for a negative exponent shows that a−n = = = 1 ÷ a÷ ...÷ a 1 1 × a× ···× a 1 an (n times) (4.11) (n times) This means that a minus sign in the exponent is just another way of writing that the whole exponential number is to be divided instead of multiplied. For example, 2−7 = = 1 2×2×2×2×2×2×2 1 27 Exercise: Application using Exponential Law 3: a−n = 1. 2−2 = 2. 3. 1 22 = 1 4 1 1 2−2 32 = 22 .32 = 36 ( 23 )−3 = ( 32 )3 = 27 8 31 1 an , a 6= 0 4.3 4.3.4 CHAPTER 4. EXPONENTIALS - GRADE 10 = mn4 4. m n−4 5. a−3 .x4 a5 .x−2 = x4 .x2 a3 .a5 x6 a8 = Exponential Law 4: am ÷ an = am−n We already realised with law 3 that a minus sign is another way of saying that the exponential number is to be divided instead of multiplied. Law 4 is just a more general way of saying the same thing. We can get this law by just multiplying law 3 by am on both sides and using law 2. am an = am a−n = am−n (4.12) For example, 27 ÷ 23 2×2×2×2×2×2×2 2×2×2 = 2×2×2×2 = = 24 = 27−3 Exercise: Exponential Law 4: am ÷ an = am−n 1. a6 a2 = a6−2 = a4 2. 32 36 = 32−6 = 3−4 = 3. 4. 4.3.5 2 32a 4a8 3x a a4 = 8a −6 = 1 34 [Always give final answer with positive index] 8 a6 = a3x−4 Exponential Law 5: (ab)n = an bn The order in which two real numbers are multiplied together does not matter. Therefore, (ab)n = = a× b × a × b × ...× a× b (n times) a× a × ...× a (n times) ×b × b × . . . × b = an b n (n times) For example, (2 · 3)4 = (2 · 3) × (2 · 3) × (2 · 3) × (2 · 3) = (2 × 2 × 2 × 2) × (3 × 3 × 3 × 3) = (24 ) × (34 ) = 24 34 32 (4.13) CHAPTER 4. EXPONENTIALS - GRADE 10 4.3 Exercise: Exponential Law 5: (ab)n = an bn 1. (2x2 y)3 = 23 x2×3 y 5 = 8x6 y 5 49a2 b6 n−4 3 2 2. ( 7a b3 ) = 3. (5a 4.3.6 ) = 125a3n−12 Exponential Law 6: (am )n = amn We can find the exponential of an exponential just as well as we can for a number. After all, an exponential number is a real number. (am )n = = = am × am × . . . × am (n times) a× a× ... × a (m × n times) (4.14) amn For example, (22 )3 = (22 ) × (22 ) × (22 ) = = (2 × 2) × (2 × 2) × (2 × 2) (26 ) = 2(2×3) Exercise: Exponential Law 6: (am )n = amn 1. (x3 )4 = x12 2. [(a4 )3 ]2 = a24 3. (3n+3 )2 = 32n+6 Worked Example 1: Simlifying indices 2x−1 x−2 .9 Question: Simplify: 5 152x−3 Answer Step 1 : Factorise all bases into prime factors: = = 52x−1 .(32 )x−2 (5.3)2x−3 52x−1 .32x−4 52x−3 .32x−3 Step 2 : Add and subtract the indices of the sam bases as per laws 2 and 4: 33 4.4 CHAPTER 4. EXPONENTIALS - GRADE 10 = 52x−1−2x−3 .32x−4−2x+3 = 52 .3−1 Step 3 : Write simplified answer with positive indices: = 25 3 Activity :: Investigation : Exponential Numbers Match the answers to the questions, by filling in the correct answer into the Answer column. Possible answers are: 32 , 1, -1, -3, 8. Question 23 73−3 ( 23 )−1 87−6 (−3)−1 (−1)23 4.4 Answer End of Chapter Exercises 1. Simplify as far as possible: (a) 3020 (e) (2x)3 (b) 10 (f) (−2x)3 (c) (xyz)0 (g) (2x)4 (d) [(3x4 y 7 z 12 )5 (−5x9 y 3 z 4 )2 ]0 (h) (−2x)4 2. Simplify without using a calculator. Leave your answers with positive exponents. (a) 3x−3 (3x)2 (b) 5x0 + 8−2 − ( 12 )−2 .1x (c) 5b−3 5b+1 3. Simplify, showing all steps: (a) 2a−2 .3a+3 6a (b) a2m+n+p am+n+p .am (c) 3n .9n−3 27n−1 2a 3 (d) ( 2x y −b ) 34 CHAPTER 4. EXPONENTIALS - GRADE 10 (e) 23x−1 .8x+1 42x−2 (f) 62x .112x 222x−1 .32x 4. Simplify, without using a calculator: (a) (−3)−3 .(−3)2 . (−3)−4 (b) (3−1 + 2−1 )−1 (c) 9n−1 .273−2n 812−n (d) 23n+2 .8n−3 43n−2 35 4.4 4.4 CHAPTER 4. EXPONENTIALS - GRADE 10 36 Chapter 5 Estimating Surds - Grade 10 5.1 Introduction You should known by now what the nth root of a number means. If the √ nth root√of a number cannot be simplified to a rational number, we call it a surd. For example, 2 and 3 6 are surds, √ but 4 is not a surd because it can be simplified to the rational number 2. √ In this chapter we will only look at surds that look like n a, where a is any positive √ √ √ number, for example 7 or 3 5. It√ is very common for n to be 2, so we usually do not write 2 a. Instead we write the surd as just a, which is much easier to read. It is sometimes useful to know the approximate value of a surd√without having to use a calculator. For example, we want to be able to guess where a surd like 3 is on the number√line. So how do we know where surds lie on the √ number line? From a calculator we know that 3 is equal to 1,73205.... It is easy to see that 3 is above 1 and below 2. But to see this for other surds like √ 18 without using a calculator, you must first understand the following fact: teresting If a and b are positive whole numbers, and a < b, then √n a < √n b. (Challenge: Interesting Fact Fact Can you explain why?) If you don’t believe this fact, check it for a few numbers to convince yourself it is true. √ How do we use this fact to help us √ guess what 18 is? Well, you can easily see that √ 18 < 25? √ 2 18 < 25. But we know that 5 = 25 so that 25 = 5. Using our rule, we also know that √ √ Now it is easy to simplify to get 18 < 5. Now we have a better idea of what 18 is. √ Now we know that 18 is less than 5, but this is only half the story. We can use the same trick again, but this time with 18 on side. You will agree that 16 < 18. Using our √ √ the right-hand rule again, we also know that 16 < 18. But we know that 16 is a perfect square, so we can √ √ simplify 16 to 4, and so we get 4 < 18! √ Can you see now that we now √ have shown that 18 is between 4 and 5? If we check on our calculator, we can see that 18 = 4,1231..., and we see that our idea was right! You will notice that our idea used perfect squares that were close to the number 18. We found the closest perfect square underneath 18, which was 42 = 16, and the closest perfect square above 18, which was 52 = 25. Here is a quick summary of what a perfect square or cube is: teresting A perfect square is the number obtained when an integer is squared. For example, Interesting Fact Fact 9 is a perfect square since 32 = 9. Similarly, a perfect cube is a number which is the cube of an integer. For example, 27 is a perfect cube, because 33 = 27. 37 5.2 CHAPTER 5. ESTIMATING SURDS - GRADE 10 To make it easier to use our idea, we will create a list of some of the perfect squares and perfect cubes. The list is shown in Table 5.1. Table 5.1: Some perfect squares and perfect cubes Integer Perfect Square Perfect Cube 0 0 0 1 1 1 2 4 8 3 9 27 4 16 64 5 25 125 6 36 216 7 49 343 8 64 512 9 81 729 10 100 1000 √ 3 Similarly, when given the surd √ 52 you should be able to tell that it lies somewhere √ √ between 3 3 3 and 4, because 27 = 3 and 64 = 4 and 52 is between 27 and 64. In fact 3 52 = 3,73 . . . which is indeed between 3 and 4. 5.2 Drawing Surds on the Number Line (Optional) √ How can√we accurately draw a surd like 5 on the number line? Well, we could use a calculator to find 5 = 2,2360... and measure the distance along the number line using a ruler. But for some surds, there is a much easier way. √ Let us call the surd we are working with a. Sometimes, we can write a as sum of two √ the √ perfect squares, so a = b2 + c2 . We know from Pythagoras’ theorem that a = b2 + c2 is the length of the hypotenuse of a triangle that has sides that have lengths of b and c. So if we draw a triangle on the number line with sides of length b and c, we can use a compass to draw a circle √ from the top of the hypotenuse down to the number line. The intersection will be the point b on the number line! teresting Not all numbers can be written as the sum of two squares. See if you can find a Interesting Fact Fact pattern of the numbers that can. Worked Example 2: Estimating Surds √ Question: Find the two consecutive integers such that 26 lies between them. (Remember that consecutive numbers that are two numbers one after the other, like 5 and 6 or 8 and 9.) Answer Step 1 : From the table find √ the largest perfect square below 26 This is 52 = 25. Therefore 5 < 26. 38 CHAPTER 5. ESTIMATING SURDS - GRADE 10 5.3 Step 2 : From the table find √ smallest perfect square above 26 This is 62 = 36. Therefore 26 < 6. Step 3 : Put the √ inequalities together Our answer is 5 < 26 < 6. Worked Example 3: Estimating Surds √ Question: 3 49 lies between: (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 (d) 4 and 5 Answer Step 1 :√Consider (a) as the solution If 1 < 3 49 < 2 then cubing√ all terms gives 1 < 49 < 23 . Simplifying gives 1 < 49 < 8 which is false. So 3 49 does not lie between 1 and 2. Step 2 :√Consider (b) as the solution If 2 < 3 49 < 3 then cubing √ all terms gives 23 < 49 < 33 . Simplifying gives 3 8 < 49 < 27 which is false. So 49 does not lie between 2 and 3. Step 3 :√Consider (c) as the solution If 3 < 3 49 < 4 then cubing all terms gives 33 < 49 < 43 . Simplifying gives √ 3 27 < 49 < 64 which is true. So 49 lies between 3 and 4. 5.3 1. 2. 3. 4. 5. 6. 7. 8. End of Chapter Excercises √ √5 lies between √10 lies between √20 lies between 30 lies between √ 3 5 lies between √ 3 10 lies between √ 3 20 lies between √ 3 30 lies between (a) (a) (a) (a) (a) (a) (a) (a) 1 1 2 3 1 1 2 3 and and and and and and and and 2 2 3 4 2 2 3 4 (b) (b) (b) (b) (b) (b) (b) (b) 2 2 3 4 2 2 3 4 and and and and and and and and 39 3 3 4 5 3 3 4 5 (c) (c) (c) (c) (c) (c) (c) (c) 3 3 4 5 3 3 4 5 and and and and and and and and 4 4 5 6 4 4 5 6 (d) (d) (d) (d) (d) (d) (d) (d) 4 4 5 6 4 4 5 6 and and and and and and and and 5 5 6 7 5 5 6 7 5.3 CHAPTER 5. ESTIMATING SURDS - GRADE 10 40 Chapter 6 Irrational Numbers and Rounding Off - Grade 10 6.1 Introduction You have seen that repeating decimals may take a lot of paper and ink to write out. Not only is that impossible, but writing numbers out to many decimal places or a high accuracy is very inconvenient and rarely gives better answers. For this reason we often estimate the number to a certain number of decimal places or to a given number of significant figures, which is even better. 6.2 Irrational Numbers Activity :: Investigation : Irrational Numbers Which of the following cannot be written as a rational number? Remember: A rational number is a fraction with numerator and denominator as integers. Terminating decimal numbers or repeating decimal numbers are rational. 1. π = 3,14159265358979323846264338327950288419716939937510 . . . 2. 1,4 3. 1,618 033 989 . . . 4. 100 Irrational numbers are numbers that cannot be written as a rational number. You should know that a rational number can be written as a fraction with the numerator and denominator as integers. This means that any number that is not a terminating decimal number or a repeating decimal number are irrational. Examples of irrational numbers are: √ √ √ 3 2, 3, 4, π, √ 1+ 5 ≈ 1,618 033 989 2 Important: When irrational numbers are written in decimal form, they go on forever and there is no repeated pattern of digits. 41 6.3 CHAPTER 6. IRRATIONAL NUMBERS AND ROUNDING OFF - GRADE 10 Important: Irrational Numbers If you are asked to identify whether a number is rational or irrational, first write the number in decimal form. If the number is terminated then it is rational. If it goes on forever, then look for a repeated pattern of digits. If there is no repeated pattern, then the number is irrational. When you write irrational numbers in decimal form, you may (if you have a lot of time and paper!) continue writing them for many, many decimal places. However, this is not convenient and it is often necessary to round off. 6.3 Rounding Off Rounding off or approximating a decimal number to a given number of decimal places is the quickest way to approximate a number. For example, if you wanted to round-off 2,6525272 to three decimal places then you would first count three places after the decimal. 2,652|5272 All numbers to the right of | are ignored after you determine whether the number in the third decimal place must be rounded up or rounded down. You round up the final digit if the first digit after the | was greater or equal to 5 and round down (leave the digit alone) otherwise. So, since the first digit after the | is a 5, we must round up the digit in the third decimal place to a 3 and the final answer of 2,6525272 rounded to three decimal places is 2,653 Worked Example 4: Rounding-Off Question: Round-off the following numbers to the indicated number of decimal places: 1. 120 99 = 1,2121212121̇2̇ to 3 decimal places 2. π = 3,141592654 . . . to 4 decimal places √ 3. 3 = 1,7320508 . . . to 4 decimal places Answer Step 1 : Determine the last digit that is kept and mark the cut-off point with |. 1. 120 99 = 1,212|1212121̇2̇ 2. π = 3,1415|92654 . . . √ 3. 3 = 1,7320|508 . . . Step 2 : Determine whether the last digit is rounded up or down. 1. The last digit of 120 99 = 1,212|1212121̇2̇ must be rounded-down. 2. The last digit of π = 3,1415|92654 . . . must be rounded-up. √ 3. The last digit of 3 = 1,7320|508 . . . must be rounded-up. Step 3 : Write the final answer. 1. 120 99 = 1,212 rounded to 3 decimal places 2. π = 3,1416 rounded to 4 decimal places √ 3. 3 = 1,7321 rounded to 4 decimal places 42 CHAPTER 6. IRRATIONAL NUMBERS AND ROUNDING OFF - GRADE 10 6.4 6.4 End of Chapter Exercises 1. Write the following rational numbers to 2 decimal places: (a) 1 2 (b) 1 (c) 0,111111 (d) 0,999991 2. Write the following irrational numbers to 2 decimal places: (a) 3,141592654 . . . (b) 1,618 033 989 . . . (c) 1,41421356 . . . (d) 2,71828182845904523536 . . . 3. Use your calculator and write the following irrational numbers to 3 decimal places: √ (a) 2 √ (b) 3 √ (c) 5 √ (d) 6 4. Use your calculator (where necessary) and write the following irrational numbers to 5 decimal places: √ (a) 8 √ (b) 768 √ (c) 100 √ (d) 0,49 √ (e) 0,0016 √ (f) 0,25 √ (g) 36 √ (h) 1960 √ (i) 0,0036 √ (j) −8 0,04 √ (k) 5 80 5. Write the following irrational numbers to 3 decimal places and then write √ them as a rational number to get an approximation to the irrational number. For example, √ 3 = 1,73205 . . .. √ 732 183 = 1 250 . Therefore, 3 is approximately To 3 decimal places, 3 = 1,732. 1,732 = 1 1000 183 . 1 250 (a) 3,141592654 . . . (b) 1,618 033 989 . . . (c) 1,41421356 . . . (d) 2,71828182845904523536 . . . 43 6.4 CHAPTER 6. IRRATIONAL NUMBERS AND ROUNDING OFF - GRADE 10 44 Chapter 7 Number Patterns - Grade 10 In earlier grades you saw patterns in the form of pictures and numbers. In this chapter we learn more about the mathematics of patterns.Patterns are recognisable regularities in situations such as in nature, shapes, events, sets of numbers. For example, spirals on a pineapple, snowflakes, geometric designs on quilts or tiles, the number sequence 0, 4, 8, 12, 16,.... Activity :: Investigation : Patterns Can you spot any patterns in the following lists of numbers? 1. 2. 3. 4. 7.1 2; 1; 1; 5; 4; 6; 8; 10; . . . 2; 4; 7; 11; . . . 4; 9; 16; 25; . . . 10; 20; 40; 80; . . . Common Number Patterns Numbers can have interesting patterns. Here we list the most common patterns and how they are made. Examples: 1. 1, 4, 7, 10, 13, 16, 19, 22, 25, ... This sequence has a difference of 3 between each number. The pattern is continued by adding 3 to the last number each time. 2. 3, 8, 13, 18, 23, 28, 33, 38, ... This sequence has a difference of 5 between each number. The pattern is continued by adding 5 to the last number each time. 3. 2, 4, 8, 16, 32, 64, 128, 256, ... This sequence has a factor of 2 between each number. The pattern is continued by multiplying the last number by 2 each time. 4. 3, 9, 27, 81, 243, 729, 2187, ... This sequence has a factor of 3 between each number. The pattern is continued by multiplying the last number by 3 each time. 45 7.2 7.1.1 CHAPTER 7. NUMBER PATTERNS - GRADE 10 Special Sequences Triangular Numbers 1, 3, 6, 10, 15, 21, 28, 36, 45, ... This sequence is generated from a pattern of dots which form a triangle. By adding another row of dots and counting all the dots we can find the next number of the sequence. Square Numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, ... The next number is made by squaring where it is in the pattern. The second number is 2 squared (22 or 2 × 2) The seventh number is 7 squared (72 or 7 × 7) etc Cube Numbers 1, 8, 27, 64, 125, 216, 343, 512, 729, ... The next number is made by cubing where it is in the pattern. The second number is 2 cubed (23 or 2 × 2 × 2) The seventh number is 7 cubed (73 or 7 × 7 × 7) etc Fibonacci Numbers 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... The next number is found by adding the two numbers before it together. The 2 is found by adding the two numbers in front of it (1 + 1) The 21 is found by adding the two numbers in front of it (8 + 13) The next number in the sequence above would be 55 (21 + 34) Can you figure out the next few numbers? 7.2 Make your own Number Patterns You can make your own number patterns using coins or matchsticks. Here is an example using dots: 3 Pattern 1 4 2 b 1 dot b b b b b 6 dots b b b b b b b b b b b b b b b b b 15 dots b b b b b b b b b b b b b b b b b b b b b b b b b b b 28 dots b 46 CHAPTER 7. NUMBER PATTERNS - GRADE 10 7.3 How many dots would you need for pattern 5 ? Can you make a formula that will tell you how many coins are needed for any size pattern? For example if the pattern 20? The formula may look something like dots = pattern × pattern + ... Worked Example 5: Study Table Question: Say you and 3 friends decide to study for Maths, and you are seated at a square table. A few minutes later, 2 other friends join you and would like to sit at your table and help you study. Naturally, you move another table and add it to the existing one. Now six of you sit at the table. Another two of your friends join your table, and you take a third table and add it to the existing tables. Now 8 of you can sit comfortably. Figure 7.1: Two more people can be seated for each table added. Examine how the number of people sitting is related to the number of tables. Answer Step 1 : Tabulate a few terms to see if there is a pattern Number of Tables, n 1 2 3 4 .. . n Number of people seated 4=4 4+2=6 4+2+2=8 4 + 2 + 2 + 2 = 10 .. . 4 + 2 + 2 + 2 + ... + 2 Step 2 : Describe the pattern We can see for 3 tables we can seat 8 people, for 4 tables we can seat 10 people and so on. We started out with 4 people and added two the whole time. Thus, for each table added, the number of persons increases by two. 7.3 Notation A sequence does not have to follow a pattern but when it does we can often write down a formula to calculate the nth -term, an . In the sequence 1; 4; 9; 16; 25; . . . where the sequence consists of the squares of integers, the formula for the nth -term is an = n 2 47 (7.1) 7.3 CHAPTER 7. NUMBER PATTERNS - GRADE 10 You can check this by looking at: a1 = 12 = 1 a2 a3 = = 22 = 4 32 = 9 a4 a5 = = 42 = 16 52 = 25 ... Therefore, using (7.1), we can generate a pattern, namely squares of integers. Worked Example 6: Study Table continued .... Question: As before, you and 3 friends are studying for Maths, and you are seated at a square table. A few minutes later, 2 other friends join you move another table and add it to the existing one. Now six of you sit at the table. Another two of your friends join your table, and you take a third table and add it to the existing tables. Now 8 of you sit comfortably as illustrated: Figure 7.2: Two more people can be seated for each table added. Find the expression for the number of people seated at n tables. Then, use the general formula to determine how many people can sit around 12 tables and how many tables are needed for 20 people. Answer Step 1 : Tabulate a few terms to see if there is a pattern Number of Tables, n 1 2 3 4 .. . n Number of people seated 4=4 4+2=6 4+2+2=8 4 + 2 + 2 + 2 = 10 .. . Formula = 4 + 2 · (0) = 4 + 2 · (1) = 4 + 2 · (2) = 4 + 2 · (3) .. . 4 + 2 + 2 + 2 + ...+ 2 = 4 + 2 · (n − 1) Step 2 : Describe the pattern The number of people seated at n tables is: an = 4 + 2 · (n − 1) Step 3 : Calculate the 12th term Using the general formula (36.1) and considering the example from the previous section, how many people can sit around, say, 12 tables? We are looking for a12 , that is, where n = 12: an = a12 = = = = a1 + d · (n − 1) 4 + 2 · (12 − 1) 4 + 2(11) 4 + 22 26 48 CHAPTER 7. NUMBER PATTERNS - GRADE 10 7.3 Step 4 : Calculate the number of terms if an = 20 an = a1 + d · (n − 1) 20 = 20 − 4 = 4 + 2 · (n − 1) 2 · (n − 1) 16 ÷ 2 = 8+1 = n n−1 n = 9 Step 5 : Final Answer 26 people can be seated at 12 tables and 9 tables are needed to seat 20 people. It is also important to note the difference between n and an . n can be compared to a place holder, while an is the value at the place “held” by n. Like our “Study Table”-example above, the first table (Table 1) holds 4 people. Thus, at place n = 1, the value of a1 = 4, and so on: n an 1 4 2 6 3 8 4 10 ... ... Activity :: Investigation : General Formula 1. Find the general formula for the following sequences and then find a10 , a50 and a100 : (a) 2, 5, 8, 11, 14, . . . (b) 0, 4, 8, 12, 16, . . . (c) 2, −1, −4, −7, −10, . . . 2. The general term has been given for each sequence below. Work out the missing terms. (a) 0; 3; ...; 15; 24 (b) 3; 2; 1; 0; ...; 2 (c) 11; ...; 7; ...; 3 7.3.1 n2 − 1 −n + 4 −13 + 2n Patterns and Conjecture In mathematics, a conjecture is a mathematical statement which appears to be true, but has not been formally proven to be true under the rules of mathematics. Other words that have a similar in meaning to conjecture are: hypothesis, theory, assumption and premise. For example: Make a conjecture about the next number based on the pattern 2; 6; 11; 17 : ... The numbers increase by 4, 5, and 6. Conjecture: The next number will increase by 7. So, it will be 17 + 7 or 24. 