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Precalculus
DeMYSTiFieD®
Rhonda Huettenmueller
Second Edition
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Precalculus DeMYSTiFieD®, Second Edition
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Contents
Introduction
Lines and Their Slopes
The Slope of a Line
Linear Equations
Parallel and Perpendicular Lines
CHAPTER 1 Applications of Lines and Slopes
Summary
Quiz
Introduction to Functions
The Definition of a Function
Evaluating a Function
More on Evaluating Functions
CHAPTER 2
Domain and Range
Summary
Quiz
Functions and Their Graphs
Evaluating a Function
The Domain and Range
Average Rate of Change
CHAPTER 3 Even and Odd Functions
Introduction to Graphing Calculators
Summary
Quiz
Combinations of Functions
Arithmetic Combinations
Function Composition
CHAPTER 4 Inverse Functions
Summary
Quiz
Transformations of Functions
Shifting Graphs
Stretching and Compressing Graphs
Reflecting Graphs
CHAPTER 5
Function Transformations and Graphing Calculators
Summary
Quiz
Quadratic Functions
The Graph of a Quadratic Function
Finding a Quadratic Function
Finding the Vertex
CHAPTER 6
Optimizing Quadratic Functions
Summary
Quiz
Polynomial Functions
Introduction to Polynomial Functions and Their Graphs
The Real Zeros of a Polynomial Function
Polynomial Division
The Rational Zero Theorem
CHAPTER 7
Complex Numbers
The Fundamental Theorem of Algebra
Summary
Quiz
Rational Functions
Introduction to Rational Functions
Sketching the Graph of a Rational Function
CHAPTER 8
Summary
Quiz
Exponential and Logarithmic Functions
Compound Growth
The Exponential Function
Logarithms
The Logarithmic Function
CHAPTER 9
Equations Involving Exponents and Logarithms
Three Important Logarithm Properties
Summary
Quiz
Systems of Equations and Inequalities
Systems of Linear Equations
Other Systems of Equations
CHAPTER 10 Systems of Inequalities
Summary
Quiz
Matrices
Matrix Arithmetic
Row Operations and Inverses
Matrices and Systems of Equations
CHAPTER 11
The Determinant of a Matrix
Summary
Quiz
Conic Sections
Parabolas
Ellipses
Hyperbolas
CHAPTER 12
Equations in Other Forms
Summary
Quiz
Trigonometry
The Unit Circle
Angles
Trigonometric Functions
Graphs of Trigonometric Functions
Right Triangle Trigonometry
CHAPTER 13 Inverse Trigonometric Functions
The Law of Sines and the Law of Cosines
Miscellaneous Formulas
Introduction to Proving Trigonometric Identities
Summary
Quiz
Sequences
The nth Term of a Sequence
Partial Sums for a Sequence
Arithmetic Sequences
CHAPTER 14
Geometric Sequences
Payments and Geometric Sequences
Summary
Quiz
Final Exam
Answers to Quizzes and Final Exam
Appendix
Index
Introduction
The purpose of this book is to prepare you for calculus. To succeed in calculus, you need strong
algebra skills as well as an understanding of functions and their graphs. Most of the book is dedicated
to covering the major families of functions (and their graphs): linear, quadratic, polynomial, rational,
exponential, logarithmic, and trigonometric functions. Important calculus concepts are introduced in
Chapters 2 and 3: the average rate of change of a function, the difference quotient, and
increasing/decreasing intervals for a function. Chapter 6 introduces another important calculus
concept: optimizing a function.
Before beginning your study of precalculus, you should have basic algebra skills: the ability to
factor expressions, simplify fractions, solve equations and inequalities, and work with exponents and
square roots. The Appendix contains a brief review of some of these topics. You should also have a
basic knowledge of the xy plane, how to plot points and how to read graphs.
You will get the most from this book if you study the examples before working the Practice
problems. If you get a problem wrong, make sure you understand where you went wrong (the solutions
are worked out) before moving to the next topic. Study the Chapter Summary at the end of each
chapter before taking the quiz and then take it as if you were in a classroom—without using notes or
looking at the chapter.
Once you have worked through the chapters, you can take the Final Exam. Rather than try all 90
problems at once, treat the exam as three separate 30problem tests. Try to improve your score with
each attempt. If you do reasonably well on these exams, then you are ready for calculus.
I wish you the best of luck!
