Wefwef

Review Notes for IB Standard Level Math
© 2015-2016, Steve Muench steve.muench@gmail.com @stevemuench
Please feel free to share the link to these notes http://bit.ly/ib-sl-maths-review-notes or my worked solutions to the November 2014 exam http://bit.ly/ib-sl-maths-nov-2014 or my worked solutions to the May 2015 (Timezone 2) exam http://bit.ly/ib-sl-maths-may-2015-tz2 or my worked solutions to the November 2015 exam https://bit.ly/ib-sl-maths-nov-2015 with any student you believe might beneﬁt from them.
If you downloaded these notes from a source other than the bit.ly link above, please check there to make sure you are reading the latest version. It may contain additional content and important corrections!
April 8, 2016

1

Contents
1 Algebra
1.1 Rules of Basic Operations . . . . . . . . . . . . . . . . . . . . .
1.2 Rules of Roots . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Rules of Exponents . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Allowed and Disallowed Calculator Functions During the Exam
1.5 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Arithmetic Sequences and Series . . . . . . . . . . . . . . . . .
1.7 Sum of Finite Arithmetic Series (u1 + · · · + un ) . . . . . . . . .
1.8 Partial Sum of Finite Arithmetic Series (uj + · · · + un ) . . . . .
1.9 Geometric Sequences and Series . . . . . . . . . . . . . . . . . .
1.10 Sum of Finite Geometric Series . . . . . . . . . . . . . . . . . .
1.11 Sum of Inﬁnite Geometric Series . . . . . . . . . . . . . . . . .
1.11.1 Example Involving Sum of Inﬁnite Geometric Series . .
1.12 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.12.1 Sigma Notation for Arithmetic Series . . . . . . . . . . .
1.12.2 Sigma Notation for Geometric Series . . . . . . . . . . .
1.12.3 Sigma Notation for Inﬁnite Geometric Series . . . . . .
1.12.4 Deﬁning Functions Using Sigma Notation . . . . . . . .
1.13 Applications: Compound Interest . . . . . . . . . . . . . . . . .
1.14 Applications: Population Growth . . . . . . . . . . . . . . . . .
1.15 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.16 Using Logarithms to Solve Equations . . . . . . . . . . . . . . .
1.17 Using Exponentiation to Solve Equations . . . . . . . . . . . .
1.18 Logarithm Facts Involving 0 and 1 . . . . . . . . . . . . . . . .
1.19 Laws of Exponents and Logarithms . . . . . . . . . . . . . . . .
1.20 Change of Base . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.21 Powers of Binomials and Pascal’s Triangle . . . . . . . . . . . .
1.22 Expansion of (a + b)n . . . . . . . . . . . . . . . . . . . . . . .
1.23 The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . .
1.23.1 Using The Binomial Theorem for a Single Term . . . . .
1.23.2 Example of Using Binomial Theorem . . . . . . . . . . .
1.24 Solving Systems of Three Linear Equations Using Substitution
1.25 Solving Systems of Three Linear Equations Using Technology .

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2 Functions and Equations
2.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Union and Intersection of Sets . . . . . . . . . . . . . .
2.3 Common Sets of Numbers . . . . . . . . . . . . . . . .
2.4 Intervals of Real Numbers . . . . . . . . . . . . . . . .
2.5 Concept of Function . . . . . . . . . . . . . . . . . . .
2.6 Graph of a Function . . . . . . . . . . . . . . . . . . .
2.7 Domain of a Function . . . . . . . . . . . . . . . . . .
2.8 Range of a Function . . . . . . . . . . . . . . . . . . .
2.9 Composing One Function with Another . . . . . . . .
2.10 Identity Function . . . . . . . . . . . . . . . . . . . . .
2.11 Inverse Function . . . . . . . . . . . . . . . . . . . . .
2.12 Determining the Inverse Function as Reﬂection in Line
2.13 Determining the Inverse Function Analytically . . . .
2.14 Drawing and Analyzing Graphs with Your Calculator
2.14.1 Drawing the Graph of a Function . . . . . . . .

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2.15

2.16

2.17
2.18
2.19
2.20
2.21
2.22

2.14.2 Restricting the Domain of a Graph . . . . . . . . . . . . . . . . . . .
2.14.3 Zooming Graph to See Exactly What You Want . . . . . . . . . . .
2.14.4 Finding a Maximum Value in an Interval . . . . . . . . . . . . . . .
2.14.5 Finding a Minimum Value Value in an Interval . . . . . . . . . . . .
2.14.6 Finding the x-Intercepts or “Zeros” of a Graph in an Interval . . . .
2.14.7 Finding the y-Intercept of a Graph . . . . . . . . . . . . . . . . . . .
2.14.8 Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14.9 Graphing Vertical Lines . . . . . . . . . . . . . . . . . . . . . . . . .
2.14.10 Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14.11 Tips to Compute Horizontal Asymptotes of Rational Functions . . .
2.14.12 Graphing Horizontal Lines . . . . . . . . . . . . . . . . . . . . . . . .
2.14.13 Symmetry: Odd Functions . . . . . . . . . . . . . . . . . . . . . . .
2.14.14 Symmetry: Even Functions . . . . . . . . . . . . . . . . . . . . . . .
2.14.15 Solving Equations Graphically . . . . . . . . . . . . . . . . . . . . .
Transformations of Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.15.1 Horizontal and Vertical Translations . . . . . . . . . . . . . . . . . .
2.15.2 Vertical Reﬂection . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.15.3 Horizontal Reﬂection . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.15.4 Vertical Stretch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.15.5 Horizontal Stretch . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.15.6 Order Matters When Doing Multiple Transformations in Sequence .
2.15.7 Graphing the Result of a Sequence of Transformations . . . . . . . .
2.15.8 Determining Point Movement Under a Sequence of Transformations
2.15.9 Vector Notation for Function Translation . . . . . . . . . . . . . . .
Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.16.1 Using the Quadratic Formula to Find Zeros of Quadratic Function .
2.16.2 Finding the Vertex If You Know the Zeros . . . . . . . . . . . . . . .
2.16.3 Graph and Axis of Symmetry . . . . . . . . . . . . . . . . . . . . . .
2.16.4 Computing the Vertex From the Coeﬃcients . . . . . . . . . . . . .
2.16.5 Using the Discriminant to Find the Number of Zeros . . . . . . . . .
2.16.6 Y-Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.16.7 X-Intercept Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.16.8 Completing the Square to Get Binomial Squared Form . . . . . . . .
2.16.9 Vertex (h, k) Form . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Reciprocal Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Continuously Compounded Interest . . . . . . . . . . . . . . . . . . . . . . .
Continuous Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . .
Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Circular Functions and Trigonometry
3.1 Understanding Radians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 Degrees Represent a Part of a Circular Path . . . . . . . . . . . . . . . .
3.1.2 Computing the Fraction of a Complete Revolution an Angle Represents
3.1.3 Attempting to Measure an Angle Using Distance . . . . . . . . . . . . .
3.1.4 Arc Distance on the Unit Circle Uniquely Identiﬁes an Angle θ . . . . .
3.1.5 Computing the Fraction of a Complete Revolution for Angle in Radians
3.2 Converting Between Radians and Degrees . . . . . . . . . . . . . . . . . . . . .
3.2.1 Converting from Degrees to Radians . . . . . . . . . . . . . . . . . . . .
3.2.2 Converting from Radians to Degrees . . . . . . . . . . . . . . . . . . . .

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3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
3.11
3.12
3.13

3.14
3.15

3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23

Length of an Arc Subtended by an Angle . . . . . . . . . . . .
Inscribed and Central Angles that Subtend the Same Arc . . .
Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . .
Deﬁnition of cos θ and sin θ . . . . . . . . . . . . . . . . . . . .
Interpreting cos θ and sin θ on the Unit Circle . . . . . . . . . .
Radian Angle Measures Can Be Both Positive and Negative . .
Remembering the Exact Values of Key Angles on Unit Circle .
The Pythagorean Identity . . . . . . . . . . . . . . . . . . . . .
Double Angle Identities . . . . . . . . . . . . . . . . . . . . . .
Deﬁnition of tan θ . . . . . . . . . . . . . . . . . . . . . . . . .
Using a Right Triangle to Solve Trigonometric Problems . . . .
3.13.1 Using Right Triangle with an Acute Angle . . . . . . . .
3.13.2 Using Right Triangle with an Obtuse Angle . . . . . . .
Using Inverse Trigonometric Functions on Your Calculator . . .
Circular Functions sin, cos, and tan . . . . . . . . . . . . . . .
3.15.1 The Graph of sin x . . . . . . . . . . . . . . . . . . . . .
3.15.2 The Graph of cos x . . . . . . . . . . . . . . . . . . . . .
3.15.3 The Graph of tan x . . . . . . . . . . . . . . . . . . . . .
3.15.4 Transformations of Circular Functions . . . . . . . . . .
3.15.5 Using Transformation to Highlight Additional Identities
3.15.6 Determining Period from Minimum and Maximum . . .
Applications of the sin Function: Tide Example . . . . . . . . .
Applications of the cos Function: Ferris Wheel Example . . . .
Solving Trigonometric Equations in a Finite Interval . . . . . .
Solving Quadratic Equations in sin, cos, and tan . . . . . . . .
Solutions of Right Triangles . . . . . . . . . . . . . . . . . . . .
The Cosine Rule . . . . . . . . . . . . . . . . . . . . . . . . . .
The Sine Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Area of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . .

4 Vectors
4.1 Vectors as Displacements in the Plane . . . . . . . . . . .
4.2 Vectors as Displacements in Three Dimensions . . . . . .
4.3 Terminology: Tip and Tail . . . . . . . . . . . . . . . . . .
4.4 Representation of Vectors . . . . . . . . . . . . . . . . . .
4.5 Magnitude of a Vector . . . . . . . . . . . . . . . . . . . .
4.6 Multiplication of a Vector by a Scalar . . . . . . . . . . .
4.7 Negating a Vector . . . . . . . . . . . . . . . . . . . . . .
4.8 Sum of Vectors . . . . . . . . . . . . . . . . . . . . . . . .
4.9 Diﬀerence of Vectors . . . . . . . . . . . . . . . . . . . . .
4.10 Unit Vectors . . . . . . . . . . . . . . . . . . . . . . . . .
4.11 Scaling Any Vector to Produce a Parallel Unit Vector . .
4.12 Position Vectors . . . . . . . . . . . . . . . . . . . . . . .
4.13 Determining Whether Vectors are Parallel . . . . . . . . .
4.14 Finding Parallel Vector with Certain Fixed Length . . . .
4.15 Scalar (or “Dot”) Product of Two Vectors . . . . . . . . .
4.16 Perpendicular Vectors . . . . . . . . . . . . . . . . . . . .
4.17 Base Vectors for Two Dimensions . . . . . . . . . . . . . .
4.18 Base Vectors for Three Dimensions . . . . . . . . . . . . .
4.19 The Angle Between Two Vectors . . . . . . . . . . . . . .
4.20 Vector Equation of a Line in Two and Three Dimensions .

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62
63
63
64
65
65
66
67
68
68
68
68
69
70
71
71
71
72
72
73
74
74
76
78
78
79
80
80
81

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82
82
82
82
83
83
85
85
86
86
88
88
89
89
90
90
90
90
91
91
92

4.21
4.22
4.23
4.24
4.25

Vector Equation of Line Passing Through Two Points .
Finding the Cartesian Equation from a Vector Line . . .
The Angle Between Two Vector Lines . . . . . . . . . .
Distinguishing Between Coincident and Parallel Lines .
Finding the Point of Intersection of Two Lines . . . . .
4.25.1 Finding Intersection Between a Line and an Axis

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93
94
94
95
95
96

5 Statistics
5.1 Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.1 Population versus Sample . . . . . . . . . . . . . . . . . .
5.1.2 Discrete Data versus Continuous Data . . . . . . . . . . .
5.2 Presentation of Data . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Frequency Distribution Tables . . . . . . . . . . . . . . .
5.2.2 Frequency Histograms with Equal Class Intervals . . . . .
5.3 Mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Median . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6 Cumulative Frequency . . . . . . . . . . . . . . . . . . . . . . . .
5.6.1 Cumulative Frequency Table . . . . . . . . . . . . . . . .
5.6.2 Cumulative Frequency Graphs . . . . . . . . . . . . . . .
5.7 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7.1 Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7.2 Quartiles . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7.3 Interquartile Range . . . . . . . . . . . . . . . . . . . . . .
5.7.4 Box and Whiskers Plots . . . . . . . . . . . . . . . . . . .
5.7.5 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7.6 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7.7 Standard Deviation . . . . . . . . . . . . . . . . . . . . .
5.8 Use Cumulative Frequency Graph to Find Median and Quartiles
5.9 Computing Statistics Using Your Calculator . . . . . . . . . . . .
5.9.1 Calculating Statistics for a Single List . . . . . . . . . . .
5.9.2 Calculating Statistics for a List with Frequency . . . . . .
5.10 Linear Correlation of Bivariate Data . . . . . . . . . . . . . . . .
5.10.1 Scatter Diagrams . . . . . . . . . . . . . . . . . . . . . . .
5.10.2 Pearson’s Product-Moment Correlation Coeﬃcient r . . .
5.10.3 Lines of Best Fit . . . . . . . . . . . . . . . . . . . . . . .
5.10.4 Equation of the Regression Line . . . . . . . . . . . . . .

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98
98
98
98
98
98
99
99
100
101
101
101
101
103
103
103
103
103
104
104
105
105
105
106
106
107
107
108
108
108

6 Probability
6.1 Probability Concepts . . . . . . . . . . . . . . . . . . . . .
6.2 Probability of an Event . . . . . . . . . . . . . . . . . . .
6.3 Probability of an Event’s Not Occurring . . . . . . . . . .
6.4 Independent, Dependent, and Mutually Exclusive Events
6.5 Probability of A and B for Independent Events . . . . . .
6.6 Probability of A or B for Mutually Exclusive Events . . .
6.7 Probability of Mutually Exclusive Events . . . . . . . . .
6.8 Probability of A or B for Non-Mutually Exclusive Events
6.9 Lists and Tables of Outcomes . . . . . . . . . . . . . . . .
6.10 Conditional Probability of Dependent Events . . . . . . .
6.11 Testing for Independent Events . . . . . . . . . . . . . . .
6.12 Venn Diagrams . . . . . . . . . . . . . . . . . . . . . . . .
6.13 Tree Diagrams . . . . . . . . . . . . . . . . . . . . . . . .

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110
110
110
110
110
111
112
112
112
112
113
113
113
113

5

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6.14 Probabilities With and Without Replacement . . . . . . . . . . . . . . . . .
6.15 Discrete Random Variables and Their Probability Distributions . . . . . . .
6.15.1 Explicitly Listed Probabilities . . . . . . . . . . . . . . . . . . . . . .
6.15.2 Probability Distribution Given by a Function . . . . . . . . . . . . .
6.15.3 Explicit Probability Distribution Involving an Unknown . . . . . . .
6.16 Expected Value (Mean), E(X) for Discrete Data . . . . . . . . . . . . . . .
6.16.1 Expected Value for a “Fair” Game . . . . . . . . . . . . . . . . . . .
6.17 Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.17.1 Using the Calculator for Binomial Distribution Problems . . . . . .
6.17.2 Tip for Calculators Whose binomCdf Does Not Have Lower Bound .
6.18 Mean and Variance of the Binomial Distribution . . . . . . . . . . . . . . .
6.18.1 Example of Mean and Variance of Binomial Distribution . . . . . . .
6.19 Normal Distribution and Curves . . . . . . . . . . . . . . . . . . . . . . . .
6.20 Graphing the Normal Distribution and Computing Bounded Area . . . . .
6.21 Standardizing Normal Variables to Get z-values (z-scores) . . . . . . . . . .
6.22 Using the Calculator for Normal Distribution Problems . . . . . . . . . . .
6.22.1 Tip for Calculators Whose normalCdf Does Not Have Lower Bound
6.23 Using Inverse Normal Cumulative Density Function . . . . . . . . . . . . .
6.24 Determining z Value and σ from the Probability . . . . . . . . . . . . . . .

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7 Calculus
7.1 Overview of Concepts for Diﬀerential Calculus . . . . . . . . . . . . . . . . .
7.1.1 Rate of Change in Distance . . . . . . . . . . . . . . . . . . . . . . . .
7.1.2 Derivative Gives Instantaneous Rate of Change Using Tangent Lines .
7.1.3 Relationships Between Function and Derivative . . . . . . . . . . . . .
7.1.4 Summary of Derivative Concepts . . . . . . . . . . . . . . . . . . . . .
7.2 Equations of Tangents and Normals . . . . . . . . . . . . . . . . . . . . . . .
7.3 Notation for Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4 Graphing Derivatives with Your Calculator . . . . . . . . . . . . . . . . . . .
7.4.1 Graphing Derivatives on the TI-Nspire CX . . . . . . . . . . . . . . .
7.4.2 Graphing Derivatives on the TI-84 Silver Edition . . . . . . . . . . . .
7.5 Computing Derivatives at a Point with Your Calculator . . . . . . . . . . . .
7.5.1 Computing Derivative at a Point on the TI-NSpire CX . . . . . . . . .
7.5.2 Computing Derivative at a Point on the TI-84 Silver Edition . . . . .
7.6 Rules for Computing Derivatives . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.1 Derivative of xn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.2 Derivative of sin x . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.3 Derivative of cos x . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.4 Derivative of tan x . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.5 Derivative of ex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6.6 Derivative of ln x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.7 Diﬀerentiating a Scalar Multiple . . . . . . . . . . . . . . . . . . . . . . . . .
7.8 Diﬀerentiating a Sum or Diﬀerence . . . . . . . . . . . . . . . . . . . . . . . .
7.9 Example Using Derivation Rules . . . . . . . . . . . . . . . . . . . . . . . . .
7.10 “Chain Rule”: Diﬀerentiating Composed Functions . . . . . . . . . . . . . . .
7.11 “Product Rule”: Diﬀerentiating Product of Functions . . . . . . . . . . . . .
7.12 “Quotient Rule”: Diﬀerentiating Quotient of Functions . . . . . . . . . . . . .
7.13 Using the First Derivative to Find Local Maxima and Minima . . . . . . . . .
7.14 Analyzing Zeros of Derivative Graph to Find Maxima and Minima . . . . . .
7.15 Using the Second Derivative to Determine Concavity and Points of Inﬂection

6

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114
115
115
115
116
116
118
118
119
120
121
121
122
122
123
124
125
125
125

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127
127
127
129
130
132
132
133
133
134
134
135
135
136
136
136
137
137
137
137
137
137
137
138
138
140
143
145
146
147

7.16 Overview of Concepts for Integral Calculus . . . . . . . . . . . . . . . . . . .
7.16.1 Area Under Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.16.2 Relationship Between Derivation and Integration . . . . . . . . . . . .
7.17 Anti-Diﬀerentiation to Compute Integrals . . . . . . . . . . . . . . . . . . . .
7.17.1 Indeﬁnite Integration as Anti-Diﬀerentiation . . . . . . . . . . . . . .
7.18 Rules for Computing the Indeﬁnite Integral (Antiderivative) . . . . . . . . . .
7.18.1 Indeﬁnite Integral of xn (n ∈ Q) . . . . . . . . . . . . . . . . . . . . .
7.18.2 Indeﬁnite Integral of sin x . . . . . . . . . . . . . . . . . . . . . . . . .
7.18.3 Indeﬁnite Integral of cos x . . . . . . . . . . . . . . . . . . . . . . . . .
1
7.18.4 Indeﬁnite Integral of x . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.18.5 Indeﬁnite Integral of ex . . . . . . . . . . . . . . . . . . . . . . . . . .
7.19 Integrating Constant Multiple of a Function . . . . . . . . . . . . . . . . . . .
7.20 Integrating Sums and Diﬀerences of Functions . . . . . . . . . . . . . . . . . .
7.21 Integration Using Substitution . . . . . . . . . . . . . . . . . . . . . . . . . .
7.21.1 Simple Example Involving Linear Factor . . . . . . . . . . . . . . . . .
7.21.2 Example Involving Additional Factor of x . . . . . . . . . . . . . . . .
7.21.3 Example Involving Other Additional Factors . . . . . . . . . . . . . .
7.21.4 Summary of Strategy for Integration by Substitution . . . . . . . . . .
7.22 Using Additional Information to Determine Constant of Integration . . . . .
7.23 Computing the Deﬁnite Integral . . . . . . . . . . . . . . . . . . . . . . . . .
7.23.1 Computing the Deﬁnite Integral Analytically . . . . . . . . . . . . . .
7.23.2 Computing the Deﬁnite Integral Using Technology . . . . . . . . . . .
7.23.3 Computing Normal Distribution Probability Using a Deﬁnite Integral
7.24 Displacement s, Velocity v, and Acceleration a . . . . . . . . . . . . . . . . .
7.25 Determining Position Function from Acceleration . . . . . . . . . . . . . . . .
7.26 Total Area Under a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.27 Area Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.27.1 Area Between Curves Using Calculator . . . . . . . . . . . . . . . . .
7.27.2 Area Between Curves Analytically (Without the Calculator) . . . . .
7.28 Net Change in Displacement versus Total Distance Traveled . . . . . . . . . .
7.28.1 Diﬀerence Between Distance and Displacement . . . . . . . . . . . . .
7.28.2 Computing Total Distance Traveled . . . . . . . . . . . . . . . . . . .