49 7.4 CHAPTER 7. NUMBER PATTERNS - GRADE 10 Worked Example 7: Number patterns Question: Consider the following pattern. 12 + 1 = 22 + 2 32 + 3 = = 42 + 4 = 22 − 2 32 − 3 42 − 4 52 − 5 1. Add another two rows to the end of the pattern. 2. Make a conjecture about this pattern. Write your conjecture in words. 3. Generalise your conjecture for this pattern (in other words, write your conjecture algebraically). 4. Prove that your conjecture is true. Answer Step 1 : The next two rows 52 + 5 = 62 − 6 62 + 6 = 72 − 7 Step 2 : Conjecture Squaring a number and adding the same number gives the same result as squaring the next number and subtracting that number. Step 3 : Generalise We have chosen to use x here. You could choose any letter to generalise the pattern. x2 + x = (x + 1)2 − (x + 1) Step 4 : Proof Lef t side : x2 + x Right side : (x + 1)2 − (x + 1) Right side = x2 + 2x + 1 − x − 1 = x2 + x = lef t side T heref ore x2 + x = (x + 1)2 − (x + 1) 7.4 Exercises 1. Find the nth term for: 3, 7, 11, 15, . . . 2. Find the general term of the following sequences: 50 CHAPTER 7. NUMBER PATTERNS - GRADE 10 (a) (b) (c) (d) (e) 7.4 −2,1,4,7, . . . 11, 15, 19, 23, . . . x − 1,2x + 5,5x + 1, . . . sequence with a3 = 7 and a8 = 15 sequence with a4 = −8 and a10 = 10 3. The seating in a section of a sports stadium can be arranged so the first row has 15 seats, the second row has 19 seats, the third row has 23 seats and so on. Calculate how many seats are in the row 25. 4. Consider the following pattern: 22 + 2 2 3 +3 42 + 4 = 32 − 3 = 42 − 4 = 52 − 5 (a) Add at least two more rows to the pattern and check whether or not the pattern continues to work. (b) Describe in words any patterns that you have noticed. (c) Try to generalise a rule using algebra i.e. find the general term for the pattern. (d) Prove or disprove that this rule works for all values. 5. The profits of a small company for the last four years has been: R10 000, R15 000, R19 000 and R23 000. If the pattern continues, what is the expected profit in the 10 years (i.e. in the 14th year of the company being in business)? 6. A single square is made from 4 matchsticks. Two squares in a row needs 7 matchsticks and 3 squares in a row needs 10 matchsticks. Determine: (a) (b) (c) (d) the first term the common difference the formula for the general term how many matchsticks are in a row of 25 squares 7. You would like to start saving some money, but because you have never tried to save money before, you have decided to start slowly. At the end of the first week you deposit R5 into your bank account. Then at the end of the second week you deposit R10 into your bank account. At the end of the third week you deposit R15. After how many weeks, do you deposit R50 into your bank account? 8. A horizontal line intersects a piece of string at four points and divides it into five parts, as shown below. 1 3 b b 2 5 b b 4 51 7.4 CHAPTER 7. NUMBER PATTERNS - GRADE 10 If the piece of string is intersected in this way by 19 parallel lines, each of which intersects it at four points, find the number of parts into which the string will be divided. 52 Chapter 8 Finance - Grade 10 8.1 Introduction Should you ever find yourself stuck with a mathematics question on a television quiz show, you will probably wish you had remembered the how many even prime numbers there are between 1 and 100 for the sake of R1 000 000. And who does not want to be a millionaire, right? Welcome to the Grade 10 Finance Chapter, where we apply maths skills to everyday financial situations that you are likely to face both now and along your journey to purchasing your first private jet. If you master the techniques in this chapter, you will grasp the concept of compound interest, and how it can ruin your fortunes if you have credit card debt, or make you millions if you successfully invest your hard-earned money. You will also understand the effects of fluctuating exchange rates, and its impact on your spending power during your overseas holidays! 8.2 Foreign Exchange Rates Is $500 (”500 US dollars”) per person per night a good deal on a hotel in New York City? The first question you will ask is “How much is that worth in Rands?”. A quick call to the local bank or a search on the Internet (for example on http://www.x-rates.com/) for the Dollar/Rand exchange rate will give you a basis for assessing the price. A foreign exchange rate is nothing more than the price of one currency in terms of another. For example, the exchange rate of 6,18 Rands/US Dollars means that $1 costs R6,18. In other words, if you have $1 you could sell it for R6,18 - or if you wanted $1 you would have to pay R6,18 for it. But what drives exchange rates, and what causes exchange rates to change? And how does this affect you anyway? This section looks at answering these questions. 8.2.1 How much is R1 really worth? We can quote the price of a currency in terms of any other currency, but the US Dollar, British Pounds Sterling or even the Euro are often used as a market standard. You will notice that the financial news will report the South African Rand exchange rate in terms of these three major currencies. So the South African Rand could be quoted on a certain date as 6,7040 ZAR per USD (i.e. $1,00 costs R6,7040), or 12,2374 ZAR per GBP. So if I wanted to spend $1 000 on a holiday in the United States of America, this would cost me R6 704,00; and if I wanted £1 000 for a weekend in London it would cost me R12 237,40. This seems obvious, but let us see how we calculated that: The rate is given as ZAR per USD, or ZAR/USD such that $1,00 buys R6,7040. Therefore, we need to multiply by 1 000 to get the 53 8.2 CHAPTER 8. FINANCE - GRADE 10 Table 8.1: Abbreviations and symbols for some common currencies. Currency Abbreviation Symbol South African Rand ZAR R United States Dollar USD $ British Pounds Sterling GBP £ Euro EUR e number of Rands per $1 000. Mathematically, $1,00 = ∴ 1 000 × $1,00 = = R6,0740 1 000 × R6,0740 R6 074,00 as expected. What if you have saved R10 000 for spending money for the same trip and you wanted to use this to buy USD? How much USD could you get for this? Our rate is in ZAR/USD but we want to know how many USD we can get for our ZAR. This is easy. We know how much $1,00 costs in terms of Rands. $1,00 = $1,00 = ∴ 6,0740 1,00 $ = 6,0740 R1,00 = = R6,0740 R6,0740 6,0740 R1,00 1,00 6,0740 $0,164636 $ As we can see, the final answer is simply the reciprocal of the ZAR/USD rate. Therefore, R10 000 will get: R1,00 = ∴ 10 000 × R1,00 = = $ 1,00 6,0740 10 000 × $ 1,00 6,0740 $1 646,36 We can check the answer as follows: ∴ $1,00 = 1 646,36 × $1,00 = = R6,0740 1 646,36 × R6,0740 R10 000,00 Six of one and half a dozen of the other So we have two different ways of expressing the same exchange rate: Rands per Dollar (ZAR/USD) and Dollar per Rands (USD/ZAR). Both exchange rates mean the same thing and express the value of one currency in terms of another. You can easily work out one from the other - they are just the reciprocals of the other. If the South African Rand is our Domestic (or home) Currency, we call the ZAR/USD rate a “direct” rate, and we call a USD/ZAR rate an “indirect” rate. In general, a direct rate is an exchange rate that is expressed as units of Home Currency per 54 CHAPTER 8. FINANCE - GRADE 10 8.2 units of Foreign Currency, i.e., Domestic Currency / Foreign Currency. The Rand exchange rates that we see on the news are usually expressed as Direct Rates, for example you might see: Table 8.2: Examples of Currency Abbreviation 1 USD 1 GBP 1 EUR exchange rates Exchange Rates R6,9556 R13,6628 R9,1954 The exchange rate is just the price of each of the Foreign Currencies (USD, GBP and EUR) in terms of our Domestic Currency, Rands. An indirect rate is an exchange rate expressed as units of Foreign Currency per units of Home Currency, i.e. Foreign Currency / Domestic Currency Defining exchange rates as direct or indirect depends on which currency is defined as the Domestic Currency. The Domestic Currency for an American investor would be USD which is the South African investor’s Foreign Currency. So direct rates from the perspective of the American investor (USD/ZAR) would be the same as the indirect rate from the perspective of the South Africa investor. Terminology Since exchange rates are simple prices of currencies, movements in exchange rates means that the price or value of the currency changed. The price of petrol changes all the time, so does the price of gold, and currency prices also move up and down all the time. If the Rand exchange rate moved from say R6,71 per USD to R6,50 per USD, what does this mean? Well, it means that $1 would now cost only R6,50 instead of R6,71. The Dollar is now cheaper to buy, and we say that the Dollar has depreciated (or weakened) against the Rand. Alternatively we could say that the Rand has appreciated (or strengthened) against the Dollar. What if we were looking at indirect exchange rates, and the exchange rate moved from $0,149 1 1 per ZAR (= 6,71 ) to $0,1538 per ZAR (= 6,50 ). Well now we can see that the R1,00 cost $0,149 at the start, and then cost $0,1538 at the end. The Rand has become more expensive (in terms of Dollars), and again we can say that the Rand has appreciated. Regardless of which exchange rate is used, we still come to the same conclusions. In general, • for direct exchange rates, the home currency will appreciate (depreciate) if the exchange rate falls (rises) • For indirect exchange rates, the home currency will appreciate (depreciate) if the exchange rate rises (falls) As with just about everything in this chapter, do not get caught up in memorising these formulae - that is only going to get confusing. Think about what you have and what you want - and it should be quite clear how to get the correct answer. Activity :: Discussion : Foreign Exchange Rates In groups of 5, discuss: 1. Why might we need to know exchange rates? 2. What happens if one countries currency falls drastically vs another countries currency? 55 8.2 CHAPTER 8. FINANCE - GRADE 10 3. When might you use exchange rates? 8.2.2 Cross Currency Exchange Rates We know that the exchange rates are the value of one currency expressed in terms of another currency, and we can quote exchange rates against any other currency. The Rand exchange rates we see on the news are usually expressed against the major currency, USD, GBP and EUR. So if for example, the Rand exchange rates were given as 6,71 ZAR/USD and 12,71 ZAR/GBP, does this tell us anything about the exchange rate between USD and GBP? Well I know that if $1 will buy me R6,71, and if £1.00 will buy me R12,71, then surely the GBP is stronger than the USD because you will get more Rands for one unit of the currency, and we can work out the USD/GBP exchange rate as follows: Before we plug in any numbers, how can we get a USD/GBP exchange rate from the ZAR/USD and ZAR/GBP exchange rates? Well, USD/GBP = USD/ZAR × ZAR/GBP. Note that the ZAR in the numerator will cancel out with the ZAR in the denominator, and we are left with the USD/GBP exchange rate. Although we do not have the USD/ZAR exchange rate, we know that this is just the reciprocal of the ZAR/USD exchange rate. USD/ZAR = 1 ZAR/USD Now plugging in the numbers, we get: USD/GBP = = = = USD/ZAR × ZAR/GBP 1 × ZAR/GBP ZAR/USD 1 × 12,71 6,71 1,894 Important: Sometimes you will see exchange rates in the real world that do not appear to work exactly like this. This is usually because some financial institutions add other costs to the exchange rates, which alter the results. However, if you could remove the effect of those extra costs, the numbers would balance again. Worked Example 8: Cross Exchange Rates Question: If $1 = R 6,40, and £1 = R11,58 what is the $/£ exchange rate (i.e. the number of US$ per £)? Answer Step 1 : Determine what is given and what is required The following are given: • ZAR/USD rate = R6,40 • ZAR/GBP rate = R11,58 56 CHAPTER 8. FINANCE - GRADE 10 8.2 The following is required: • USD/GBP rate Step 2 : Determine how to approach the problem We know that: USD/GBP = USD/ZAR × ZAR/GBP. Step 3 : Solve the problem USD/GBP = = = = USD/ZAR × ZAR/GBP 1 × ZAR/GBP ZAR/USD 1 × 11,58 6,40 1,8094 Step 4 : Write the final answer $1,8094 can be bought for £1. Activity :: Investigation : Cross Exchange Rates - Alternate Method If $1 = R 6,40, and £1 = R11,58 what is the $/£ exchange rate (i.e. the number of US$ per £)? Overview of problem You need the $/£ exchange rate, in other words how many dollars must you pay for a pound. So you need £1. From the given information we know that it would cost you R11,58 to buy £1 and that $ 1 = R6,40. Use this information to: 1. calculate how much R1 is worth in $. 2. calculate how much R11,58 is worth in $. Do you get the same answer as in the worked example? 8.2.3 Enrichment: Fluctuating exchange rates If everyone wants to buy houses in a certain suburb, then house prices are going to go up - because the buyers will be competing to buy those houses. If there is a suburb where all residents want to move out, then there are lots of sellers and this will cause house prices in the area to fall because the buyers would not have to struggle as much to find an eager seller. This is all about supply and demand, which is a very important section in the study of Economics. You can think about this is many different contexts, like stamp-collecting for example. If there is a stamp that lots of people want (high demand) and few people own (low supply) then that stamp is going to be expensive. And if you are starting to wonder why this is relevant - think about currencies. If you are going to visit London, then you have Rands but you need to “buy” Pounds. The exchange rate is the price you have to pay to buy those Pounds. Think about a time where lots of South Africans are visiting the United Kingdom, and other South Africans are importing goods from the United Kingdom. That means there are lots of Rands (high supply) trying to buy Pounds. Pounds will start to become more expensive (compare this to the house price example at the start of this section if you are not convinced), and the 57 8.3 CHAPTER 8. FINANCE - GRADE 10 exchange rate will change. In other words, for R1 000 you will get fewer Pounds than you would have before the exchange rate moved. Another context which might be useful for you to understand this: consider what would happen if people in other countries felt that South Africa was becoming a great place to live, and that more people were wanting to invest in South Africa - whether in properties, businesses - or just buying more goods from South Africa. There would be a greater demand for Rands - and the “price of the Rand” would go up. In other words, people would need to use more Dollars, or Pounds, or Euros ... to buy the same amount of Rands. This is seen as a movement in exchange rates. Although it really does come down to supply and demand, it is interesting to think about what factors might affect the supply (people wanting to “sell” a particular currency) and the demand (people trying to “buy” another currency). This is covered in detail in the study of Economics, but let us look at some of the basic issues here. There are various factors affect exchange rates, some of which have more economic rationale than others: • economic factors (such as inflation figures, interest rates, trade deficit information, monetary policy and fiscal policy) • political factors (such as uncertain political environment, or political unrest) • market sentiments and market behaviour (for example if foreign exchange markets perceived a currency to be overvalued and starting selling the currency, this would cause the currency to fall in value - a self fulfilling expectation). Exercise: Foreign Exchange 1. I want to buy an IPOD that costs £100, with the exchange rate currently at £1 = R14. I believe the exchange rate will reach R12 in a month. (a) How much will the MP3 player cost in Rands, if I buy it now? (b) How much will I save if the exchange rate drops to R12? (c) How much will I lose if the exchange rate moves to R15? 2. Study the following exchange rate table: Country Currency United Kingdom (UK) Pounds(£) United States (USA) Dollars ($) Exchange Rate R14,13 R7,04 (a) In South Africa the cost of a new Honda Civic is R173 400. In England the same vehicle costs £12 200 and in the USA $ 21 900. In which country is the car the cheapest if you compare it to the South African Rand ? (b) Sollie and Arinda are waiters in a South African Restaurant attracting many tourists from abroad. Sollie gets a £6 tip from a tourist and Arinda gets $ 12. How many South African Rand did each one get ? 8.3 Being Interested in Interest If you had R1 000, you could either keep it in your wallet, or deposit it in a bank account. If it stayed in your wallet, you could spend it any time you wanted. If the bank looked after it for you, then they could spend it, with the plan of making profit off it. The bank usually “pays” you to deposit it into an account, as a way of encouraging you to bank it with them, This payment is like a reward, which provides you with a reason to leave it with the bank for a while, rather than keeping the money in your wallet. 58 CHAPTER 8. FINANCE - GRADE 10 8.4 We call this reward ”interest”. If you deposit money into a bank account, you are effectively lending money to the bank - and you can expect to receive interest in return. Similarly, if you borrow money from a bank (or from a department store, or a car dealership, for example) then you can expect to have to pay interest on the loan. That is the price of borrowing money. The concept is simple, yet it is core to the world of finance. Accountants, actuaries and bankers, for example, could spend their entire working career dealing with the effects of interest on financial matters. In this chapter you will be introduced to the concept of financial mathematics - and given the tools to cope with even advanced concepts and problems. Important: Interest The concepts in this chapter are simple - we are just looking at the same idea, but from many different angles. The best way to learn from this chapter is to do the examples yourself, as you work your way through. Do not just take our word for it! 8.4 Simple Interest Definition: Simple Interest Simple interest is where you earn interest on the initial amount that you invested, but not interest on interest. As an easy example of simple interest, consider how much you will get by investing R1 000 for 1 year with a bank that pays you 5% simple interest. At the end of the year, you will get an interest of: Interest = R1 000 × 5% 5 = R1 000 × 100 = R1 000 × 0,05 = R50 So, with an “opening balance” of R1 000 at the start of the year, your “closing balance” at the end of the year will therefore be: Closing Balance = = = Opening Balance + Interest R1 000 + R50 R1 050 We sometimes call the opening balance in financial calculations Principal, which is abbreviated as P (R1 000 in the example). The interest rate is usually labelled i (5% in the example), and the interest amount (in Rand terms) is labelled I (R50 in the example). So we can see that: I =P ×i and Closing Balance = = = = Opening Balance + Interest P +I P + (P × i) P (1 + i) 59 (8.1) 8.4 CHAPTER 8. FINANCE - GRADE 10 This is how you calculate simple interest. It is not a complicated formula, which is just as well because you are going to see a lot of it! Not Just One You might be wondering to yourself: 1. how much interest will you be paid if you only leave the money in the account for 3 months, or 2. what if you leave it there for 3 years? It is actually quite simple - which is why they call it Simple Interest. 1. Three months is 1/4 of a year, so you would only get 1/4 of a full year’s interest, which is: 1/4 × (P × i). The closing balance would therefore be: Closing Balance = P + 1/4 × (P × i) = P (1 + (1/4)i) 2. For 3 years, you would get three years’ worth of interest, being: 3 × (P × i). The closing balance at the end of the three year period would be: Closing Balance = = P + 3 × (P × i) P × (1 + (3)i) If you look carefully at the similarities between the two answers above, we can generalise the result. In other words, if you invest your money (P ) in an account which pays a rate of interest (i) for a period of time (n years), then, using the symbol (A) for the Closing Balance: Closing Balance,(A) = P (1 + i · n) (8.2) As we have seen, this works when n is a fraction of a year and also when n covers several years. Important: Interest Calculation Annual Rates means Yearly rates. and p.a.