Rhonda Huettenmueller
About the Author
Rhonda Huettenmueller has taught mathematics at the college level for 20 years. She earned a PhD
in mathematics from the University of North Texas and is the author of several books in the
Demystified series, including the bestselling Algebra Demystified.
chapter 1
Lines and Their Slopes
Most of an introductory course in calculus is concerned with the rate of change, that is, how fast a
variable is changing. That rate of change is measured by the slope of a line, so we begin our
preparation for calculus by studying lines and their slopes. We will learn an application from calculus
for the slope of a line in Chapter 2. We will also use linear equations to help us solve applications
(word problems) in this chapter.
CHAPTER OBJECTIVES
In this chapter, you will
Find the slope of a line
Work with horizontal and vertical lines
Work with two lines that are parallel or perpendicular
Find an equation of a line from two of its points
Interpret the slope of a line
Work with applications of lines
The Slope of a Line
The slope of a line measures its tilt. The sign of the slope tells us whether the line tilts up (if the slope
is positive) or tilts down (if the slope is negative), see Figure 1-1. The larger the number, the steeper
the slope (see Figure 1-2).
FIGURE 1-1
FIGURE 1-2
We can find the slope of a line by putting the coordinates of any two points on the line, ( x1, y1) and
(x2, y2), in the slope formula.
For example, (0, 3), (−2, 2), (6, 6), and (−1, ) are all points on the same line. We can pick any pair of
points to compute the slope.
A slope of means that if we increase the x-value by 2, then we need to increase the y-value by 1 to
get another point on the line. For example, knowing that (0, 3) is on the line means that we know (0 +
2, 3 + 1) = (2, 4) is also on the line (see Figure 1-3).
FIGURE 1-3
Because a horizontal line is not tilted at all, we expect the slope to be zero. As we can see from
Figure 1-4 (on page 4), (−4, −2) and (1, −2) are two points on a horizontal line. When we put these
points in the slope formula, we will see that the slope really is zero.
FIGURE 1-4
The y-values on a horizontal line do not change but the x-values do, so the numerator of the slope is
always zero for a horizontal line. What happens to the slope formula for two points on a vertical line?
The points (3, 2) and (3, −1) are on the vertical line in Figure 1-5 (on page 4). Let’s see what
happens when we put them in the slope formula.
FIGURE 1-5
This is not a number, so the slope of a vertical line does not exist (we also say that it is undefined).
The x-values on a vertical line do not change but the y-values do.
Linear Equations
Every line in the plane is the graph of a linear equation. The equation of a horizontal line is y = a
(where a is the y-value of every point on the line). Some examples of horizontal lines are y = 4, y = 1,
and y = −5. These lines are sketched in Figure 1-6. The equation of a vertical line is x = a (where a is
the x-value of every point on the line). Some examples are x = −3, x = 2, and x = 4 (see Figure 1-7).
FIGURE 1-6
FIGURE 1-7
Other equations normally come in one of two forms: Ax + By = C and y = mx + b. We will usually
use the form y = mx + b in this book. An equation in this form gives us two important pieces of
information. The first is m, the slope. The second is b, the y-intercept (where the line crosses the yaxis). For this reason, this form is called the slope-intercept form. In the line y = x + 4, the slope of
the line is and the y-intercept is (0, 4), or simply 4.
We can find an equation of a line from its slope and any point on the line. There are two common
methods for finding this equation. One is to put m, x, and y (x and y are the coordinates of the point we
know) in y = mx + b and use algebra to find b. The other is to put these same numbers in the pointslope form of the line, y − y1 = m(x − x1), and then use algebra to solve for y. We will use both
methods in the next example.
EXAMPLE 1-1
_______________________________________
Find an equation of the line with slope − containing the point (8, −2).
We let m = − , x = 8, and y = −2 in y = mx + b to find b.
The line is y = − x + 4.
Now we let m = − , x1 = 8, and y1 = −2 in y − y1 = m(x − x1).
Find an equation of the line with slope 4, containing the point (0, 3).
We know the slope is 4 and the y-intercept is 3 [because (0, 3) is on the line], so we can write the
equation without having to do any work: y = 4x + 3.
Find an equation of the horizontal line that contains the point (5, −6).
Because the y-values are the same on a horizontal line, we know that this equation is y = −6. We
can still find the equation algebraically using the fact that m = 0, x = 5, and y = −6. Then y = mx +
b becomes −6 = 0(5) + b. From here we can see that b = −6, so y = 0x − 6, or simply y = −6.
Find an equation of the vertical line containing the point (10, −1).
Because the x-values are the same on a vertical line, we know that the equation is x = 10. We
cannot find this equation algebraically because m does not exist.
We can find an equation of a line if we know any two points on the line. First we need to use the
slope formula to find m, and then we pick one of the points to put into y = mx + b.
EXAMPLE 1-2
_______________________________________
Find an equation of the line containing the given points.