7

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147
148
149
150
151
151
152
152
153
153
153
153
154
154
155
156
157
157
157
159
159
160
160
161
161
162
163
163
163
164
164
165

1 Algebra
1.1 Rules of Basic Operations
You need to be familiar with the rules of basic operations shown in Table 1. a b

a (b + c) = ab + ac a b

c d +

=

a b ·

d d +

c d ·

b b =

ad+cb bd ·

c d =

ac bd − (a + b) = −a − b

− (a − b) = b − a

(a + b)2 = a2 + 2ab + b2

(a − b)2 = a2 − 2ab + b2

(a + b) (a − b) = a2 − b2

(a + b) (c + d) = ac + ad + bc + bd
Table 1: Basic Operations

1.2 Rules of Roots
You need to be familiar with the rules of roots shown in Table 2.
√

ab =

−p q a

=

√ √ a b
1
√

( q a)

p
√
a

p

·

√ a √a b = b
√
√
√a = p a a a

1

aq =

√ q a

√ 2
( a) = a

p
√ p a q = ( q a)

√ q
( q a) = a

Table 2: Rules of Roots

1.3 Rules of Exponents
You need to be familiar with the rules of exponents shown in Table 3. ab · ac = ab+c a c d ab ac =

ab

ac dc = a(b−c)

c

= ab·c

a(−c) = ab de

·

dc af =

ab af ·

dc de 1 ac = ab−f · dc−e =

ab−f de−c Table 3: Rules of Exponents

1.4 Allowed and Disallowed Calculator Functions During the Exam
The International Baccalaureate documentation for examiners dated 18/11/2015, provides the following guidance about the functions of the TI NSpire calculator during an exam. The following options in “Press to Test” mode must be ticked and therefore blocked: Limit geometry functions; Disable function and conic grab and move; Disable vector functions including eigenvectors and eigenvalues; Disable isPrime function; Disable diﬀerential equation graphing; Disable 3D graphing; Disable implicit graphing, conic templates, conic analysis, and geometric conics. The following options in “Press to Test” mode must be unticked and therefore allowed: Disable inequality graphing; Limit trigonometric functions; logb x template and summation functions;
Disable Polynomial Root Finder and Simultaneous Equation Solver.

8

1.5 Sequences and Series
A sequence is an ordered list of numbers. The list can be ﬁnite or inﬁnite. We name the sequence with a letter like u, and refer to an individual element (or “term”) of this ordered list using a subscript that represents the ordinal position the particular element occupies in the list.
The ﬁrst term in the sequence u is denoted by u1 , the second term by u2 , and so on. The nth term in the sequence is un .
You describe a sequence by giving a formula for its nth term. This could be something simple like the sequence each of whose terms is the number 1: un = 1
More typically, the formula for the nth term involves the variable n in some way. It might be involved in calculating the value of the term like: un = n2

or

un =

1 n The n can also appear in the subscript to reference the values of previous terms. For example, after deﬁning the value of the ﬁrst and second term, this sequence deﬁnes the nth term to be the sum of the previous two terms: u1 = 1 u2 = 1 un = un−1 + un−2
A series is the sum of the terms in a sequence.

1.6 Arithmetic Sequences and Series
A sequence is arithmetic if you compute its nth term by adding a constant value d to the previous element: un = un−1 + d
This value d is known as the common diﬀerence because if you take any term in the sequence and subtract the value of the preceding one, the diﬀerence is d. This is a result of rearranging the previous equation to get: un − un−1 = d
Given the above deﬁnition for un involving d, we can also represent the nth element of an arithmetic sequence as the ﬁrst element to which the common diﬀerence d has been added n − 1 times. un = u1 + (n − 1)d
An arithmetic series is the sum of an arithmetic sequence.

1.7 Sum of Finite Arithmetic Series (u1 + · · · + un )
It only makes sense to talk about the sum of a ﬁnite arithmetic sequence. Since the elements of an inﬁnite arithmetic sequence continue growing by the common diﬀerence, we don’t try to compute the sum of the corresponding inﬁnite arithmetic series because its value would always be ∞ (or −∞ if d is a negative number).

9

To calculate the sum of the ﬁrst n terms of a ﬁnite arithmetic sequence we use the formula:
Sn =

n
(u1 + un )
2

This formula comes from observing that the sum of the pair of u1 + un (the ﬁrst and last terms) is the same as the sum of u2 + un−1 (the second and penultimate terms), which is the same as the sum of u3 + un−2 (the third and antepenultimate terms), etc. Since we’re pairing up the n terms, we end up with pairs having the same sum as u1 + un . By multiplying this sum by
2
the number of pairs, we get the sum of all of the terms of the series up to n. Since above we saw that un = u1 + (n − 1)d, we can also write this formula as:
Sn =

n n (u1 + (u1 + (n − 1)d)) = (2u1 + (n − 1)d)
2
2

1.8 Partial Sum of Finite Arithmetic Series (uj + · · · + un )
Sometimes you might need to calculate the sum of a series starting at a jth term other than the ﬁrst term, that is where j > 1. These kind of problems are easy once you understand that the sum of the ﬁrst n terms equals the sum of the ﬁrst j − 1 terms plus the sum of the terms from the jth position to the nth. In other words,
Sn = (u1 + · · · + uj−1 ) + (uj + · · · + un )
Sn = Sj−1 + Sj...n
Rearranging the equation we have:
Sj...n = Sn − Sj−1
Therefore, to ﬁnd the sum of the series from the jth term through the nth term (inclusive), calculate Sn and subtract from it the value of Sj−1 .

1.9 Geometric Sequences and Series
A sequence is geometric if you compute its nth term by multiplying the previous element by a constant value r: un = un−1 · r
This value r is known as the common ratio because if you take any term in the sequence and divide by the value of the preceding one, the ratio is r. This is a result of rearranging the previous equation to get: un =r un−1 Given the above deﬁnition for un involving r, we can also represent the nth element of an arithmetic sequence as the ﬁrst element by which the common ratio r has been multiplied n − 1 times. un = u1 · rn−1
A geometric series is the sum of a geometric sequence.

10

1.10 Sum of Finite Geometric Series
To calculate the sum of the ﬁrst n terms of a geometric series, use the formula:
Sn =

u1 (rn − 1)
,
r−1

r=1

1.11 Sum of Inﬁnite Geometric Series
If the absolute value of the common ratio is less than one, then each time we compute the next term of a geometric sequence it is some fraction of the previous term. In other words, the new term is smaller in absolute value than the previous term. Another way to think of this is that the new term is closer to zero than the previous term was. This means we can calculate the sum of the inﬁnite geometric series because the terms tend to zero as n gets larger and larger.
To calculate the sum of an inﬁnite geometric series, use the formula:
S∞ =

u1
,
1−r

|r| < 1

1.11.1 Example Involving Sum of Inﬁnite Geometric Series
As shown in Figure 1, the sides of an outermost square are 16 units. To create a smaller square, connect midpoints of the original square. Consider the sequence An of the areas of the shaded triangles in the lower-left corner of each square, formed when the next smaller square is drawn.
Find the sum of the areas of these shaded triangles when this process continues inﬁnitely.

Figure 1: Progressively Smaller Triangle Areas
Since the original side length is 16, the legs of the ﬁrst shaded triangle are 8, and the area
1
A1 = 2 · 8 · 8 = 32. Using Pythagorean Theorem, we can see that for a square of side-length a,
√
√
√
its hypotenuse will always be a2 + a2 = a 2, so the hypotenuse of the ﬁrst triangle is 8 2.
√
This means that the legs of the √ second triangle are half of this, or 4 2. Therefore the area of
√
the second triangle is A2 = 1 · 4 2 · 4 2 = 16. The hypotenuse of the second shaded triangle
2
√ √ is 4 2 2 = 8, so the legs of the third triangle are half of this: 4. Hence, the area of the third
√
shaded triangle is A3 = 1 · 4 · 4 = 8. The hypotenuse of the third shaded triangle is 4 2, so
2
√ the legs of √ third triangle are half of this: 2 2. So, the area of the fourth shaded triangle the √ is A4 = 1 ·2 2·2 2 = 4. So, the pattern that is emerging is that A = 32, 16, 8, 4, . . . , 26−n , . . . .
2
Considering the sequence A of the triangle areas, we see that each subsequent term is obtained by multiplying the previous term by 1 , so the sequence of the areas is a geometric sequence
2
with common ratio r = 1 . Since the common ratio is less than one in absolute value, it makes
2

11

sense to compute the iniﬁte sum, which is given by the formula in Section 1.11, “Sum of Inﬁnite
Geometric Series”:
∞

An = n=1 A1
32
= 64
=
1−r
1− 1
2

1.12 Sigma Notation
Sigma notation is another convenient way to represent a series. Instead of saying an English phrase like “the sum of the ﬁrst 5 terms of the sequence deﬁned by un = 2n2 + 1”, we can write the same expression using the much more compact formula:
5

2n2 + 1 n=1 The symbol is the capital Greek letter sigma, the ﬁrst letter in the word “sum”. The n = 1 below the indicates the variable whose value will be changing to compute the terms in the series, and it also indicates the starting value of this variable. The 5 above the symbol provides the ending value of the variable n speciﬁed below the symbol. You compute each term in the sequence by substituting the current value of the changing variable into the expression the follows the symbol. The series is the sum of these terms.
In the example above, the variable is n and its value starts at 1 and goes to 5, incrementing by 1 each time. This means that we can “expand” the compact sigma notation in the example into this expression:

 
 
 
 

5

2n2 + 1 = 2 (1)2 + 1 + 2 (2)2 + 1 + 2 (3)2 + 1 + 2 (4)2 + 1 + 2 (5)2 + 1 n=1 n=1

n=2

n=3

n=4

n=5

= (2 · 1 + 1) + (2 · 4 + 1) + (2 · 9 + 1) + (2 · 16 + 1) + (2 · 25 + 1)
= (3) + (9) + (19) + (33) + (51)
= 115
In the example above, by computing the diﬀerence between two pairs of consecutive terms, we can observe that the sequence (and hence the series) is not arithmetic because the terms do not share a common diﬀerence: u2 − u1 = 9 − 3 = 6 u3 − u2 = 19 − 9 = 10 = 6
By computing the ratio between two pairs of consecutive terms, we can also observe that the sequence (and hence the series) is not geometric because the terms do not share a common ratio: u2
9
= =3 u1 3 u3 19
=
=3 u2 9
Of course there is nothing special about the letter n and the value of the variable being incremented from the start index to the end index need not start with the value 1. The sigma notation can be used with any letter for the variable and any integer values for the index start

12

and end. Other common variable letters that get used are i, j, k, and m. The following series are equivalent:
9

9

9

2n2 + 1 = n=3 2i2 + 1 = i=3 9

2j 2 + 1 =

2m2 + 1 m=3 j=3

1.12.1 Sigma Notation for Arithmetic Series
The series that the sigma notation represents might be an arithmetic series. Consider the example:  
 
 
 


5

2n + 1 = 2 (1) + 1 + 2 (2) + 1 + 2 (3) + 1 + 2 (4) + 1 + 2 (5) + 1 n=1 n=1

n=2

n=3

n=4

n=5

= (2 · 1 + 1) + (2 · 2 + 1) + (2 · 3 + 1) + (2 · 4 + 1) + (2 · 5 + 1)
= (3) + (5) + (7) + (9) + (11)
= 35
If we compute the diﬀerence between two pairs of consecutive terms, we see that they both have a common diﬀerence of 2, so this series is arithmetic: u2 − u1 = 5 − 3 = 2 u3 − u2 = 7 − 5 = 2
Computing the ratio between two pairs of consecutive terms, we see that there is no common ratio, so the series is not geometric. u2 5
=
u1
3
u3
7
5
= = u2 5
3
1.12.2 Sigma Notation for Geometric Series
The series that the sigma notation represents might be a geometric series. Consider the example:

 
 
 

4 k=1  1   1   1   1 
1
=  2·1  +  2·2  +  2·3  +  2·4 
3  3  3  3 
2k
3 k=1 k=2

k=3

k=4

1
1
1
1
= +
+
+
9 81 729 6561
820
=
6561
If we compute the diﬀerence between two pairs of consecutive terms, we see that they do not have a common diﬀerence, so this series is not arithmetic:
1
1
8
− =−
81 9
81
1
1
8
8
u3 − u2 =
−
=−
=−
729 81
729
81 u2 − u1 =

13

Computing the ratio between two pairs of consecutive terms, we see that there is common ratio
1
of 9 , so the series is geometric. u2 = u1 u3
=
u2

1
1 9
81
1 = 81 · 1
9
1
1
729
1 = 729 ·
81

=

1
9

81
1
=
1
9

1.12.3 Sigma Notation for Inﬁnite Geometric Series
The sigma notation can also represent an inﬁnite sum by having ∞ as the end value of the variable. The inﬁnite sum of the geometric series from the previous section would therefore be written like:
∞
k=1

1
32k

Since we determined above that this series is geometric with a common ratio of r = 1 and since
9
this respects the requirement for inﬁnite series to have |r| < 1, we can use the formula from section 1.11 to calculate the inﬁnite sum.
∞

1 u1 1
2·1
=
= 3 1 =
1−r
32k
1− 9 k=1 1
9
8
9

=

1 9
1
· =
9 8
8

1.12.4 Deﬁning Functions Using Sigma Notation
You can deﬁne a function using sigma notation by including an additional variable in the expression other than the so-called “index variable” that is counting from the start to end value. For example, consider the following function f (x): f (x) = 1 + x + x2 + x3 + x4
To write the function more compactly, you could use sigma notation like this:
4

xk

f (x) = k=0 When expanding the sigma notation, each term replaces the value of the index variable with the current integer value (from 0 to 4 in this example). Any other variables, like the x here, remains as they are in the expression. It even makes sense to use this notation for functions deﬁned by inﬁnite sums. We could write:
∞

xk

f (x) = k=0 This would be the function with an inﬁnite number of terms: f (x) = 1 + x + x2 + x3 + x4 + x5 + . . .

14

1.13 Applications: Compound Interest
Interest is a percentage paid on an original amount of money, the principal, after a certain interval of time. During subsequent periods, the interest earned in one period begins earning interest (as part of the new balance) during the next period. In this way, the interest compounds
(or accumulates) period after period.
If P is the initial principal, r is the annual interest rate expressed as a decimal, and we are compounding annually, then after one year the amount of money will be P + P · r = P · (1 + r).
That 1 inside the parenthesis ensures that the original principal amount P is included in the new total, added to the fraction r of the principal that represents the interest. After the second year, the amount from the end of the ﬁrst year is multiplied by the same (1 + r) quantity to produce the new balance of [P · (1 + r)] (1 + r) = P · (1 + r)2 . After three years, the balance is
P · (1 + r)3 . Observe that the yearly balance of the bank account is a geometric sequence with ﬁrst (zeroth) term (u0) equal to P and a common ratio of (1 + r). The t-th term in the sequence represents the balance after t years of compounding annually.
P · (1 + r)t
The period of compounding can be more frequent than once per year. It could be twelve times a year (monthly), or ﬁfty-two times a year (weekly), or 365 times a year (daily). In these cases, the annual interest percentage r is divided over the number of compounding periods a r r r year, 12 , 52 , or 365 respectively. The number of compounding periods will increase as well in these cases. In t years, there would be 12t monthly periods, 52t weekly periods, or 365t daily periods. In general, if there are n compounding periods per year, then the per-period fractional r amount of the annual interest rate will be n and the number of compounding periods will be n · t. Therefore, the formula for calculating the balance of an initial investment P at r percent annual interest over t years, compounding n times a year is: r n·t
P · 1+ n 1.14 Applications: Population Growth
The same formula for annual compounding of interest can be used to calculate population growth as well. If a population is initially P people, and it grows at the rate of r percent per year (expressed as a decimal), then the total number of people after t years is:
P · (1 + r)t

1.15 Logarithms
We know intuitively when we look at an equation like 5x = 125 what the value of x has to be to make the equality true. What is x here? It is the power to which we need to raise the base 5 to get the result 125. In this example, it must equal 3 since 5 · 5 · 5 = 53 = 125. Similarly, if we see the equation 5x = 25 we know x = 2. But how do we solve an equation like 5x = 100? From our previous two observations, we understand that 52 is less than 100 and 53 is greater than 100.
In other words, we know that 52 < 5x < 53 . Recall that the value of an exponent is not limited to integer numbers like 2 or 3. Intuitively, the value of x must lie somewhere between 2 and 3.
We could start guessing at the value by picking a number like 2.5 and using our calculator to compute 52.5 = 55.9017, which is still too small. Next we could try a value like 52.9 = 106.417, which is too large. So, now we know that 52.5 < 5x < 52.9 . Trying other values of x between
2.5 and 2.9 we could eventually converge on a value for x like 2.86135 after ﬁnding that:
52.8613 < 5x < 52.8614
99.9915

100

15

100.008

In fact, 52.86135 = 99.9995 ≈ 100. So, using this manual procedure and some intuition we’ve calculated an approximate value for the power to which we would need to raise the base 5 to get the value 100.
What if we were not happy with an approximation, but instead wanted the exact value of x that solves the equation 5x = 100? This is where logarithms come into play. They provide a mathematical notation to represent the exact value of the exponent required on a given base like 5 to equal a desired value like 100. So rather than being contented by saying that x is approximately equal to 2.86135, instead we can say that x equals “log base 5 of 100”, which we write like this: x = log5 100
√
The expression on the right hand side of the equals sign is an exact number (like π or 2) that has a place on the number line close to, but ever so slightly to the right of 2.86135. We conclude this because we saw that 52.86135 = 99.9995 was just a bit smaller than 100. What is the value of this expression?
5log5 100
The number log5 100 is exactly the right number that, when used as a power on the base 5, gives the number 100. Here we are using it as a power on the base 5, so the result of this calculation is exactly 100.
5log5 100 = 100
So, we see that the action of “raising base 5 to a power” and “taking the log base 5” are inverse operations. One operation “undoes” the other. Consider the following expression. What does it equal? log5 53
We know from above that log5 53 is a number that is exactly the right exponent to use on the base 5 to get the value 53 . So what is that exponent? log5 53 = 3
The exponent we need to use on the base 5 to get the result of 53 is just the number 3. So again, we see that the actions of “raising base 5 to a power” and “taking the log base 5” are inverse operations. One operation “undoes” the other.

1.16 Using Logarithms to Solve Equations
Given that the actions of “raising base b to a power” and “taking the log base b” are inverse operations, we can use these actions as tools when solving equations. If the unknown variable x is “trapped” in the position of the exponent like this:
92x = 6561
We can “take the log of both sides” to free the expression involving x from its trapped position up in the exponent. In order to have the inverse eﬀect, we need to pick the correct base of the logarithm to match the base that is trapping our expression. In this case, the base is 9, so we’ll take the log9 of both sides to keep the equation equal:
92x = 6561 log9 92x = log9 6561

16

The expression log9 92x represents the power we need to put on the base 9 to get 92x , so that is just 2x. As before the operations of “taking the log base 9” and “raising the base 9 to a power” are inverses, so they cancel or neutralize each other: log9 92x = log9 6561
2x = log9 6561
4
log9 6561
= =2 x= 2
2

1.17 Using Exponentiation to Solve Equations
Given that the actions of “taking the log base b” and “raising base b to a power” are inverse operations, we can use these actions as tools when solving equations. If the unknown variable x is “trapped” in the expression of a logarithm like this: log9 (3x) = 4
We can “raise a base to a power” on both sides to free the expression involving x from its trapped position inside the logarithm. In order to have the inverse eﬀect, we need to pick the correct base of the logarithm to match the base that is trapping our expression. In this case, the base is 9, so we’ll use 9 as the base on both sides of the equals sign, and use the expression on both sides as the exponent on this base. Since those expressions were equal before, putting equal exponents on the same base 9 will produce results that are also equal: log9 (3x) = 4
9log9 (3x) = 94
The expression log9 (3x) represents the power we need to put on the base 9 to get 3x, and we have used it as power on exactly that base 9 here, so it must produce 3x. As before the operations of “raising the base 9 to a power” and “taking the log base 9” are inverses, so they cancel or neutralize each other:
9log9 (3x) = 94
3x = 6561
6561
x=
= 2187
3

1.18 Logarithm Facts Involving 0 and 1
Two facts that stem from the deﬁnition of logarithm are the following (for any value of b > 0): logb 1 = 0

(since b0 = 1)

logb b = 1

(since b1 = b)

logb 0 = undeﬁned

(since no value x makes bx = 0)

1.19 Laws of Exponents and Logarithms
There are three laws to remember to help simplify expressions involving logarithms: logb a + logb c = logb (a · c) a logb a − logb c = logb c logb (ac ) = c · logb (a)

17

1.20 Change of Base
A fourth law related to logarithms is the “Change of Base” law. This allows you to change a logarithm in one base p to a new base q. logp a =

logq a logq p

Consider the expression log12 100. To compute a decimal value for this using the change of base rule, you can re-express it in terms of base 10 as follows: log12 100 =

log10 100
2
=
= 1.85326 log10 12
1.07918

1.21 Powers of Binomials and Pascal’s Triangle
A polynomial is a mathematical expression involving a sum of powers of one or more variables multiplied by coeﬃcients. For example, the following is a polynomial in variables x and y with integer coeﬃcients:
15x2 y 3 + 7xy 2 − 6xy
The individual terms in the sum are called monomials. A polynomial that is the sum of two monomials is known as a binomial. For example, all of the following are examples of binomials:
15x2 y 3 + 7xy 2
15x2 − 7x
15x + 7y x+1 When you multiply one binomial by another you expand the product one step at a time using the distributive property of multiplication. For example, consider:
(15x + 7y)(x + 2) = (15x + 7y) · x + (15x + 7y) · 2
= 15x · x + 7y · x + 15x · 2 + 7y · 2
= 15x2 + 7xy + 30x + 14y
The same approach works, of course, when you multiply a binomial by itself:
(x + 1)(x + 1) = (x + 1) · x + (x + 1) · 1
=x·x+1·x+x·1+1·1
= x2 + x + x + 1
= x2 + 2x + 1
Using shorthand notation, we see then that:
(x + 1)2 = (x + 1)(x + 1) = x2 + 2x + 1
We can use the same approach to multiply a binomial by itself to higher powers. For example, consider: (x + 1)3 = (x + 1)2 (x + 1) = (x + 1)2 · x + (x + 1)2 · 1
= (x2 + 2x + 1) · x + (x2 + 2x + 1) · 1
= x2 · x + 2x · x + 1 · x + x2 · 1 + 2x · 1 + 1 · 1
= x3 + 2x2 + x + x2 + 2x + 1
= x3 + 3x2 + 3x + 1