(per annum) = per year Worked Example 9: Simple Interest Question: If I deposit R1 000 into a special bank account which pays a Simple Interest of 7% for 3 years, how much will I get back at the end? Answer Step 1 : Determine what is given and what is required • opening balance, P = R1 000 • interest rate, i = 7% • period of time, n = 3 years We are required to find the closing balance (A). Step 2 : Determine how to approach the problem We know from (8.2) that: Closing Balance,(A) = P (1 + i · n) 60 CHAPTER 8. FINANCE - GRADE 10 8.4 Step 3 : Solve the problem A = = = P (1 + i · n) R1 000(1 + 3 × 7%) R1 210 Step 4 : Write the final answer The closing balance after 3 years of saving R1 000 at an interest rate of 7% is R1 210. Worked Example 10: Calculating n Question: If I deposit R30 000 into a special bank account which pays a Simple Interest of 7.5% ,for how many years must I invest this amount to generate R45 000 Answer Step 1 : Determine what is given and what is required • opening balance, P = R30 000 • interest rate, i = 7,5% • closing balance, A = R45 000 We are required to find the number of years. Step 2 : Determine how to approach the problem We know from (8.2) that: Closing Balance (A) = P (1 + i · n) Step 3 : Solve the problem Closing Balance (A) = R45 000 = (1 + 0,075 × n) = 0,075 × n = n = n = P (1 + i · n) R30 000(1 + n × 7,5%) 45000 30000 1,5 − 1 0,5 0,075 6,6666667 Step 4 : Write the final answer n has to be a whole number, therefore n = 7. The period is 7 years for R30 000 to generate R45 000 at a simple interest rate of 7,5%. 8.4.1 Other Applications of the Simple Interest Formula 61 8.4 CHAPTER 8. FINANCE - GRADE 10 Worked Example 11: Hire-Purchase Question: Troy is keen to buy an addisional hard drive for his laptop advertised for R 2 500 on the internet. There is an option of paying a 10% deposit then making 24 monthly payments using a hire-purchase agreement where interest is calculated at 7,5% p.a. simple interest. Calculate what Troy’s monthly payments will be. Answer Step 1 : Determine what is given and what is required A new opening balance is required, as the 10% deposit is paid in cash. • • • • 10% of R 2 500 = R250 new opening balance, P = R2 500 − R250 = R2 250 interest rate, i = 7,5% = 0,075pa period of time, n = 2 years We are required to find the closing balance (A) and then the montly payments. Step 2 : Determine how to approach the problem We know from (8.2) that: Closing Balance,(A) = P (1 + i · n) Step 3 : Solve the problem A = = = Monthly payment = = P (1 + i · n) R2 250(1 + 2 × 7,5%) R2 587,50 2587,50 ÷ 24 R107,81 Step 4 : Write the final answer Troy’s monthly payments = R 107,81 Worked Example 12: Depreciation Question: Seven years ago, Tjad’s drum kit cost him R12 500. It has now been valued at R2 300. What rate of simple depreciation does this represent ? Answer Step 1 : Determine what is given and what is required • opening balance, P = R12 500 • period of time, n = 7 years • closing balance, A = R2 300 We are required to find the rate(i). Step 2 : Determine how to approach the problem We know from (8.2) that: Closing Balance,(A) = P (1 + i · n) Therefore, for depreciation the formula will change to: Closing Balance,(A) = P (1 − i · n) 62 CHAPTER 8. FINANCE - GRADE 10 8.5 Step 3 : Solve the problem A = P (1 − i · n) R2 300 = R12 500(1 − 7 × i) i = 0,11657... Step 4 : Write the final answer Therefore the rate of depreciation is 11,66% Exercise: Simple Interest 1. An amount of R3 500 is invested in a savings account which pays simple interest at a rate of 7,5% per annum. Calculate the balance accumulated by the end of 2 years. 2. Calculate the simple interest for the following problems. (a) A loan of R300 at a rate of 8% for l year. (b) An investment of R225 at a rate of 12,5% for 6 years. 3. I made a deposit of R5 000 in the bank for my 5 year old son’s 21st birthday. I have given him the amount of R 18 000 on his birtday. At what rate was the money invested, if simple interest was calculated ? 4. Bongani buys a dining room table costing R 8 500 on Hire Purchase. He is charged simple interest at 17,5% per annum over 3 years. (a) How much will Bongani pay in total ? (b) How much interest does he pay ? (c) What is his montly installment ? 8.5 Compound Interest To explain the concept of compound interest, the following example is discussed: Worked Example 13: Using Simple Interest to lead to the concept Com- pound Interest Question: If I deposit R1 000 into a special bank account which pays a Simple Interest of 7%. What if I empty the bank account after a year, and then take the principal and the interest and invest it back into the same account again. Then I take it all out at the end of the second year, and then put it all back in again? And then I take it all out at the end of 3 years? Answer Step 1 : Determine what is given and what is required • opening balance, P = R1 000 • interest rate, i = 7% 63 8.5 CHAPTER 8. FINANCE - GRADE 10 • period of time, 1 year at a time, for 3 years We are required to find the closing balance at the end of three years. Step 2 : Determine how to approach the problem We know that: Closing Balance = P (1 + i · n) Step 3 : Determine the closing balance at the end of the first year Closing Balance = P (1 + i · n) = R1 000(1 + 1 × 7%) = R1 070 Step 4 : Determine the closing balance at the end of the second year After the first year, we withdraw all the money and re-deposit it. The opening balance for the second year is therefore R1 070, because this is the balance after the first year. Closing Balance = P (1 + i · n) = R1 070(1 + 1 × 7%) = R1 144,90 Step 5 : Determine the closing balance at the end of the third year After the second year, we withdraw all the money and re-deposit it. The opening balance for the third year is therefore R1 144,90, because this is the balance after the first year. Closing Balance = P (1 + i · n) = R1 144,90(1 + 1 × 7%) = R1 225,04 Step 6 : Write the final answer The closing balance after withdrawing the all the money and re-depositing each year for 3 years of saving R1 000 at an interest rate of 7% is R1 225,04. In the two worked examples using simple interest, we have basically the same problem because P =R1 000, i=7% and n=3 years for both problems. Except in the second situation, we end up with R1 225,04 which is more than R1 210 from the first example. What has changed? In the first example I earned R70 interest each year - the same in the first, second and third year. But in the second situation, when I took the money out and then re-invested it, I was actually earning interest in the second year on my interest (R70) from the first year. (And interest on the interest on my interest in the third year!) This more realistically reflects what happens in the real world, and is known as Compound Interest. It is this concept which underlies just about everything we do - so we will look at more closely next. Definition: Compound Interest Compound interest is the interest payable on the principal and its accumulated interest. Compound interest is a double edged sword, though - great if you are earning interest on cash you have invested, but crippling if you are stuck having to pay interest on money you have borrowed! In the same way that we developed a formula for Simple Interest, let us find one for Compound Interest. 64 CHAPTER 8. FINANCE - GRADE 10 8.5 If our opening balance is P and we have an interest rate of i then, the closing balance at the end of the first year is: Closing Balance after 1 year = P (1 + i) This is the same as Simple Interest because it only covers a single year. Then, if we take that out and re-invest it for another year - just as you saw us doing in the worked example above then the balance after the second year will be: Closing Balance after 2 years = = [P (1 + i)] × (1 + i) P (1 + i)2 And if we take that money out, then invest it for another year, the balance becomes: = [P (1 + i)2 ] × (1 + i) = P (1 + i)3 Closing Balance after 3 years We can see that the power of the term (1 + i) is the same as the number of years. Therefore, Closing Balance after n years = P (1 + i)n 8.5.1 (8.3) Fractions add up to the Whole It is easy to show that this formula works even when n is a fraction of a year. For example, let us invest the money for 1 month, then for 4 months, then for 7 months. Closing Balance after 1 month = Closing Balance after 5 months = = = = Closing Balance after 12 months = 1 P (1 + i) 12 Closing Balance after 1 month invested for 4 months more 4 1 [P (1 + i) 12 ] 12 1 4 P (1 + i) 12 + 12 5 P (1 + i) 12 Closing Balance after 5 month invested for 7 months more 5 7 = [P (1 + i) 12 ] 12 = P (1 + i) 12 + 12 = = P (1 + i) 12 P (1 + i)1 5 7 12 which is the same as investing the money for a year. Look carefully at the long equation above. It is not as complicated as it looks! All we are doing is taking the opening amount (P ), then adding interest for just 1 month. Then we are taking that new balance and adding interest for a further 4 months, and then finally we are taking the new balance after a total of 5 months, and adding interest for 7 more months. Take a look again, and check how easy it really is. Does the final formula look familiar? Correct - it is the same result as you would get for simply investing P for one full year. This is exactly what we would expect, because: 1 month + 4 months + 7 months = 12 months, which is a year. Can you see that? Do not move on until you have understood this point. 8.5.2 The Power of Compound Interest To see how important this “interest on interest” is, we shall compare the difference in closing balances for money earning simple interest and money earning compound interest. Consider an amount of R10 000 that you have to invest for 10 years, and assume we can earn interest of 9%. How much would that be worth after 10 years? 65 8.5 CHAPTER 8. FINANCE - GRADE 10 The closing balance for the money earning simple interest is: Closing Balance = = = P (1 + i · n) R10 000(1 + 9% × 10) R19 000 The closing balance for the money earning compound interest is: Closing Balance = = = P (1 + i)n ) R10 000(1 + 9%)10 R23 673,64 So next time someone talks about the “magic of compound interest”, not only will you know what they mean - but you will be able to prove it mathematically yourself! Again, keep in mind that this is good news and bad news. When you are earning interest on money you have invested, compound interest helps that amount to increase exponentially. But if you have borrowed money, the build up of the amount you owe will grow exponentially too. Worked Example 14: Taking out a Loan Question: Mr Lowe wants to take out a loan of R 350 000. He does not want to pay back more than R625 000 altogether on the loan. If the interest rate he is offered is 13%, over what period should he take the loan Answer Step 1 : Determine what has been provided and what is required • opening balance, P = R350 000 • closing balance, A = R625 000 • interest rate, i = 13% peryear We are required to find the time period(n). Step 2 : Determine how to approach the problem We know from (8.3) that: Closing Balance,(A) = P (1 + i)n We need to find n. Therefore we covert the formula to: A = (1 + i)n P and then find n by trial and error. Step 3 : Solve the problem A = (1 + i)n P 625000 = (1 + 0,13)n 350000 1,785... = (1,13)n Try n = 3 : (1,13)3 = 1,44... Try n = 4 : Try n = 5 : (1,13)4 = 1,63... (1,13)5 = 1,84... Step 4 : Write the final answer Mr Lowe should take the loan over four years 66 CHAPTER 8. FINANCE - GRADE 10 8.5.3 8.5 Other Applications of Compound Growth Worked Example 15: Population Growth Question: South Africa’s population is increasing by 2,5% per year. If the current population is 43 million, how many more people will there be in South Africa in two year’s time ? Answer Step 1 : Determine what has been provided and what is required • opening balance, P = 43 000 000 • period of time, n = 2 year • interest rate, i = 2,5% peryear We are required to find the closing balance(A). Step 2 : Determine how to approach the problem We know from (8.3) that: Closing Balance,(A) = P (1 + i)n Step 3 : Solve the problem A = = = P (1 + i)n 43 000 000(1 + 0,025)2 45 176 875 Step 4 : Write the final answer There are 45 176 875 − 43 000 000 = 2 176 875 more people in 2 year’s time Worked Example 16: Compound Decrease Question: A swimming pool is being treated for a build-up of algae. Initially, 50m2 of the pool is covered by algae. With each day of treatment, the algae reduces by 5%. What area is covered by algae after 30 days of treatment ? Answer Step 1 : Determine what has been provided and what is required • opening balance, P = 50m2 • period of time, n = 30 days • interest rate, i = 5% perday We are required to find the closing balance(A). Step 2 : Determine how to approach the problem We know from (8.3) that: Closing Balance,(A) = P (1 + i)n 67 8.6 CHAPTER 8. FINANCE - GRADE 10 But this is compound decrease so we can use the formula: Closing Balance,(A) = P (1 − i)n Step 3 : Solve the problem A = = = P (1 − i)n 50(1 − 0,05)30 10,73m2 Step 4 : Write the final answer Therefore the area still covered with algae is 10,73m2 Exercise: Compound Interest 1. An amount of R3 500 is invested in a savings account which pays compound interest at a rate of 7,5% per annum. Calculate the balance accumulated by the end of 2 years. 2. If the average rate of inflation for the past few years was 7,3% and your water and electricity account is R 1 425 on average, what would you expect to pay in 6 years time ? 3. Shrek wants to invest some money at 11% per annum compound interest. How much money (to the nearest rand) should he invest if he wants to reach a sum of R 100 000 in five year’s time ? 8.6 Summary As an easy reference, here are the key formulae that we derived and used during this chapter. While memorising them is nice (there are not many), it is the application that is useful. Financial experts are not paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve financial problems. 8.6.1 P i n Definitions Principal (the amount of money at the starting point of the calculation) interest rate, normally the effective rate per annum period for which the investment is made 8.6.2 Equations Closing Balance - simple interest Solve for i = P (1 + i · n) Solve for n 68 CHAPTER 8. FINANCE - GRADE 10 8.7 Closing Balance - compound interest Solve for i = P (1 + i)n Solve for n Important: Always keep the interest and the time period in the same units of time (e.g. both in years, or both in months etc.). 8.7 End of Chapter Exercises 1. You are going on holiday to Europe. Your hotel will cost e200 per night. How much will you need in Rands to cover your hotel bill, if the exchange rate is e1 = R9,20. 2. Calculate how much you will earn if you invested R500 for 1 year at the following interest rates: (a) 6,85% simple interest (b) 4,00% compound interest 3. Bianca has R1 450 to invest for 3 years. Bank A offers a savings account which pays simple interest at a rate of 11% per annum, whereas Bank B offers a savings account paying compound interest at a rate of 10,5% per annum. Which account would leave Bianca with the highest accumulated balance at the end of the 3 year period? 4. How much simple interest is payable on a loan of R2 000 for a year, if the interest rate is 10%? 5. How much compound interest is payable on a loan of R2 000 for a year, if the interest rate is 10%? 6. Discuss: (a) Which type of interest would you like to use if you are the borrower? (b) Which type of interest would you like to use if you were the banker? 7. Calculate the compound interest for the following problems. (a) A R2 000 loan for 2 years at 5%. (b) A R1 500 investment for 3 years at 6%. (c) An R800 loan for l year at 16%. 8. If the exchange rate 100 Yen = R 6,2287 and 1 AUD = R 5,1094 , determine the exchange rate between the Australian Dollar and the Japanese Yen. 9. Bonnie bought a stove for R 3 750. After 3 years she paid for it and the R 956,25 interest that was charged for hire-purchase. Determine the simple rate of interest that was charged. 69 8.7 CHAPTER 8. FINANCE - GRADE 10 70 Chapter 9 Products and Factors - Grade 10 9.1 Introduction In this chapter you will learn how to work with algebraic expressions. You will recap some of the work on factorisation and multiplying out expressions that you learnt in earlier grades. This work will then be extended upon for Grade 10. 9.2 Recap of Earlier Work The following should be familiar. Examples are given as reminders. 9.2.1 Parts of an Expression Mathematical expressions are just like sentences and their parts have special names. You should be familiar with the following names used to describe the parts of an mathematical expression. a · xk + b · x + cm = 0 p d·y +e·y+f ≤0 Name term expression coefficient exponent (or index) base constant variable equation inequality binomial trinomial 9.2.2 (9.1) (9.2) Examples (separated by commas) a · xk ,b · x, cm , d · y p , e · y, f a · xk + b · x + cm , d · y p + e · y + f a, b, d, e k, p x, y, c a, b, c, d, e, f x, y a · xk + b · x + cm = 0 d · yp + e · y + f ≤ 0 expression with two terms expression with three terms Product of Two Binomials A binomial is a mathematical expression with two terms, e.g. (ax + b) and (cx + d). If these two binomials are multiplied, the following is the result: 71 9.2 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 (a · x + b)(c · x + d) = (ax)(c · x + d) + b(c · x + d) = (ax)(cx) + (ax)d + b(cx) + b · d Worked Example 17: Product of two Binomials Question: Find the product of (3x − 2)(5x + 8) Answer (3x − 2)(5x + 8) = = (3x)(5x) + (3x)(8) + (−2)(5x) + (−2)(8) 15x2 + 24x − 10x − 16 15x2 + 14x − 16 = The product of two identical binomials is known as the square of the binomials and is written as: (ax + b)2 = a2 x2 + 2abx + b2 If the two terms are ax + b and ax − b then their product is: (ax + b)(ax − b) = a2 x2 − b2 This is known as the difference of squares. 9.2.3 Factorisation Factorisation is the opposite of expanding brackets. For example expanding brackets would require 2(x + 1) to be written as 2x + 2. Factorisation would be to start with 2x + 2 and to end up with 2(x + 1). In previous grades you factorised based on common factors and on difference of squares. Common Factors Factorising based on common factors relies on there being common factors between your terms. For example, 2x − 6x2 can be factorised as follows: 2x − 6x2 = 2x(1 − 3x) Activity :: Investigation : Common Factors Find the highest common factors of the following pairs of terms: (a) 6y; 18x (f) 2xy; 4xyz (b) 12mn; 8n (g) 3uv; 6u (c) 3st; 4su (h) 9xy; 15xz 72 (d) 18kl; 9kp (i) 24xyz; 16yz (e) abc; ac (j) 3m; 45n CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 9.2 Difference of Squares We have seen that: (ax + b)(ax − b) = a2 x2 − b2 (9.3) Since 9.3 is an equation, both sides are always equal. This means that an expression of the form: a2 x2 − b2 can be factorised to (ax + b)(ax − b) Therefore, a2 x2 − b2 = (ax + b)(ax − b) For example, x2 − 16 can be written as (x2 − 42 ) which is a difference of squares. Therefore the factors of x2 − 16 are (x − 4) and (x + 4). Worked Example 18: Factorisation Question: Factorise completely: b2 y 5 − 3aby 3 Answer b2 y 5 − 3aby 3 = by 3 (by 2 − 3a) Worked Example 19: Factorising binomials with a common bracket Question: Factorise completely: 3a(a − 4) − 7(a − 4) Answer Step 1 : bracket (a − 4) is the common factor 3a(a − 4) − 7(a − 4) = (a − 4)(3a − 7) Worked Example 20: Factorising using a switch around in brackets Question: Factorise 5(a − 2) − b(2 − a) Answer Step 1 : Note that (2 − a) = −(a − 2) 5(a − 2) − b(2 − a) = 5(a − 2) − [−b(a − 2)] = 5(a − 2) + b(a − 2) = (a − 2)(5 + b) 73 9.3 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 Exercise: Recap 1. Find the products of: (a) 2y(y + 4) (d) (y + 8)(y + 4) (b) (y + 5)(y + 2) (e) (2y + 9)(3y + 1) (c) (y + 2)(2y + 1) (f) (3y − 2)(y + 6) (b) 20a − 10 (e) 16k 2 − 4k (h) −2ab − 8a (k) 12k 2 j + 24k 2 j 2 (n) a(a − 1) − 5(a − 1) (q) 3b(b − 4) − 7(4 − b) (c) 18ab − 3bc (f) 3a2 + 6a − 18 (i) 24kj − 16k 2 j (l) 72b2q − 18b3 q 2 (o) bm(b+4)−6m(b+4) (r) a2 b2 c2 − 1 2. Factorise: (a) (b) (c) (d) (e) 2l + 2w 12x + 32y 6x2 + 2x + 10x3 2xy 2 + xy 2 z + 3xy −2ab2 − 4a2 b 3. Factorise completely: (a) 7a + 4 (d) 12kj + 18kq (g) −6a − 24 (j) −a2 b − b2 a (m) 4(y − 3) + k(3 − y) (p) a2 (a + 7) + a(a + 7) 9.3 More Products We have seen how to multiply two binomials in section 9.2.2. In this section we learn how to multiply a binomial (expression with two terms) by a trinomial (expression with three terms). Fortunately, we use the same methods we used to multiply two binomials to multiply a binomial and a trinomial. For example, multiply 2x + 1 by x2 + 2x + 1. = (2x + 1)(x2 + 2x + 1) 2x(x2 + 2x + 1) + 1(x2 + 2x + 1) (apply distributive law) = = [2x(x2 ) + 2x(2x) + 2x(1)] + [1(x2 ) + 1(2x) + 1(1)] 4x3 + 4x2 + 2x + x2 + 2x + 1 (expand the brackets) = = 4x3 + (4x2 + x2 ) + (2x + 2x) + 1 (group like terms to simplify) 4x3 + 5x2 + 4x + 1 (simplify to get final answer) Important: Multiplication of Binomial with Trinomial If the binomial is A + B and the trinomial is C + D + E, then the very first step is to apply the distributive law: (A + B)(C + D + E) = A(C + D + E) + B(C + D + E) If you remember this, you will never go wrong! 