(−2, 3) and (10, 15)
First, we find the slope.
We now use x = −2, y = 3, and m = 1 in y = mx + b to find b.
The equation is y = 1x + 5, or simply y = x + 5.
( , −1) and (4, 3)
Again, we begin by finding the slope.
Using x = 4, y = 3, and m = in y = mx + b, we have
The equation is
(0, 1) and (12, 1)
The y-values are the same, making this a horizontal line. The equation is y = 1.
If a graph is clear enough, we can find two points on the line or even its slope. In fact, if the slope
and y-intercept are easy enough to see on the graph, we know right away what the equation is.
EXAMPLE 1-3
_______________________________________
The line in Figure 1-8 crosses the y-axis at 1, so b = 1. From this point, we can go right 2 and up 3 to
reach the point (2, 4) on the line. “Right 2” means that the denominator of the slope is 2. “Up 3”
means that the numerator of the slope is 3. The slope is , so the equation of the line is y = x + 1.
FIGURE 1-8
The y-intercept for the line in Figure 1-9 is not easy to determine, but we do have two points. We
can either find the slope by using the slope formula, or visually (as we did above). We can find the
slope visually by asking how we can go from (−4, 3) to (2, −1) : Down 4 (making the numerator of the
slope −4) and right 6 (making the denominator 6). If we use the slope formula, we have
FIGURE 1-9
Using x = 2 and y = −1 in y = mx + b, we have −1 = − (2) + b. From this, we have b = . The equation is
y=− x+ .
The line in Figure 1-10 is vertical, so it has the form x = a. All of the x-values are −2, so the
equation is x = −2.
FIGURE 1-10
When an equation for a line is in the form Ax + By = C, we can find the slope by solving the
equation for y. This puts the equation in the form y = mx + b.
EXAMPLE 1-4
_______________________________________
Find the slope of the line 6x − 2y = 3.
The slope is 3 (or ).
Parallel and Perpendicular Lines
Two lines are parallel if their slopes are equal (or if both lines are vertical), see Figure 1-11.
FIGURE 1-11
Two lines are perpendicular if their slopes are negative reciprocals of each other (or if one line is
horizontal and the other is vertical), see Figure 1-12.
FIGURE 1-12
Two numbers are negative reciprocals of each other if one is positive and the other is negative and
inverting one gets the other (if we ignore the sign).
EXAMPLE 1-5
_______________________________________
and − are negative reciprocals
− and are negative reciprocals
−2 and are negative reciprocals
1 and −1 are negative reciprocals
We can decide whether two lines are parallel or perpendicular or neither by writing them in the form y
= mx + b and then comparing their slopes.
EXAMPLE 1-6
_______________________________________
Determine whether the lines are parallel or perpendicular or neither.
4x − 3y = −15 and 4x − 3y = 6
We begin by solving each equation for y, and then we compare their slopes.
The lines have the same slope, so they are parallel.
3x − 5y = 20 and 5x − 3y = −15
The slopes are reciprocals of each other but not negative reciprocals, so they are not
perpendicular. They are not parallel, either.
x − y = 2 and x + y = −8
The slope of the first line is 1 and that of the second is −1. Because 1 and −1 are negative
reciprocals, these lines are perpendicular.
y = 10 and x = 3
The line y = 10 is horizontal, and the line x = 3 is vertical. They are perpendicular.
Sometimes we need to find an equation of a line when we know only a point on the line and an
equation of another line that is either parallel or perpendicular to it. We need to find the slope of
the line whose equation we have and use this to find the equation of the line we are looking for.
EXAMPLE 1-7
_______________________________________
Find an equation of the line containing the point (−4, 5) that is parallel to the line y = 2x + 1.
The slope of y = 2x + 1 is 2. This is the same as the line we want, so we let x = −4, y = 5, and m =
2 in y = mx + b. We get 5 = 2(−4) + b, so b = 13. The equation of the line we want is y = 2x + 13.
Find an equation of the line with x-intercept 4 that is perpendicular to x − 3y = 12.
The x-intercept is 4 means that the point (4, 0) is on the line. The slope of the line we want is the
negative reciprocal of the slope of the line x − 3y = 12. We find the slope of x − 3y = 12 by
solving for y.
The slope we want is −3, which is the negative reciprocal of . When we let x = 4, y = 0, and m =
−3 in y = mx + b, we have 0 = −3(4) + b, which gives us b = 12. The line is y = −3x + 12.
Find an equation of the line containing the point (3, −8), perpendicular to the line y = 9.
The line y = 9 is horizontal, so the line we want is vertical. The vertical line passing through (3,
−8) is x = 3.