18

Continuing the pattern, we can calculate:
(x + 1)4 = (x + 1)3 (x + 1) = (x + 1)3 · x + (x + 1)3 · 1
= (x3 + 3x2 + 3x + 1) · x + (x3 + 3x2 + 3x + 1) · 1
= x3 · x + 3x2 · x + 3x · x + 1 · x + x3 · 1 + 3x2 · 1 + 3x · 1 + 1 · 1
= x4 + 4x3 + 6x2 + 4x + 1
If you adopt the convention of always ordering the powers of x from highest to lowest going left to right, then as you continue expanding the binomial (x + 1) to higher and higher powers, you would notice that the coeﬃcients of the terms follow the pattern in Pascal’s Triangle shown in Figure 2. If you study a row in the triangle like Row 6, you can see that each row begins and ends with 1 and that the coeﬃcients in the inside positions are symmetric. Each of these
“middle coeﬃcients” in Row 6 is obtained by adding the coeﬃcient above and to its left with the one above and to its right from the previous row. So 6 = 1 + 5, 15 = 5 + 10, and 20 = 10 + 10.
Knowing this pattern, it would be easy to extend the triangle to any number of rows.
1

Row 0:
1

Row 1:
1

Row 2:
1

Row 3:
1

Row 4:
1

Row 5:
Row 6:
Row 7: 1

1

2

4

3
6

10
15

21

1

3

5
6

7

1

4
10

20
35

1
1
5
15

35

1
6

21

1
7

1

Figure 2: Row n in Pascal’s Triangle are coeﬃcients of binomial expansion of (x + 1)n

1.22 Expansion of (a + b)n
In practice, the binomial raised to a power that you will need to expand will be more complex than the simple (x + 1)n we studied in the previous section. So it won’t be possible, in general, to just read the numbers oﬀ of Pascal’s Triangle and be done with it. For example, consider a slightly more complicated binomial expansion like this:
(3x + 2y)6
We can still use Pascal’s Triangle to ﬁnd the coeﬃcients, but we need to be more careful about following the pattern. Since we’re raising the binomial to the sixth power, we’ll use the coeﬃcients from Row 6 in the triangle:
(3x + 2y)6 = 1 · (. . . ) + 6 · (. . . ) + 15 · (. . . ) + 20 · (. . . ) + 15 · (. . . ) + 6 · (. . . ) + 1 · (. . . )
From the simpler (x + 1)n examples in the previous section, we saw that the highest power of x started with the same power to which the binomial was being raised and then proceeded to get smaller by one in each successive term. The same pattern works when the ﬁrst term in the binomial is 3x...
(3x + 2y)6 = 1 · (3x)6 (. . . ) + 6 · (3x)5 (. . . ) + 15 · (3x)4 (. . . )
+ 20 · (3x)3 (. . . ) + 15 · (3x)2 (. . . ) + 6 · (3x)1 (. . . ) + 1 · (3x)0 (. . . )
While it was hard to observe when the second term in the binomial was 1, there was a similar pattern occurring for the powers of the second term. However, the powers of the second term

19

start at 6 in the ﬁnal term and count down to 0 in each preceding term (going right to left).
Alternatively, you can think of them starting with 0 and counting up to 6 going left to right, like this...
(3x + 2y)6 = 1 · (3x)6 (2y)0 + 6 · (3x)5 (2y)1 + 15 · (3x)4 (2y)2
+ 20 · (3x)3 (2y)3 + 15 · (3x)2 (2y)4 + 6 · (3x)1 (2y)5 + 1 · (3x)0 (2y)6
Notice in each term of the expansion, the powers of the two monomials add up to 6, which is the power to which the original binomial is being raised. If we simplify the expression by raising each monomial to the indicated power, we get:
(3x + 2y)6 = 1 · 36 · x6 · 20 · y 0 + 6 · 35 · x5 · 21 · y 1 + 15 · 34 · x4 · 22 · y 2
+ 20 · 33 · x3 · 23 · y 3 + 15 · 32 · x2 · 24 · y 4 + 6 · 31 · x1 · 25 · y 5
+ 1 · 30 · x0 · 26 · y 6
= 729x6 + 6 · 243x5 · 2y + 15 · 81x4 · 4y 2 + 20 · 27x3 · 8y 3
+ 15 · 9x2 · 16y 4 + 6 · 3x · 32y 5 + 64y 6
= 729x6 + 2916x5 y + 4860x4 y 2 + 4320x3 y 3 + 2160x2 y 4 + 576xy 5 + 64y 6

1.23 The Binomial Theorem
We have seen how to compute (a + b)n using the ”brute force” procedure in the previous section, using the n-th row from Pascal’s Triangle for the coeﬃcients. However, this process would quickly become tedious if n were a larger number like 17 or 37, especially if you only need to calculate one term in the expansion.
The Binomial Theorem gives a more concise formula to calculate any particular term in the expansion of (a + b)n for any value of n. It relies on the observation that the r-th entry in n-th row of Pascal’s Triangle is precisely the number of ways there are to choose r items out of a set of n items (where the order of the items does not matter). This is a calculation from combinatorics that has a special notation named “n choose r” that looks like this: n r

=
=
=
=
=
=

Permutations of r elements without repetition from set of n elements
Permutations of r elements n · (n − 1) · · · (n − r + 1) r! 1
· (n · (n − 1) · · · (n − r + 1)) r! $
$
$
$
$
(n − r) (n − r −
1 n · (n − 1) · · · (n − r + 1) · $$$ · $$$$ 1) · $$$$ 2) · · · 2 · 1
(n − r −
¡ ¡
·
$
$
$
$ · (n$$$$ · (n$$$$ · · · 2 · 1 r! (n −
¡ ¡
$$ r) $ − r − 1) $ − r − 2)
1
n!
·
r! (n − r)! n! (r!)(n − r)!

Here the exclamation point stands for the “factorial” of a number, which is the product of all natural numbers from n down to 1. For example, 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720. If you have 6 items and you want to know how many diﬀerent ways there are to pick 4 items out of that set (where order does not matter), then you can use this “6 choose 4” formula to calculate it like this:
6
4

=

6!
6!
6·5·4·3·2·1
6·5
=
=
=
= 15
(4!)(6 − 4)!
(4!)(2)!
(4 · 3 · 2 · 1)(2 · 1)
2

20

The binomial theorem uses this “n choose r” notation and the sigma notation to write the expansion of the binomial (a + b) to the n-th power: n n
(a)n−r (b)r r (a + b)n = r=0 Using this notation we can recompute the example of (3x + 2y)6 from the previous section:
6
6

(3x + 2y) = r=0 =

6
(3x)6−r (2y)r r 6
(3x)6−0 (2y)0 +
0
6
(3x)6−3 (2y)3 +
3
6
(3x)6−6 (2y)6
6

6
(3x)6−1 (2y)1 +
1
6
(3x)6−4 (2y)4 +
4

6
(3x)6−2 (2y)2
2
6
(3x)6−5 (2y)5
5

Using your calculator’s nCr(n,r) function (in the Probability category under Combinations)
6
to calculate the value of each expression, expanding the power of each term, and simplifying k you’ll end up with the same answer we calculated above.
(3x + 2y)6 = 1 · (3x)6 + 6 · (3x)5 (2y)1 + 15 · (3x)4 (2y)2
20(3x)3 (2y)3 + 15(3x)2 (2y)4 + 6(3x)6−5 (2y)5
1 · (2y)6
= 729x6 + 2916x5 y + 4860x4 y 2 + 4320x3 y 3 + 2160x2 y 4 + 576xy 5 + 64y 6
1.23.1 Using The Binomial Theorem for a Single Term
1
A question like, “What is the coeﬃcient of the x2 term in (3x2 + )7 ?” is easy to answer using x the Binomial Theorem, without having to do all the work to expand every term.
It tells us that for any r between 0 and 7, the r-th term will be:
7
1
(3x2 )7−r ( )r r x
We can rewrite this r-th term by simplifying the exponents and writing
7
1
(3x2 )7−r ( )r = r x
=
=

1 x as x−1 like this:

7 7−r 2(7−r) −1 r
3 x
(x ) r 7 7−r 14−2r −r
3 x
(x)
r
7
(3)7−r (x)14−3r r So, we can ﬁnd which value of r will produce the term with x2 in the expansion by solving the simple equation based on the exponent 14 − 3r in the r-th term:
14 − 3r = 2
12 = 3r
12
=4 r= 3

21

Therefore, the r = 4 term will be the one that has x2 . Substituting r = 4 into the Binomial
Theorem, we can compute the desired term of the expansion. N.B. Since the ﬁrst term has r = 0, the r = 4 term is the ﬁfth monomial in the expansion:
7
(3x2 )7−4
4

1 x 4

= 35 · (3x2 )3
= 35 · 33 · x6

1 x 4

1 x4 = 35 · 27 · x2
= 945x2
Therefore, without expanding all of the terms, we have determined that 945 is the answer to
1
the question “What is the coeﬃcient of the x2 term in (3x2 + )7 ?” x 1.23.2 Example of Using Binomial Theorem
Consider the following problem:
Given that

2
1+ x
3

n

(3 + nx)2 = 9 + 84x + · · · , ﬁnd the value of n

Since the right side of the equation only provides us information about the units (9) and the x term (84x), we will focus our attention only on the terms with no x and x to the ﬁrst power.
We can ignore terms with x2 or higher powers of x.
Using the Binomial Theorem, the ﬁrst (r = 0) term and second (r = 1) term of the expansion is: 2
1+ x
3

n

n
2 0 n 2 1
(1)n−0
x +
(1)n−1
x + ···
0
3
1
3 n 2 n 2
=1·1·1+
· 1 · x + ··· = 1 +
· x + ···
1
1
3
3
=

So, we can substitute this expansion into the left-hand side of the equation in the problem, and simplify. Any time we get a power of x2 while multiplying we can ignore that term since we only care about the units and the x term:
2
1+ x
3

n

n 2 x + · · · (3 + nx)2
1 3 n 2
= 1 · (3 + nx)2 + x(3 + nx)2 + · · ·
1 3 n 2
= 9 + 6nx + n2 x2 + x 9 + 6nx + n2 x2 + · · ·
1 3 n 2
= (9 + 6nx + · · · ) + x (9 + 6nx + · · · ) + · · ·
1 3 n 2 n 2
= 9 + 6nx + x·9+ x · 6nx + · · ·
1 3
1 3 n n
= 9 + 6nx + 6 x+4 nx2 + · · ·
1
1 n = 9 + 6n + 6 x + ···
1

(3 + nx)2 =

1+

22

At this point, we can set the coeﬃcient on x equal to 84 and solve the problem:
6n + 6

n
1

n
1
n n+ 1

6 n+

Since

n
1

=

= 84
= 84
=

84
= 14
6

n! n(n$$$ − 1)!
= $ $$ = n, we have:
(1!)(n − 1)!
(1)$$ 1)!
(n −
84
= 14
6
84
= 14
(n + n) =
6
2n = 14

n+

n
1

=

n=7
Out of curiosity, Mathematica tells me that the complete expansion works out to:
6272x9 23744x8 39328x7 4144x6 22232x5 8596x4
+
+
+
+
+
+ 714x3 + 329x2 + 84x + 9
2187
729
243
9
27
9

1.24 Solving Systems of Three Linear Equations Using Substitution
Consider the function f (x) = px3 + qx2 + rx. The graph passes through the origin and the points A(−2, −8), B(1, −2), and C(2, 0). Find three linear equations in p, q, and r, then solve the system of linear equations to ﬁnd the values of p, q, and r.
By substituting in the x and y values of the three points into f (x), we produce the following system of three linear equations in three variables:

−8p + 4q − 2r = −8

Equation using A(−2, −8)

p + q + r = −2

Equation using B(1, −2)

8p + 4q + 2r = 0

Equation using C(2, 0)

One way to solve this problem is to use the substitution method. Adding the ﬁrst and third equations together gives:
−8p + 4q − 2r = −8
+ 8p + 4q + 2r = 0
8q = −8 q = −1
Then, once ﬁnding the value of q, you can substitute it into any two of the three equations to end up with two equations in two variables p and r. Picking the ﬁrst and second equations and substituting in our value of q we get:

23

−8p − 2r = −4 p + r = −1

We can solve for p using the second equation to get p = −1 − r and then substitute that into the other equation to solve for r:
−8 (−1 − r) − 2r = −4
8 + 8r − 2r = −4
6r = −12 r = −2
So, now we know that q = −1 and r = −2, we substitute those into any of the three equations to ﬁnd the remaining value p. Using the second equation of the original three, and substituting in the values of q and r that we know, we have: p − 1 − 2 = −2 p=1 So the answer is p = 1, q = −1, and r = −2.

1.25 Solving Systems of Three Linear Equations Using Technology
You can use the TI Nspire CX calculator to easily solve a system of three linear equations like the one in the previous section. To access this functionality, use the “Menu” button while in a calculator document to pick the “Algebra →Solve Systems of Linear Equations...” option. As shown in Figure 3, enter the number of equations and a comma-separated list of the variables involved, then press the “OK” button to continue.

Figure 3: Entering Number of Equations and Variables Involved
Enter the three equations as shown in Figure 3, and press “Enter” to compute the solution.

Figure 4: Entering System of Three Linear Equations in p, q, and r variables
In the time it takes to type in the equations, your calculator produces the answer {1, −1, 2}.
The order of the solutions matches the order of the variables you entered into the “Variables” ﬁeld of the dialog box in the ﬁrst step. Therefore, since you typed in p, q, r into that box earlier, the solution is p = 1, q = −1, and r = 2.

24

2 Functions and Equations
2.1 Sets
A set is a collection of elements. The set can contain no elements, a ﬁnite (i.e. countable) number of elements, or an inﬁnite number of elements. If the set is empty, then we write:
S=∅
If the set is ﬁnite, then its elements can be listed explicitly:
S = {2, 3, 5, 7, 11}
If there is an obvious pattern to the elements in the set, then rather than listing every one, you can provide an example of the list and use dots to show where the pattern continues. For example, the set T below contains all odd numbers between 1 and 101:
T = {1, 3, 5, 7, 9, . . . , 101}
If the set contains an inﬁnite number of elements, the dots can indicate that the pattern goes on forever. For example, the set D below are all even numbers greater than or equal to 2:
D = {2, 4, 6, 8, 10, . . . }
A set can be described in words like the example below that deﬁnes the set P to be the set of all prime numbers:
P = {n ∈ N | n is prime}
We read this deﬁnition, “P is the set of all natural numbers n such that n is prime.” To the left of the vertical “such that” bar, we provide a variable and indicate the set of numbers to which it belongs. On the right of the vertical bar, we provide a condition that indicates the “rule” that deﬁnes which elements of that set belong to the current set we are deﬁning.
The vertical “such that” bar can be used with other kind of conditions to help deﬁne the elements of a set. For example, the set W below is deﬁned to be the set of all real numbers between −2π and 2π (inclusive).
W = {x ∈ R | −2π ≤ x ≤ 2π}

2.2 Union and Intersection of Sets
Addition and multiplication deﬁne two operations on numbers that we use to produce new numbers as a result. Analogously, we frequently use two operations on sets called union and intersection to produce new sets from existing ones. The union of two sets A and B is deﬁned as the set that contains the elements that are either in A or in B (or in both). It is written using the “cup” symbol that looks like the letter u in the word union. For example:
{1, 3, 5, 7} ∪ {2, 3, 6} = {1, 2, 3, 5, 6, 7}
Notice that even though the number 3 appears in both sets, it only occurs once in the set that is their union. Sets never contain repeated occurrences of the same element.
The intersection of two sets A and B is deﬁned as the set that contains the elements that are in both A and B. It is written using the “cap” symbol that looks like an upside-down letter
u. For example:
{1, 3, 5, 7} ∩ {2, 3, 6} = {3}
If the sets contain no common elements, then their intersection is the empty set:
{1, 3, 5, 7} ∩ {2, 6} = ∅

25

2.3 Common Sets of Numbers
In IB Maths Standard we work with four important sets of numbers. The set of natural numbers N includes zero as well as the “counting numbers”, the positive whole numbers 1, 2, 3, etc. The set of integers Z includes all of the natural numbers along with their negatives. The blackboard letter Z used to denote this set comes from the German word for numbers zahlen.
The set of rational numbers Q contains elements formed by a ratio of two integers, that is, one number divided by another: in short, all fractions. By thinking of the fractions formed by an integer in the numerator and the number one (1) in the denominator, it’s easy to see that rationals also include all integers. The blackboard letter Q naming this set comes from the word
“quotient”, another word for one number divided by another. Finally, the set of real numbers R
√
includes all of the rational numbers, along with irrational numbers like π and 2 which cannot be represented exactly as a fraction.
When all the elements of one set A are contained in another set B, we say that A is a subset of B and write A ⊂ B. Given the deﬁnitions above, we have the following subset relationships between the common sets of numbers we will work with:
N⊂Z⊂Q⊂R

2.4 Intervals of Real Numbers
When working with the real numbers R we often will work with subsets deﬁned by the set of reals contained between a lower bound and an upper bound. For example, the set W of all real numbers between 0 and π , which above we learned could be written as:
4
W = x∈R 0≤x≤

π
4

If we change the ≤ symbol in the set’s deﬁnition to use < instead, then we can deﬁne a subset
J of W that does not include the lower bound 0 and does not include the upper bound π :
4
x∈R 0 7}
√
√
(−∞, 2] = x ∈ R x ≤ 2
(−∞, π) = {x ∈ R | x < π}
(−∞, ∞) = {x ∈ R}

2.5 Concept of Function
A function is an operation that uniquely associates each element of an “input” set with a member of another “output” set. In IB Standard Level Maths, both the input and output sets are typically the set of all real numbers R, or some subset of it like the positive real numbers or the interval [0, 2π], for example.
We give each function a name like f or g and say that the function “maps” a value in the input set to a value in the output set. Using a variable like x to represent an arbitrary element of the input set, we can describe a function by the operation it performs on this arbitrary element. For example, a function named f that takes an arbitrary value x from its input set and raises the base 2 to the power x would be written: f (x) = 2x
We read the expression f (x) as “f of x”. Figure 5 shows the function, its input set, and its output set in a pictorial form.

Figure 5: Function f associates elements of input set with output set

2.6 Graph of a Function
For functions that map elements of the real numbers to other real numbers (R → R) , we can visualize the operation the function performs by considering the set of all points (x, y) in the
Cartesian plane (written in shorthand as R × R or R2 ) where the x coordinate is in the input set of the function and the y coordinate is the value f (x) to which the function maps x. Consider the
√
x+4 function f (x) =
+ 1. Figure 6 shows the visualization the Texas Instruments Nspire-CX x−4 calculator produces for the set of points deﬁned by:
√
x+4
2
(x, y) ∈ R y =
+1
x−4

27

This visualization is known as the graph of the function f (x). Since the calculator supports graphing multiple functions on the same graph, it names the ﬁrst function f1(x), the second function f2(x), and so on.

√

Figure 6: Graph of f (x) =

x+4 x−4 +1

2.7 Domain of a Function
A function f ’s input set is called its domain. On some occasions, a function’s domain may be limited on purpose by the problem statement to focus your attention on a particular interval such as {x ∈ R | −2π ≤ x ≤ 2π}. However, more typically a function’s domain will consist of the set of all real numbers √ except where the function is not deﬁned. Consider again the
R,
x+4 function in Figure 6 f (x) =
+ 1. The value 4 would cause the x − 4 in the denominator x−4 to be zero, which would cause trouble for the function f because we cannot divide by zero.
Furthermore, the numerator is only well-deﬁned if the value under the square root is nonnegative. In other words, it must be the case that x + 4 ≥ 0, so x ≥ −4. So the domain of f (x) is the set {x ∈ R | x ≥ −4 and x = 4}. Another way of writing this using interval notation is
[−4, 4) ∪ (4, ∞). Notice in Figure 6 that the graph of this function does not exist to the left of x = −4 and that it is undeﬁned at x = 4 where there is a vertical asymptote. We can see the domain of the function f in its graph by scanning across the x-axis and noticing which values of x the function takes on along its graph.

2.8 Range of a Function
A function f ’s output set is called its range. It consists of all elements that result by applying the function f (x) to all elements x in the domain. We can see the range of the function f in its graph by scanning up and down the y-axis and noticing which values of y the function takes on along its graph. Looking again at Figure 6, we see that around the vertical asymptote at x = 4, the function’s value goes down to negative inﬁnity (−∞) on the left of the asymptote and up to positive inﬁnity (∞) on the right of the asymptote. So, by choosing a value for x that is appropriately close (but not equal to) 4, we can produce function values that are as large as we want in either positive or negative direction. So, it would appear that the range is all of the real numbers (R). However, the function has a horizontal asymptote at y = 1. As the value of x gets larger and larger in the positive direction, the value of the function gets closer and closer to 1 from above, but never actually reaches the value 1. So, on second thought it would appear that the range of the function f from Figure 6 is all real numbers except 1, or written in set notation {y ∈ R | y = 1}. However, notice that the function is deﬁned at x = −4 and that
√
−4 + 4
0
f (−4) =
+1 =
+ 1 = 1. Therefore, even though the function never takes on the
−4 − 4
−8
value 1 as the value of x gets larger and larger, it does equal 1 right at x = −4 which is part of

28

the domain. Therefore, the range is all real numbers (R).
Consider another example of the function f (x) = x2 shown in Figure 7.

Figure 7: Graph of f (x) = x2
Here there are no values in R where the function is undeﬁned, so the domain is all real numbers
(R). Scanning up and down along the y-axis, we see that the function’s y values never go below zero. When the function is applied to any negative numbers in the domain, the result is a positive number. So, the range of this function is the set of all non-negative real numbers
{y ∈ R | y ≥ 0}.