74 (9.4) CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 Worked Example 21: Multiplication of Binomial with Trinomial Question: Multiply x − 1 with x2 − 2x + 1. Answer Step 1 : Determine what is given and what is required We are given two expressions: a binomial, x − 1, and a trinomial, x2 − 2x + 1. We need to multiply them together. Step 2 : Determine how to approach the problem Apply the distributive law and then simplify the resulting expression. Step 3 : Solve the problem (x − 1)(x2 − 2x + 1) x(x2 − 2x + 1) − 1(x2 − 2x + 1) (apply distributive law) [x(x2 ) + x(−2x) + x(1)] + [−1(x2 ) − 1(−2x) − 1(1)] = = x3 − 2x2 + x − x2 + 2x − 1 (expand the brackets) x3 + (−2x2 − x2 ) + (x + 2x) − 1 (group like terms to simplify) = = x3 − 3x2 + 3x − 1 = (simplify to get final answer) Step 4 : Write the final answer The product of x − 1 and x2 − 2x + 1 is x3 − 3x2 + 3x − 1. Worked Example 22: Sum of Cubes Question: Find the product of x + y and x2 − xy + y 2 . Answer Step 1 : Determine what is given and what is required We are given two expressions: a binomial, x + y, and a trinomial, x2 − xy + y 2 . We need to multiply them together. Step 2 : Determine how to approach the problem Apply the distributive law and then simplify the resulting expression. Step 3 : Solve the problem (x + y)(x2 − xy + y 2 ) = = x(x2 − xy + y 2 ) + y(x2 − xy + y 2 ) (apply distributive law) [x(x2 ) + x(−xy) + x(y 2 )] + [y(x2 ) + y(−xy) + y(y 2 )] = x3 − x2 y + xy 2 + yx2 − xy 2 + y 3 = = 3 2 2 2 (expand the brackets) 2 x + (−x y + yx ) + (xy − xy ) + y 3 x3 + y 3 (simplify to get final answer) Step 4 : Write the final answer The product of x + y and x2 − xy + y 2 is x3 + y 3 . 75 (group like terms to simplify) 9.3 9.4 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 Important: We have seen that: (x + y)(x2 − xy + y 2 ) = x3 + y 3 This is known as a sum of cubes. Activity :: Investigation : Difference of Cubes Show that the difference of cubes (x3 − y 3 ) is given by the product of x − y and 2 x + xy + y 2 . Exercise: Products 1. Find the products of: (a) (−2y 2 − 4y + 11)(5y − 12) (c) (4y 2 + 12y + 10)(−9y 2 + 8y + 2) (e) (10y 5 + 3)(−2y 2 − 11y + 2) (g) (−10)(2y 2 + 8y + 3) (i) (6y 7 − 8y 2 + 7)(−4y − 3)(−6y 2 − 7y − 11) (k) (8y 5 + 3y 4 + 2y 3 )(5y + 10)(12y 2 + 6y + 6) (m) (4y 3 + 5y 2 − 12y)(−12y − 2)(7y 2 − 9y + 12) (o) (9)(8y 2 − 2y + 3) (q) (−6y 4 + 11y 2 + 3y)(10y + 4)(4y − 4) (s) (−11y 5 + 11y 4 + 11)(9y 3 − 7y 2 − 4y + 6) (b) (−11y + 3)(−10y 2 − 7y − 9) (d) (7y 2 − 6y − 8)(−2y + 2) (f) (−12y − 3)(12y 2 − 11y + 3) (h) (2y 6 + 3y 5 )(−5y − 12) (j) (−9y 2 + 11y + 2)(8y 2 + 6y − 7) (l) (−7y + 11)(−12y + 3) (n) (7y + 3)(7y 2 + 3y + 10) (p) (−12y + 12)(4y 2 − 11y + 11) (r) (−3y 6 − 6y 3 )(11y − 6)(10y − 10) (t) (−3y + 8)(−4y 3 + 8y 2 − 2y + 12) 2. Remove the brackets and simplify:(2h + 3)(4h2 − 6h + 9) 9.4 Factorising a Quadratic Finding the factors of a quadratic is quite easy, and some are easier than others. The simplest quadratic has the form ax2 , which factorises to (x)(ax). For example, 25x2 factorises to (5x)(5x) and 2x2 factorises to (2x)(x). The second simplest quadratic is of the form ax2 + bx. We can see here that x is a common factor of both terms. Therefore, ax2 + bx factorises to x(ax + b). For example, 8y 2 + 4y factorises to 4y(2y + 1). The third simplest quadratic is made up of the difference of squares. We know that: (a + b)(a − b) = a2 − b2 . This is true for any values of a and b, and more importantly since it is an equality, we can also write: a2 − b2 = (a + b)(a − b). This means that if we ever come across a quadratic that is made up of a difference of squares, we can immediately write down what the factors are. 76 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 9.4 Worked Example 23: Difference of Squares Question: Find the factors of 9x2 − 25. Answer Step 1 : Examine the quadratic We see that the quadratic is a difference of squares because: (3x)2 = 9x2 and 52 = 25. Step 2 : Write the quadratic as the difference of squares 9x2 − 25 = (3x)2 − 52 Step 3 : Write the factors (3x)2 − 52 = (3x − 5)(3x + 5) Step 4 : Write the final answer The factors of 9x2 − 25 are (3x − 5)(3x + 5). The three types of quadratic that we have seen are very simple to factorise. However, many quadratics do not fall into these categories, and we need a more general method to factorise quadratics like x2 − x − 2? We can learn about how to factorise quadratics by looking at how two binomials are multiplied to get a quadratic. For example, (x + 2)(x + 3) is multiplied out as: (x + 2)(x + 3) = x(x + 3) + 2(x + 3) = (x)(x) + 3x + 2x + (2)(3) = x2 + 5x + 6. We see that the x2 term in the quadratic is the product of the x-terms in each bracket. Similarly, the 6 in the quadratic is the product of the 2 and 3 the brackets. Finally, the middle term is the sum of two terms. So, how do we use this information to factorise the quadratic? Let us start with factorising x2 + 5x+ 6 and see if we can decide upon some general rules. Firstly, write down two brackets with an x in each bracket and space for the remaining terms. ( x )( x ) Next decide upon the factors of 6. Since the 6 is positive, these are: Factors of 6 1 6 2 3 -1 -6 -2 -3 Therefore, we have four possibilities: Option 1 (x + 1)(x + 6) Option 2 (x − 1)(x − 6) Option 3 (x + 2)(x + 3) 77 Option 4 (x − 2)(x − 3) 9.4 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 Next we expand each set of brackets to see which option gives us the correct middle term. Option 1 (x + 1)(x + 6) x2 + 7x + 6 Option 2 (x − 1)(x − 6) x2 − 7x + 6 Option 3 (x + 2)(x + 3) x2 + 5x + 6 Option 4 (x − 2)(x − 3) x2 − 5x + 6 We see that Option 3 (x+2)(x+3) is the correct solution. As you have seen that the process of factorising a quadratic is mostly trial and error, however the is some information that can be used to simplify the process. Method: Factorising a Quadratic 1. First divide the entire equation by any common factor of the coefficients, so as to obtain an equation of the form ax2 + bx + c = 0 where a, b and c have no common factors and a is positive. 2. Write down two brackets with an x in each bracket and space for the remaining terms. ( x )( x ) (9.5) 3. Write down a set of factors for a and c. 4. Write down a set of options for the possible factors for the quadratic using the factors of a and c. 5. Expand all options to see which one gives you the correct answer. There are some tips that you can keep in mind: • If c is positive, then the factors of c must be either both positive or both negative. The factors are both negative if b is negative, and are both positive if b is positive. If c is negative, it means only one of the factors of c is negative, the other one being positive. • Once you get an answer, multiply out your brackets again just to make sure it really works. Worked Example 24: Factorising a Quadratic Question: Find the factors of 3x2 + 2x − 1. Answer Step 1 : Check whether the quadratic is in the form ax2 + bx + c = 0 with a positive. The quadratic is in the required form. Step 2 : Write down two brackets with an x in each bracket and space for the remaining terms. ( x )( x ) (9.6) Write down a set of factors for a and c. The possible factors for a are: (1,3). The possible factors for c are: (-1,1) or (1,-1). Write down a set of options for the possible factors for the quadratic using the factors of a and c. Therefore, there are two possible options. Option 1 (x − 1)(3x + 1) 3x2 − 2x − 1 78 Option 2 (x + 1)(3x − 1) 3x2 + 2x − 1 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 9.5 Step 3 : Check your answer (x + 1)(3x − 1) = = = = x(3x − 1) + 1(3x − 1) (x)(3x) + (x)(−1) + (1)(3x) + (1)(−1) 3x2 − x + 3x − 1 x2 + 2x − 1. Step 4 : Write the final answer The factors of 3x2 + 2x − 1 are (x + 1) and (3x − 1). Exercise: Factorising a Trinomial 1. Factorise the following: (a) x2 + 8x + 15 (d) x2 + 9x + 14 (b) x2 + 10x + 24 (e) x2 + 15x + 36 (c) x2 + 9x + 8 (f) x2 + 13x + 36 2. Factorise the following: (a) x2 − 2x − 15 (b) x2 + 2x − 3 (c) x2 + 2x − 8 (d) x2 + x − 20 (e) x2 − x − 20 3. Find the factors for the following quadratic expressions: (a) 2x2 + 11x + 5 (b) 3x2 + 19x + 6 (c) 6x2 + 7x + 2 (d) 12x2 + 7x + 1 (e) 8x2 + 6x + 1 4. Find the factors for the following trinomials: (a) 3x2 + 17x − 6 (b) 7x2 − 6x − 1 (c) 8x2 − 6x + 1 (d) 2x2 − 5x − 3 9.5 Factorisation by Grouping One other method of factorisation involves the use of common factors. We know that the factors of 3x + 3 are 3 and (x + 1). Similarly, the factors of 2x2 + 2x are 2x and (x + 1). Therefore, if we have an expression: 2x2 + 2x + 3x + 3 then we can factorise as: 2x(x + 1) + 3(x + 1). You can see that there is another common factor: x + 1. Therefore, we can now write: (x + 1)(2x + 3). We get this by taking out the x + 1 and see what is left over. We have a +2x from the first term and a +3 from the second term. This is called factorisation by grouping. 79 9.6 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 Worked Example 25: Factorisation by Grouping Question: Find the factors of 7x + 14y + bx + 2by by grouping Answer Step 1 : Determine if there are common factors to all terms There are no factors that are common to all terms. Step 2 : Determine if there are factors in common between some terms 7 is a common factor of the first two terms and b is a common factor of the second two terms. Step 3 : Re-write expression taking the factors into account 7x + 14y + bx + 2by = 7(x + 2y) + b(x + 2y) Step 4 : Determine if there are more common factors x + 2y is a common factor. Step 5 : Re-write expression taking the factors into account 7(x + 2y) + b(x + 2y) = (x + 2y)(7 + b) Step 6 : Write the final answer The factors of 7x + 14y + bx + 2by are (7 + b) and (x + 2y). Exercise: Factorisation by Grouping 1. Factorise by grouping: 6x + 9 + 2ax + 3 2. Factorise by grouping: x2 − 6x + 5x − 30 3. Factorise by grouping: 5x + 10y − ax − 2ay 4. Factorise by grouping: a2 − 2a − ax + 2x 5. Factorise by grouping: 5xy − 3y + 10x − 6 9.6 Simplification of Fractions In some cases of simplifying an algebraic expression, the expression will be a fraction. For example, x2 + 3x x+3 has a quadratic in the numerator and a binomial in the denominator. You can apply the different factorisation methods to simplify the expression. = = x2 + 3x x+3 x(x + 3) x+3 x provided x 6= −3 80 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 Worked Example 26: Simplification of Fractions Question: Simplify: 2x−b+x−ab ax2 −abx Answer Step 1 : Factorise numerator and denominator Use grouping for numerator and common factor for denominator in this example. (ax − ab) + (x − b) ax2 − abx a(x − b) + (x − b) ax(x − b) (x − b)(a + 1) ax(x − b) = = = Step 2 : Cancel out same factors The simplified answer is: = a+1 ax Worked Example 27: Simplification of Fractions 2 2 +x Question: Simplify: x x−x−2 ÷ xx2 +2x 2 −4 Answer Step 1 : Factorise numerators and denominators = (x + 1)(x − 2) x(x + 1) ÷ (x + 2)(x − 2) x(x + 2) Step 2 : Multiply by factorised reciprocal = (x + 1)(x − 2) x(x + 2) × (x + 2)(x − 2) x(x + 1) Step 3 : Cancel out same factors The simplified answer is = Exercise: Simplification of Fractions 1. Simplify: 81 1 9.6 9.7 CHAPTER 9. PRODUCTS AND FACTORS - GRADE 10 2a+10 4 a2 −4a a−4 9a+27 9a+18 2 y−8xy (h) 16x12x−6 3a+9 (j) 14 ÷ 7a+21 a+3 12p2 (l) 3xp+4p ÷ 8p 3x+4 (y) 24a−8 ÷ 9a−3 12 6 2 +pq (p) p 7p ÷ 8p+8q 21q 2 a2 (r) f fa−f −a 3a 15 5a+20 a+4 3a2 −9a 2a−6 (g) 6ab+2a 2b (i) 4xyp−8xp 12xy 2 −5a (k) a2a+10 ÷ 3a+15 4a 2 +8x 16 ÷ 6x 12 (x) 2xp+4x 2 ÷ 2a+4 (o) a +2a 5 20 6b2 ÷ (q) 5ab−15b 4a−12 a+b (b) (d) (f) (a) (c) (e) 2. Simplify: 9.7 x2 −1 3 × 1 x−1 − 1 2 End of Chapter Exercises 1. Factorise: (a) a2 − 9 (d) 16b6 − 25a2 (g) 16ba4 − 81b (j) 2a2 − 12ab + 18b2 (m) 125a3 + b3 (p) 64b3 + 1 (b) m2 − 36 (e) m2 − (1/9) (h) a2 − 10a + 25 (k) −4b2 − 144b8 + 48b5 (n) 128b7 − 250ba6 (q) 5a3 − 40c3 (c) 9b2 − 81 (f) 5 − 5a2 b6 (i) 16b2 + 56b + 49 (l) a3 − 27 (o) c3 + 27 (r) 2b4 − 128b 2. Show that (2x − 1)2 − (x − 3)2 can be simplified to (x + 2)(3x − 4) 3. What must be added to x2 − x + 4 to make it equal to (x + 2)2 82 Chapter 10 Equations and Inequalities - Grade 10 10.1 Strategy for Solving Equations This chapter is all about solving different types of equations for one or two variables. In general, we want to get the unknown variable alone on the left hand side of the equation with all the constants on the right hand side of the equation. For example, in the equation x − 1 = 0, we want to be able to write the equation as x = 1. As we saw in section 2.9 (page 13), an equation is like a set of weighing scales, that must always be balanced. When we solve equations, we need to keep in mind that what is done to one side must be done to the other. Method: Rearranging Equations You can add, subtract, multiply or divide both sides of an equation by any number you want, as long as you always do it to both sides. For example, in the equation x + 5 − 1 = −6, we want to get x alone on the left hand side of the equation. This means we need to subtract 5 and add 1 on the left hand side. However, because we need to keep the equation balanced, we also need to subtract 5 and add 1 on the right hand side. x+5−1 = −6 x+0+0 x = −11 + 1 = −10 x+5−5−1+1 = −6 − 5 + 1 In another example, 23 x = 8, we must divide by 2 and multiply by 3 on the left hand side in order to get x alone. However, in order to keep the equation balanced, we must also divide by 2 and multiply by 3 on the right hand side. 2 x 3 2 x÷2×3 3 2 3 × ×x 2 3 1×1×x x = 8 = 8÷2×3 8×3 2 = 12 = 12 = These are the basic rules to apply when simplifying any equation. In most cases, these rules have to be applied more than once, before we have the unknown variable on the left hand side 83 10.2 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 of the equation. We are now ready to solve some equations! Important: The following must also be kept in mind: 1. Division by 0 is undefined. 2. If x y = 0, then x = 0 and y 6= 0, because division by 0 is undefined. Activity :: Investigation : Strategy for Solving Equations In the following, identify what is wrong. 4x − 8 4(x − 2) 4(x − 2) (x − 2) 4 10.2 = = = = 3(x − 2) 3(x − 2) 3(x − 2) (x − 2) 3 Solving Linear Equations The simplest equation to solve is a linear equation. A linear equation is an equation where the power on the variable(letter, e.g. x) is 1(one). The following are examples of linear equations. 2x + 2 2−x 3x + 1 4 x−6 3 = 1 = 2 = 7x + 2 In this section, we will learn how to find the value of the variable that makes both sides of the linear equation true. For example, what value of x makes both sides of the very simple equation, x + 1 = 1 true. Since the highest power on the variable is one(1) in a linear equation, there is at most one solution or root for the equation. This section relies on all the methods we have already discussed: multiplying out expressions, grouping terms and factorisation. Make sure that you are comfortable with these methods, before trying out the work in the rest of this chapter. 2x + 2 = 2x = 2x = 1 1 − 2 (like terms together) −1 (simplified as much a possible) Now we see that 2x = −1. This means if we divide both sides by 2, we will get: x=− 84 1 2 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.2 If we substitute x = − 21 , back into the original equation, we get: = = = 2x + 2 1 2(− ) + 2 2 −1 + 2 1 That is all that there is to solving linear equations. Important: Solving Equations When you have found the solution to an equation, substitute the solution into the original equation, to check your answer. Method: Solving Equations The general steps to solve equations are: 1. Expand(Remove) all brackets. 2. ”Move” all terms with the variable to the left hand side of equation, and all constant terms (the numbers) to the right hand side of the equal to-sign. Bearing in mind that the sign of the terms will chance(from (+) to (-) or vice versa, as they ”cross over” the equal to sign. 3. Group all like terms together and simplify as much as possible. 4. Factorise if necessary. 5. Find the solution. 6. Substitute solution into original equation to check answer. Worked Example 28: Solving Linear Equations Question: Solve for x: 4 − x = 4 Answer Step 1 : Determine what is given and what is required We are given 4 − x = 4 and are required to solve for x. Step 2 : Determine how to approach the problem Since there are no brackets, we can start with grouping like terms and then simplifying. Step 3 : Solve the problem 4−x = −x = −x = −x = −x = ∴ x = 4 4 − 4 (move all constant terms (numbers) to the RHS (right hand side)) 0 (group like terms together) 0 (simplify grouped terms) 0 0 Step 4 : Check the answer 85 10.2 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Substitute solution into original equation: 4−0=4 4=4 Since both sides are equal, the answer is correct. Step 5 : Write the Final Answer The solution of 4 − x = 4 is x = 0. Worked Example 29: Solving Linear Equations Question: Solve for x: 4(2x − 9) − 4x = 4 − 6x Answer Step 1 : Determine what is given and what is required We are given 4(2x − 9) − 4x = 4 − 6x and are required to solve for x. Step 2 : Determine how to approach the problem We start with expanding the brackets, then grouping like terms and then simplifying. Step 3 : Solve the problem 4(2x − 9) − 4x = 4 − 6x 8x − 36 − 4x = 4 − 6x (expand the brackets) 8x − 4x + 6x = 4 + 36 (move all terms with x to the LHS and all constant terms to the RHS of the =) (8x − 4x + 6x) = (4 + 36) (group like terms together) 10x = 40 10 40 x = 10 10 x = 4 (simplify grouped terms) (divide both sides by 10) Step 4 : Check the answer Substitute solution into original equation: 4(2(4) − 9) − 4(4) = 4(8 − 9) − 16 = 4(−1) − 16 = −4 − 16 = −20 = 4 − 6(4) 4 − 24 −20 −20 −20 Since both sides are equal to −20, the answer is correct. Step 5 : Write the Final Answer The solution of 4(2x − 9) − 4x = 4 − 6x is x = 4. Worked Example 30: Solving Linear Equations Question: Solve for x: Answer 2−x 3x+1 =2 86 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.2 Step 1 : Determine what is given and what is required 2−x We are given 3x+1 = 2 and are required to solve for x. Step 2 : Determine how to approach the problem Since there is a denominator of (3x+1), we can start by multiplying both sides of the equation by (3x+1). But because division by 0 is not permissible, there is a restriction on a value for x. (x 6= −1 3 ) Step 3 : Solve the problem 2−x 3x + 1 (2 − x) 2−x = 2 = 2(3x + 1) = 6x + 2 (remove/expand brackets) −x − 6x = 2 − 2 −7x = 0 theref ore x x (move all terms containing x to the LHS and all constant terms (numbers) to the RHS.) (simplify grouped terms) = 0 ÷ (−7) = 0 zero divide by any number is 0 Step 4 : Check the answer Substitute solution into original equation: 2 − (0) 3(0) + 1 2 1 = 2 = 2 Since both sides are equal to 2, the answer is correct. Step 5 : Write the Final Answer 2−x = 2 is x = 0. The solution of 3x+1 Worked Example 31: Solving Linear Equations Question: Solve for x: 34 x − 6 = 7x + 2 Answer Step 1 : Determine what is given and what is required We are given 34 x − 6 = 7x + 2 and are required to solve for x. Step 2 : Determine how to approach the problem We start with multiplying each of the terms in the equation by 3, then grouping like terms and then simplifying. Step 3 : Solve the problem 4 x−6 = 3 4x − 18 = 4x − 21x = −17x = −17 x = −17 x = 7x + 2 21x + 6 (each term is multiplied by 3 6 + 18 (move all terms with x to the LHS and all constant terms to the RHS of the =) 24 24 −17 −24 17 (simplify grouped terms) (divide both sides by -17) 87 10.2 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Step 4 : Check the answer Substitute solution into original equation: 4 −24 × −6 3 17 4 × (−8) −6 (17) (−32) −6 17 −32 − 102 17 −134 17 Since both sides are equal to −134 17 , = = = = = −24 +2 17 7 × (−24) +2 17 −168 +2 17 (−168) + 34 17 −134 17 7× the answer is correct. Step 5 : Write the Final Answer The solution of 34 x − 6 = 7x + 2 is, x = −24 17 . Exercise: Solving Linear Equations 1. Solve for y: 2y − 3 = 7 2. Solve for w: −3w = 0 3. Solve for z: 4z = 16 4. Solve for t: 12t + 0 = 144 5. Solve for x: 7 + 5x = 62 6. Solve for y: 55 = 5y + 3 4 7. Solve for z: 5z = 3z + 45 8. Solve for a: 23a − 12 = 6 + 2a 9. Solve for b: 12 − 6b + 34b = 2b − 24 − 64 10. Solve for c: 6c + 3c = 4 − 5(2c − 3). 11. Solve for p: 18 − 2p = p + 9 12. Solve for q: 4 q = 16 24 13. Solve for q: 4 1 = q 2 14. Solve for r: −(−16 − r) = 13r − 1 15. Solve for d: 6d − 2 + 2d = −2 + 4d + 8 16. Solve for f : 3f − 10 = 10 17. Solve for v: 3v + 16 = 4v − 10 18. Solve for k: 10k + 5 + 0 = −2k + −3k + 80 19. Solve for j: 8(j − 4) = 5(j − 4) 20. Solve for m: 6 = 6(m + 7) + 5m 88 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.3 10.3 Solving Quadratic Equations A quadratic equation is an equation where the power on the variable is at most 2. The following are examples of quadratic equations. 2x2 + 2x = 1 2−x = 2x 3x + 1 4 x − 6 = 7x2 + 2 3 Quadratic equations differ from linear equations by the fact that a linear equation only has one solution, while a quadratic equation has at most two solutions. There are some special situations when a quadratic equation only has one solution. We solve quadratic equations by factorisation, that is writing the quadratic as a product of two expressions in brackets. For example, we know that: (x + 1)(2x − 3) = 2x2 − x − 3. In order to solve: 2x2 − x − 3 = 0 we need to be able to write 2x2 − x − 3 as (x + 1)(2x − 3), which we already know how to do. Activity :: Investigation : Factorising a Quadratic Factorise the following quadratic expressions: 1. x + x2 2. x2 + 1 + 2x 3. x2 − 4x + 5 4. 16x2 − 9 5. 