2.9 Composing One Function with Another
Given a value x in the domain of a function f , if you take the result of applying f to x and then apply another function g to that result, you end up with g(f (x)), which we read “g of f of x”. We call this “chaining” of functions composition. Consider two functions f and g deﬁned as follows: f (x) = 2x

g(x) = x2 + 1

Figure 8 shows what happens if we compose function g with function f , which means ﬁrst applying f (x) and then “chaining” that output into the input of g(x) to produce g(f (x)). We write the function “g composed with f ” using a circle symbol between them: g ◦ f . We see that the result of composing g with f is the function: g ◦ f (x) = (2x )2 + 1

Figure 8: Composite function g ◦ f (x)
On the other hand, as shown in Figure 9 if we ﬁrst apply g to x and then “chain” that output g(x) into the input of the function f , we produce a diﬀerent resulting composed function f ◦g(x): f ◦ g(x) = 2x

29

2 +1

Figure 9: Composite function f ◦ g(x)
So we see from this simple example that by changing the order of function composition, we produce diﬀerent resulting functions. This tells us that the order of composition is not commutative like addition or multiplication is. In other words, in most cases: f ◦ g(x) = g ◦ f (x)

2.10 Identity Function
The identity function maps a value in the domain to itself and is represented by the function f (x) = x.

2.11 Inverse Function
Given a function f , its inverse function f −1 “reverses” or “inverts” the mapping that f performs on its input x. If you ﬁrst put a value x into the function f and then put that result into the function f −1 , then you get back the original value x. In other words, f composed with f −1 is the identity function. f ◦ f −1 (x) = f −1 ◦ f (x) = x

2.12 Determining the Inverse Function as Reﬂection in Line y = x
Figure 10 shows how the ex and ln x functions, which are inverses of each other, are a reﬂection in the line y = x. If you only need to graph the inverse, simply draw its reﬂection in the line y = x.

Figure 10: Inverse Functions ex and ln x are Reﬂections in y = x
Another important fact to remember about inverse functions is that if the point (p, q) lies on the curve of function f (x), then the point (q, p) lies on the curve of the inverse f −1 (x). This is a direct corollary of the fact that any inverses are reﬂections in the line y = x.

30

Figure 11: If (p, q) Lies on Graph of f (x), then (q, p) Lies on Graph of Inverse f −1 (x) ln x
Figure 11 shows the graphs of two functions which are inverses f 1(x) = e0.25x and f 2(x) = 0.25 .
Notice how, as expected, they are each others reﬂection in the line y = x. But also notice how the points (4, 2.72) and (6, 4.48) lie on the curve f 1(x) and the points with inverted x and y coordinates (2.72, 4) and (4.48, 6) lie on the inverse curve f 2(x). This makes sense because the deﬁnition of an inverse function is that it “undoes” the mapping that the other function did.
If f 1(x) maps the number 4 in the domain (x-axis) to the number 2.72 in the range (y-axis)
– which is what saying “the point (4, 2.72) is on curve f 1” means – then the inverse function must map 2.72 back to 6 in order that the original operation be “undone” to get back to the point you started with.

2.13 Determining the Inverse Function Analytically
Given a function f (x) to determine its inverse function f −1 (x) analytically, follow these steps:
1. Write the function using y = · · · instead of f (x) = · · ·
2. Switch all occurrences of x and y
3. Solve the new equation for y
4. Write the function using f −1 (x) = · · · instead of y = · · ·
For example, assume that f (x) =

x+1 for x > −2. To ﬁnd the inverse f −1 (x), do these steps. x+2 Start by replacing f (x) with y: y= x+1 x+2 Then swap the x and y variables and solve for y: y+1 y+2 x(y + 2) = y + 1 x= xy + 2x = y + 1
2x − 1 = y − xy
2x − 1 = y(1 − x)
2x − 1
=y
1−x
Finally, replace the y with f −1 (x). f −1 (x) =

31

2x − 1
1−x

Figure 12 shows the graphs of f and f −1 . Notice again that the graphs of inverse functions are reﬂections in the line y = x.

Figure 12: Inverse Functions are Reﬂections in y = x

2.14 Drawing and Analyzing Graphs with Your Calculator
Your graphing calculator can graph multiple functions and analyze the graphs to ﬁnd minimum, maximum, x-intercept, y-intercept, where functions intersect, and where the function has
“zeros” (which means where it intersects the x-axis, in other words, the line y = 0).
2.14.1 Drawing the Graph of a Function
On the TI-Nspire calculator, to draw a graph create a new document of type “Graph”. You can type one or more functions that the calculator names f 1(x), f 2(x), etc. The checkbox next to the function name indicates whether you want to show or hide the function in question. When entering the equation, it is best to use the dedicated buttons on the calculator to enter common constructs like ln(x), ex , or sin x to ensure the calculator understands what you mean.
2.14.2 Restricting the Domain of a Graph
When typing in a function to graph on the TI-Nspire, by using the “such that” symbol (vertical bar), you can restrict the domain to limit the values of x for which the graph will be drawn.
Figure 13 shows entering the function x2 − x − 1 but restricting the domain to only between -2 and 2.

Figure 13: Restricting the Domain of a Graph Using the “Such That” Vertical Bar

32

2.14.3 Zooming Graph to See Exactly What You Want
After you’ve graphed a function, use the “Window/Zoom” option to scale the graph to see the portion you care to work with. The easiest option to use is the combination of “Zoom - Out” and then “Zoom - Box”. Zoom out until you see all of the graph you care about, then zoom to the box that you want to ﬁll the screen. You click in the upper left and lower right corners to deﬁne the bounding box of what you’d like to see in the window. Figure 14 shows the graph x3 − 16x of 2 after ﬁrst zooming out, and then zooming to a box to deﬁne the exact area to display. x +1

NOTE: If you select a rectangular box with diﬀerent width and height, then the calculator will scale the axes diﬀerently to make it ﬁt the window. For this reason, if possible, it’s best to select a box that is roughly square in shape so the axes will maintain the same scale.

Figure 14: Graph After Zooming Out and then Zooming to Box
2.14.4 Finding a Maximum Value in an Interval
To ﬁnd the maximum value of a graph in a given interval, use the “Menu” button to pick the
“Analyze Graph →Maximum” option. You need to pick a lower bound and a higher bound of an interval in which the calculator should determine the maximum value. When the calculator prompts to you to pick a “lower bound?” you click to position the slider just to the left of the point that visually appears to be the local minimum. Next, as shown in Figure 15, when it prompts you to pick an “upper bound?” you click to position the slider just to the right of that minimum. Once you’ve done this, the calculator adds the (x, y) coordinates of the maximum point in that interval to the graph.

33

Figure 15: Selecting the Upper Bound in Which to Determine the Maximum
2.14.5 Finding a Minimum Value Value in an Interval
To ﬁnd the minimum value of a graph in a given interval, use the “Menu” button to pick the
“Analyze Graph →Minimum” option. The steps after that are similar to those in Section 2.14.4,
“Finding a Maximum Value in an Interval”. Figure 16 shows the graph from the previous section after analyzing it to ﬁnd both the maximum and the minimum values

Figure 16: Graph Showing Minimum and Maximum Points
2.14.6 Finding the x-Intercepts or “Zeros” of a Graph in an Interval
To ﬁnd the points where a graph crosses (or intercepts) the x-axis, for each x-intercept you see in the graph, repeat the following steps. Use the “Menu” button to pick the “Analyze Graph
→Zero” option. Then click to deﬁne the lower and upper bounds of the interval around the zero you want to identify. The calculator will add that zero to the graph. Figure 17 shows the graph from the previous section after having repeated the “Analyze Graph →Zero” option three times.

34

Figure 17: Graph Showing Zero Points
2.14.7 Finding the y-Intercept of a Graph
To ﬁnd the points where a graph crosses (or intercepts) the y-axis, use the “Trace →Graph
Trace” option. You could use the arrow keys and press [enter] to place points on the graph, but it’s even easier to just type the value of x whose function value you want to see in the trace. Since we are looking for the y-intercept, that will be when the value of x = 0. Therefore, after picking the trace option, type the [0] key and press [enter] twice, followed by [esc] to signal you’re done adding trace points. As shown in Figure 18, which uses a slightly diﬀerent equation for the graph to have a more interesting y-intercept, it places the (x, y) coordinate of the y-intercept on the graph. In your own work, when using the “Graph Trace” feature to add points to a curve, keep in mind that if you have multiple functions on your graph that you can use the up and down arrow keys to select which graph you want to trace before entering a particular x value for which to trace/plot a point.

Figure 18: Graph Showing Y-Intercept Point
2.14.8 Vertical Asymptotes
To ﬁnd vertical asymptotes, analyze where the function would be undeﬁned. Typically this will be the value(s) of x that will make the expression in the denominator equal zero.
2x2 − 3x − 6
. To ﬁnd the values of x that make the denominator zero x2 − 4 and cause the function to be undeﬁned, solve the equation:
Consider the function

x2 − 4 = 0
(x + 2)(x − 2) = 0

35

So x = 2 or x = −2. As shown in Figure 19 the graph of this function has vertical asymptotes as x = 2 and x = −2.

Figure 19: Graph with Two Vertical Asymptotes at −2 and 2
2.14.9 Graphing Vertical Lines
If you want to add a vertical line to a graph (perhaps to illustrate a vertical asymptote), the way you do it on the Nspire calculator is to choose:
“Graph Entry/Edit →Equation →Line →Vertical Line x = c”
Then you enter the value of x that you want the vertical line to pass through. After repeating this for the values x = 2 and x = −2 and using the graph attributes to make the line dotted, you end up with the graph in Figure 20.

Figure 20: Dotted Vertical Lines Showing Vertical Asymptotes of Another Graph
2.14.10 Horizontal Asymptotes
Horizontal asymptotes occur when the function value approaches a ﬁxed number as the value of x increases without bound, or decreases without bound. Said another way, as x → ∞ if f (x) gets closer and closer to some value a but never reaches it, then say that the function f (x) has a horizontal asymptote at y = a. The way we determine whether such a value a exists is to consider the limit of the function as x → ∞ and as x → −∞.
2x2 − 3x − 6
. When we take the limit of this function x2 − 4 as x → ∞, the highest powers of x in the numerator and denominator make a diﬀerence. So,

Consider the function in Figure 20,

36

we can ignore the lower powers of x when thinking about the limit.
2x2 − 3x − 6
2x2
= lim 2 = 2 x→∞ x→∞ x x2 − 4 lim The same limit occurs if we let x → −∞ because the even powers of x cause the result to be positive like before. So in both cases, the limit as x → ±∞ is 2. So the graph of this function has a horizontal asymptote at y = 2.
2.14.11 Tips to Compute Horizontal Asymptotes of Rational Functions
While calculating the limit for x → ±∞ in in Section 2.14.10, “Horizontal Asymptotes”, we looked only at the highest power of x in the numerator and the denominator, since its value will dominate as x gets larger and larger. In that section’s example, the highest power in the numerator and denominator was the same (i.e. 2) so these canceled out and the value of the limit was given by the ratio of the coeﬃcients. However, if the numerator’s highest power of x is larger than the denominator’s highest power of x, then there is no limit. That is, the limit will be ∞ because the numerator gets larger faster than the denominator. In contrast, if the numerator’s highest power of x is smaller than the denominator’s highest power of x, then the limit is zero (0) since the denominator gets larger faster than the numerator.
2.14.12 Graphing Horizontal Lines
If you want to add a horizontal line to a graph (perhaps to illustrate a horizontal asymptote), the way you do it on the Nspire calculator is to choose:
“Graph Entry/Edit →Equation →Line →Line Slope Intercept x = c”
Then you enter the value of zero (0) in the ﬁrst box as the coeﬃcient for x and the value a that you want the y-intercept to be. After doing this for the values y = 0 · x + 2 and using the graph attributes to make the line dotted, you end up with the graph in Figure 21.

Figure 21: Dotted Vertical Lines Showing Vertical Asymptotes of Another Graph
2.14.13 Symmetry: Odd Functions
A function f that is symmetric about the origin is called an odd function. For each x in the domain of f , an odd function satisﬁes the condition that f (−x) = −f (x). That is, if you put in the value −5 for example, you get the opposite (i.e. negative) of the value you would get if you put positive 5 into the function. For an odd function, this happens for any value of x.
The functions are called “odd” because this eﬀect occurs for some functions where the highest

37

power of x is an odd number. Not all such equations are odd functions, however. A line that passes through the origin is an odd function. The power on the x in a linear equation is the odd number 1. A cubic equation that passes through the origin is also an odd function. The power on the x in a linear equation is the odd number 3. Figure 22 shows examples of a linear and cubic equation that are odd functions because they are symmetric about the origin.

Figure 22: Linear and Cubic Equations Through the Origin are Odd Functions
2.14.14 Symmetry: Even Functions
A function f that is symmetric about the y-axis is called an even function. For each x in the domain of f , an even function satisﬁes the condition that f (−x) = f (x). That is, if you put in the value −5 for example, you get the same value you would get as if you put positive 5 into the function. For an even function, this happens for any value of x.
The functions are called “even” because this symmetry eﬀect occurs for some functions where highest power of x is an even number. Not all such equations are even functions, however. An absolute value function can be an even function, as well as a quadratic equation if its vertex lies on the y-axis. Figure 23 shows examples of a linear and cubic equation that are odd functions because they are symmetric about the origin.

Figure 23: Absolute Value and Quadratic Examples of Even Functions
2.14.15 Solving Equations Graphically
Very often we are presented with two functions and asked to ﬁnd the value x that makes both functions equal. Of course, for some situations, we have existing techniques to ﬁnd the solution.
If both functions are lines, we can use substitution to solve the system of linear equations. Even

38

if both functions are quadratic like: f 1(x) = x2 − x − 4 f 2(x) = 6x2 + x − 10 we can create an equation by setting the function f 1(x) equal to the function f 2(x) and then solve this equation like any quadratic we’re familiar with: x2 − x − 4 = 6x2 + x − 10
0 = 5x2 + 2x − 6
Then we use the quadratic formula (with a = 5, b = 2, and c = −6) to ﬁnd the two values of x that solve the equation:
22 − 4 · 5 · (−6)
√ 2·5
−2 ± 124
=
10
−2 ± 11.136
=
10
= −1.314 or 0.914

x=

−2 ±

However, if you don’t have an easy analytic way to solve the equation involving the two functions, then fall back to using a sureﬁre visual technique using your graphing calculator to get the solution. When presented with two functions f 1(x) and f 2(x), of course their graphs may or may not intersect. If they do intersect, then this means that those points of intersection belong to both graphs. In other words, when function f 1(x) and f 2(x) intersect at the point
(p, q), that means that f 1(p) = q = f 2(p). Therefore, ﬁnding where the two functions intersect is a graphical way to solve the equation f 1(x) = f 2(x).
Consider the two non-trivial equations below:
1
f 1(x) = x5 − 4x3 + x + 7
2
f 2(x) = ex + 5
Even if we set these two equations equal, we don’t know any analytical techniques to solve for x. Here, the graphing calculator comes to the rescue. We simply graph f 1(x) and f 2(x) and use the calculator’s “Analyze Graph →Intersection” feature to compute the coordinates of the points where the two functions intersect. The x coordinate values of these point we can ﬁnd visually in this way are the solutions to the equation f 1(x) = f 2(x). The feature works similarly to ﬁnding the minimum, maximum, or zero points. The calculator asks you to indicate the lower bound and upper bound of an interval in which you want its help in computing the intersection. You can repeat the process for each of the places that your eyes see the graphs cross. Figure 24 shows the result of following this procedure to compute the coordinates of the three places where f 1(x) = f 2(x). The solutions to this equation are the x values from the three coordinates: x ∈ {−2.81, 0.59, 3.02}.

39

Figure 24: Finding Points where f 1(x) = f 2(x) Graphically

NOTE: If your two curves intersect in multiple points, instead of repeatedly using “Analyze
Graph →Intersection” try the “Geometry →Points & Lines →Intersection Point(s)” instead.
After clicking on two curves on your graph, this alternative ﬁnds all their points of intersection in a single command.

2.15 Transformations of Graphs
There are four main kinds of transformations you can do to a graph:
1. Translate it a given number of units horizontally and/or vertically
2. Reﬂect it in the x-axis or the y-axis
3. Vertically Stretch it by a given factor
4. Horizontally Stretch is by a given factor
Each of these transformations corresponds to an analytical change that you make to the original function’s deﬁnition to achieve the modiﬁcation to the graph. The following sections describe each one in turn.
2.15.1 Horizontal and Vertical Translations
Given a function f (x), to translate it b units to the right, use f (x − b). To translate it b units to the left, use f (x − (−b)) = f (x + b). Figure 25 shows these two kinds of horizontal translation.

40

Figure 25: Horizontal Translation of f 1(x) to Left and Right
Given a function f (x), to translate it b units up, use f (x) + b. To translate it b units down, use f (x) + (−b) = f (x) − b. Figure 26 shows both kinds of this vertical translation.

Figure 26: Vertical Translation of f 1(x) Up and Down
2.15.2 Vertical Reﬂection
Given a function f (x), to reﬂect it vertically – causing what was “up” to become ”down” – multiply the function by negative one. In other words, use −f (x). The line of symmetry around which you reﬂect the function will be a horizontal line of symmetry, parallel (or in some cases, exactly equal to) the x-axis.

Figure 27: Reﬂection of f 1(x) in the x-Axis

41

2.15.3 Horizontal Reﬂection
Given a function f (x), to reﬂect it horizontally – causing what was “left” to become ”right”
– multiply the input to the function by negative one. In other words, use f (−x). The line of symmetry around which you reﬂect the function will be a vertical line of symmetry, parallel (or in some cases, exactly equal to) the y-axis.

Figure 28: Reﬂection of f 1(x) in the y-Axis
2.15.4 Vertical Stretch
Given a function f (x) to stretch it by scale factor b in the vertical direction, use b · f (x). Notice in the example in Figure 29 that if |b| > 1 then the deformation is a stretch. If |b| < 1 then the deformation is a shrink (or squash).

Figure 29: Vertical Stretch of f 1(x) by factor of 3 and

1
4

2.15.5 Horizontal Stretch
1
Given a function f (x) to stretch it by scale factor b in the horizontal direction, use f ( · x). b Notice in the example in Figure 30 that if |b| > 1 (meaning 1 < 1) the deformation is a stretch. b If |b| < 1 (meaning 1 > 1)the deformation is a shrink (or squash). b 42

Figure 30: Horizontal Stretch of f 1(x) by factor of 3 and

1
4

2.15.6 Order Matters When Doing Multiple Transformations in Sequence
You can apply multiple transformation to a graph by applying the operations one after another.
As we’ll see, even in this very simple example, the order in which you carry out the operations is important because changing the order can produce diﬀerent results. For this example, we’ll start with a function f (x) = x3 + 3x2 − 3 and try two diﬀerent experiments:
1. Translate f (x) up by 5 units, then reﬂect the result in the x-axis
2. Reﬂect f (x) in the x axis, then translate the result up by 5 units
To carry out the ﬁrst experiment, use your graphing calculator to deﬁne f 1(x) = x3 + 3x2 − 3.
Then deﬁne f 2(x) = f 1(x) + 5. This represents a vertical translation by 5 units and the new function f 2(x) represents the result of this transformation operation. Finally, deﬁne f 3(x) =
−f 2(x). This reﬂects the result of the ﬁrst transformation in the x-axis. Figure 31 shows the f 3(x) graph that results from these two consecutive transformations in this order.

Figure 31: First Translate Up by 5, Then Reﬂect in x-Axis
To carry out the second experiment, use your graphing calculator to change f 2(x) = −f 1(x).
This reﬂects the initial f 1(x) function in the x-axis as the ﬁrst transformation. Next, change f 3(x) = f 2(x) + 5. This translates the result of the ﬁrst transformation up by 5 units. Figure 32 shows the diﬀerent resulting f 3(x) function graph obtained by these two consecutive transformations in this diﬀerent order.

43

Figure 32: First Reﬂect in x-Axis, Then Translate Up by 5
2.15.7 Graphing the Result of a Sequence of Transformations
In this section we study an example of doing six operations in sequence to understand how to graph the result of this combined transformation. Starting with the original function f (x) =
1 2
2 x − 4x + 6, we’ll do the following six transformation operations in this order:
1. Translate left by 1 unit
2. Reﬂect in the y-axis
3. Stretch horizontally by factor of 2
4. Translate down by 2 units
5. Reﬂect in the x-axis
6. Stretch vertically by factor of

1
3

(resulting in a shrink/squash)

To carry out the work, we start by graphing the function f 1(x) = 1 x2 − 4x + 6, then we deﬁne
2
the following additional functions to correspond to the six operations in order:
1. f 2(x) = f 1(x + 1)

Translate left by 1 unit

2. f 3(x) = f 2(−x)

Reﬂect in the y-axis

3. f 4(x) =

1 f 3( 2 x)

Stretch horizontally by factor of 2

4. f 5(x) = f 4(x) − 2

Translate down by 2 units

5. f 6(x) = −f 5(x)

Reﬂect in the x-axis

6. f 7(x) =

1
3 f 6(x)

Stretch vertically by factor of

1
3

(resulting in a shrink/squash)

As you enter each equation, by default the calculator will graph them, but you can hide the graphs of the interim functions f 2, f 3, f 4, f 5, and f 6 to leave only the original graph f 1(x) and the ﬁnal f 7(x). The easiest way to hide the interim graphs is to revisit the function entry area and uncheck the ones you don’t want to see. Figure 33 shows the result of graphing the initial f 1(x) function as well as the ﬁnal f 7 function.

44

Figure 33: Function f 7(x) Results After Six Transformation Operations to Function f 1(x)
If you are required to state the resulting, simpliﬁed equation for the ﬁnal transformed graph, then you need to work backwards and in each step substitute the function with its deﬁnition until you get back to an expression in x that you can simplify completely. The steps involved to get the transformed equation in the above example go like this:
1
f 7(x) = f 6 (x)
3
1
= (−f 5 (x))
3
1
= (− (f 4 (x) − 2))
3
1
1
=
− f3 x −2
3
2
1
1
=
− f2 − x − 2
3
2
1
1
=
− f1 − x + 1 − 2
3
2
=

1
3

−

1
2

1
− x+1
2

1 2 x −
8
1 x f 7(x) = − x2 + −
24
2
=

1
3

−

2

1
−4 − x+1 +6−2
2

x 1
+ + 2x
2 2
1
6

2.15.8 Determining Point Movement Under a Sequence of Transformations
In this section we continue with the example from Section 2.15.7, “Graphing the Result of a
Sequence of Transformations”. To determine exactly where a sequence of transformations moves a particular point on the original graph, we cannot simply take the x coordinate of that original point and put it in into the resulting, new function. To see why that does not work, let’s look at an example. Figure 34 shows the vertex (4, 2) of the original f 1(x) function. Intuitively by translating and reﬂecting and stretching, the vertex of the original parabola should have moved to the vertex of the resulting parabola. However, if we use the calculator’s “Graph Trace” to add a point corresponding to (4, f 7(4)), we see that the point on the resulting curve with x-coordinate of 4 is not the new curve’s vertex. So, we need a diﬀerent technique to get the right answer.

45

Figure 34: Visualizing Why Plugging in Original Point’s x-Coordinate Doesn’t Work
The correct way to determine where a particular point moves under sequence of transformations is to consider what each transformation in the sequence does to the x-coordinate or the ycoordinate of that point. The approach works for any point, but let’s use our original parabola’s vertex (4, −2) as the example whose progress we follow through each step of the sequence of transformations. In each step below, we adjust either the x-coordinate or the y coordinate based on what the current transformation would do to it. Horizontal translations or stretches only aﬀect the x-coordinate. Vertical translations or stretches only aﬀect the y-coordinate. A reﬂection in the x-axis ﬂips the sign of the y-coordinate. A reﬂection in the y-axis ﬂips the sign of the x-coordinate.
Starting with our vertex point (4, −2), we proceed to record that each transformation step does to its coordinates:
1. (4, −2) → ( 3 , −2)

Translate left by 1 unit

2. (3, −2) → ( −3 , −2)

Reﬂect in the y-axis

3. (−3, −2) → ( −6 , −2)

Stretch horizontally by factor of 2

4. (−6, −2) → (−6, −4 )

Translate down by 2 units

5. (−6, −4) → (−6, 4 )

Reﬂect in the x-axis

6. (−6, 4) → (−6,

4
3

)

Stretch vertically by factor of

1
3

(resulting in a squash)

After repeating the same steps for the graph’s y-intercept point (0, 6) and the graph’s two zeros at (2, 0) and (6, 0), Figure 35 shows the result which looks a lot more intuitive. The original y-intercept of f 1 at (0, 6) is mapped by the six-step transformation to the point (2, − 4 ) on the
3
resulting f 7 curve. The original zeros of f 1 at (2, 0) and (6, 0) are mapped to points on the resulting curve that are no longer on the x-axis.

46

Figure 35: Vertex, y-intercept, and Zeros of f 1 Plotted on Transformed f 7
2.15.9 Vector Notation for Function Translation
As an alternative to expressing a translation in words like “function f is tranformed p units to the right and q units up” sometimes you’ll see the translation expressed in vector notation like this:  
p
function f undergoes translation   q The p is in the x-coordinate of the vector, so it represents the horizontal translation. A positive number p in this notation means “translated p units to the right”. The q is in the y-coordinate of the vector, so it represents the vertical translation. A positive number q in this notation means “translated q units up”.
Let’s consider an example that uses this vector notation and asks to follow a speciﬁc point through a transformation of functions. Let f and g be functions such that g(x) = 2f (x + 1) + 5.
The graph of f is mapped to the graph of g under the following transformations:
 
p
vertical stretch by a factor of k, followed by a translation  . q First, write down the values of k, p, and q. Then, consider the function h(x) = −g(3x). The point A(6, 5) on the graph of g is mapped to the point A on the graph of h. Find A .
Looking at the function g(x), we see that starting with the original f (x) that it has been:
1. translated horizontally left by 1

→

f (x − (−1))

2. stretched vertically by 2

→

2f (x − (−1))

3. translated vertically up by 5

→

2f (x − (−1)) + 5

So this tells us that k = 2, p = −1, and q = 5. Next, we look closely at the deﬁnition of h(x) in terms of g(x). We see that it has been:
1. g(3x)

→

stretched horizontally by

2. −g(3x)

→

reﬂected in the x-axis

47

1
3

NOTE: In step 1 above, recall from Section 2.15.5, “Horizontal Stretch” that to stretch f (x) horizontally by a factor of b, we multiply the argument of the function by the reciprocal of b.
That is, f ( 1 x). Therefore if we see h(x) = g(3x), this means that 1 = 3 and therefore b = 1 . b b
3
In other words, if the function h(x) = g(3x) then in words we say that “h(x) has horizontally
1
stretched g(x) by a factor of 3 . When you “stretch” something by a factor less than one, it is sometimes called a “squash”.

We now use this information to follow what happens to our original point A(6, 5) on g(x) under the ordered steps of this transformation that produces the new function h(x).
1. (6, 5) → ( 2 , 5)

stretched horizontally by

2. (2, 5) → (2, −5 )

1
3

reﬂected in the x-axis

Figure 36 illustrates three possible functions f , g, and h that satisfy this example, calling them f 1, f 2, and f 3, respectively. It also depicts the original point A and the resulting point A .

Figure 36: Original Function f 1 and Transformed Functions f 2 and f 3

2.16 Quadratic Functions
Quadratic functions have the form f (x) = ax2 + bx + c where a, b, and c are real number constants. The coeﬃcient of the x2 term needs to be non-zero or else the function would be a
1
linear function, not quadratic. Figure 37 shows the graph of the quadratic function x2 −4x+6.
2

Figure 37: Example of a Quadratic Function

48

2.16.1 Using the Quadratic Formula to Find Zeros of Quadratic Function
The quadratic formula gives an explicit way to compute the zeros of (i.e. solutions to) any quadratic equation ax2 + bx + c = 0. The zeros are always a formula of the coeﬃcients a, b, and c as follows:
√
−b ± b2 − 4ac x= 2a
1
In the example function shown in Figure 38, we have a = , b = −4, and c = 6, so the values
2
of its zeros using the quadratic formula are:
−(−4) ± x= 1
2

·6

21
2

16 − 12
√1
=4± 4

=

4±

√

(−4)2 − 4 ·

=4±2
Therefore, x = 4 − 2 = 2 or x = 4 + 2 = 6.

Figure 38: Two Distinct Zeros of Quadratic Function Found Using Quadratic Formula
2.16.2 Finding the Vertex If You Know the Zeros
Suppose that you already know the zeros x1 and x2 of a quadratic equation, by any of the following means:
• You factored it and determined which values of x make each factor zero
• You used the quadratic equation
• You read oﬀ the values because the function was already in x-intercept form
Then, it is very easy to ﬁnd the x-coordinate of the vertex.
• If there is only a single root (also known as “two equal roots”), then that root is the x coordinate of the vertex.
• If there are two distinct roots, then the vertex is halfway between the two roots. To ﬁnd the point that is halfway between the roots, just take the average of the roots: add them together and divide by two.
Once you have the x coordinate of the vertex, call it h, compute the y coordinate k of the vertex by computing f (h).

49

2.16.3 Graph and Axis of Symmetry
The graph of a quadratic function forms a parabola, and it is symmetric about the vertical line x = h that passes through its vertex (h, k). We can derive information about the symmetry of the function by rearranging the quadratic formula from Section 2.16.1, “Using the Quadratic
Formula to Find Zeros of Quadratic Function”. If we separate the two terms in the numerator by writing each as a fraction over the common denominator of 2a, we get:
√
−b ± b2 − 4ac x= 2a
√
b2 − 4ac
−b
x=
±
2a
2a
−b
This tells us that the zeros are always computed by taking the quantity and adding to it
2a
√
2 − 4ac b . This means that the axis of symmetry for or subtracting from it the same quantity
2a
−b any quadratic function will always be the line x =
. Figure 39 shows the axis of symmetry
2a
1 of the function x2 − 4x + 6 lies on the line:
2
x=

4
−b
−(−4)
=
=
1
2a
1
2( 2 )

x=4

Figure 39: Quadratic Function’s Axis of Symmetry x =

−b
2a

2.16.4 Computing the Vertex From the Coeﬃcients
Since the parabola is symmetric about its vertex, and in Section 2.16.3, “Graph and Axis of
−b
,
Symmetry” we learned that a quadratic function’s axis of symmetry is always the line x =
2a
we can infer an easy way to calculate the coordinates of its vertex. The vertex lies on the line
−b
. But the vertex lies on the function f (x)’s graph, of symmetry, so its x coordinate must be
2a
so that means that the y coordinate of the vertex must be the function f ’s value at this point
−b
x=
.
2a
−b
−b
Therefore, for any quadratic function f (x), its vertex is at
,f
.
2a
2a

50

b
Since for any quadratic function we have f (x) = ax2 + bx + c, if we evaluate f (− ), we
2a
get: f (−

b 2 b b
)=a −
+b· −
2a
2a
2a
2
2
b b =a· 2 −· +c
4a
2a b2 b2
−
+c
=
4a 2a b2 − 2b2
=
+c
4a
b2
=c−
4a

So, the vertex of any quadratic function is at the coordinates

+c

−

b b2 .
,c −
2a
4a

2.16.5 Using the Discriminant to Find the Number of Zeros
In Section 2.16.3 we rewrote the quadratic equation from Section 2.16.1, “Using the Quadratic
Formula to Find Zeros of Quadratic Function” in the equivalent form:
√
−b b2 − 4ac x= ±
2a
2a
The second term of this expression (i.e. after the ± sign) includes a square root with the expression b2 − 4ac under it. Since you can’t take the square root of a negative number, we know that the square root function is only well-deﬁned when b2 − 4ac ≥ 0. This quantity under the square root of the quadratic formula is called the discriminant, because as we’ll see in this section, its value helps us discriminate between three possible situations for the zeros of a quadratic function:
1. There are two distinct zeros
2. There is a single zero (vertex lies on x-axis), also known as having “two equal zeros”
3. There is no zero (parabola never intersects x-axis)
If the discriminant (b2 − 4ac) is greater than zero, then the second term of the expression above is non-zero, and therefore the quadratic function has two distinct zeros that are obtained by tak√
−b
b2 − 4ac ing the quantity and adding to it and subtracting from it, the non-zero quantity
.
2a
2a
This is the case in Figure 38.
If the discriminant (b2 − 4ac) equals zero, then the second term of this expression evaluates to zero, which means then that the zeros of the function are:
−b
±0
2a
−b x= 2a x= In this case, the parabola only has a single zero. Another way you will see this stated is that that parabola has “two equal zeros” or “two equal roots”. This corresponds to the graph where

51

the vertex lies right on the x-axis. Figure 40 shows an example of a quadratic function with a single zero. Its discriminant is: b2 − 4ac = 182 − 4 · 1 · 81 = 324 − 324 = 0

Figure 40: Quadratic with Discriminant = 0 has Single Zero (or “Two Equal Zeros/Roots”)
Finally, if the discriminant (b2 − 4ac) is less than zero, then the square root in the second term of the expression above is undeﬁned, so there are no zeros. As shown in Figure 41, its graph never intersects the x-axis.

Figure 41: Quadratic Function with Negative Discriminant has No Zeros
2.16.6 Y-Intercept Form
The standard way of writing a quadratic function, f (x) = ax2 +bx+c, is called the “Y-Intercept
Form” because in this form you can easily read oﬀ the value of y where the function intercepts
(i.e. crosses) the y axis. Since the y intercept is the point where x = 0, if we evaluate f (0) = a · 02 + b · 0 + c = c we see that the y intercept is just the coeﬃcient c of the units term. By following the steps explained in Section 2.14.7, “Finding the y-Intercept of a Graph”, we can easily plot the y-intercept of the example quadratic function in Figure 42. We see that the y coordinate of the y-intercept is just the coeﬃcient 6 from its units term, as expected.

52

Figure 42: Quadratic Function with y Intercept Point
2.16.7 X-Intercept Form
In order to put a quadratic equation into a form that makes it easy to read oﬀ the x-intercepts of the function, we need to factor the original expression ax2 + bx + c into a form that looks like a(x − p)(x − q), where p and q are the two roots (i.e. “zeros” or solutions) of the quadratic equation. We can do this mechanically by ﬁrst using the quadratic formula to ﬁnd the roots p and q, and then rewriting the original quadratic equation as a(x − p)(x − q). Another possibility is to factor out the constant a that is the coeﬃcient of the x2 term, and then try to factor the remaining quadratic expression that will now start with x2 .
The reverse is of course also true. If you see a quadratic equation that is already in the x-intercept form, you can just read oﬀ the values of the zeros p and q. For example, if f (x) = 17(x − 10)(x + 3), then to ﬁnd the zeros consider 17(x − 10)(x + 3) = 0 which can only be zero if either (x − 10) = 0 or else if (x + 3) = 0. Therefore, x = 10 or x = −3.
Those were numbers you can immediately see when the quadratic expression is written in the x-intercept format.
1
Given the example function f (x) = x2 − 4x + 6 from Figure 42, let’s rewrite it in x-intercept
2
form for practice:
1 2
1 2 x − 4x + 6 = x − 8x + 12
2
2
1
= (x − 6) (x − 2)
2

Start by factoring out the coeﬃcient of the x2 term
Factoring is straightforward, so we don’t need quadratic formula

So, the x intercepts are 6 and 2 and the equivalent x-intercept form of the original function is therefore: 1 2
1
x − 4x + 6 = (x − 6) (x − 2)
2
2 y-Intercept Form

x-Intercept Form

When written in this x-intercept format, we can easily read oﬀ that the zeros of the function are
6 and 2, which matches the results we found using the quadratic formula as shown in Figure 38.
2.16.8 Completing the Square to Get Binomial Squared Form
“Completing the square” is a procedure that you can follow to rewrite any quadratic function in a form that includes a squared binomial (x − h)2 . The procedure goes like this:

53

ax2 + bx + c

Start with the equation in standard y-intercept form

ax2 + bx + c

Group the x2 and x terms with parentheses

b a x2 + x + c a b a x2 + x+ a 2

b

+ c−a

2a

a x+ a x− −

b
2a

b
2a

2

Factor out a from the group b 2a

2

Add constant inside, subtract constant outside

+ c−

b2
4a

Factor into square binomial and simplify

+ c−

b2
4a

Rewrite to see x = −

2

In the fourth step above, the constant

b
2a

b coordinate of vertex
2a

2

that we add inside of the x2 and x group is the

b
. We add this strategic value because in a general, when you square the binomial (x + q), you get the expression (x + q)2 = x2 + 2q · x + q 2 .
The units term q 2 is the square of one half of the coeﬃcient of the x term (= 2q). So the number we add inside the group is the one that must be there in order to use this squaring-a-binomial rule in reverse to factor it back to the binomial square form. However, if we introduce a term being added somewhere in the original expression, we also have to subtract that same quantity somewhere else in the expression to avoid inadvertently changing the meaning of the original expression. Notice that the a has been factored out of the x2 and x group parenthesis so anything added inside the parenthesis is really being multiplied by a as well. This is why we have b 2 from the constant c outside of the group to avoid changing to subtract the quantity a
2a
the original expression. square of one half of the coeﬃcient of the x term

So, for any quadratic equation, we have the following equality with standard y-intercept form on the left of the equals sign and “binomial squared” form on the right: ax2 + bx + c = a x − −

b

2

2a

+ c−

b2
4a

2.16.9 Vertex (h, k) Form
We saw in Section 2.16.4, “Computing the Vertex From the Coeﬃcients” that the vertex of a b b2 quadratic function was at the point − , c −
. We also saw in Section 2.16.8, “Completing
2a
4a the Square to Get Binomial Squared Form” that a quadratic function in standard y-intercept form can be written as: a x− −

2

b
2a

+ c−

b2
4a

b
The quantity being subtracted from x in the expression above − is none other than the x2a coordinate of the vertex we see just above. Likewise, the quantity being added to the binomial

54

squared term

c−

b2
4a

is exactly the y-coordinate of the vertex. So, if we let:

b
2a
b2 k =c−
4a
then we can write any quadratic equation in the following form, where h and k are the x and y coordinates respectively of the function’s vertex: h=− ax2 + bx + c = a (x − h)2 + k
This form is called the vertex form of a quadratic function.
1
Given the example function f (x) = x2 − 4x + 6 from Figure 42, let’s rewrite it in vertex
2
form for practice. Start by identifying the coeﬃcients:
1
a=
2
b = −4 c=6 Next we compute the h and k coordinates of the vertex using the formula we learned above:
−4
b
=− 1 =4 h=− 2a
2· 2 k =c−

b2
(−4)2
16
=6−
1 = 6 − 2 = 6 − 8 = −2
4a
4· 2

So, the vertex is at point (4, −2) and the equivalent vertex form of the original function is therefore: 1 2
1
x − 4x + 6 = (x − 4)2 − 2
2
2 y-Intercept Form

Vertex Form

2.17 Reciprocal Functions k Reciprocal functions are of the form f (x) = x where k is some constant. Their graph is a hyperbola with two curving branches having both a horizontal and vertical asymptote. They are symmetric around the line y = x, which makes them their own inverse function. If k > 0 then the curving branches of the function are in quadrants I and III as shown in Figure 43. If k < 0, then they are instead in quadrants II and IV.

Figure 43: Reciprocal Function is a Hyperbola

55

2.18 Rational Functions ax + b
. Like cx + d reciprocal functions, they have a horizontal asymptote and a vertical asymptote. Following the rules we learned in Section 2.14.8, “Vertical Asymptotes”, we can compute that the vertical asymptote must exist where the denominator cx + d = 0. Therefore the vertical asymptote is d the line x = − . Following the rules we learned in Section 2.14.10, “Horizontal Asymptotes”, c a we can determine that the a horizontal asymptote must be the line y = . The x intercept will c b occur when the numerator ax + b = 0, so we can compute that the x intercept is at x = − . a b
By evaluating the function at x = 0 we can ﬁnd that the y-intercept is . Figure 44 shows an d example of a rational function.
Rational functions are the quotient of two linear functions and have the form

Figure 44: Rational Function is a Hyperbola with Asymptotes Away from the Axes

2.19 Exponential Functions
Exponential functions have the form f (x) = kamx for some constants a (the base) and coeﬃcients k and m. Very frequently the base e is used, since it occurs in formulas relating to the growth and decay of processes in nature and in the business world. As shown in Figure 45, for positive values of m, the graph models growth. The values of the function get larger and larger as x gets larger. When m > 1, the larger the value, the more steep the growth. When values
0 < m < 1, then the function shows slower, ﬂatter growth. When the value of m is negative, this models decay instead of growth since the values of the graph get smaller and smaller as x increases. Figure 45: Examples of Exponential Functions Having Base e

56

Recall from Section 1.16, “Using Logarithms to Solve Equations” that we can exploit the fact that these the ex and ln functions are inverses to help solve problems. Consider an example equation like:
1234e0.19x = 20000
The variable x whose value we seek to solve for is “trapped” up in the exponent of the base
e. We can free it from that position by applying the appropriate inverse function, the natural logarithm ln(x) in this case, to both sides of the equation:
1234e0.19x = 20000
20000
≈ 16.2075 e0.19x =
1234
ln e0.19x = ln 16.2075
0.19x = 2.7855 x = 14.6604

2.20 Continuously Compounded Interest
In section Section 1.13, “Applications: Compound Interest” you saw how to calculate compound interest A(t) after t years, at a rate of r for a discrete number of periods n in a year:

A(t) = P · 1 +

r n n·t

If you let the number of compounding periods n in a year increase without bound, you arrive at the concept of continuously compounded interest. In this model, your money is continuously earning tiny amounts of additional interest as time moves forward. These types of continually compounded growth (or decay, in the case of losing money) are modeled by exponential functions using the base of the natural logarithm e. The formula for what sounds like an even more complex concept, continuosly compounded interest, is actually even simpler than the formula for the discrete compounding case. It is:
A(t) = P · ert where r is again the rate of interest and t is the time in years.

2.21 Continuous Growth and Decay
Many natural processes exhibit continuous growth or decay. Two examples are population growth and the decay of radioactivity in a given material. These processes of continuous growth or decay are modeled using essentially the same exponential equation as for continuously compounded interest. You start out with some initial quantity C of some entity (people, radioactive particles, etc.), and a rate r, then the amount of the entity after t units of time is given by:
A(t) = C · ert

2.22 Logarithmic Functions
Logarithmic functions have the form f (x) = k · logb x. For a given base b, the function f (x) = logb x is the inverse function of the exponential function f (x) = bx . So, y = logb x if and only if by = x. Many natural processes model their continuous growth or decay based on the exponential function in Section 2.21, “Continuous Growth and Decay”. Since this uses the base

57

e, the natural logarithm (loge x usually written ln x) is it’s inverse. Figure 46 shows how the graphs of ex and ln(x) are reﬂections in the line y = x, which means they are inverse functions.

Figure 46: Exponential Base ex and Natural Logarithm ln(x) are Inverse Functions
Recall from Section 1.17, “Using Exponentiation to Solve Equations” that if we have an equation involving a logarithm, where the variable x whose value we seek is “trapped” inside the log function, then we can free it from that position by raising an appropriate base to the power represented by each side of the equation. If the two sides were equal before, then so will be the same base raised to equal powers. Consider the example:

25 ln (0.35x) = 200
First we isolate the ln(0.35x) by dividing both sides by 25:

25 ln (0.35x) = 200
200
ln (0.35x) =
=8
25
Next, we write a new equation involving the base e of the natural logarithm on both sides:

e

···

=e

···

Since we know that the two expressions that we will write into the boxes as exponents on the base e are equal already, we can conclude that result of raising e to those equal powers will also be equal. Therefore, we can write: eln(0.35x) = e8
0.35x = e8 ≈ 2980.958
2980.958
= 8517.023 x= 0.35

58

3 Circular Functions and Trigonometry
3.1 Understanding Radians
This section motivates how radians provide an alternative to degrees in measuring the part of a circular path that an angle represents. It explains how to compute the fraction of the complete circular revolution to which this part corresponds. Finally, it shows how to convert back and forth between radians and degrees.
3.1.1 Degrees Represent a Part of a Circular Path
The angle θ in Figure 47 represents the amount of rotation that would be required to bring one intersecting line (or segment) in line with the other. About 2400 BC, the Sumerians observed the planets and noticed that the Sun’s path across the sky took about 360 days to make one year’s cycle. So, they divided this circular path into 360 degrees to track each day’s passage of the Sun’s whole journey. These 360 degrees (written 360°) are still in use today as one way to measure the amount of rotation in an angle. The degrees measuring the angle θ’s rotation are some part of the total journey around the circle.