4x2 + 4x + 1 Being able to factorise a quadratic means that you are one step away from solving a quadratic equation. For example, x2 − 3x − 2 = 0 can be written as (x − 1)(x − 2) = 0. This means that both x − 1 = 0 and x − 2 = 0, which gives x = 1 and x = 2 as the two solutions to the quadratic equation x2 − 3x − 2 = 0. Method: Solving Quadratic Equations 1. First divide the entire equation by any common factor of the coefficients, so as to obtain an equation of the form ax2 + bx + c = 0 where a, b and c have no common factors. For example, 2x2 + 4x + 2 = 0 can be written as x2 + 2x + 1 = 0 by dividing by 2. 2. Write ax2 + bx + c in terms of its factors (rx + s)(ux + v). This means (rx + s)(ux + v) = 0. 3. Once writing the equation in the form (rx + s)(ux + v) = 0, it then follows that the two solutions are x = − rs or x = − uv . Extension: Solutions of Quadratic Equations There are two solutions to a quadratic equation, because any one of the values can solve the equation. 89 10.3 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Worked Example 32: Solving Quadratic Equations Question: Solve for x: 3x2 + 2x − 1 = 0 Answer Step 1 : Find the factors of 3x2 + 2x − 1 As we have seen the factors of 3x2 + 2x − 1 are (x + 1) and (3x − 1). Step 2 : Write the equation with the factors (x + 1)(3x − 1) = 0 Step 3 : Determine the two solutions We have x+1=0 or 3x − 1 = 0 1 3. Therefore, x = −1 or x = Step 4 : Write the final answer 3x2 + 2x − 1 = 0 for x = −1 or x = 13 . Worked Example 33: Solving Quadratic Equations √ Question: Solve for x: x + 2 = x Answer Step 1 : Square both sides of the equation Both sides of the equation should be squared to remove the square root sign. x + 2 = x2 Step 2 : Write equation in the form ax2 + bx + c = 0 x2 (subtract x2 to both sides) = = 0 0 (divide both sides by -1) x2 − x + 2 = 0 x+2 = x + 2 − x2 −x − 2 + x2 Step 3 : Factorise the quadratic x2 − x + 2 The factors of x2 − x + 2 are (x − 2)(x + 1). Step 4 : Write the equation with the factors (x − 2)(x + 1) = 0 Step 5 : Determine the two solutions We have x+1=0 or x−2=0 90 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.3 Therefore, x = −1 or x = 2. Step 6 : Check whether solutions are valid √ Substitute x = −1into the original equation x + 2 = x: LHS p (−1) + 2 √ 1 1 = = = but RHS = (−1) Therefore LHS6=RHS Therefore x 6= −1 √ Now substitute x = 2 into original equation x + 2 = x: LHS √ 2+2 = √ 4 = = 2 and RHS = 2 Therefore LHS = RHS Therefore x = 2 is the only valid solution Step 7 : Write the final answer √ x + 2 = x for x = 2 only. Worked Example 34: Solving Quadratic Equations Question: Solve the equation: x2 + 3x − 4 = 0. Answer Step 1 : Check if the equation is in the form ax2 + bx + c = 0 The equation is in the required form, with a = 1. Step 2 : Factorise the quadratic You need the factors of 1 and 4 so that the middle term is +3 So the factors are: (x − 1)(x + 4) Step 3 : Solve the quadratic equation x2 + 3x − 4 = (x − 1)(x + 4) = 0 Therefore x = 1 or x = −4. Step 4 : Write the final solution Therefore the solutions are x = 1 or x = −4. 91 (10.1) 10.3 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Worked Example 35: Solving Quadratic Equations Question: Find the roots of the quadratic equation 0 = −2x2 + 4x − 2. Answer Step 1 : Determine whether the equation is in the form ax2 + bx + c = 0, with no common factors. There is a common factor: -2. Therefore, divide both sides of the equation by -2. −2x2 + 4x − 2 = x2 − 2x + 1 = 0 0 Step 2 : Factorise x2 − 2x + 1 The middle term is negative. Therefore, the factors are (x − 1)(x − 1) If we multiply out (x − 1)(x − 1), we get x2 − 2x + 1. Step 3 : Solve the quadratic equation x2 − 2x + 1 = (x − 1)(x − 1) = 0 In this case, the quadratic is a perfect square, so there is only one solution for x: x = 1. Step 4 : Write the final solution The root of 0 = −2x2 + 4x − 2 is x = 1. Exercise: Solving Quadratic Equations 1. Solve for x: (3x + 2)(3x − 4) = 0 2. Solve for a: (5a − 9)(a + 6) = 0 3. Solve for x: (2x + 3)(2x − 3) = 0 4. Solve for x: (2x + 1)(2x − 9) = 0 5. Solve for x: (2x − 3)(2x − 3) = 0 6. Solve for x: 20x + 25x2 = 0 7. Solve for a: 4a2 − 17a − 77 = 0 8. Solve for x: 2x2 − 5x − 12 = 0 9. Solve for b: −75b2 + 290b − 240 = 0 10. Solve for y: 2y = 13 y 2 − 3y + 14 32 11. Solve for θ: θ2 − 4θ = −4 12. Solve for q: −q 2 + 4q − 6 = 4q 2 − 5q + 3 13. Solve for t: t2 = 3t 14. Solve for w: 3w2 + 10w − 25 = 0 15. Solve for v: v 2 − v + 3 16. Solve for x: x2 − 4x + 4 = 0 17. Solve for t: t2 − 6t = 7 18. Solve for x: 14x2 + 5x = 6 19. Solve for t: 2t2 − 2t = 12 20. Solve for y: 3y 2 + 2y − 6 = y 2 − y + 2 92 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.4 10.4 Exponential Equations of the form ka(x+p) = m examples solved by trial and error) Exponential equations generally have the unknown variable as the power. The following are examples of exponential equations: 2x 2 3x+1 4 −6 3 −x = 1 = 2 = 7x + 2 You should already be familiar with exponential notation. Solving exponential equations are simple, if we remember how to apply the laws of exponentials. Activity :: Investigation : Solving Exponential Equations Solve the following equations by completing the table: 2x = 2 -3 -2 -1 x 0 1 2 3 -3 -2 -1 x 0 1 2 3 2x 3x = 9 3x 2x+1 = 8 -3 -2 -1 x 0 1 2 3 2x+1 10.4.1 Algebraic Solution Definition: Equality for Exponential Functions If a is a positive number such that a > 0, then: ax = ay if and only if: x=y . This means that if we can write all terms in an equation with the same base, we can solve the exponential equations by equating the indices. For example take the equation 3x+1 = 9. This can be written as: 3x+1 = 32 . Since the bases are equal (to 3), we know that the exponents must also be equal. Therefore we can write: x + 1 = 2. This gives: x = 1. 93 10.4 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Method: Solving Exponential Equations Try to write all terms with the same base. Activity :: Investigation : Exponential Numbers Write the following with the same base. The base is the first in the list. For example, in the list 2, 4, 8, the base is two and we can write 4 as 22 . 1. 2,4,8,16,32,64,128,512,1024 2. 3,9,27,81,243 3. 5,25,125,625 4. 13,169 5. 2x, 4x2 , 8x3 , 49x8 Worked Example 36: Solving Exponential Equations Question: Solve for x: 2x = 2 Answer Step 1 : Try to write all terms with the same base. All terms are written with the same base. 2x = 21 Step 2 : Equate the indices x=1 Step 3 : Check your answer 2x = 2(1) = 21 Since both sides are equal, the answer is correct. Step 4 : Write the final answer x=1 x is the solution to 2 = 2. Worked Example 37: Solving Exponential Equations Question: Solve: 2x+4 = 42x 94 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.4 Answer Step 1 : Try to write all terms with the same base. 2x+4 = 42x 2x+4 = 22(2x) 2x+4 = 24x Step 2 : Equate the indices x + 4 = 4x Step 3 : Solve for x x+4 = 4x x − 4x = −4 −3x = −4 −4 x = −3 4 x = 3 Step 4 : Check your answer LHS = = RHS 2x+4 4 2( 3 +4) 16 = 23 = = (216 ) 3 42x = = 42( 3 ) 8 43 = (48 ) 3 = ((22 )8 ) 3 = (216 ) 3 = LHS 1 4 1 1 1 Since both sides are equal, the answer is correct. Step 5 : Write the final answer x= 4 3 is the solution to 2x+4 = 42x . Exercise: Solving Exponential Equations 1. Solve the following exponential equations. a. 2x+5 = 25 d. 65−x = 612 b. 32x+1 = 33 e. 64x+1 = 162x+5 2. Solve: 39x−2 = 27 95 c. 52x+2 = 53 f. 125x = 5 10.5 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 3. Solve for k: 81k+2 = 27k+4 4. The growth of an algae in a pond is can be modeled by the function f (t) = 2t . Find the value of t such that f (t) = 128? 5. Solve for x: 25(1−2x) = 54 6. Solve for x: 27x × 9x−2 = 1 10.5 Linear Inequalities graphically; Activity :: Investigation : Inequalities on a Number Line Represent the following on number lines: 1. x = 4 2. x < 4 3. x ≤ 4 4. x ≥ 4 5. x > 4 A linear inequality is similar to a linear equation and has the power on the variable is equal to 1. The following are examples of linear inequalities. 2x + 2 2−x 3x + 1 4 x−6 3 ≤ 1 ≥ 2 < 7x + 2 The methods used to solve linear inequalities are identical to those used to solve linear equations. The only difference occurs when there is a multiplication or a division that involves a minus sign. For example, we know that 8 > 6. If both sides of the inequality are divided by −2, −4 is not greater than −3. Therefore, the inequality must switch around, making −4 < −3. Important: When you divide or multiply both sides of an inequality by any number with a minus sign, the direction of the inequality changes. For example, if x < 1, then −x > −1. In order to compare am inequality to a normal equation, we shall solve an equation first. Solve 2x + 2 = 1. 2x + 2 = 1 2x = 1 − 2 2x = −1 1 x = − 2 If we represent this answer on a number line, we get 96 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 -3 x = − 12 b -1 0 -2 1 10.5 2 3 2 3 Now let us solve the inequality 2x + 2 ≤ 1. 2x + 2 ≤ 1 2x ≤ 1 − 2 2x ≤ −1 1 x ≤ − 2 If we represent this answer on a number line, we get x ≤ − 21 -3 -2 b -1 0 1 As you can see, for the equation, there is only a single value of x for which the equation is true. However, for the inequality, there is a range of values for which the inequality is true. This is the main difference between an equation and an inequality. Worked Example 38: Linear Inequalities Question: Solve for r: 6 − r > 2 Answer Step 1 : Move all constants to the RHS −r > 2 − 6 −r > −4 Step 2 : Multiply both sides by -1 When you multiply by a minus sign, the direction of the inequality changes. r<4 Step 3 : Represent answer graphically r<4 0 1 2 3 bc 4 5 Worked Example 39: Linear Inequalities Question: Solve for q: 4q + 3 < 2(q + 3) and represent solution on a number line. Answer Step 1 : Expand all brackets 4q + 3 4q + 3 < 2(q + 3) < 2q + 6 97 10.5 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Step 2 : Move all constants to the RHS and all unknowns to the LHS 4q + 3 4q − 2q 2q < < < 2q + 6 6−3 3 Step 3 : Solve inequality 2q q < 3 Divide both sides by 2 3 < 2 Step 4 : Represent answer graphically q< 0 3 2 1 c b 2 3 4 5 Worked Example 40: Compound Linear Inequalities Question: Solve for x: 5 ≤ x + 3 < 8 and represent solution on a number line. Answer Step 1 : Subtract 3 from Left, middle and right of inequalities 5−3≤ x+3−3 2≤ x <8−3 <5 Step 2 : Represent answer graphically 0 b 2 1 2≤x<5 3 4 bc 5 Exercise: Linear Inequalities 1. Solve for x and represent the solution graphically: (a) 3x + 4 > 5x + 8 (b) 3(x − 1) − 2 ≤ 6x + 4 2x−3 (c) x−7 3 > 2 (d) −4(x − 1) < x + 2 (e) 21 x + 13 (x − 1) ≥ 65 x − 1 3 2. Solve the following inequalities. Illustrate your answer on a number line if x is a real number. (a) −2 ≤ x − 1 < 3 98 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.6 (b) −5 < 2x − 3 ≤ 7 3. Solve for x: 7(3x + 2) − 5(2x − 3) > 7. Illustrate this answer on a number line. 10.6 Linear Simultaneous Equations Thus far, all equations that have been encountered have one unknown variable, that must be solved for. When two unknown variables need to be solved for, two equations are required and these equations are known as simultaneous equations. The solutions to the system of simultaneous equations, are the values of the unknown variables which satisfy the system of equations simultaneously, that means at the same time. In general, if there are n unknown variables, then n equations are required to obtain a solution for each of the n variables. An example of a system of simultaneous equations is: 2x + 2y = 1 2−x =2 3y + 1 10.6.1 (10.2) Finding solutions In order to find a numerical value for an unknown variable, one must have at least as many independent equations as variables. We solve simultaneous equations graphically and algebraically/ 10.6.2 Graphical Solution Simultaneous equations can also be solved graphically. If the graphs corresponding to each equation is drawn, then the solution to the system of simultaneous equations is the co-ordinate of the point at which both graphs intersect. x = 2y (10.3) y = 2x − 3 1 1 −1 2 y= 1 x 2 3 2x − −1 b y= −2 (2,1) 3 Draw the graphs of the two equations in (10.3). The intersection of the two graphs is (2,1). So the solution to the system of simultaneous equations in (10.3) is y = 1 and x = 2. 99 10.6 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 This can be shown algebraically as: x = ∴ y y − 4y −3y y = = = = Substitute into the first equation: x = = 2y 2(2y) − 3 −3 −3 1 2(1) 2 Worked Example 41: Simultaneous Equations Question: Solve the following system of simultaneous equations graphically. 4y + 3x = 4y − 19x = 100 12 Answer Step 1 : Draw the graphs corresponding to each equation. For the first equation: 4y + 3x = 4y = y = 100 100 − 3x 3 25 − x 4 and for the second equation: 4y − 19x = 4y = y = 12 19x + 12 19 x+3 4 = x 19 40 4y 4y + 3x = 100 12 − 30 20 10 −8 −6 −4 −2 2 4 6 8 Step 2 : Find the intersection of the graphs. The graphs intersect at (4,22). Step 3 : Write the solution of the system of simultaneous equations as given by the intersection of the graphs. x = y = 100 4 22 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.6.3 10.6 Solution by Substitution A common algebraic technique is the substitution method: try to solve one of the equations for one of the variables and substitute the result into the other equations, thereby reducing the number of equations and the number of variables by 1. Continue until you reach a single equation with a single variable, which (hopefully) can be solved; back substitution then yields the values for the other variables. In the example (??), we first solve the first equation for x: x= 1 −y 2 and substitute this result into the second equation: 2−x 3y + 1 2 − ( 21 − y) 3y + 1 1 2 − ( − y) 2 1 2− +y 2 = 2 = 2 = 2(3y + 1) = 6y + 2 y − 6y = −2 + −5y ∴ x 1 2 = y = − = = = = 1 +2 2 1 10 1 −y 2 1 1 − (− ) 2 10 6 10 3 5 The solution for the system of simultaneous equations (??) is: 3 5 x = y = − 1 10 Worked Example 42: Simultaneous Equations Question: Solve the following system of simultaneous equations: 4y + 3x = 100 4y − 19x = 12 Answer Step 1 : If the question, does not explicitly ask for a graphical solution, then the system of equations should be solved algebraically. 101 10.6 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Step 2 : Make x the subject of the first equation. 4y + 3x = 3x = x = 100 100 − 4y 100 − 4y 3 Step 3 : Substitute the value obtained for x into the second equation. 100 − 4y ) = 3 12y − 19(100 − 4y) = 4y − 19( 12 36 12y − 1900 + 76y 88y = = 36 1936 y = 22 Step 4 : Substitute into the equation for x. x = = = = 100 − 4(22) 3 100 − 88 3 12 3 4 Step 5 : Substitute the values for x and y into both equations to check the solution. 4(22) + 3(4) = 88 + 12 = 100 X 4(22) − 19(4) = 88 − 76 = 12 X Worked Example 43: Bicycles and Tricycles Question: A shop sells bicycles and tricycles. In total there are 7 cycles and 19 wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels. Answer Step 1 : Identify what is required The number of bicycles and the number of tricycles are required. Step 2 : Set up the necessary equations If b is the number of bicycles and t is the number of tricycles, then: b+t = 2b + 3t = 7 19 Step 3 : Solve the system of simultaneous equations using substitution. 102 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 b = Into second equation: 2(7 − t) + 3t 14 − 2t + 3t = = t Into first equation: b = = = 10.7 7−t 19 19 5 7−5 2 Step 4 : Check solution by substituting into original system of equations. 2+5 = 2(2) + 3(5) = 4 + 15 = 7 X 19 X Exercise: Simultaneous Equations 1. Solve graphically and confirm your answer algebraically: 3a − 2b7 = 0 , a − 4b + 1 = 0 2. Solve algebraically: 15c + 11d − 132 = 0, 2c + 3d − 59 = 0 3. Solve algebraically: −18e − 18 + 3f = 0, e − 4f + 47 = 0 4. Solve graphically: x + 2y = 7, x + y = 0 10.7 Mathematical Models 10.7.1 Introduction Tom and Jane are friends. Tom picked up Jane’s Physics test paper, but will not tell Jane what her marks are. He knows that Jane hates maths so he decided to tease her. Tom says: “I have 2 marks more than you do and the sum of both our marks is equal to 14. How much did we get?” Let’s help Jane find out what her marks are. We have two unknowns, Tom’s mark (which we shall call t) and Jane’s mark (which we shall call j). Tom has 2 more marks than Jane. Therefore, t=j+2 Also, both marks add up to 14. Therefore, t + j = 14 The two equations make up a set of linear (because the highest power is one) simultaneous 103 10.7 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 equations, which we know how to solve! Substitute for t in the second equation to get: t+j = 14 j+2+j 2j + 2 = 14 = 14 2(j + 1) = 14 j+1 = 7 j = 7−1 = 6 Then, t = j+2 = 6+2 = 8 So, we see that Tom scored 8 on his test and Jane scored 6. This problem is an example of a simple mathematical model. We took a problem and we able to write a set of equations that represented the problem, mathematically. The solution of the equations then gave the solution to the problem. 10.7.2 Problem Solving Strategy The purpose of this section is to teach you the skills that you need to be able to take a problem and formulate it mathematically, in order to solve it. The general steps to follow are: 1. Read ALL of it ! 2. Find out what is requested. 3. Let the requested be a variable e.g. x. 4. Rewrite the information given in terms of x. That is, translate the words into algebraic language. This is the reponse 5. Set up an equation (i.e. a mathematical sentence or model) to solve the required variable. 6. Solve the equation algebraically to find the result. Important: Follow the three R’s and solve the problem... Request - Response - Result 10.7.3 Application of Mathematical Modelling Worked Example 44: Mathematical Modelling: One variable Question: A fruit shake costs R2,00 more than a chocolate milkshake. If three fruit shakes and 5 chocolate milkshakes cost R78,00, determine the individual prices. Answer Step 1 : Summarise the information in a table Fruit Chocolate Price number x+2 3 x 5 104 Total 3(x + 2) 5x CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.7 Step 2 : Set up an algebraic equation 3(x + 2) + 5x = 78 Step 3 : Solve the equation 3x + 6 + 5x = 78 8x = 72 x = 9 Step 4 : Present the final answer Chocolate milkshake costs R 9,00 and the Fruitshake costs R 11,00 Worked Example 45: Mathematical Modelling: Two variables Question: Three rulers and two pens cost R 21,00. One ruler and one pen cost R 8,00. Find the cost of one ruler and one pen Answer Step 1 : Translate the problem using variables Let the cost of one ruler be x rand and the cost of one pen be y rand. Step 2 : Rewrite the information in terms of the variables 3x + 2y x+y = = 21 8 Step 3 : Solve the equations simultaneously First solve the second equation for y: y =8−x and substitute the result into the first equation: 3x + 2(8 − x) = 21 3x + 16 − 2x = 21 x = 5 therefore y y = 8−5 = 3 Step 4 : Present the final answers one Ruler costs R 5,00 and one Pen costs R 3,00 105 (10.4) (10.5) 10.7 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Exercise: Mathematical Models 1. Stephen has 1 l of a mixture containing 69% of salt. How much water must Stephen add to make the mixture 50% salt? Write your answer as a fraction. 2. The diagonal of a rectangle is 25 cm more than its width. The length of the rectangle is 17 cm more than its width. What are the dimensions of the rectangle? 3. The sum of 27 and 12 is 73 more than an unknown number. Find the unknown number. 4. The two smaller angles in a right-angled triangle are in the ratio of 1:2. What are the sizes of the two angles? 5. George owns a bakery that specialises in wedding cakes. For each wedding cake, it costs George R150 for ingredients, R50 for overhead, and R5 for advertising. George’s wedding cakes cost R400 each. As a percentage of George’s costs, how much profit does he make for each cake sold? 6. If 4 times a number is increased by 7, the result is 15 less than the square of the number. Find the numbers that satisfy this statement, by formulating an equation and then solving it. 7. The length of a rectangle is 2 cm more than the width of the rectangle. The perimeter of the rectangle is 20 cm. Find the length and the width of the rectangle. 10.7.4 End of Chapter Exercises 1. What are the roots of the quadratic equation x2 − 3x + 2 = 0? 2. What are the solutions to the equation x2 + x = 6? 3. In the equation y = 2x2 − 5x − 18, which is a value of x when y = 0? 4. Manuel has 5 more CDs than Pedro has. Bob has twice as many CDs as Manuel has. Altogether the boys have 63 CDs. Find how many CDs each person has. 5. Seven-eighths of a certain number is 5 more than one-third of the number. Find the number. 6. A man runs to a telephone and back in 15 minutes. His speed on the way to the telephone is 5 m/s and his speed on the way back is 4 m/s. Find the distance to the telephone. 7. Solve the inequality and then answer the questions: x x 3 − 14 > 14 − 4 (a) If xǫR, write the solution in interval notation. (b) if xǫZ and x < 51, write the solution as a set of integers. 