Figure 47: Angle is Rotation to Bring One Intersecting Segment in Line with the Other
3.1.2 Computing the Fraction of a Complete Revolution an Angle Represents
Knowing the total number of units that represent a complete circular trip, for example 360 degrees, you can use the part of the total journey divided by the total to calculate the fraction of the overall journey the angle represents. For example, if θ measures 45°, then the fraction of part 45°
1
a total revolution that this rotation represents is
=
= of the circular journey. Put total 360°
8
1 another way, after 45 days the Sumerian’s knew that their 360-day year was over. So, while
8
the number 360 is a historical quantity, the key concept involved is that the angle represents a part of a circular path and you can use this to calculate the fraction of a complete revolution. 3.1.3 Attempting to Measure an Angle Using Distance
Another way to describe the angle θ would be to explain how far someone would have to walk along a circular path to “push” one segment to be in line with another. Figure 48 shows this idea. Imagine the same angle θ from Figure 47 with its vertex sitting at the center of a circle and a man standing some small distance rs away from the vertex. Let this distance rs be the radius of the circle. Suppose the segments of the angle are ﬁxed with a pin at the center of this circle that allows them to move like hands of a clock. If the man’s goal is to push the segment closest to him to be in line with the other segment forming the angle, how far would he need to walk around the circumference of this circular path? The answer is, “It depends on the radius of the circle!” We know the formula for the circumference of a circle of radius rs is 2πrs , so a walk around the entire circumference would be 2πrs units. But clearly the man wouldn’t need to walk all the way around the circle to line up the segments, so he only needs to walk some part of the circular path to get the job done.

59

Figure 48: Rotation θ in Terms of Distance Walked Around a Circumference
If the man measures the distance ds that he walks, how can he determine what fraction of the circular path he has traveled? A fraction always needs a part quantity in the numerator and a total or “whole” amount in the denominator. In this example, the part value for the numerator is the distance ds the man walked. The quantity for the denominator’s total distance is the full circumference of the circle 2πrs . Therefore, the fraction of the circular path that the man has part ds traveled is
=
. total 2πrs
What happens if the man moves a bigger distance rb (for “bigger”) away from the center of the circle and repeats his walk? This time he will have to walk a greater distance db along the circumference of the larger circle of radius rb . Using the same formula as above, we would part db calculate the fraction of the circular path to be
=
. Figure 49 shows the man’s longer total 2πrb walk on his second attempt.

Figure 49: Man Walks Longer Along Circle with Larger Radius
Figure 49 also depicts his third attempt. That time he positions himself exactly one unit away from the vertex, and walks a distance of du (for “unit”). On this third attempt, the fraction of part du du the circular path would be
=
=
.
total
2π · 1
2π
Therefore, using an arc distance alone is an ambiguous way to represent an angle because the same angle θ can correspond to many diﬀerent arc distances (du , ds , db ), depending on how far away from the vertex you measured the “walk” along the circular path.

60

3.1.4 Arc Distance on the Unit Circle Uniquely Identiﬁes an Angle θ
In Figure 49, since the angle θ hasn’t changed between the ﬁrst walk of length ds , the longer walk of length db , and the third walk of length du , the fraction of the circular path is still the same as it was before. Therefore, we must have: db du ds =
=
2πrs
2πrb
2π · 1
We can multiple through by 2π to eliminate the common factor in all the denominators, in order to reveal a very interesting observation. ds du db = 2π ·
= 2π ·
2πr
2πrb
2π
ds db d
¨
¨
¨ u
¨ ·
2π ¨ = ¨ · ¨ = ¨ · ¨
2π
2π
¨
¨ b
¨
2πr
2πr
2π db ds
=
= du rs rb
2π ·

In other words, for any radius other than 1, the ratio of the arc distance to the radius of its circular path is equal to the arc distance du on the unit circle. This means that if we ﬁx our sights on only using the unit circle to measure angles, then the arc distance itself uniquely identiﬁes the angle θ. We deﬁne the radian measure of any angle θ to be this unambiguous arc distance on the unit circle.

3.1.5 Computing the Fraction of a Complete Revolution for Angle in Radians
Radians measure the part of a circular path – speciﬁcally of the unit circular path – just like degrees do. To compute the fraction of the complete revolution this represents, just divide the number of radians by 2π, which represents the total distance around the circumference of the unit circle. If you are given an arc distance for an angle and the radius is not equal to one, then in order to compute the radian measure of the angle just divide the arc distance by the radius.
This normalizes the arc distance by scaling it to correspond to the unambiguous distance on the unit circle.

3.2 Converting Between Radians and Degrees
For a given angle θ, θrad radians and θdeg degrees are diﬀerent numbers that each corresponds to θrad θdeg the same part of a unit circular path. Correspondingly, and both represent the same
2π
360 fraction of a complete revolution that the angle represents. We can use this fact to convert between them.
3.2.1 Converting from Degrees to Radians
If an angle θ measures θdeg degrees, to convert it to radians:
• Determine the fraction of a complete revolution that θdeg represents by dividing it by 360.
• The measure in radians will be the same fraction of the total unit circle distance of 2π, so multiply the fraction determined in the ﬁrst step by 2π.

61

Therefore to convert from degrees to radians, we have: θrad =

θdeg
360

×

= θdeg ·

2π

2π
=
360

θdeg ·

π
180

Fraction of

Circumference of

Radian measure

complete

complete revolu-

of partial distance

revolution

tion of unit circle

around unit circle

π
In short, to convert an angle θ in degrees to radians, multiply it by
. To remember the
180
right fraction to multiply by, think about the fact that you are “introducing π” when going from degrees to radians, so the π goes in the numerator.
3.2.2 Converting from Radians to Degrees
If an angle θ measures θrad radians, to convert it to degrees:
• Determine the fraction of a complete revolution that θrad represents by dividing it by 2π.
• The measure in radians will be the same fraction of the total number of degrees 360, so multiply the fraction determined in the ﬁrst step by 360.
Therefore to convert from degrees to radians, we have: θdeg =

θrad
2π

×

= θrad ·

360

360
=
2π

θrad ·

180 π Fraction of

Total degrees in

Degree measure

complete

complete revolu-

of partial rotation

revolution

tion of unit circle

around unit circle

180
In short, to convert an angle θ in radians to degrees, multiply it by
. To remember the π right fraction to multiply by, think about the fact that you are “removing π” when going from radians to degrees, so the π goes in the denominator.

3.3 Length of an Arc Subtended by an Angle
When an angle θ has its vertex at the center of a circle, the angle determines (or “subtends”) an arc on the circle as shown in Figure 50.

Figure 50: If Angle θ is in radians, Arc Length L = θ · r
Given a circle of radius r and an angle θ in radians, to compute the length of the arc subtended

62

by the angle, do the following:
• Determine the fraction of a complete revolution that θrad represents by dividing it by 2π.
• The arc length L will be the same fraction of the circle’s total circumference, so multiply the fraction determined in the ﬁrst step by the circumference 2πr. This gives:

L=

θrad
2π

×

2πr

Fraction of

Circumference

complete

= θrad ·

¨
¨ ·r
2π
=
¨
¨
2π

θrad · r
Arc length

of the circle

revolution

In short, given a circle of radius r and an angle θ in radians, the length of the arc subtended by the angle is θr. If the angle θ is in degrees, then ﬁrst convert it to radians by multiplying π by then proceed to multiply that by the radius to calculate the arc length.
180

3.4 Inscribed and Central Angles that Subtend the Same Arc
The “Central Angle Theorem” is an important geometrical fact worth remembering that pertains to circles, angles, and arcs subtended by a central angle. Figure 51 illustrates that if a central angle ∠AOB subtends an arc AB, then its measure θ is double the angle of any inscribed angle ∠ACB or ∠ADB that subtends the same arc. Or, vice versa, the inscribed angle’s measure is half of the measure of the central angle when they subtend the same arc of the circle.

Figure 51: Central Angle is Double the Inscribed Angle that Subtends the Same Arc

3.5 Area of a Sector
When an angle θ has its vertex at the center of a circle, the angle’s segments and the arc it subtends form a sector as shown in Figure 52.

1
Figure 52: If Angle θ is in radians, Area of Sector A = 2 θ · r2

63

Given a circle of radius r and an angle θ in radians, to compute the area of the sector formed by the angle, do the following:
• Determine the fraction of a complete revolution that θrad represents by dividing it by 2π.
• The area A of the sector will be the same fraction of the circle’s total area, so multiply the fraction determined in the ﬁrst step by the area πr2 . This gives:

A=

θrad
2π
Fraction of

πr2

×

2

= θrad ·

π
&· r
=
2& π 1 θrad · r2
2
Sector area

Area of the circle

complete revolution In short, given a circle of radius r and an angle θ in radians, the area of the sector formed
1
by the angle is θr2 . If the angle θ is in degrees, then ﬁrst convert it to radians by multiplying
2
π by then proceed to multiply that by one half of the radius squared to calculate the arc
180
length.

3.6 Deﬁnition of cos θ and sin θ
Given the right triangle in Figure 53 with angles A and B and corresponding opposite side lengths a and b, we deﬁne the sine function of an angle θ (written sin θ) as the ratio of the length of the side opposite the angle to the hypotenuse and the cosine function of an angle θ
(written cos θ) as the ratio of the length of the side adjacent the angle to the hypotenuse.

Figure 53: Right Triangle with Angles A and B
Given this deﬁnition and the angles in Figure 53, we have: a opp
=
hyp c opp b sin B =
=
hyp c adj b = hyp c adj a cos B =
=
hyp c cos A =

sin A =

These sin and cos functions are two of the three so-called trigonometric functions that we’ll cover in this program. The word “trigonometry” comes from the Latin words “trigon” for threesided ﬁgure and “metry” for measurement. Trigonometric functions help measure interesting relationships between the sides of a three-sided ﬁgure, the triangle. We’ll deﬁne the third trigonometric function in Section 3.12, “Deﬁnition of tan θ”.

64

3.7 Interpreting cos θ and sin θ on the Unit Circle
If we imagine the unit circle centered at the origin of the Cartesian plane, at any point (x, y) on the circle we can drop a perpendicular line to the x-axis to form a right triangle. As shown in Figure 54, the hypotenuse of this right triangle is always 1 unit because it is always a radius of our unit circle. Notice that the two legs of the triangle are exactly the distances x and y that are given by the point’s (x, y) coordinates. From angle θ’s point of view, the side of the triangle that sits opposite it has length y, and the side that sits adjacent to it has length x. Using the deﬁnition of the sine function from Section 3.6, putting these facts together we can derive the following fact about the angle theta related to the unit circle:

sin θ =

opp y = =y hyp 1

cos θ =

adj x = =x hyp 1

So as shown in Figure 54, any point on the unit circle has coordinates (cos θ, sin θ).

Figure 54: Right Triangle in Unit Circle with Angle θ

3.8 Radian Angle Measures Can Be Both Positive and Negative
From Section 3.1.4, “Arc Distance on the Unit Circle Uniquely Identiﬁes an Angle θ”, we know that radians are a measure of arc distance on the unit circle that corresponds to an amount of rotation. By convention, a positive radian angle measure is a distance measured starting at the point (1, 0) on the unit circle (and on the x-axis) and traveling in a counter-clockwise direction.
If instead the rotation is in a clockwise direction, then the radian angle measure is a negative distance. Figure 55 shows a θpos angle with a positive radian measure representing a certain amount of counter-clockwise rotation. It also shows a θneg angle with a negative radian measure representing a diﬀerent amount of clockwise rotation.

Figure 55: Right Triangle in Unit Circle with Angle θ

65

3.9 Remembering the Exact Values of Key Angles on Unit Circle
It’s important to remember the cos and sin values for a number of key angles on the unit circle because they are not in the formula booklet and you are expected to know them on the non-calculator part of your exam. Figure 56 shows a visual approach using your left hand that you can use to memorize the cos and sin values for ﬁve key angles in the ﬁrst quadrant. First, remember that the values are listed as cos, sin. That is, for each pair of values the cosine value of the angle is ﬁrst and the sine value of the angle is second. To keep it simple, remember that all values have 2 in the denominator. The values in the numerator range from 0 to 4 and they are all under a square root.√Going√ counter-clockwise order, the values for the cosine
√ √
√ in numerators √ √ down ( 4, 3, 2, 1, 0 ). The values √ the sine numerators count up count √ √ √
√ for
√
( 0, 1, 2, 3, 4 ). Of course, some of the values like 4, 1, and 0 simplify to 2, 1, and 0 respectively, but remembering the counting up and down pattern makes it easier to remember.

Figure 56: Using Your Hand to Remember Key cos and sin Values in First Quadrant
Once you know the values of the cos and sin functions for the ﬁve key angles in the ﬁrst quadrant, you can easily remember the values of the other key angles in the other three quadrants because their cos and sin values are the same except for a strategically placed minus sign. Figure 57 shows the pattern of where the cos and sin function values are positive and negative. To calculate the cos and sin values of key angles in quadrants II, III, and IV , just determine which angle in the ﬁrst quadrant should be its reﬂection in the x-axis or y axis, remember the value from the
“hand” and then apply the minus sign where appropriate based on which quadrant the angle falls in.

Figure 57: Positive/Negative Sign of cos and sin Values in Four Quadrants
Figure 58 shows cos and sin values for all key angles. Using the “hand” and “quadrant signs” makes remembering all of the values as easy as counting on your ﬁngers.

66

y

(0, 1)
1
−2,

√

√
3
1
2, 2

3
2

√

√
2
2
2 , 2

−

√
2π
3

√

3 1
2 ,2

−

3π
4

120◦

5π
6

√
2
2
2 , 2

π
2

90◦

π
3

60◦

150◦
(−1, 0)

30◦

210◦

√

−

4π
3

√

2
2
2 ,− 2

−1, −
2

x

2π

330◦
240◦

5π
4

3
1
2 , −2

(1, 0)

360◦
0◦

7π
6
√

3 1
2 ,2

π
6

180◦

π

−

√ π 4

270◦
3π
2

11π
6

300◦

√

7π
4
5π
3

3
1
2 , −2

√
2
2
2 ,− 2

√
√
3
1
2, − 2

√

3
2

(0, −1)

Figure 58: Values of cos and sin for All Key Unit Circle Angles

3.10 The Pythagorean Identity
As shown in Figure 59, the right triangle formed in the unit circle with left-hand angle θ has legs that measure cos θ and sin θ. Since it’s a right trianle, Pythagorean Theorem tells us that the sum of the squares of the legs equals the square of the hypotenuse. This gives rise to the
Pythagorean Identity for cos and sin that cos2 θ + sin2 θ = 1.

Figure 59: Pythagorean Identity for cos and sin

67

3.11 Double Angle Identities
The formula booklet lists two additional handy identities that can be used to calculate the cos or sin of a “double angle” 2θ if you already know the cos θ and sin θ. You will use these identities to substitute one expression for another while solving equations involving sin and cos functions. cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ sin 2θ = 2 cos θ sin θ

3.12 Deﬁnition of tan θ
In addition to the two functions deﬁned in Section 3.6, “Deﬁnition of cos θ and sin θ”, the tangent function is a third important trigonometric ratio that captures interesting information about an angle θ in a right triangle. It is written tan θ and is deﬁned as the ratio between the length of the side opposite the angle and the length of the side adjacent to it.

Figure 60: Right Triangle with Angles A and B
Given this deﬁnition and the angles in Figure 60, we can derive the following information. Note that in addition to being the ratio of opposite to adjacent sides, the tan θ is also the ratio of sin θ to cos θ. tan A =

opp a = = adj b

tan B =

opp b = = adj a

a c b c b c a c =

sin A cos A

=

sin B cos B

3.13 Using a Right Triangle to Solve Trigonometric Problems
We have talked in earlier sections about sin, cos, and tan in the context of the unit circle, but their deﬁnitions do not depend on having a hypotenuse of unit length. Any right triangle will do. The values of sin and cos range from −1 to 1, so any value in that range is a legal value for the sin or cos of some angle θ. The value of the tan function can be any real number, so any value is legal to be the tan of some angle θ.
If a problem tells you the value of the sin, cos, or tan of an angle θ and that value does not correspond to one of the familiar “key angles” from the unit circle, then draw a right triangle and label the sides with the information you know. Using the Pythagorean Theorem, you can then ﬁgure out the remaining side, and then subsequently you can read oﬀ the values of the other trigonometric functions you may need to calculate.
3.13.1 Using Right Triangle with an Acute Angle
For example, a typical problem of this kind might take this form:

68

• Given cos x =

3
4

and that x is acute 0 < x <

π
2

, ﬁnd sin 2x

We know that x is an angle whose cosine is 3 . This value does not look like any of the small set
4
√

√

of values (0, 1, 1 , 22 , and 23 ) for the key angles of the unit circle, so we cannot immediately
2
know the measure of the angle x in radians. Using the triangle technique here, we’ll see that to solve this particular problem we don’t really even need to ﬁnd the value of x to get the ﬁnal answer we’re being asked to calculate.
Since the angle x is not one of the key angles, we draw a triangle with the information we
3
know. As shown in Figure 61, we place the angle x in one corner. Then, since the cos x = 4 we adjacent that 3 can be the length of the adjacent side know from the deﬁnition of cosine hypotenuse 3 while 4 is the hypotenuse. This is by deﬁnition what it means that cos x = 4 .

Figure 61: Using Right Triangle and Pythagorean Theorem for Trig Problems
Recall from Section 3.11, “Double Angle Identities”, that the sin 2x we’ve been asked to ﬁnd can be rewritten via an identity as: sin 2x = 2 cos x sin x
We already know the value of cos x since the problem statement gave that to us. To complete the problem we need to determine the value of sin x. From the deﬁnition of the sine function, its value is the ratio of the opposite side to the hypotenuse. The problem didn’t provide us the length of the opposite side, but now that we’ve drawn a right triangle, we can use the
Pythagorean√
Theorem to ﬁgure it out. As shown in the ﬁgure, the value of the opposite side is
√
√
2 − 32 =
4
16 − 9√ 7. So, the ratio of the opposite side to the hypotenuse that provides
=
7 the value of sin x =
. Finally, to calculate the value of sin 2x we just need to substitute the
4
values we now know for both cos x and sin x into the above double-angle formula: sin 2x = 2 cos x sin x
√
√
√
7
3 7
3 7
3
=2· ·
=2
=
4 4
16
8
3.13.2 Using Right Triangle with an Obtuse Angle
Another typical problem will involve an obtuse angle. Consider this question:
• Given sin x =

2
3

and that x is obtuse

π
2

< x < π , ﬁnd cos x and cos 2x

As shown in Figure 62, we setup our triangle with the obtuse angle x, putting the 2 for the opposite side and 3 for the hypotenuse since the deﬁnition of the sin function is the ratio of the opposite side to the hypotenuse.

69

Figure 62: Using Right Triangle and Pythagorean Theorem for Obtuse Angle Problem
√
Using the Pythagorean Theorem, we determine that √ adjacent side is 5, so the cos x should the 5 be the ratio of the adjacent side to the hypotenuse
. However, given that this obtuse angle
3
x is in the second quadrant, recall from Section 3.9, “Remembering the Exact Values of Key
Angles on Unit Circle” that the value of its cosine should be a negative number. So the correct
√
5 answer is cos x = −
.
3
To ﬁnd the value of cos 2x, we can use any of the identities from Section 3.11, “Double Angle
Identities” and then ﬁll in the values of the sin x and cos x that we already have. The three
1
equations below show how all three identities allow us to conclude that cos 2x = :
9
cos 2x = 1 − 2 sin2 x = 1 − 2 ·
√

5
= 2 cos x − 1 = 2 −
3
√
2

2

2

= cos x − sin x =

2

2
3

5
−
3

=1−2·
2

−1=2·
2

−

2
3

4
8
1
=1− =
9
9
9

5
10
1
−1=
−1=
9
9
9

2

=

5 4
1
− =
9 9
9

3.14 Using Inverse Trigonometric Functions on Your Calculator
Suppose a problem tells you, or as part of your work you conclude, that sin x = 0.156. Since the value is not one of the “key angles” from the unit circle, you need to use the inverse sine function (sin−1 ) on your calculator to solve for x. Since the functions are periodic, however, there are inﬁnitely many angles x that would have a sin value of 0.156. For this reason, the range of the sin−1 function is restricted to the interval − π , π so that there is an unambiguous
2 2 answer when you type it into your calculator. This interval represents quadrants I and IV .
Depending on the problem you are solving, you may have to add some integer multiple of 2π to the answer your calculator gives you.
Similar inverse functions exist for cosine and tangent, and are written cos−1 and tan−1 . For similar reasons as noted above for the inverse sine function, the range of the cos−1 is restricted to the interval [0, π]. This interval represents quadrants I and II. Depending on the problem you are solving, you may have to add some integer multiple of 2π to the answer your calculator gives you.
The tan−1 function’s domain accepts any real number, but its range is like sin’s − π , π but
2 2 omitting the endpoints. Depending on the problem you are solving, you may have to add some integer multiple of π to the answer your calculator gives you.

70

3.15 Circular Functions sin, cos, and tan
We’ve seen the trigonometric functions sin, cos, and tan in earlier sections. In addition to being related to right-triangles, since the functions also relate to the unit circle they are sometimes know synonymously as circular functions. In this section we’ll study the graphs of the circular functions sin, cos, and tan. These are periodic functions whose values repeat after a given period. The value of the sin and cos functions cycle back and forth between a minimum value and a maximum value, whereas the value of tan is unbounded. The amplitude of a periodic function is one half of the diﬀerence between its maximum value and its minimum value. For each graph we’ll list the domain, range, amplitude, period, minimum, maximum, and zeros.
3.15.1 The Graph of sin x
Figure 63 shows a portion of the graph of the sin x function. The domain of the function is all real numbers (R), and its range is {x ∈ R | −1 ≤ x ≤ 1}. It cycles smoothly between its maximum value at 1 and its minimum value at −1 over a period of 2π. That is, every positive and negative multiple of 2π the graph repeats itself. Its maximum value occurs at x = π , where
2
it equals one, and its minimum value occurs at x = 3π , where its value is −1. These minima
2
and maxima recur at every integer multiple of 2π of these values as well. Its amplitude is one
1
1 half of the diﬀerence between the maximum and minimum, so that is (1 − (−1)) = · 2 = 1.
2
2
The zeros of the graph occur at x = 0, x = π, and at x = 2π as well as at every positive and negative multiple of 2π of those. We can express this by saying that the zeros are at x = 2πk and x = π + 2πk for k ∈ Z. Since in general a periodic function like sin will have inﬁnitely-many zeros, most problems involving a sin function will limit the domain to an interval like [0, 2π] to focus your attention on a small, ﬁnite set of zeros.