8. Solve for a: 1−a 2 − 2−a 3 9. Solve for x: x − 1 = >1 42 x 10. Solve for x and y: 7x + 3y = 13 and 2x − 3y = −4 chapterFunctions and Graphs - Grade 10 106 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.8 10.8 Introduction to Functions and Graphs Functions are mathematical building blocks for designing machines, predicting natural disasters, curing diseases, understanding world economies and for keeping aeroplanes in the air. Functions can take input from many variables, but always give the same answer, unique to that function. It is the fact that you always get the same answer from a set of inputs, which is what makes functions special. A major advantage of functions is that they allow us to visualise equations in terms of a graph. A graph is an accurate drawing of a function and is much easier to read than lists of numbers. In this chapter we will learn how to understand and create real valued functions, how to read graphs and how to draw them. Despite their use in the problems facing humanity, functions also appear on a day-to-day level, so they are worth learning about. A function is always dependent on one or more variables, like time, distance or a more abstract quantity. 10.9 Functions and Graphs in the Real-World Some typical examples of functions you may already have met include:• how much money you have, as a function of time. You never have more than one amount of money at any time because you can always add everything to give one number. By understanding how your money changes over time, you can plan to spend your money sensibly. Businesses find it very useful to plot the graph of their money over time so that they can see when they are spending too much. Such observations are not always obvious from looking at the numbers alone. • the temperature is a very complicated function because it has so many inputs, including; the time of day, the season, the amount of clouds in the sky, the strength of the wind, where you are and many more. But the important thing is that there is only one temperature when you measure it. By understanding how the temperature is effected by these things, you can plan for the day. • where you are is a function of time, because you cannot be in two places at once! If you were to plot the graphs of where two people are as a function of time, if the lines cross it means that the two people meet each other at that time. This idea is used in logistics, an area of mathematics that tries to plan where people and items are for businesses. • your weight is a function of how much you eat and how much exercise you do, but everybody has a different function so that is why people are all different sizes. 10.10 Recap The following should be familiar. 10.10.1 Variables and Constants In section 2.4 (page 8), we were introduced to variables and constants. To recap, a variable can take any value in some set of numbers, so long is the equation is consistent. Most often, a variable will be written as a letter. A constant has a fixed value. The number 1 is a constant. Sometimes letters are used to represent constants, as its easier to work with. Activity :: Investigation : Variables and Constants In the following expressions, identify the variables and the constants: 107 10.10 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 1. 2x2 = 1 2. 3x + 4y = 7 3. y = −5 x x 4. y = 7 − 2 10.10.2 Relations and Functions In earlier grades, you saw that variables can be related to each other. For example, Alan is two years older than Nathan. Therefore the relationship between the ages of Alan and Nathan can be written as A = N + 2, where A is Alan’s age and N is Nathan’s age. In general, a relation is an equation which relates two variables. For example, y = 5x and y 2 + x2 = 5 are relations. In both examples x and y are variables and 5 is a constant, but for a given value of x the value of y will be very different in each relation. Besides writing relations as equations, they can also be represented as words, tables and graphs. Instead of writing y = 5x, we could also say “y is always five times as big as x”. We could also give the following table: x 2 6 8 13 15 y = 5x 10 30 40 65 75 Activity :: Investigation : Relations and Functions Complete the following table for the given functions: x y = x y = 2x y = x + 2 1 2 3 50 100 10.10.3 The Cartesian Plane When working with real valued functions, our major tool is drawing graphs. In the first place, if we have two real variables, x and y, then we can assign values to them simultaneously. That is, we can say “let x be 5 and y be 3”. Just as we write “let x = 5” for “let x be 5”, we have the shorthand notation “let (x, y) = (5, 3)” for “let x be 5 and y be 3”. We usually think of the real numbers as an infinitely long line, and picking a number as putting a dot on that line. If we want to pick two numbers at the same time, we can do something similar, but now we must use two dimensions. What we do is use two lines, one for x and one for y, and rotate the one for y, as in Figure 10.1. We call this the Cartesian plane. 108 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 y 10.10 4 3 (−3,2) (2,2) b b 2 1 x −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 Figure 10.1: The Cartesian plane is made up of an x−axis (horizontal) and a y−axis (vertical). 10.10.4 Drawing Graphs In order to draw the graph of a function, we need to calculate a few points. Then we plot the points on the Cartesian Plane and join the points with a smooth line. The great beauty of doing this is that it allows us to “draw” functions, in a very abstract way. Assume that we were investigating the properties of the function f (x) = 2x. We could then consider all the points (x, y) such that y = f (x), i.e. y = 2x. For example, (1, 2), (2.5, 5), and (3, 6) would all be such points, whereas (3, 5) would not since 5 6= 2 × 3. If we put a dot at each of those points, and then at every similar one for all possible values of x, we would obtain the graph shown in 5 b 4 b 3 b 2 b b 1 b 1 2 3 4 5 Figure 10.2: Graph of f (x) = 2x The form of this graph is very pleasing – it is a simple straight line through the middle of 109 10.10 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 the plane. The technique of “plotting”, which we have followed here, is the key element in understanding functions. Activity :: Investigation : Drawing Graphs and the Cartesian Plane Plot the following points and draw a smooth line through them. (-6; -8),(-2; 0), (2; 8), (6; 16) 10.10.5 Notation used for Functions Thus far you would have seen that we can use y = 2x to represent a function. This notation however gets confusing when you are working with more than one function. A more general form of writing a function is to write the function as f (x), where f is the function name and x is the independent variable. For example, f (x) = 2x and g(t) = 2t + 1 are two functions. Both notations will be used in this book. Worked Example 46: Function notation Question: If f (n) = n2 − 6n + 9, find f (k − 1) in terms of k. Answer Step 1 : Replace n with k − 1 f (n) = f (k − 1) = n2 − 6n + 9 (k − 1)2 − 6(k − 1) + 9 Step 2 : Remove brackets on RHS and simplify = = k 2 − 2k + 1 − 6k + 6 + 9 k 2 − 8k + 16 Worked Example 47: Function notation Question: If f (x) = x2 − 4, calculate b if f (b) = 45. Answer Step 1 : Replace x with b f (b) = butf (b) = 110 b2 − 4 45 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.10 Step 2 : f(b) = f(b) b2 − 4 = 2 b − 49 = b = 45 0 +7or − 7 {ExerciseRecap 1. Guess the function in the form y = . . . that has the values listed in the table. x y 1 1 2 2 3 3 40 40 50 50 600 600 700 700 800 800 900 900 1000 1000 2. Guess the function in the form y = . . . that has the values listed in the table. x y 1 2 2 4 3 6 40 80 50 100 600 1200 700 1400 800 1600 900 1800 1000 2000 3. Guess the function in the form y = . . . that has the values listed in the table. x y 1 10 2 20 3 30 40 400 50 500 600 6000 700 7000 800 8000 900 9000 1000 10000 4. On a Cartesian plane, plot the following points: (1,2), (2,4), (3,6), (4,8), (5,10). Join the points. Do you get a straight-line? 5. If f (x) = x + x2 , write out: (a) f (t) (b) f (a) (c) f (1) (d) f (3) 6. If g(x) = x and f (x) = 2x, write out: (a) f (t) + g(t) (b) f (a) − g(a) (c) f (1) + g(2) (d) f (3) + g(s) 7. A car drives by you on a straight highway. The car is travelling 10 m every second. Complete the table below by filling in how far the car has travelled away from you after 5, 10 and 20 seconds. Time (s) Distance (m) 0 0 1 10 2 20 5 10 20 Use the values in the table and draw a graph of distance on the y-axis and time on the x-axis. 111 10.11 10.11 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Characteristics of Functions - All Grades There are many characteristics of graphs that help describe the graph of any function. These properties are: 1. dependent and independent variables 2. domain and range 3. intercepts with axes 4. turning points 5. asymptotes 6. lines of symmetry 7. intervals on which the function increases/decreases 8. continuous nature of the function Some of these words may be unfamiliar to you, but each will be clearly described. Examples of these properties are shown in Figure 10.3. bE f (x) 3 2 1 Bb g(x) 3 2 1 bA bF −3 −2 −1 −1 bC 1 −2 2 3 h −3 −2 −1 −1 2 3 −2 b D −3 1 −3 (a) A B, C, F D, E (b) y-intercept x-intercept turning points Figure 10.3: (a) Example graphs showing the characteristics of a function. (b) Example graph showing asymptotes of a function. 10.11.1 Dependent and Independent Variables Thus far, all the graphs you have drawn have needed two values, an x-value and a y-value. The y-value is usually determined from some relation based on a given or chosen x-value. These values are given special names in mathematics. The given or chosen x-value is known as the independent variable, because its value can be chosen freely. The calculated y-value is known as the dependent variable, because its value depends on the chosen x-value. 112 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.11.2 10.11 Domain and Range The domain of a relation is the set of all the x values for which there exists at least one y value according to that relation. The range is the set of all the y values, which can be obtained using at least one x value. If the relation is of height to people, then the domain is all living people, while the range would be about 0.1 to 3 metres — no living person can have a height of 0m, and while strictly its not impossible to be taller than 3 metres, no one alive is. An important aspect of this range is that it does not contain all the numbers between 0.1 and 3, but only six billion of them (as many as there are people). As another example, suppose x and y are real valued variables, and we have the relation y = 2x . Then for any value of x, there is a value of y, so the domain of this relation is the whole set of real numbers. However, we know that no matter what value of x we choose, 2x can never be less than or equal to 0. Hence the range of this function is all the real numbers strictly greater than zero. These are two ways of writing the domain and range of a function, set notation and interval notation. Both notations are used in mathematics, so you should be familiar with each. Set Notation A set of certain x values has the following form: {x : conditions, more conditions} (10.6) We read this notation as “the set of all x values where all the conditions are satisfied”. For example, the set of all positive real numbers can be written as {x : x ∈ R, x > 0} which reads as “the set of all x values where x is a real number and is greater than zero”. Interval Notation Here we write an interval in the form ’lower bracket, lower number, comma, upper number, upper bracket’. We can use two types of brackets, square ones [, ] or round ones (, ). A square bracket means including the number at the end of the interval whereas a round bracket means excluding the number at the end of the interval. It is important to note that this notation can only be used for all real numbers in an interval. It cannot be used to describe integers in an interval or rational numbers in an interval. So if x is a real number greater than 2 and less than or equal to 8, then x is any number in the interval (2,8] (10.7) It is obvious that 2 is the lower number and 8 the upper number. The round bracket means ’excluding 2’, since x is greater than 2, and the square bracket means ’including 8’ as x is less than or equal to 8. 10.11.3 Intercepts with the Axes The intercept is the point at which a graph intersects an axis. The x-intercepts are the points at which the graph cuts the x-axis and the y-intercepts are the points at which the graph cuts the y-axis. In Figure 10.3(a), the A is the y-intercept and B, C and F are x-intercepts. You will usually need to calculate the intercepts. The two most important things to remember is that at the x-intercept, y = 0 and at the y-intercept, x = 0. For example, calculate the intercepts of y = 3x + 5. For the y-intercept, x = 0. Therefore the y-intercept is yint = 3(0) + 5 = 5. For the x-intercept, y = 0. Therefore the x-intercept is found from 0 = 3xint + 5, giving xint = − 35 . 113 10.11 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.11.4 Turning Points Turning points only occur for graphs of functions that whose highest power is greater than 1. For example, graphs of the following functions will have turning points. f (x) g(x) h(x) = 2x2 − 2 = x3 − 2x2 + x − 2 2 4 x −2 = 3 There are two types of turning points: a minimal turning point and a maximal turning point. A minimal turning point is a point on the graph where the graph stops decreasing in value and starts increasing in value and a maximal turning point is a point on the graph where the graph stops increasing in value and starts decreasing. These are shown in Figure 10.4. y y b 2 y-values decreasing 1 y-values increasing 1 0 −1 minimal turning point b 2 maximal turning point 0 y-values increasing −1 −2 (a) y-values decreasing −2 (b) Figure 10.4: (a) Maximal turning point. (b) Minimal turning point. In Figure 10.3(a), E is a maximal turning point and D is a minimal turning point. 10.11.5 Asymptotes An asymptote is a straight or curved line, which the graph of a function will approach, but never touch. In Figure 10.3(b), the y-axis and line h are both asymptotes as the graph approaches both these lines, but never touches them. 10.11.6 Lines of Symmetry Graphs look the same on either side of lines of symmetry. These lines include the x- and yaxes. For example, in Figure 10.5 is symmetric about the y-axis. This is described as the axis of symmetry. 10.11.7 Intervals on which the Function Increases/Decreases In the discussion of turning points, we saw that the graph of a function can start or stop increasing or decreasing at a turning point. If the graph in Figure 10.3(a) is examined, we find that the values of the graph increase and decrease over different intervals. We see that the graph increases (i.e. that the y-values increase) from -∞ to point E, then it decreases (i.e. the y-values decrease) from point E to point D and then it increases from point D to +∞. 10.11.8 Discrete or Continuous Nature of the Graph A graph is said to be continuous if there are no breaks in the graph. For example, the graph in Figure 10.3(a) can be described as a continuous graph, while the graph in Figure 10.3(b) has a 114 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.11 1 −2 −1 1 2 −1 Figure 10.5: Demonstration of axis of symmetry. The y-axis is an axis of symmetry, because the graph looks the same on both sides of the y-axis. break around the asymptotes. In Figure 10.3(b), it is clear that the graph does have a break in it around the asymptote. Exercise: Domain and Range 1. The domain of the function f (x) = 2x + 5 is -3; -3; -3; 0. Determine the range of f . 2. If g(x) = −x2 + 5 and x is between - 3 and 3, determine: (a) the domain of g(x) (b) the range of g(x) 3. Label, on the following graph: (a) (b) (c) (d) the x-intercept(s) the y-intercept(s) regions where the graph is increasing regions where the graph is decreasing 4 3 2 1 −3 −2 −1 1 −1 −2 −3 4. Label, on the following graph: (a) the x-intercept(s) (b) the y-intercept(s) 115 2 3 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 (c) regions where the graph is increasing (d) regions where the graph is decreasing 3 2 1 −2 −1 1 2 −1 −2 10.12 Graphs of Functions 10.12.1 Functions of the form y = ax + q Functions with a general form of y = ax + q are called straight line functions. In the equation, y = ax + q, a and q are constants and have different effects on the graph of the function. The general shape of the graph of functions of this form is shown in Figure 10.6 for the function f (x) = 2x + 3. 12 b b 9 b 6 3 − 32 −5 −4 −3 b b b −2 b b b −1 1 2 3 −3 −6 Figure 10.6: Graph of f (x) = 2x + 3 Activity :: Investigation : Functions of the Form y = ax + q 1. On the same set of axes, plot the following graphs: (a) a(x) = x − 2 116 4 5 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 (b) (c) (d) (e) 10.12 b(x) = x − 1 c(x) = x d(x) = x + 1 e(x) = x + 2 Use your results to deduce the effect of q. 2. On the same set of axes, plot the following graphs: (a) (b) (c) (d) (e) f (x) = −2 · x g(x) = −1 · x h(x) = 0 · x j(x) = 1 · x k(x) = 2 · x Use your results to deduce the effect of a. You should have found that the value of a affects the slope of the graph. As a increases, the slope of the graph increases. If a > 0 then the graph increases from left to right (slopes upwards). If a < 0 then the graph increases from right to left (slopes downwards). For this reason, a is referred to as the slope or gradient of a straight-line function. You should have also found that the value of q affects where the graph passes through the y-axis. For this reason, q is known as the y-intercept. These different properties are summarised in Table 10.1. Table 10.1: Table summarising general shapes and positions of graphs of functions of the form y = ax + q. a>0 a<0 q>0 q<0 Domain and Range For f (x) = ax + q, the domain is {x : x ∈ R} because there is no value of x ∈ R for which f (x) is undefined. The range of f (x) = ax + q is also {f (x) : f (x) ∈ R} because there is no value of f (x) ∈ R for which f (x) is undefined. For example, the domain of g(x) = x − 1 is {x : x ∈ R} because there is no value of x ∈ R for which g(x) is undefined. The range of g(x) is {g(x) : g(x) ∈ R}. Intercepts For functions of the form, y = ax + q, the details of calculating the intercepts with the x and y axis is given. 117 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 The y-intercept is calculated as follows: y = ax + q yint = = a(0) + q q (10.8) (10.9) (10.10) For example, the y-intercept of g(x) = x − 1 is given by setting x = 0 to get: g(x) = yint = = x−1 0−1 −1 The x-intercepts are calculated as follows: y = 0 = a · xint = xint = ax + q (10.11) a · xint + q (10.12) −q q − a (10.13) (10.14) For example, the x-intercepts of g(x) = x − 1 is given by setting y = 0 to get: g(x) = 0 = xint = x−1 xint − 1 1 Turning Points The graphs of straight line functions do not have any turning points. Axes of Symmetry The graphs of straight-line functions do not, generally, have any axes of symmetry. Sketching Graphs of the Form f (x) = ax + q In order to sketch graphs of the form, f (x) = ax + q, we need to determine three characteristics: 1. sign of a 2. y-intercept 3. x-intercept Only two points are needed to plot a straight line graph. The easiest points to use are the x-intercept (where the line cuts the x-axis) and the y-intercept. For example, sketch the graph of g(x) = x − 1. Mark the intercepts. Firstly, we determine that a > 0. This means that the graph will have an upward slope. The y-intercept is obtained by setting x = 0 and was calculated earlier to be yint = −1. The x-intercept is obtained by setting y = 0 and was calculated earlier to be xint = 1. Worked Example 48: Drawing a straight line graph Question: Draw the graph of y = 2x + 2 118 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.12 3 2 1 (1,0) −4 −3 −2 −1 −1 b 1 b (0,-1) 2 3 4 −2 −3 −4 Figure 10.7: Graph of the function g(x) = x − 1 Answer Step 1 : Find the y-intercept For the intercept on the y-axis, let x = 0 y = 2(0) + 2 = 2 Step 2 : Find the x-intercept For the intercept on the x-axis, let y = 0 0 = 2x = 2x + 2 −2 x = −1 Step 3 : Draw the graph 3 2 1 2x +2 −2 y= −3 −1 1 2 −1 −2 119 3 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Exercise: Intercepts 1. List the y-intercepts for the following straight-line graphs: (a) (b) (c) (d) y=x y =x−1 y = 2x − 1 y + 1 = 2x 2. Give the equation of the illustrated graph below: y (0;3) (4;0) x 3. Sketch the following relations on the same set of axes, clearly indicating the intercepts with the axes as well as the co-ordinates of the point of interception on the graph: x + 2y − 5 = 0 and 3x − y − 1 = 0 10.12.2 Functions of the Form y = ax2 + q The general shape and position of the graph of the function of the form f (x) = ax2 + q is shown in Figure 10.8. Activity :: Investigation : Functions of the Form y = ax2 + q 1. On the same set of axes, plot the following graphs: (a) (b) (c) (d) (e) a(x) = −2 · x2 + 1 b(x) = −1 · x2 + 1 c(x) = 0 · x2 + 1 d(x) = 1 · x2 + 1 e(x) = 2 · x2 + 1 Use your results to deduce the effect of a. 2. On the same set of axes, plot the following graphs: (a) f (x) = x2 − 2 (b) g(x) = x2 − 1 120 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.12 9 8 7 6 5 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 Figure 10.8: Graph of the f (x) = x2 − 1. (c) h(x) = x2 + 0 (d) j(x) = x2 + 1 (e) k(x) = x2 + 2 Use your results to deduce the effect of q. Complete the following table of values for the functions a to k to help with drawing the required graphs in this activity: x −2 −1 0 1 2 a(x) b(x) c(x) d(x) e(x) f (x) g(x) h(x) j(x) k(x) From your graphs, you should have found that a affects whether the graph makes a smile or a frown. If a < 0, the graph makes a frown and if a > 0 then the graph makes a smile. This is shown in Figure 10.9. b b a > 0 (a positive smile) b b a < 0 (a negative frown) Figure 10.9: Distinctive shape of graphs of a parabola if a > 0 and a < 0. You should have also found that the value of q affects whether the turning point is to the left of the y-axis (q > 0) or to the right of the y-axis (q < 0). These different properties are summarised in Table ??. 121 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 Table 10.2: Table summarising general shapes and positions of functions of the form y = ax2 +q. a>0 a<0 q>0 q<0 Domain and Range For f (x) = ax2 + q, the domain is {x : x ∈ R} because there is no value of x ∈ R for which f (x) is undefined. The range of f (x) = ax2 + q depends on whether the value for a is positive or negative. We will consider these two cases separately. If a > 0 then we have: x2 ax2 ax2 + q f (x) ≥ 0 ≥ 0 (The square of an expression is always positive) (Multiplication by a positive number maintains the nature of the inequality) ≥ q ≥ q This tells us that for all values of x, f (x) is always greater than q. Therefore if a > 0, the range of f (x) = ax2 + q is {f (x) : f (x) ∈ [q,∞)}. Similarly, it can be shown that if a < 0 that the range of f (x) = ax2 + q is {f (x) : f (x) ∈ (−∞,q]}. This is left as an exercise. For example, the domain of g(x) = x2 + 2 is {x : x ∈ R} because there is no value of x ∈ R for which g(x) is undefined. The range of g(x) can be calculated as follows: x2 2 x +2 g(x) ≥ 0 ≥ ≥ 2 2 Therefore the range is {g(x) : g(x) ∈ [2,∞)}. Intercepts For functions of the form, y = ax2 + q, the details of calculating the intercepts with the x and y axis is given. The y-intercept is calculated as follows: y yint = ax2 + q 2 = a(0) + q = q 122 (10.15) (10.16) (10.17) CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.12 For example, the y-intercept of g(x) = x2 + 2 is given by setting x = 0 to get: g(x) = x2 + 2 yint 02 + 2 2 = = The x-intercepts are calculated as follows: y = 0 = ax2int = xint = ax2 + q (10.18) ax2int (10.19) (10.20) +q −q r q ± − a (10.21) However, (10.21) is only valid if − aq > 0 which means that either q < 0 or a < 0. This is consistent with what we expect, since if q > 0 and a > 0 then − aq is negative and in this case the graph lies above the x-axis and therefore does not intersect the x-axis. If however, q > 0 and a < 0, then − aq is positive and the graph is hat shaped and should have two x-intercepts. Similarly, if q < 0 and a > 0 then − aq is also positive, and the graph should intersect with the x-axis. For example, the x-intercepts of g(x) = x2 + 2 is given by setting y = 0 to get: g(x) = 0 = −2 = x2 + 2 x2int + 2 x2int which is not real. Therefore, the graph of g(x) = x2 + 2 does not have any x-intercepts. Turning Points The turning point of the function of the form f (x) = ax2 + q is given by examining the range of the function. We know that if a > 0 then the range of f (x) = ax2 + q is {f (x) : f (x) ∈ [q,∞)} and if a < 0 then the range of f (x) = ax2 + q is {f (x) : f (x) ∈ (−∞,q]}. So, if a > 0, then the lowest value that f (x) can take on is q. Solving for the value of x at which f (x) = q gives: q = ax2tp + q 0 = ax2tp 0 = x2tp xtp = 0 ∴ x = 0 at f (x) = q. The co-ordinates of the (minimal) turning point is therefore (0; q). Similarly, if a < 0, then the highest value that f (x) can take on is q and the co-ordinates of the (maximal) turning point is (0; q). Axes of Symmetry There is one axis of symmetry for the function of the form f (x) = ax2 + q that passes through the turning point. Since the turning point lies on the y-axis, the axis of symmetry is the y-axis. Sketching Graphs of the Form f (x) = ax2 + q In order to sketch graphs of the form, f (x) = ax2 + q, we need to calculate determine four characteristics: 1. sign of a 123 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 2. domain and range 3. turning point 4. y-intercept 5. x-intercept For example, sketch the graph of g(x) = − 21 x2 − 3. Mark the intercepts, turning point and axis of symmetry. Firstly, we determine that a < 0. This means that the graph will have a maximal turning point. The domain of the graph is {x : x ∈ R} because f (x) is defined for all x ∈ R. The range of the graph is determined as follows: x2 1 − x2 2 1 − x2 − 3 2 ∴ f (x) ≥ 0 ≤ 0 ≤ −3 ≤ −3 Therefore the range of the graph is {f (x) : f (x) ∈ (−∞, − 3]}. Using the fact that the maximum value that f (x) achieves is -3, then the y-coordinate of the turning point is -3. The x-coordinate is determined as follows: 1 − x2 − 3 = 2 −3 1 − x2 − 3 + 3 = 2 1 − x2 = 2 Divide both sides by − 21 : x2 = 0 Take square root of both sides: x = 0 x = 0 ∴ The coordinates of the turning point are: (0, − 3). The y-intercept is obtained by setting x = 0. This gives: yint = = = 1 − (0)2 − 3 2 1 − (0) − 3 2 −3 The x-intercept is obtained by setting y = 0. This gives: 0 3 −3 · 2 −6 1 = − x2int − 3 2 1 = − x2int 2 = x2int = x2int which is not real. Therefore, there are no x-intercepts. We also know that the axis of symmetry is the y-axis. 124 0 0 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 −4 −3 −2 −1 −1 1 2 3 10.12 4 −2 −3 b (0,-3) −4 −5 −6 Figure 10.10: Graph of the function f (x) = − 12 x2 − 3 Exercise: Parabolas 1. Show that if a < 0 that the range of f (x) = ax2 +q is {f (x) : f (x) ∈ (−∞,q]}. 2. Draw the graph of the function y = −x2 + 4 showing all intercepts with the axes. 3. Two parabolas are drawn: g : y = ax2 + p and h : y = bx2 + q. y g 23 (-4; 7) (4; 7) x 3 h -9 (a) (b) (c) (d) 10.12.3 Find the values of a and p. Find the values of b and q. Find the values of x for which ax2 + p ≥ bx2 + q. For what values of x is g increasing ? Functions of the Form y = a x +q Functions of the form y = xa + q are known as hyperbolic functions. The general form of the graph of this function is shown in Figure 10.11. Activity :: Investigation : Functions of the Form y = 1. On the same set of axes, plot the following graphs: (a) a(x) = −2 x +1 (b) b(x) = −1 x +1 125 a x +q 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 5 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 −2 Figure 10.11: General shape and position of the graph of a function of the form f (x) = (c) c(x) = (d) d(x) = (e) e(x) = a x + q. 0 x +1 +1 x +1 +2 x +1 Use your results to deduce the effect of a. 2. On the same set of axes, plot the following graphs: (a) f (x) = x1 − 2 (b) g(x) = x1 − 1 (c) h(x) = x1 + 0 (d) j(x) = x1 + 1 (e) k(x) = x1 + 2 Use your results to deduce the effect of q. You should have found that the value of a affects whether the graph is located in the first and third quadrants of Cartesian plane. You should have also found that the value of q affects whether the graph lies above the x-axis (q > 0) or below the x-axis (q < 0). These different properties are summarised in Table 10.3. The axes of symmetry for each graph are shown as a dashed line. Domain and Range For y = a x + q, the function is undefined for x = 0. The domain is therefore {x : x ∈ R,x 6= 0}. We see that y = a x + q can be re-written as: y = y−q = If x 6= 0 then: (y − q)(x) = x = a +q x a x a a y−q This shows that the function is undefined at y = q. Therefore the range of f (x) = {f (x) : f (x) ∈ (−∞,q) ∪ (q,∞)}. 126 a x + q is CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.12 Table 10.3: Table summarising general shapes and positions of functions of the form y = The axes of symmetry are shown as dashed lines. a>0 a<0 a x + q. q>0 q<0 For example, the domain of g(x) = x = 0. 2 x + 2 is {x : x ∈ R, x 6= 0} because g(x) is undefined at y = (y − 2) = If x 6= 0 then: x(y − 2) = x = 2 +2 x 2 x 2 2 y−2 We see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. Intercepts For functions of the form, y = xa + q, the intercepts with the x and y axis is calculated by setting x = 0 for the y-intercept and by setting y = 0 for the x-intercept. The y-intercept is calculated as follows: y = yint = a +q x a +q 0 which is undefined. Therefore there is no y-intercept. For example, the y-intercept of g(x) = 2 x + 2 is given by setting x = 0 to get: y = yint = which is undefined. 127 2 +2 x 2 +2 0 (10.22) (10.23) 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 The x-intercepts are calculated by setting y = 0 as follows: y a +q x a +q xint = 0 = a xint a xint For example, the x-intercept of g(x) = 2 x (10.24) (10.25) = −q (10.26) = −q(xint ) a −q (10.27) = (10.28) (10.29) + 2 is given by setting x = 0 to get: y = 0 = −2 = −2(xint ) = xint = xint = 2 +2 x 2 +2 xint 2 xint 2 2 −2 −1 Asymptotes There are two asymptotes for functions of the form y = the domain and range. a x +q. They are determined by examining We saw that the function was undefined at x = 0 and for y = q. Therefore the asymptotes are x = 0 and y = q. For example, the domain of g(x) = x2 + 2 is {x : x ∈ R, x 6= 0} because g(x) is undefined at x = 0. We also see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. From this we deduce that the asymptotes are at x = 0 and y = 2. Sketching Graphs of the Form f (x) = a x +q In order to sketch graphs of functions of the form, f (x) = four characteristics: a x + q, we need to calculate determine 1. domain and range 2. asymptotes 3. y-intercept 4. x-intercept For example, sketch the graph of g(x) = 2 x + 2. Mark the intercepts and asymptotes. We have determined the domain to be {x : x ∈ R, x 6= 0} and the range to be {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. Therefore the asymptotes are at x = 0 and y = 2. There is no y-intercept and the x-intercept is xint = −1. 128 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.12 6 5 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 −2 −3 Figure 10.12: Graph of g(x) = 2 x + 2. Exercise: Graphs 1. Using grid paper, draw the graph of xy = −6. (a) Does the point (-2; 3) lie on the graph ? Give a reason for your answer. (b) Why is the point (-2; -3) not on the graph ? (c) If the x-value of a point on the drawn graph is 0,25, what is the corresponding y-value ? (d) What happens to the y-values as the x-values become very large ? (e) With the line y = −x as line of symmetry, what is the point symmetrical to (-2; 3) ? 2. Draw the graph of xy = 8. (a) How would the graph y = 83 + 3 compare with that of xy = 8? Explain your answer fully. (b) Draw the graph of y = 38 + 3 on the same set of axes. 10.12.4 Functions of the Form y = ab(x) + q Functions of the form y = ab(x) + q are known as exponential functions. The general shape of a graph of a function of this form is shown in Figure 10.13. Activity :: Investigation : Functions of the Form y = ab(x) + q 1. On the same set of axes, plot the following graphs: (a) a(x) = −2 · b(x) + 1 (b) b(x) = −1 · b(x) + 1 (c) c(x) = −0 · b(x) + 1 129 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 4 3 2 1 −4 −3 −2 −1 1 2 3 4 Figure 10.13: General shape and position of the graph of a function of the form f (x) = ab(x) +q. (d) d(x) = −1 · b(x) + 1 (e) e(x) = −2 · b(x) + 1 Use your results to deduce the effect of a. 2. On the same set of axes, plot the following graphs: (a) (b) (c) (d) (e) f (x) = 1 · b(x) − 2 g(x) = 1 · b(x) − 1 h(x) = 1 · b(x) + 0 j(x) = 1 · b(x) + 1 k(x) = 1 · b(x) + 2 Use your results to deduce the effect of q. You should have found that the value of a affects whether the graph curves upwards (a > 0) or curves downwards (a < 0). You should have also found that the value of q affects the position of the y-intercept. These different properties are summarised in Table 10.4. Table 10.4: Table summarising general shapes and positions of functions of the form y = ab(x) + q. a>0 a<0 q>0 q<0 Domain and Range For y = ab(x) + q, the function is defined for all real values of x. Therefore, the domain is {x : x ∈ R}. 130 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.12 The range of y = ab(x) + q is dependent on the sign of a. If a > 0 then: b(x) a · b(x) a · b(x) + q f (x) ≥ 0 ≥ 0 ≥ q ≥ q Therefore, if a > 0, then the range is {f (x) : f (x) ∈ [q,∞)}. If a < 0 then: b(x) (x) a·b a · b(x) + q f (x) ≤ 0 ≤ 0 ≤ q ≤ q Therefore, if a < 0, then the range is {f (x) : f (x) ∈ (−∞,q]}. For example, the domain of g(x) = 3 · 2x + 2 is {x : x ∈ R}. For the range, 2x 3 · 2x ≥ ≥ 3 · 2x + 2 ≥ 0 0 2 Therefore the range is {g(x) : g(x) ∈ [2,∞)}. Intercepts For functions of the form, y = ab(x) + q, the intercepts with the x and y axis is calculated by setting x = 0 for the y-intercept and by setting y = 0 for the x-intercept. The y-intercept is calculated as follows: y yint = ab(x) + q = ab(0) + q (10.30) (10.31) = a(1) + q = a+q (10.32) (10.33) For example, the y-intercept of g(x) = 3 · 2x + 2 is given by setting x = 0 to get: y yint = = = = 3 · 2x + 2 3 · 20 + 2 3+2 5 The x-intercepts are calculated by setting y = 0 as follows: y = 0 = ab(xint ) = (xint ) = b ab(x) + q ab(xint ) + q (10.34) (10.35) −q q − a (10.36) (10.37) Which only has a real solution if either a < 0 or q < 0. Otherwise, the graph of the function of form y = ab(x) + q does not have any x-intercepts. 131 10.12 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 For example, the x-intercept of g(x) = 3 · 2x + 2 is given by setting y = 0 to get: y = 3 · 2x + 2 0 = 3 · 2xint + 2 −2 = 3 · 2xint −2 2xint = 3 which has no real solution. Therefore, the graph of g(x) = 3 · 2x + 2 does not have any x-intercepts. Asymptotes There are two asymptotes for functions of the form y = ab(x) + q. They are determined by examining the domain and range. We saw that the function was undefined at x = 0 and for y = q. Therefore the asymptotes are x = 0 and y = q. For example, the domain of g(x) = 3 · 2x + 2 is {x : x ∈ R, x 6= 0} because g(x) is undefined at x = 0. We also see that g(x) is undefined at y = 2. Therefore the range is {g(x) : g(x) ∈ (−∞,2) ∪ (2,∞)}. From this we deduce that the asymptotes are at x = 0 and y = 2. Sketching Graphs of the Form f (x) = ab(x) + q In order to sketch graphs of functions of the form, f (x) = ab(x) + q, we need to calculate determine four characteristics: 1. domain and range 2. y-intercept 3. x-intercept For example, sketch the graph of g(x) = 3 · 2x + 2. Mark the intercepts. We have determined the domain to be {x : x ∈ R} and the range to be {g(x) : g(x) ∈ [2,∞)}. The y-intercept is yint = 5 and there are no x-intercepts. 6 5 4 3 2 1 −4 −3 −2 −1 1 2 3 4 Figure 10.14: Graph of g(x) = 3 · 2x + 2. 132 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 10.13 Exercise: Exponential Functions and Graphs 1. Draw the graphs of y = 2x and y = ( 12 )x on the same set of axes. (a) Is the x-axis and asymptote or and axis of symmetry to both graphs ? Explain your answer. (b) Which graph is represented by the equation y = 2−x ? Explain your answer. (c) Solve the equation 2x = ( 12 )x graphically and check that your answer is correct by using substitution. (d) Predict how the graph y = 2.2x will compare to y = 2x and then draw the graph of y = 2.2x on the same set of axes. 2. The curve of the exponential function f in the accompanying diagram cuts the f b 4 B(2,4) 3 2 b 1 y-axis at the point A(0; 1) and B(2; 4) is on f . A(0,1) 0 1 2 (a) Determine the equation of the function f . (b) Determine the equation of h, the function of which the curve is the reflection of the curve of f in the x-axis. (c) Determine the range of h. 10.13 End of Chapter Exercises 1. Given the functions f (x) = −2x2 − 18 and g(x) = −2x + 6 (a) Draw f and g on the same set of axes. (b) Calculate the points of intersection of f and g. (c) Hence use your graphs and the points of intersection to solve for x when: i. f (x) > 0 ii. f (x) g(x) ≤0 (d) Give the equation of the reflection of f in the x-axis. 2. After a ball is dropped, the rebound height of each bounce decreases. The equation y = 5(0.8)x shows the relationship between x, the number of bounces, and y, the height of the bounce, for a certain ball. What is the approximate height of the fifth bounce of this ball to the nearest tenth of a unit ? 3. Marc had 15 coins in five rand and two rand pieces. He had 3 more R2-coins than R5coins. He wrote a system of equations to represent this situation, letting x represent the number of five rand coins and y represent the number of two rand coins. Then he solved the system by graphing. 133 x 10.13 CHAPTER 10. EQUATIONS AND INEQUALITIES - GRADE 10 (a) Write down the system of equations. (b) Draw their graphs on the same set of axes. (c) What is the solution? 134 Chapter 11 Average Gradient - Grade 10 Extension 11.1 Introduction In chapter 10.7.4, we saw that the gradient of a straight line graph is calculated as: y2 − y1 x2 − x1 (11.1) for two points (x1 ,y1 ) and (x2 ,y2 ) on the graph. We can now define the average gradient between any two points, (x1 ,y1 ) and (x2 ,y2 ) as: y2 − y1 . x2 − x1 (11.2) This is the same as (11.1). 11.2 Straight-Line Functions Activity :: Investigation : Average Gradient - Straight Line Function Fill in the table by calculating the average gradient over the indicated intervals for the function f (x) = 2x − 2: x1 A-B A-C B-C x2 y1 y2 y2 −y1 x2 −x1 y 2 C(2,2) b 1 b −1 1 −1 −2 −3 A(-1,-4) 135 b −4 B(1,0) x 11.3 CHAPTER 11. AVERAGE GRADIENT - GRADE 10 EXTENSION What do you notice about the gradients over each interval? The average gradient of a straight-line function is the same over any two intervals on the function. 11.3 Parabolic Functions Activity :: Investigation : Average Gradient - Parabolic Function Fill in the table by calculating the average gradient over the indicated intervals for the function f (x) = 2x − 2: x1 x2 y1 y2 b A(-3,7) y2 −y1 x2 −x1 b G(3,7) y A-B B-C C-D D-E E-F F-G What do you notice about the average gradient over each interval? What can you say about the average gradients between A and D compared to the average gradients between D and G? B(-2,2) F(2,2) b b C(-1,-1) x b b E(1,-1) b D(0,-2) The average gradient of a parabolic function depends on the interval and is the gradient of a straight line that passes through the points on the interval. For example, in Figure 11.1 the various points have been joined by straight-lines. The average gradients between the joined points are then the gradients of the straight lines that pass through the points. b A(-3,7) B(-2,2) b G(3,7) y F(2,2) b C(-1,-1) b b b E(1,-1) x b D(0,-2) Figure 11.1: The average gradient between two points on a curve is the gradient of the straight line that passes through the points. Method: Average Gradient Given the equation of a curve and two points (x1 , x2 ): 136 CHAPTER 11. AVERAGE GRADIENT - GRADE 10 EXTENSION 1. Write the equation of the curve in the form y = . . .. 2. Calculate y1 by substituting x1 into the equation for the curve. 3. Calculate y2 by substituting x2 into the equation for the curve. 4. Calculate the average gradient using: y2 − y1 x2 − x1 Worked Example 49: Average Gradient Question: Find the average gradient of the curve y = 5x2 − 4 between the points x = −3 and x = 3 Answer Step 1 : Label points Label the points as follows: x1 = −3 x2 = 3 to make it easier to calculate the gradient. Step 2 : Calculate the y coordinates We use the equation for the curve to calculate the y-value at x1 and x2 . y1 = 5x21 − 4 = 5(−3)2 − 4 = 5(9) − 4 = 41 y2 = = = = 5x22 − 4 5(3)2 − 4 5(9) − 4 41 Step 3 : Calculate the average gradient y2 − y1 x2 − x1 41 − 41 3 − (−3) 0 = 3+3 0 = 6 = 0 = Step 4 : Write the final answer The average gradient between x = −3 and x = 3 on the curve y = 5x2 − 4 is 0. 137 11.3 11.4 11.4 CHAPTER 11. AVERAGE GRADIENT - GRADE 10 EXTENSION End of Chapter Exercises 1. An object moves according to the function d = 2t2 + 1 , where d is the distance in metres and t the time in seconds. Calculate the average speed of the object between 2 and 3 seconds. 2. Given: f (x) = x3 − 6x. Determine the average gradient between the points where x = 1 and x = 4. 138 Chapter 12 Geometry Basics 12.1 Introduction The purpose of this chapter is to recap some of the ideas that you learned in geometry and trigonometry in earlier grades. You should feel comfortable with the work covered in this chapter before attempting to move onto the Grade 10 Geometry Chapter (Chapter 13) or the Grade 10 Trigonometry Chapter (Chapter 14). This chapter revises: 1. Terminology: quadrilaterals, vertices, sides, angles, parallel lines, perpendicular lines, diagonals, bisectors, transversals 2. Similarities and differences between quadrilaterals 3. Properties of triangles and quadrilaterals 4. Congruence 5. Classification of angles into acute, right, obtuse, straight, reflex or revolution 6. Theorem of Pythagoras which is used to calculate the lengths of sides of a right-angled triangle 12.2 Points and Lines The two simplest objects in geometry are points and lines. A point is something that is not very wide or high and is usually used in geometry as a marker of a position. Points are usually labelled with a capital letter. Some examples of points are shown in Figure 12.1. A line is formed when many points are placed next to each other. Lines can be straight or curved, but are always continuous. This means that there are never any breaks in the lines. The endpoints of lines are labelled with capital letters. Examples of two lines are shown in Figure 12.1. bS b b R Q P Some points b B C D E Some lines Figure 12.1: Examples of some points (labelled P , Q, R and S) and some lines (labelled BC and DE). 