Figure 63: Portion of the Graph of sin x
3.15.2 The Graph of cos x
Figure 63 shows a portion of the graph of the cos x function. The domain of the function is all real numbers (R), and its range is {x ∈ R | −1 ≤ x ≤ 1}. It cycles smoothly between its maximum value at 1 and its minimum value at −1 over a period of 2π. That is, every positive and negative multiple of 2π the graph repeats itself. Its maximum value occurs at x = 0, where it equals one, and its minimum value occurs at x = π, where its value is −1. These minima and maxima recur at every integer multiple of 2π of these values as well. Its amplitude is one
1
1 half of the diﬀerence between the maximum and minimum, so that is (1 − (−1)) = · 2 = 1.
2
2
The zeros of the graph occur at x = π and x = 3π , as well as at every positive and negative
2
2 multiple of 2π of those. We can express this by saying that the zeros are at x = π + 2πk and
2
x = 3π + 2πk for k ∈ Z. Since in general a periodic function like cos will have inﬁnitely-many
2
zeros, most problems involving a cos function will limit the domain to an interval like [0, 2π] to focus your attention on a small, ﬁnite set of zeros.

71

Figure 64: Portion of the Graph of cos x
3.15.3 The Graph of tan x
Figure 63 shows a portion of the graph of the tan x function. The domain of the function is all real numbers except integer multiples of π , where it has vertical asymptotes. Its range is
2
all real numbers (R). It cycles between −∞ and ∞ repeating every π, so its period is π. As it has no minimum or maximum value, it does not make sense to talk about the amplitude. The zeros of the graph occur at x = 0 as well as at every positive and negative multiple of π. We can express this by saying that the zeros are at x = πk for k ∈ Z. Since in general a periodic function like tan will have inﬁnitely-many zeros, most problems involving a tan function will limit the domain to an interval like − π , π to focus your attention on a small, ﬁnite set of
2 2 zeros. Figure 65: Portion of the Graph of tan x
3.15.4 Transformations of Circular Functions
As described more generally in Section 2.15, “Transformations of Graphs”, the graphs of sin, cos, and tan can be transformed in all the same ways. In its most general form, the graph f (x) of a transformed sine function can be written with four coeﬃcients that control the transformations of the original sin x graph:

f (x) = a sin(b(x − c)) + d

72

From Section 2.15, “Transformations of Graphs”, you recognize that:
• a is a vertical stretch: it changes the amplitude from 1 to a
2π
• b is a horizontal stretch: it changes the period from 2π to b • c is a horizontal translation: it shifts the graph c units to the right
• d is a vertical translation: it shifts the graph d units up
Of course if a < 0 then you can think of that as a reﬂection in the x-axis, followed by a vertical stretch of |a|. Similarly, if b < 0, then you can think of that as a reﬂection in the y-axis, followed by a horizontal stretch of |b|.
Notice that when written in this form, any coeﬃcient in front of the x inside the sin function is factored out so that what is left inside the parenthesis has the x with a coeﬃcient of 1.
For example, given a function like 6 sin (3x − 2), we cannot immediately “read oﬀ” the horizon2 tal translation. We must ﬁrst factor out the coeﬃcient in front of x to get: 6 sin 3 x − 3 .
Only when the x has a coeﬃcient of 1 in front of it can we read oﬀ the horizontal translation.
In this example, it is a shift to the right by 2 (and not by 2!).
3
3.15.5 Using Transformation to Highlight Additional Identities
Figure 66 shows the graphs of both sin x and cos x. You can see that the graphs’ shapes are identical, but that they are “out of phase” by π . It is visually clear that if we...
2
1. Shift cos x to the right by
2. Shift cos x to the left by

π
2

π
2,

we get sin x.

and reﬂect in the x-axis, we also get sin x.

3. Shift the graph of sin x to the left by
4. Shift sin x to the right by

π
2

π
2

we get cos x.

and reﬂect in the x-axis, we also get cos x.

Figure 66: Portion of the Graphs of sin x and cos x
These observations above explain the four handy identities below (for x in radians):
1. cos x −

π
2

2. −cos x +
3. sin x +

= sin x π 2

π
2

4. −sin x −

= sin x
= cos x

π
2

= cos x

73

3.15.6 Determining Period from Minimum and Maximum
Sometimes a problem will tell you the minimum and maximum points on a sin or cos curve and ask you to ﬁnd the period. We know that on both of these functions in one complete cycle, the minimum occurs only once and the maximum occurs only once. In the simplest version of the sin function with period 2π, the maximum occurs at π and the minimum occurs at 3π . This is
2
2
3
1 the way around the circle and 4 the way around the circle, respectively. Said another way,
4
if you go from the minimum to the maximum you have gone 3 − 1 = 1 of the way around, or
4
4
2
half of the period.
Therefore, if a problem tells you the minimum and the maximum points, subtract their x coordinates and that distance will be one half of the period. Double that value to get the period. To ﬁnd out the value of the constant b that’s needed to produce a transformed sin function that has the computed period value, use the equation:
2π
= period b So, for example, if the minimum occurs occurs at point (-2,0) and the maximum occurs at point
(16,4), then the distance between their x coordinates is 16 − (−2) = 18 and that represents half of the period. The period is therefore 2 × 18 = 36, and the value of b that is needed in the general a sin (b (x − c)) + d form of the circular function would be:
2π
= 36 b 2π
=b
36 π b=
18

3.16 Applications of the sin Function: Tide Example
During high tide, the water depth in a harbor is 22 m, and during low tide it is 10 m. Assuming a 12 h cycle:
• Find an expression for water depth t hours after low tide
• draw a graph of the function for a 48 h period
• State the times at which the water level is at a maximum, a minimum, and a mean sea level. We need to ﬁnd the appropriate values of a, b, c, and d to “adjust” the general form of the sin function (a sin(b(x − c)) + d) to ﬁt our data. We start by noting that the period is 12 hours.
2π
π
This means that we need
= 12. Solving this simple equation gives b = . b 6
Next, we determine the amplitude of the function. This is one half of the diﬀerence between
1
the maximum water depth of 22 m and the minimum water depth of 10 m: a = (22 − 10) =
2
1
(12) = 6. After determining the period and the amplitude, we have a graph that looks like
2
what you see in Figure 67. We can see that the period is correct because the graph completes one cycle exactly after twelve hours. We can see that the amplitude is right because it goes up to 6 and down to −6.

74

Figure 67: Initial sin Graph with Period and Amplitude Correct
The last steps are translating the graph horizontally and vertically to ﬁnish the job. Currently the graph has a maximum of 6, but in our problem statement the maximum is 22 m, so the value of d that we need to shift the graph up vertically is d = 22 − 6 = 16. After doing the vertical translation, we have the graph in Figure 68.

Figure 68: Initial sin Graph Attempt After Vertical Translation
Finally, the problem asks that the graph start at time t = 0 at the minimum value. At the moment the minimum value occurs after nine hours, so we need to shift the graph 9 units to the left to translate that point to be on the y-axis where t = 0. This means that our value for c = −9. This gives a ﬁnal function of 6 sin π (x + 9) + 16, whose graph appears in Figure 69.
6

Figure 69: Final Tidal Graph Involving sin Graphed on Interval [0, 48] Hours

75

So, the expression for the water depth t hours after low tide is 6 sin π (t + 9) + 16. The graph
6
of the tide depth over 48 hours appears in Figure 69. By also graphing the line y = 16, which is the horizontal “mid-line” halfway between the maximum and minimum values, and using the calculator to analyze the graph to ﬁnd intersections, we can ﬁnd all the points the problem asks for: • The minimum tide depths occur at hours 0, 12, 24, 36, 48
• The maximum tide depths occur at hours 6, 18, 30, 42
• The mean tide depths occur at hours 3, 9, 15, 21, 27, 33, 39, 45

3.17 Applications of the cos Function: Ferris Wheel Example
A ferris wheel of diameter 122 m rotates clockwise at constant speed. The wheel completes 2.4 rotations every hour. The bottom of the wheel is 13 m above the ground. A seat starts at the bottom of the wheel. The height above the ground of the seat after t minutes is h(t) =
74 + a cos (bt).
• Find the maximum height above the ground of the seat
• Show the period is 25 minutes
• Find the exact value of a and b
• Sketch the graph of h for 0 < t < 50
At the lowest point on the bottom of the circle, we know the seat is 13 m above the ground. At the highest point, it will be an entire diameter higher than that, which is 122 m higher. So the maximum height above ground of the seat is 13 + 122 = 135 m.
We’re told that in 60 minutes, the seat does 2.4 rotations. By dividing 60 by 2.4, we can
60
ﬁnd how many minutes it takes to perform a single rotation. 2.4 = 25, so a single rotation takes
25 minutes.
To ﬁnd the value of b, we use the equation

2π b = 25. Solving for b gives b =

2π
.
25

To ﬁnd the value of a, we ﬁnd the diﬀerence between the maximum and minimum value of h, and then divide by two since the amplitude is the height above the horizontal midline of symmetry of the cos function. To compute this, we calculate 135−13 = 122 = 61.
2
2
If we use these values of a = 61 and b = 2π in the equation for h(t), then we get a graph
25
like Figure 70. However, immediately we notice something is not right. The problem says that
“A seat starts at the bottom of the wheel,” which means that the height of the seat at time zero h(0) should equal 13. However, from the graph we see that h(t) = 135 and that it does take on the value of 135 until h(12.5). So, things are backwards. We need the graph to start out “low” instead of starting “high” where it is now.

76

Figure 70: Initial Attempt at Graph of Ferris Wheel Seat Height
In order to correct this mistake, we simply need to reﬂect the graph vertically, reﬂecting it around its horizontal midline of symmetry. Recall from Section 2.15.2, “Vertical Reﬂection” that the way to reﬂect a graph vertically is to multiply the function cos by −1. This means that instead of using a = 61 we need to instead us a = (−1) · 61 = −61. This keeps the amplitude of the cos graph at the required value of 61, but ﬂip-ﬂops the graph so what was “up” becomes
”down” instead. Figure 71 shows the graph of the resulting h(t) = 74 − 61 cos( 2π t) function
25
using this new, correct value for a, and restricts the graph as requested only to the values of t between 0 and 50 minutes. Note that the calculator requires the function variable to be x instead of allow us to use t, but the result is the same. Notice that now, as demanded by the problem statement, the height of the ferris wheel seat at time zero h(0) equals 13 as it should, and one half of the way through the period of 25 minutes, the height at time 12.5 minutes h(12.5) = 135 which occurs when the seat is at its maximum height. So, the moral of the story is that sometimes after determining the amplitude, it might be necessary to use the negative value of that amplitude as the multiplier in front of the original function if the problem requires the graph to be ﬂipped vertically. Therefore, a = −61 .

Figure 71: Correct Ferris Wheel Graph with Vertically Reﬂection

NOTE: Remember from Section 2.14.2, “Restricting the Domain of a Graph” that when entering a function to graph on the TI-NSpire, it’s possible to type the vertical bar “such that” symbol and provide a range of values that constrains the function domain. In this example, I typed the function in as f 1(x) = 74 − 61 cos 2π x | 0 ≤ x ≤ 50
25

77

3.18 Solving Trigonometric Equations in a Finite Interval
When you are given an equation involving a trigonometric function that does not involve any squared functions (e.g. sin2 x, cos2 x, or tan2 x), then you can isolate the trig function on one side of the equation and then use your calculator to take the appropriate inverse function of both sides to solve the equation. The problem will always provide an interval in which to seek the solution so that the number of answers is a small, ﬁnite number.

3.19 Solving Quadratic Equations in sin, cos, and tan
When you are given an equation involving squared trigonometric functions like sin2 x, cos2 x, or tan2 x, then to solve them, follow these steps:
1. If the equation involves diﬀerent trigonometric functions (e.g. both sin x and cos x), then use an appropriate identity to rewrite the equation to use only a single trigonometric function (e.g. sin2 x + cos2 x = 1).
2. Substitute a variable u for the trigonometric function to end up with a simple quadratic equation in u2 and u.
3. Use the quadratic equation or factoring to ﬁnd the two solutions for u
4. Replace u by its trigonometric function again and solve the equations using the unit circle or inverse trig functions
Consider the example equation 6 cos2 x + 7 sin x = 8.
Since it involves both cosine and sine functions, use an identity to replace the cos2 x so that the equation ends up using only sin x. The most common identity for this purpose is one described in Section 3.10, “The Pythagorean Identity”. Since we know from this identity that sin2 x + cos2 x = 1, rearranging this equation we get cos2 x = 1 − sin2 x. Substituting the right hand side of this identity equation into the original problem, we get:
6 cos2 x + 7 sin x = 8
6 1 − sin2 x + 7 sin x − 8 = 0
6 − 6 sin2 x + 7 sin x − 8 = 0
−6 sin2 x + 7 sin x − 2 = 0
6 sin2 x − 7 sin x + 2 = 0
Next substitute u = sin x to get a simpler-looking quadratic equation to solve: 6u2 − 7u + 2 = 0.
Using the quadratic equation with a = 6, b = −7, and c = 2, we solve for u:
√
−(−7) ± (−7)2 − 4 · 6 · 2
7 ± 49 − 48
7±1
8
6
2
1
u=
=
=
=
or
= or
2·6
12
12
12
12
3
2
Now, substitute back the original expression for u so that we have:
2
3
2
sin x =
3
u=

2
3
x = 0.72973

sin−1 (sin x) = sin−1

or or or or 78

1
2
1 sin x =
2
u=

1 sin−1 (sin x) = sin−1
2
π x= 6

Notice that one of the values of sin x was easily recognizable from the “key angles” of the unit π circle, so in that case we provide the exact answer for x of . In the other case, we needed to
6
2 use the calculator to compute the inverse sine of 3 . To three signiﬁcant ﬁgures, x ∈ 0.730, π .
6
Before giving the ﬁnal answer, we need to look again at the interval stipulated by the original problem. They ask us to ﬁnd the solutions over the interval [0, 4π]. Given this, we need to expand the solution set to include any multiples of 2π that would still fall inside this interval.
As a decimal, 4π is approximately 12.57 and 2π is approximately 6.28. So, for the decimal solution x = 0.730 we can add 6.28 once and still be inside the 12.57 upper bound. This gives π us one additional decimal solution of 0.730 + 6.28 = 7.01. Adding 2π to we can get another
6
13π additional solution of that is still inside the upper bound. Therefore, the ﬁnal solution to
6
π 13π
.
the problem is: x ∈ 0.730, 7.01, ,
6 6

3.20 Solutions of Right Triangles
In addition to the Pythagorean Theorem, trigonometric functions give you a new way to solve for missing lengths in a right triangle when you know the measure of one of the angles. Figure 72 shows two diﬀerent kinds of angles that can occur in problems like this, so-called angles of depression and elevation. For a particular problem, in practice, the angles are the same. One man’s angle of depression is another man’s angle of elevation as shown in the ﬁgure.

Figure 72: Angles of Depression and Elevation are Equal
Consider the example showing in Figure 73, where a man stands 100 m away from the front door of the Chrysler Building, and measures an angle of elevation to the top of the building’s spire of 73°. How tall is the building?

Figure 73: Using the tan Function to Calculate a Building’s Height

79

If we call the unknown height of the building x, then we can use the tangent function to represent an equation that relates the angle of elevation to the ratio of the opposite over adjacent sides of the right triangle formed by the little man, the front door of the building, and the top of the spire: tan 73° =

opposite x = adjacent 100

Normally the tan function expects an angle measure in radians as its argument. On your calculator, you can enter the degree symbol after the 73 and it will automatically convert the degrees to radians and then compute the tangent. Be careful to add the degree symbol, or else the calculator will compute the tangent of 73 radians! The calculator gives 3.27 as the value of the tan 73°, so we have: x 100
100 · 3.27 = x

tan 73° = 3.27 =

x = 327
So, the building is 327 m tall.

3.21 The Cosine Rule
The Cosine Rule is a more generalized form of the Pythagorean Theorem. While the latter only is applicable to right triangles, this rule works for any triangle. The Cosine Rule is useful when you are given the lengths of two sides and the angle between them. It allows you to solve for the length of the missing side. a2 = b2 + c2 − 2bc cos A b2 = a2 + c2 − 2ac cos B c2 = a2 + b2 − 2ab cos C

3.22 The Sine Rule
The Sine Rule says that in any triangle, the ratio of the length of a side to the sin of the angle opposite that side stays the same. This rule is useful when you know any angle in a triangle as well as the length of the side opposite. Then, depending on whether you know another angle or another side, you can use the Sine Rule to compute the missing quantity. a b c =
=
sin A sin B sin C
Consider the triangle in Figure 74. Suppose you are given the length of sides a and b, as well as angle A, and are asked to ﬁnd angle B that you are told (and can see) is obtuse.

Figure 74: Sine Rule Always Finds the Acute Angle

80

Using the Sine Rule, you will therefore solve this equation for B: a b
=
sin A sin B b · sin A sin B = a b · sin A sin−1 (sin B) = sin−1 a b · sin A
B = sin−1 a Since the range of the sin−1 function is from − π to π , the value that you compute for B will
2
2 be an acute angle. You can see in the ﬁgure above that the dotted line with the same length as side a can also form another triangle with side length b and angle A that has an acute angle labeled Bacute . That is the angle whose measure you ﬁnd using the computation above. In the isosceles triangle with equals side-lengths a, we know that the two base angles are equal. So the value of the obtuse angle you are looking for Bobtuse is the complement of Bacute , which means that Bobtuse = π − Bacute (or 180° − Bacute if in degrees).

3.23 Area of a Triangle
If you know the length of two sides of a triangle, and the angle between them, then you can compute the area of the triangle with the formula:
1
Area = ab sin C
2

81

4 Vectors
4.1 Vectors as Displacements in the Plane
A vector represents the shortest path in (two-dimensional) space starting at some initial point
A and terminating at another point B. Unlike a segment, a vector is represented as an arrow, whose length indicates the distance required to move in a straight line from A to B, and whose arrow head indicates the direction of the movement. Two vectors are equal if they have the same magnitude and the same direction, irrespective of their initial point. As shown in Figure 75,
#»
the vector v indicates the directed straight path from A to B, but it is the same vector if it is translated to have its initial point coincide with the origin of the Cartesian plane. In either
#»
position, the vector v points in the same direction and has the same length (also known as magnitude). Figure 75: Vectors Represent Distance and Direction, Irrespective of Origin Point
We typically distinguish the name of a vector from other kinds of variable names with an arrow written above and a bolder type. If the initial and terminal points of a vector are labeled with letters like A and B, then an equivalent way of naming the vector uses these point labels listed
# »
#»
in the order the arrow is pointing. So, for example, Figure 75 vector v can also be written AB.

4.2 Vectors as Displacements in Three Dimensions
A vector has exactly the same meaning in three dimensions as it does in the plane. It represents the shortest path through space starting at some initial point A and terminating at another point B. Figure 76 shows such a vector with initial point along the x axis and terminating at a point above the y axis in the yz-plane. As was the case in two dimensions, we can always translate a vector so that its initial point is the origin of the x, y, and z axes without changing
#»
the vector. In other words, both vectors labeled v in Figure 76 are the same vector: same magnitude or length, same direction.

Figure 76: Vectors Also Represent Distance and Direction in 3 Dimensions

4.3 Terminology: Tip and Tail
Since vectors are drawn as arrows, the terminal point with the arrowhead is sometimes called the tip of the vector, while the initial point is called the tail. When one vector starts at exactly the place where another vector terminates, we say that they are placed “tip to tail”.

82

4.4 Representation of Vectors
Since vectors having the same length and direction are equivalent, we always describe a vector as if its initial point were at the origin. By doing this, we can use the (x, y) coordinates of its
#»
#» terminal point to describe it. Figure 77 shows two vectors v and w. The x and y coordinates
#» correspond to the fact that to get from the initial point at the of the point (4, 3) for vector v origin to that terminal point, you need to “walk” 4 units to the right and 3 units up. Similarly,
#»
for w, its terminal point (−2, 2) means that to travel from the origin to there, you need to go 2 units left, and 2 units up. Even though the length of the vector is measured along the straight path from the initial point to the terminal point, the vector with initial point at the origin is uniquely identiﬁed by these x and y coordinates that represent the distance of the legs of a right triangle having the vector as the hypotenuse.