139 12.3 CHAPTER 12. GEOMETRY BASICS Lines are labelled according to the start point and end point. We call the line that starts at a point A and ends at a point B, AB. Since the line from point B to point A is the same as the line from point A to point B, we have that AB=BA. The length of the line between points A and B is AB. So if we say AB = CD we mean that the length of the line between A and B is equal to the length of the line between C and D. In science, we sometimes talk about a vector and this is just a fancy way of saying the we are referring to the line that starts at one point and moves in the direction of the other point. We ~ referring to the vector from the point A label a vector in a similar manner to a line, with AB ~ is the line segment with length AB and in the direction from point A to point B. Similarly, BA with the same length but direction from point B to point A. Usually, vectors are only equal if ~ 6= BA. ~ they have the same length and same direction. So, usually, AB A line is measured in units of length. Some common units of length are listed in Table 12.1. Table 12.1: Some common units of length and their abbreviations. Unit of Length Abbreviation kilometre km metre m centimetre cm millimetre mm 12.3 Angles An angle is formed when two straight lines meet at a point. The point at which two lines meet is known as a vertex. Angles are labelled with a ˆ on a letter, for example, in Figure 12.3, the angle is at B̂. Angles can also be labelled according to the line segments that make up the angle. For example, in Figure 12.3, the angle is made up when line segments CB and BA meet. So, the angle can be referred to as ∠CBA or ∠ABC. The ∠ symbol is a short method of writing angle in geometry. Angles are measured in degrees which is denoted by ◦ . C B A Figure 12.2: Angle labelled as B̂, ∠CBA or ∠ABC C F A (a) B E (b) G Figure 12.3: Examples of angles. Â = Ê, even though the lines making up the angles are of different lengths. 140 CHAPTER 12. GEOMETRY BASICS 12.3.1 12.3 Measuring angles The size of an angle does not depend on the length of the lines that are joined to make up the angle, but depends only on how both the lines are placed as can be seen in Figure 12.3. This means that the idea of length cannot be used to measure angles. An angle is a rotation around the vertex. Using a Protractor A protractor is a simple tool that is used to measure angles. A picture of a protractor is shown in Figure 12.4. 90◦ ◦ 120 150◦ 60◦ 30◦ 180◦ 0◦ Figure 12.4: Diagram of a protractor. Method: Using a protractor 1. Place the bottom line of the protractor along one line of the angle. 2. Move the protractor along the line so that the centre point on the protractor is at the vertex of the two lines that make up the angle. 3. Follow the second line until it meets the marking on the protractor and read off the angle. Make sure you start measuring at 0◦ . Activity :: Measuring Angles : Use a protractor to measure the following angles: 12.3.2 Special Angles What is the smallest angle that can be drawn? The figure below shows two lines (CA and AB) making an angle at a common vertex A. If line CA is rotated around the common vertex A, down towards line AB, then the smallest angle that can be drawn occurs when the two lines are pointing in the same direction. This gives an angle of 0◦ . 141 12.3 CHAPTER 12. GEOMETRY BASICS C swing point C down towards AB 0◦ A C C A B B b 180◦ A B If line CA is now swung upwards, any other angle can be obtained. If line CA and line AB point in opposite directions (the third case in the figure) then this forms an angle of 180◦ . Important: If three points A, B and C lie on a straight line, then the angle between them is 180◦ . Conversely, if the angle between three points is 180◦ , then the points lie on a straight line. An angle of 90◦ is called a right angle. A right angle is half the size of the angle made by a straight line (180◦ ). We say CA is perpendicular to AB or CA ⊥ AB. An angle twice the size of a straight line is 360◦ . An angle measuring 360◦ looks identical to an angle of 0◦ , except for the labelling. We call this a revolution. C 90◦ A 360◦ B A b C B Figure 12.5: An angle of 90◦ is known as a right angle. Extension: Angles larger than 360◦ All angles larger than 360◦ also look like we have seen them before. If you are given an angle that is larger than 360◦ , continue subtracting 360◦ from the angle, until you get an answer that is between 0◦ and 360◦ . Angles that measure more than 360◦ are largely for mathematical convenience. Important: • Acute angle: An angle ≥ 0◦ and < 90◦ . • Right angle: An angle measuring 90◦ . • Obtuse angle: An angle > 90◦ and < 180◦ . • Straight angle: An angle measuring 180◦ . • Reflex angle: An angle > 180◦ and < 360◦ . • Revolution: An angle measuring 360◦ . These are simply labels for angles in particular ranges, shown in Figure 12.6. Once angles can be measured, they can then be compared. For example, all right angles are 90◦ , therefore all right angles are equal, and an obtuse angle will always be larger than an acute angle. 142 CHAPTER 12. GEOMETRY BASICS C 12.3 reflex A C b acute A b obtuse B A B b B C Figure 12.6: Three types of angles defined according to their ranges. 12.3.3 Special Angle Pairs In Figure 12.7, straight lines AB and CD intersect at point X, forming four angles: X̂1 , X̂2 , X̂3 and X̂4 . C 3 A B 2 X b 1 4 D Figure 12.7: Two intersecting straight lines with vertical angles X̂1 ,X̂3 and X̂2 ,X̂4 . The table summarises the special angle pairs that result. Special Angle adjacent angles linear pair (adjacent angles on a straight line) vertically opposite angles supplementary angles complementary angles Property share a common vertex and a common side adjacent angles formed by two intersecting straight lines that by definition add to 180◦ angles formed by two intersecting straight lines that share a vertex but do not share any sides Example (X̂1 ,X̂2 ), (X̂2 ,X̂3 ), (X̂3 ,X̂4 ), (X̂4 ,X̂1 ) X̂1 + X̂2 X̂2 + X̂3 X̂3 + X̂4 X̂4 + X̂1 = 180◦ = 180◦ = 180◦ = 180◦ X̂1 = X̂3 X̂2 = X̂4 two angles whose sum is 180◦ two angles whose sum is 90◦ Important: The vertically opposite angles formed by the intersection of two straight lines are equal. Adjacent angles on a straight line are supplementary. 12.3.4 Parallel Lines intersected by Transversal Lines Two lines intersect if they cross each other at a point. For example, at a traffic intersection, two or more streets intersect; the middle of the intersection is the common point between the streets. 143 12.3 CHAPTER 12. GEOMETRY BASICS Parallel lines are lines that never intersect. For example the tracks of a railway line are parallel. We wouldn’t want the tracks to intersect as that would be catastrophic for the train! All these lines are parallel to each other. Notice the arrow symbol for parallel. teresting A section of the Australian National Railways Trans-Australian line is perhaps Interesting Fact Fact one of the longest pairs of man-made parallel lines. Longest Railroad Straight (Source: www.guinnessworldrecords.com) The Australian National Railways Trans-Australian line over the Nullarbor Plain, is 478 km (297 miles) dead straight, from Mile 496, between Nurina and Loongana, Western Australia, to Mile 793, between Ooldea and Watson, South Australia. A transversal of two or more lines is a line that intersects these lines. For example in Figure 12.8, AB and CD are two parallel lines and EF is a transversal. We say AB k CD. The properties of the angles formed by these intersecting lines are summarised in the table below. F h C g D a b d A c B e f E Figure 12.8: Parallel lines intersected by a transversal Extension: Euclid’s Parallel Line Postulate If a straight line falling on two straight lines makes the two interior angles on the same side less than two right angles (180◦ ), the two straight lines, if produced indefinitely, will meet on that side. This postulate can be used to prove many identities about the angles formed when two parallel lines are cut by a transversal. 144 CHAPTER 12. GEOMETRY BASICS Name of angle interior angles exterior angles 12.3 Definition the angles that lie inside the parallel lines the angles that lie outside the parallel lines Examples Notes a, b, c and d are interior angles the word interior means inside e, f , g and h are exterior angles the word exterior means outside alternate interior angles the interior angles that lie on opposite sides of the transversal (a,c) and (b,d) are pairs of alternate interior angles, a = c, b = d co-interior angles on the same side co-interior angles that lie on the same side of the transversal (a,d) and (b,c) are interior angles on the same side. a + d = 180◦, b + c = 180◦ corresponding angles the angles on the same side of the transversal and the same side of the parallel lines (a,e), (b,f ), (c,g) and (d,h) are pairs of corresponding angles. a = e, b = f , c = g, d = h Z shape C shape F shape Important: 1. If two parallel lines are intersected by a transversal, the sum of the co-interior angles on the same side of the transversal is 180◦ . 2. If two parallel lines are intersected by a transversal, the alternate interior angles are equal. 3. If two parallel lines are intersected by a transversal, the corresponding angles are equal. 4. If two lines are intersected by a transversal such that any pair of co-interior angles on the same side is supplementary, then the two lines are parallel. 5. If two lines are intersected by a transversal such that a pair of alternate interior angles are equal, then the lines are parallel. 6. If two lines are intersected by a transversal such that a pair of alternate corresponding angles are equal, then the lines are parallel. Exercise: Angles 1. Use adjacent, corresponding, co-interior and alternate angles to fill in all the angles labeled with letters in the diagram alongside: b g 145 d f e a c 30◦ 12.3 CHAPTER 12. GEOMETRY BASICS E A 1 B 2. Find all the unknown angles in the figure alongside: 1 30◦ 2 F 3 1 2 C 2 ◦ 100 D A G 1 3 3 H 1 D X 4x 3. Find the value of x in the figure alongside: x Y x+20◦ Z B C 4. Determine whether there are pairs of parallel lines in the following figures. M O O Q S 1 A 115◦ K Q 2 3 1 3 35◦ 1 2 P N 3 T a) R b) P K T 85◦ U M 1 2 1 3 V c) Y 2 3 N 85 ◦ L C 5. If AB is parallel to CD and AB is parallel to A EF, prove that CD is parallel to EF: E 146 3 L 55◦ B1 45◦ 2 2 D B F R CHAPTER 12. GEOMETRY BASICS 12.4 12.4 Polygons If you take some lines and join them such that the end point of the first line meets the starting point of the last line, you will get a polygon. Each line that makes up the polygon is known as a side. A polygon has interior angles. These are the angles that are inside the polygon. The number of sides of a polygon equals the number of interior angles. If a polygon has equal length sides and equal interior angles then the polygon is called a regular polygon. Some examples of polygons are shown in Figure 12.9. * Figure 12.9: Examples of polygons. They are all regular, except for the one marked * 12.4.1 Triangles A triangle is a three-sided polygon. There are four types of triangles: equilateral, isosceles, right-angled and scalene. The properties of these triangles are summarised in Table 12.2. Properties of Triangles Activity :: Investigation : Sum of the angles in a triangle 1. Draw on a piece of paper a triangle of any size and shape 2. Cut it out and label the angles Â, B̂ and Ĉ on both sides of the paper 3. Draw dotted lines as shown and cut along these lines to get three pieces of paper 4. Place them along your ruler as shown to see that Â + B̂ + Ĉ = 180◦ B A C Important: The sum of the angles in a triangle is 180◦ . 147 C A B 12.4 CHAPTER 12. GEOMETRY BASICS Name Table 12.2: Types of Triangles Diagram Properties C b 60◦ equilateral A 60◦ b All three sides are equal in length and all three angles are equal. 60◦ b B C b Two sides are equal in length. The angles opposite the equal sides are equal. isosceles b A B b B b This triangle has one right angle. The side opposite this angle is called the hypotenuse. po hy te right-angled nu se b b A C All sides and angles are different. scalene (non-syllabus) 148 CHAPTER 12. GEOMETRY BASICS 12.4 B A C Figure 12.10: In any triangle, ∠A + ∠B + ∠C = 180◦ Important: Any exterior angle of a triangle is equal to the sum of the two opposite interior angles. An exterior angle is formed by extending any one of the sides. ˆ + BCA ˆ = CBD ˆ BAC D B ˆ + CBA ˆ = BCD ˆ BAC B b b C B b b A ˆ + BAC ˆ = ACD ˆ ABC A b b b C D A b b C D Figure 12.11: In any triangle, any exterior angle is equal to the sum of the two opposite interior angles. 149 12.4 CHAPTER 12. GEOMETRY BASICS Congruent Triangles Label Description RHS If the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and the respective side of another triangle then the triangles are congruent. SSS If three sides of a triangle are equal in length to the same sides of another triangle then the two triangles are congruent SAS If two sides and the included angle of one triangle are equal to the same two sides and included angle of another triangle, then the two triangles are congruent. AAS If one side and two angles of one triangle are equal to the same one side and two angles of another triangle, then the two triangles are congruent. Diagram Similar Triangles Description Diagram a If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar. If all pairs of corresponding sides of two triangles are in proportion, then the triangles are similar. a y x z q p r x p 150 c b c b = y q = z r CHAPTER 12. GEOMETRY BASICS 12.4 The theorem of Pythagoras b If △ABC is right-angled (B̂ = 90◦ ) then b 2 = a 2 + c2 A c b c C a B Converse: If b2 = a2 + c2 , then △ABC is right-angled (B̂ = 90◦ ). a Exercise: Triangles 1. Calculate the unknown variables in each of the following figures. All lengths are in mm. a) N b) N x c) x 30o y 68o 36o y 68o x O P O P O 68o P N d) N N x 19 P e) R 116 O 15 x 76 f) N R 12 P S g) 14 x 20 O N R x 5 15 9 P P O S 21 T y O 2. State whether or not the following pairs of triangles are congruent or not. Give reasons for your answers. If there is not enough information to make a descision, say why. 151 12.4 CHAPTER 12. GEOMETRY BASICS B E B b) a) C A B C A D D E B c) E d) C C D A B D A e) b A C b D 12.4.2 Quadrilaterals A quadrilateral is any polygon with four sides. The basic quadrilaterals are the trapezium, parallelogram, rectangle, rhombus, square and kite. Name of quadrilateral trapezium parallelogram rectangle rhombus square kite Figure Figure 12.12 Figure 12.13 Figure 12.14 Figure 12.15 Figure 12.16 Figure 12.17 Table 12.3: Examples of quadrilaterals. Trapezium A trapezium is a quadrilateral with one pair of parallel opposite sides. It may also be called a trapezoid. A special type of trapezium is the isosceles trapezium, where one pair of opposite sides is parallel, the other pair of sides is equal in length and the angles at the ends of each parallel side are equal. An isosceles trapezium has one line of symmetry and its diagonals are equal in length. isosceles trapezium Figure 12.12: Examples of trapeziums. 152 CHAPTER 12. GEOMETRY BASICS 12.4 Parallelogram A trapezium with both sets of opposite sides parallel is called a parallelogram. A summary of the properties of a parallelogram is: • Both pairs of opposite sides are parallel. • Both pairs of opposite sides are equal in length. • Both pairs of opposite angles are equal. • Both diagonals bisect each other (i.e. they cut each other in half). D b b / C /// b /// / A b b B Figure 12.13: An example of a parallelogram. Rectangle A rectangle is a parallelogram that has all four angles equal to 90◦ . A summary of the properties of a rectangle is: • Both pairs of opposite sides are parallel. • Both pairs of opposite sides are of equal length equal. • Both diagonals bisect each other. • Diagonals are equal in length. • All angles are right angles. D b / b C b B / b / / A b Figure 12.14: Example of a rectangle. Rhombus A rhombus is a parallelogram that has all four side of equal length. A summary of the properties of a rhombus is: • Both pairs of opposite sides are parallel. • All sides are equal in length. 153 12.4 CHAPTER 12. GEOMETRY BASICS • Both pairs of opposite angles equal. • Both diagonals bisect each other at 90◦ . • Diagonals of a rhombus bisect both pairs of opposite angles. b x • / • x /// b D C b x A b / /// • x • b B Figure 12.15: An example of a rhombus. A rhombus is a parallelogram with all sides equal. Square A square is a rhombus that has all four angles equal to 90◦ . A summary of the properties of a rhombus is: • Both pairs of opposite sides are parallel. • All sides are equal in length. • All angles are equal to 90◦ . • Both pairs of opposite angles equal. • Both diagonals bisect each other at 90◦ . • Diagonals are equal in length. • Diagonals bisect both pairs of opposite angles (ie. all 45◦ ). D b • • • b C b B • / / A • b / / b • • • Figure 12.16: An example of a square. A square is a rhombus with all angles equal to 90◦ . Kite A kite is a quadrilateral with two pairs of adjacent sides equal. A summary of the properties of a kite is: • Two pairs of adjacent sides are equal in length. 154 CHAPTER 12. GEOMETRY BASICS 12.4 • One pair of opposite angles are equal where the angles must be between unequal sides. • One diagonal bisects the other diagonal and one diagonal bisects one pair of opposite angles. • Diagonals intersect at right-angles. C b x x b / b b / A B •• b D Figure 12.17: An example of a kite. 12.4.3 Other polygons There are many other polygons, some of which are given in the table below. Sides 5 6 7 8 10 15 Name pentagon hexagon heptagon octagon decagon pentadecagon Table 12.4: Table of some polygons and their number of sides. octagon nonagon decagon pentagon hexagon heptagon Figure 12.18: Examples of other polygons. 155 12.4 12.4.4 CHAPTER 12. GEOMETRY BASICS Extra Angles of regular polygons You can calculate the size of the interior angle of a regular polygon by using: Â = n−2 × 180◦ n (12.1) where n is the number of sides and Â is any angle. Areas of Polygons 1. Area of triangle: 1 2× 2. Area of trapezium: h base × perpendicular height 1 2× h (sum of k sides) × perpendicular height h 3. Area of parallelogram and rhombus: base × perpendicular height b 4. Area of rectangle: length × breadth l s 5. Area of square: length of side × length of side s r 6. Area of circle: π x radius2 Exercise: Polygons 1. For each case below, say whether the statement is true or false. For false statements, give a counter-example to prove it: (a) All squares are rectangles (b) All rectangles are squares (c) All pentagons are similar (d) All equilateral triangles are similar (e) All pentagons are congruent (f) All equilateral triangles are congruent 2. Find the areas of each of the given figures - remember area is measured in square units (cm2 , m2 , mm2 ). 156 CHAPTER 12. GEOMETRY BASICS a) 12.5 b) c) 10cm 10cm 10cm 7cm d) b 5cm 5cm 3cm e) f) 6cm 10cm 5cm 12cm g) h) 10cm 16cm 15cm 9cm 12.5 5cm 8cm 21cm Exercises 1. Find all the pairs of parallel lines in the following figures, giving reasons in each case. A N B 62 M ◦ 137◦ 57◦ 62◦ (a) D C 123◦ G H (b) 120◦ 60◦ K 60◦ L (c) 2. Find a, b, c and d in each case, giving reasons. P b a Q d c S (a) 73◦ R 157 O P 12.5 CHAPTER 12. GEOMETRY BASICS W K ◦ 100 a A B L b C c d N d a U V b F (c) X O (b) c 50◦ D M E 45◦ T (a) Which of the following claims are true? Give a counter-example for those that are incorrect. i. All equilateral triangles are similar. ii. All regular quadrilaterals are similar. iii. In any △ABC with ∠ABC = 90◦ we have AB 3 + BC 3 = CA3 . iv. All right-angled isosceles triangles with perimeter 10 cm are congruent. v. All rectangles with the same area are similar. (b) Say which of the following pairs of triangles are congruent with reasons. D A F C E i. B G L J K I ii. H P N iii. M R Q O U R V S T iv. Q (c) For each pair of figures state whether they are similar or not. Give reasons. 158 CHAPTER 12. GEOMETRY BASICS (a) 12.5 A P √ 2 2 2 45◦ 3 45◦ C B Q (b) H J 7,5 W 5 X 5 L 12.5.1 R 3 Z K Y Challenge Problem 1. Using the figure below, show that the sum of the three angles in a triangle is 180◦ . Line DE is parallel to BC. A D E d e a B c b C 159 12.5 CHAPTER 12. GEOMETRY BASICS 160 Appendix A GNU Free Documentation License Version 1.2, November 2002 c 2000,2001,2002 Free Software Foundation, Inc. Copyright 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA Everyone is permitted to copy and distribute verbatim copies of this license document, but changing it is not allowed. PREAMBLE The purpose of this License is to make a manual, textbook, or other functional and useful document “free” in the sense of freedom: to assure everyone the effective freedom to copy and redistribute it, with or without modifying it, either commercially or non-commercially. 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