Figure 77: Coordinates of Terminal Point Uniquely Identify a Vector
To make it clear we are talking about a vector instead of just a point in the plane, we not only write the vector variable with the bold font and arrow above, we also write the coordinates of the vector’s terminal point in a column like this:
 
4
#»   v = 
3
 
−2
#»   w= 
2

In general, a vector in the plane is written in a column with its x coordinate at the top, and y coordinate at the bottom. A vector in three-dimensions extends this to list the z coordinate at the bottom, like this:
    x 4
#»     a = = 
Two-dimensional vector y 3
   
−2 x
#»    
   
Three-dimensional vector b =  2  = y
   
    z 1

4.5 Magnitude of a Vector
#»
#»
Given a vector v , its length (known as magnitude) is written | v |. As shown in Figure 78, we calculate the magnitude using the Pythagorean Theorem with the x and y coordinates as the

83

legs of the right triangle for which the vector is the hypotenuse. This gives:
√
#»
| v | = 42 + 32 = 25 =
=5
Therefore, in general we have:
 
a
  = b a2 + b2

Figure 78: Using the Pythagorean Theorem to Find (2D) Vector Magnitude
As shown in Figure 79, for a three-dimensional vector we also calculate the magnitude using the Pythagorean Theorem. First we use the theorem to ﬁnd the length of the diagonal in the xy-plane, which is (−4)2 + (−2)2 . Then we use the theorem again with the right triangle having this ﬁrst diagonal as one leg and the segment parallel to the z axis as the other leg. This gives: #»
|v | =
=

(−4)2 + (−2)2

2

+ 32
√
√
(−4)2 + (−2)2 + 32 = 16 + 4 + 9 = 29

Therefore, in general we have:
 
a
 
 
b =
 
  c a2 + b2 + c2

Figure 79: Using the Pythagorean Theorem to Find (3D) Vector Magnitude

84

4.6 Multiplication of a Vector by a Scalar
To multiply a vector by a scalar k, simply multiply each coordinate by k:
   
a ka
   
    k  b  =  kb 
   
    c kc
As shown in Figure 80, when you multiply a vector by a scalar number like 2, you stretch it 2 times in the same direction it’s already pointing. A consequence is that the magnitude of the new vector is 2 times the magnitude of the original vector.
#»
#»
|k a| = k | a|

Figure 80: Multiplying a Vector by a Scalar Stretches It By That Factor

4.7 Negating a Vector
#»
#»
Given a vector a, you can negate the vector by putting a minus sign in front of it: − a. Just as with simple numbers, putting a minus sign in front is the same as multiplying the vector by
  a #»   the scalar −1. So if a =  , we have: b  
   
a
a −a
#»
− a = −   = −1   =   b b
−b
Note that the magnitude is not changed by this operation:
#»
#»
|− a| = | a|
Figure 81 shows the visual interpretation of what negating a vector does. It ﬂips the vector around so that it’s pointing in the opposite direction, without changing its length. As with any vector, we can then translate the negated vector so that it’s initial point is at the origin without changing it.

85

Figure 81: Negating a Vector Makes It Point in the Opposite Direction

4.8 Sum of Vectors
To add vectors, simply add their coordinates:
    

a d a + d
 + =

b e b+e
 
 
4
#» 1
#» 
Figure 82 shows the visual interpretation of adding vectors a =   and b =   together.
3
5
#»
Since we can move vector b ’s initial point around without changing it, we move it to coincide
#»
#» #» with the terminal point of a as shown in the ﬁgure. The vector a + b is the result of ﬁrst
#»
#»
#»
traveling in the direction of a for the distance | a|, then traveling in the direction of b for the
#»
distance b . This is exactly the same destination you would get to by starting at the origin and traveling the distance of the diagonal of the parallelogram formed by the two vectors. For this reason vector addition is sometimes called the “parallelogram rule”.

Figure 82: Adding Vectors Produces Their Parallelogram Diagonal

4.9 Diﬀerence of Vectors
To subtract vectors, simply subtract their coordinates in the correct order:
    

a d a − d
 − =

b e b−e

86

#»
#» #»
#» #»
#»
Figure 83 shows the visual interpretation of a − b . Recall that a − b = a + (− b ), so we
#»
#» can start by negating b to ﬂip its direction around, translate the resulting − b to start at the
#»
#» origin, and then ﬁnally use the “parallelogram rule” above for adding vectors a and − b . If we
#»
#» #» translate the resulting vector a − b to start at the terminal point of b , then we can say that
#»
#»
#»
#» a − b is the vector that connects the tip of b to the tip of a.

Figure 83: Vector a − b Connects Tip of b to Tip of a
Just as subtraction of numbers is not commutative (e.g. 5 − 3 = 3 − 5), the same is true of
#» #» vector subtraction. Figure 84 shows the visual interpretation of b − a. As we did above, we
#»
#» start by negating a and translating its initial point to the origin. Then, we add the vector b
#» #»
#»
with this − a using the parallelogram rule. If we translate the resulting vector b − a to start
#» #»
#»
at the terminal point of a, then we can say that b − a is the vector that connects the tip of
#»
#» a to the tip of b . If you compare the result to Figure 83, you’ll see that it’s the same vector but with the opposite direction. In other words, we have:
#» #»
#» #» a − b = −( b − a)

Figure 84: Vector b − a Connects Tip of a to Tip of b

87

4.10 Unit Vectors
A unit vector is one whose length is equal to one unit. For example, the following three vectors all are unit vectors because their length is one:
 
√
1
  = 12 + 0 2 = 1 = 1
0
 
√
0
  = 12 + 0 2 = 1 = 1
1
 
4
5
 
42 32
42 + 3 2
16 + 9
25 √
 
+
=
=
=
= 1=1
  =
2
 
5
5
5
25
25
 
3
5

4.11 Scaling Any Vector to Produce a Parallel Unit Vector
#»
You can make any vector v into a unit vector by following these two steps:
#»
• If | v |=1, then you’re done. It’s already a unit vector
#»
#»
#»
• Otherwise, multiply v by the reciprocal of | v |. This will scale v to stretch or shrink it to have exactly the length 1 without changing its direction.
 
4
For example, consider the vector  . We start by computing:
5
 
4
  =
5

42 + 5 2 =

√

16 + 25 =

√

41

So the vector’s length is not equal to 1, therefore it is not a unit vector. To turn it into a unit vector, multiply it by the reciprocal of its magnitude:



4
√ 
  
 41 

1 4 

√  =


41


5
 5 
√
41
This vector should be a unit vector now. We can verify it by computing its magnitude:


4
 √41 


2
2


4
5
42 + 5 2
41 √

 =
√
+ √
=
=
= 1=1


41
41
41
41


 5 
√
41

88

4.12 Position Vectors
#»
Given a point A,the position vector of A (typically written a) is the vector with initial point at the origin and terminal point at A. Given another point B, it also has an associated position
#»
# » vector b that points from the origin to B. As shown in Figure 85, the vector AB is therefore
#» #» b − a.

Figure 85: Vector AB Connects Tip of Position Vector a to Tip of Position Vector b

4.13 Determining Whether Vectors are Parallel
#»
#»
Two vectors a and b are parallel if one is a scalar multiple of the other. Recall that you can
#»
#» change the initial point of a and b by translating them to the origin (without changing either
#»
#» one’s direction!). They are parallel, then, if a = k b for some scalar k. In a visual sense, it
#»
#»
#»
means that a and b are pointing in the same direction and you can stretch (or shrink) a by
#»
some amount to get b .
 
 
a1 
b1 
 
 
#»
 
 
#»
Given two vectors a = a2  and b = b2 , to determine if they are parallel compute
 
 
 
  a3 b3 b3 b1 b2
, , and . the three quotients: a1 a2 a3 #»
#»
If any two of these quotients are not equal, then you can conclude that a and b are NOT parallel. If all three of the quotients are equal, then that value is the scalar multiple k by which
#»
#» a was multiplied to get b .


 
−91
7

 
#» 


#»  
For example, consider the vectors a = −3 and b =  39 . Computing the three quotients

 



 
−65
5 we get:
−91
= −13
7

39
= −13
−3

−65
= −13
5

#»
#»
Since all three quotients equal −13, then we know that b = −13 a and this tells us that they are parallel, since one is a scalar multiple of the other.

89

4.14 Finding Parallel Vector with Certain Fixed Length
#»
#»
#»
Given a vector v of a given length | v |, you may be asked to ﬁnd a parallel vector w with a speciﬁc diﬀerent length L. The procedure is similar to that described in Section 4.11, “Scaling
Any Vector to Produce a Parallel Unit Vector”, with one additional twist. Instead of scaling the
1
original vector by multiplying by #» , scale it by that factor multiplied by the desired length L:
|v |
L #»
#»
w = #» v
|v |

4.15 Scalar (or “Dot”) Product of Two Vectors
#»
#»
#» #»
The scalar product (also known as the dot product) of vectors a and b is written a · b and is deﬁned as follows:
 
  a #» b1 
#» #»
#»  1 
If a =   and b =   , then a · b = a1 b1 + a2 b2 b2 a2
The scalar product is called that because the result is a scalar number, not a vector. It’s also called the dot product because the operation is written with a dot between the two vectors.
If the vectors are in three dimensions, the scalar product also includes the product of their z-coordinate values:
 
 
a1 
b1 
 
#»  
 
#»  
#» #»
If a = a2  and b = b2  , then a · b = a1 b1 + a2 b2 + a3 b3
 
 
 
  a3 b3

4.16 Perpendicular Vectors
#»
#»
Two vectors a and b are perpendicular if and only if their scalar product is zero:
#» #»
#» #» a · b = 0 ⇐⇒ a ⊥ b

4.17 Base Vectors for Two Dimensions
In two dimensions, base vectors are a set of two perpendicular unit vectors. Any vector in two dimensions can be expressed as the sum of appropriate scalar multiples of these two base
   


 1

  0 vectors. The canonical base vectors comprise the set   ,   . This is the unit vector


 0
1  from the origin pointing a distance of one unit along the x-axis and the unit vector from the origin pointing a distance of one unit up the y axis. By convention, the one pointing along the
#»
#» x-axis is called i and the one pointing along the y axis is called j . Figure 86 shows the   point 3
(3, 4) in the plane, and its position vector. In column notation, this vector is written  .
4
#»
#»
However, if we multiply the i unit vector by the scalar value 3 to get 3 i and we multiply the
#»
#» j unit vector by the scalar value 4 to get 4 j , then we can use the parallelogram rule from

90

#»
#»
Section 4.8, “Sum of Vectors” to add 3 i + 4 j to get the position vector for the point (3, 4).
 
#»
#»
3
When we write a vector   in the form 3 i + 4 j , we say that it is in “base vector form”.
4

Figure 86: Any Vector Can Be Expressed as Sum of Scalar Multiples of Base Vectors

4.18 Base Vectors for Three Dimensions
In three dimensions, base vectors are a set of three perpendicular unit vectors. Any vector in three dimensions can be expressed as the sum of appropriate    of these three base
 scalar multiples


 1
  0 0


     



      vectors. The canonical base vectors comprise the set 0 , 1 , 0 . This is the unit
     
     






 0
0
1  vector from the origin pointing a distance of one unit along the x-axis, the unit vector from the origin pointing a distance of one unit up the y axis, and the unit vector from the origin
#»
#» #» pointing a distance of one unit up the z axis. By convention, these are called i , j , and k respectively. In an analogous way to what we illustrated in Section 4.17, “Base Vectors for Two
Dimensions”, in three dimensions any vector can be written in base vector form. For example:
 
3
 
#»
#»
#»
 
 17  = 3 i + 17 j − 2 k
 
 
−2

4.19 The Angle Between Two Vectors
#»
#»
Given the two vectors a and b in Figure 87, recall that the cosine of the angle θ between them is: 

#» · #»
#» · #» a b  a b therefore θ = cos−1  cos θ =
#» #»
#» #»
| a| b
| a| b

91

Figure 87: Angle θ Between a and b is cos−1

a·b
|a| |b|

π
#»
#»
If vectors a and b are perpendicular, then the angle between them is . We know from the
2
# #
» » π a· b unit circle “key angles” that the cos = 0. This means that #» #» = cos π = 0, so for that
2
| a || b |
2
#» · #» = 0. This is the condition we learned about in to happen, it must be the case that a b
Section 4.16, “Perpendicular Vectors”.

4.20 Vector Equation of a Line in Two and Three Dimensions
A line is a straight, one-dimensional ﬁgure that we normally describe using the equation y = mx + b in slope-intercept form, or y − y1 = m(x − x1 ) in point-slope form. In these two ways to express the set of points (x, y) that belong to a straight line, the slope m is the value that determines the “direction” the line is pointing, and the y-intercept (0, b) or arbitrary point (x1 , y1 ) deﬁnes a point that the line passes through. In both cases, once we’ve indicated a direction for the line and one point that it passes through, we have uniquely deﬁned the line.
Using these same concepts, we can use a vector equation as an alternative way to describe a line. Figure 88 shows the technique we’ll adopt. Given a line L, we need the same two “in#» gredients” as in the two examples above: a direction vector d that indicates the direction the
#»
line is pointing, and a position vector p for some point that the line passes through.

Figure 88: Vector Line r = p + kd Has Position Vector p and Direction Vector d
#»
#»
As shown in Figure 88, if we use the “parallelogram rule” to add p and d (after translating it to the origin as well), we get a position vector for some other point on the line L. If we
#»
#» repeat the process of adding p to another multiple of the direction vector like 3 d , again we
#»
get another position vector for diﬀerent point on the line L. Trying a third time, if we add p
#»
to yet another multiple of the direction vector like 5 d , we get a third position vector for yet
#»
another point on the line L. If instead of adding the position vector p only to a few diﬀerent
#»
#» scalar multiples of the direction vector d , we added to p every possible positive or negative real number multiple k of the direction vector, then we could end up producing position vectors

92

for every point on the line L.
Since the line L is made up of inﬁnitely many position vectors, one that connects the origin to each point on the line, we can describe the set of these inﬁnitely many position vectors using the vector equation:
#»
#» = p + k d
#»
r

(k ∈ R)

#»
In other words, the line L is the set of all vectors #» that result by taking a position vector p r for some point on the line and adding to it every possible real number multiple of the direction
#»
vector d that determines the line’s direction.
Keep in mind that any position vector on the line is ﬁne to use for the position vector of the vector line equation. If you happen to know more than one position vector on the line, just pick one of them. It doesn’t matter which one you choose. Once you add all the scalar multiples of the direction vector to it, you will still end up producing the same line L.

4.21 Vector Equation of Line Passing Through Two Points
Given two points, for example (5, 6) and (1, 3), to determine the vector equation of the line that passes through both of them, do the following:
• Write the points in vector form and subtract one from the other to get the direction vector.
• Pick one of the two points (doesn’t matter which) to use as the position vector
   
 
5 1
4
So,   −   =  . This is the direction vector. Subtracting them in the other order
6
3
3
 
 
4
−4 would give   = −  , so its just pointing in the opposite direction. It still ends up
−3
3 generating the same line when we consider adding all real number multiples of the direction vector to the position vector.
 
5
For the position vector, pick either one. For example, we can use  . So, writing the
6
equation in vector line form we get:
 
 
5
4
#» =  r   + k   (k ∈ R)
6
3

93

4.22 Finding the Cartesian Equation from a Vector Line
 
 
5
4
If you’re given a line in vector line form #» =   + k  , you can easily convert it to r 6
3
Cartesian form by following these steps:
 
x
• Write the vector #» as   r y
• Write the system of two equations involving x and k as well as involving y and k
• Choose one of the two equations and solve for k
• Substitute that value of k into the other equation to eliminate k and end up with a
Cartesian equation in only x and y.
 
x
In the example above, start by replacing #» with  : r y
   
 
x 5
4
  =  +k  y 6
3
So the system of equations is:



x = 5 + 4k 


y = 6 + 3k 

Using the ﬁrst equation to solve for k we get: x = 5 + 4k x − 5 = 4k
5
1 x− =k
4
4
Substituting this value of k into the other equation in the system, we can eliminate k and end up with a Cartesian equation for the line in only x and y: y = 6 + 3k
5
1 x− 4
4
3
15
y =6+ x−
4
4
3
24 15 y = x+
−
4
4
4
3
9 y = x+
4
4 y =6+3

4.23 The Angle Between Two Vector Lines
To compute the angle between two vector lines, ﬁnd the angle between their respective direction vectors using the formula in Section 4.19, “The Angle Between Two Vectors”.

94

4.24 Distinguishing Between Coincident and Parallel Lines
Two vector lines are parallel if their respective direction vectors are parallel (equal or one is a scalar multiple of the other). See Section 4.13, “Determining Whether Vectors are Parallel” for tip on determining if two vectors are parallel.
Two vector lines are coincident if the specify the same line. This means that they must be parallel, and you can ﬁnd some point that exists on both lines.

4.25 Finding the Point of Intersection of Two Lines
To ﬁnd the point of intersection of two vector lines, set the two vectors equal and solve the resulting system of simultaneous equations. Make sure that the two diﬀerent equations use two distinct variables for the real number multiple of the direction vector. If the lines are in two-dimensions, then they will either be parallel or they will intersect. In three dimensions the lines might be parallel, intersect, or be skew (no intersection but not parallel).
 
 
 
 
4
5
1
−1
 
 
 
 
 
 
 
 
#»
#»
For example, given the two lines r1 = 7 + t  0  and r2 =  3  + s  2  ﬁnd the
 
 
 
 
 
 
 
 
2
−4
−2
4 point B where they intersect.
If the two lines intersect, then there will be some values for s and t that produce the same
#» #» position vector on both lines. So, we start by setting r1 = r2 :
 
   
 
4
5 1
−1
 
   
 
 
   
 
7 + t  0  =  3  + s  2 
 
   
 
 
   
 
2
−4
−2
4
This implies the system of linear equations:
4 + 5t = 1 − s
7 + 0t = 3 + 2s
2 − 4t = −2 + 4s
Given that the second equation has 0 × t, that one will only have variable s left.
7 + 0t = 3 + 2s
7 = 3 + 2s
2=s
So s = 2. Substituting that back into one of the other two equations in the system (let’s pick the ﬁrst one), we get:
4 + 5t = 1 − s
4 + 5t = 1 − (2)
4 + 5t = −1 t = −1

95

Now we check our work to ensure that using the values s = 2 and t = −1 produce the same vector:  
  
  
4
 5   4 − 5  −1
 
  
  
 
  
  
+ (−1)  0  =  7 − 0  =  7 
7
 
  
  
 
  
  
2
−4
2 − (−4)
6
 
  
  
1
−1  1 − 2  −1
 
  
  
 
  
  
 3  + (2)  2  =  3 + 4  =  7 
 
  
  
 
  
  
−2
4
−2 + 8
6
So, the point B where the two lines intersect is (−1, 7, 6).

NOTE: If the problem appears on a calculator test, remember from Section 1.25, “Solving
Systems of Three Linear Equations Using Technology” that you can enter the system of three linear equations in three variables into the calculator and it will give you the answer in a single step.

4.25.1 Finding Intersection Between a Line and an Axis
If you are given the equation of one vector line L1 and asked to ﬁnd where it intersects an axis like the x-axis, use a vector equation for the desired axis as your L2 and then solve the problem as described in Section 4.25, “Finding the Point of Intersection of Two Lines”. Since the origin
 
0
 
 
0 is a point on all three axes, it is the simplest vector to use for the position vector of the
 
 
0
#»
#» #» axis line equations. Then you use the base vector ( i , j , or k ) from Section 4.18, “Base Vectors for Three Dimensions” (or the two-dimensional versions if you are working in two dimensions).

96

So, the vector equations for the three axes in three dimensions are:
 
 
0
1
 
 
 
 
• x-axis: r = 0 + s 0
 
 
 
 
0
0
 
 
0
0
 
 
 
 
• y-axis: r = 0 + s 1
 
 
 
 
0
0
 
 
0
0
 
 
 
 
• z-axis: r = 0 + s 0
 
 
 
 
0
1
The analogous vector equations for the axes in two dimensions are:
 
 
0
1
• x-axis: r =   + s  
0
0
 
 
0

0
• y-axis: r =   + s  
0
1

97

5 Statistics
5.1 Concepts
5.1.1 Population versus Sample
Statistics are drawn from an overall set called a population. When you draw a subset of examples from the population, you are selecting a sample. Since it may be impractical (or impossible) to measure statistics about an entire population, we use the statistics for a sample to estimate population statistics. A random sample is a subset of the population that is chosen randomly, to increase the chances that the sample statistics fairly represent the whole population.
5.1.2 Discrete Data versus Continuous Data
Discrete data is counted, while continuous data is measured.
Discrete data typically have natural number values and have a distinct set of possibilities you can identify (and count) ahead of time. Examples include:
• dress sizes
• outcomes from rolling dice
• candies in a box
In contrast, continuous data occur somewhere in a real number interval of values, and it’s not possible to count the number of distinct values ahead of time. Examples include:
• height or weight of students in a class
• times of a swimming race
• length of leaves fallen from a tree

5.2 Presentation of Data
Numerical data is typically arranged in a table, which has one row for each group or class of data. For discrete data you usually have one row per distinct data value, whereas for continuous data each row typically represents a subinterval of the overall range of possible values. Be careful when labeling the range groups to partition the data uniquely into groups. If the ﬁrst group were labeled as 0 f (x) here

f (x)>g(x) here

Of course, in a non-calculator problem the points of intersection and the function values to compute would be something simpler than the functions and values in the previous section, but the concept of identifying the intersection points and performing the integral as separate parts added together (with the larger function written ﬁrst in the subtraction) is still the key point to remember here.

7.28 Net Change in Displacement versus Total Distance Traveled
It’s important to understand the diﬀerence between displacement and distance traveled so you know how to calculate each one when requested.
7.28.1 Diﬀerence Between Distance and Displacement
Figure 136 illustrates displacement. It is a vector quantity that tells how a body’s current position related to its starting point.

Figure 136: Displacement is Vector Measure from Initial Position to Final Position7
Figure 137 illustrates distance. It is a scalar quantity that measures how far a body has to travel along a path to get from its initial position to its current position.

Figure 137: Distance is Scalar Measure of Actual Length of Path Traveled8

7
8

Source: tutorvista.com/physics/difference-between-distance-and-displacement
Source: tutorvista.com/physics/difference-between-distance-and-displacement

164

7.28.2 Computing Total Distance Traveled
The velocity function v(t) is the derivative of the displacement function s(t), giving the rate of change in distance at any time t. Vice versa, the displacement function s(t) is the integral of the velocity function v(t). The Fundamental Theorem of Calculus tells us that the area under the velocity curve equals the change in the displacement function. Since displacement has direction and can be backwards and forwards, when you compute the integral of the velocity function, you end up determining the net change in displacement. If some of the area under the velocity function is negative, it “cancels out” the area under the velocity function that might be positive, possibly resulting in a net change in displacement that is zero or maybe even negative.
If you want to calculate the total distance traveled (regardless of direction) you need to compute the area of the absolute value of the velocity function so that any negative values don’t cancel out positive values in the distance calculation. Therefore, to compute the total distance traveled from time t1 to t 2, calculate the deﬁnite integral between t1 and t2 of the absolute value of the velocity function, with respect to time. t2 |v(t)| dt

Total Distance Traveled = t1 165

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