FURTHER ALGEBRA – A BRIDGE TO ABSTRACTION A COLLEGE ALGEBRA COURSE CONTAINING ALGEBRA II- Complete E

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FURTHER ALGEBRA – A BRIDGE TO ABSTRACTION A COLLEGE ALGEBRA COURSE CONTAINING ALGEBRA II

Mark Daniels Clinical Professor of Mathematics Assistant Chair - Mathematics Department Associate Co-Director - UTeach Natural Sciences Co-Director IBL Project - Mathematics Department OnRamps

The University of Texas at Austin c 2018 i

To the Student It is my genuine hope that the student/learner experiences this textbook in the manner in which I intend it. I am fully aware that any algebra text, in the minds of many students, is full of information and techniques that will likely not be used beyond the life of the course. For many, this has been true of various previously encountered mathematics courses. As I wrote and envisioned the topics and teaching of the Units of this text, I purposefully designed the student experience with the material to be different from most mathematics courses encountered previously.

Based on future degree paths and career choices, some may need to know the mathematics topics of this text very well, while others may not. This is why I wrote this course in such a way that it is more about thinking than about learning any individual topic of the curriculum. The skills I hope the student/learner take away from this course are broader than any single topic in the course. I mainly wish that the Exercises in this book teach one the skills of:

1. Critical Thinking 2. Creativity 3. Learning from failure 4. and Persistence.

It is also my wish that the course is taught to you in such a way as to foster collaboration and communication among your peers and your instructor as part of the process of working through the Exercises in the text.

If, indeed, the student/learner finishes the course better at the skills of critical thinking, creativity, learning from failure, and persistence, then the student’s life is enriched in a way that is helpful for any future path chosen.

Sincerely, The Author

Cover Page: Pictured is the quadratic function, z 2 + c, where c ∈ C, evaluated under iteration for all values of c under the orbit of 0. Given these circumstances the graph produced is infinitely complex in nature. The picture, created by Mark and Gunter Daniels, is named Moons Over Mandelbrot. You will learn about the complex numbers and function iteration in this course.

Dedication: This work is dedicated to the memory of professors Efraim Armendariz and Ted Odell.

Their friendship and guidance was invaluable to me.

ii

Contents 1 Essential Prerequisites

1

1.1 Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2 Thinking Like a Mathematician . . . . . . . . . . . . . . . . . . . . . . . . . .

7

1.3 Prerequisite Topics Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

2 Functions

13

2.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13

2.2 Linear and Absolute Value Functions . . . . . . . . . . . . . . . . . . . . . .

18

2.3 Composition of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

2.4 A Function Recursive Process . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

2.5 Function Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

2.6 Inverse of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

35

3 Systems

41

3.1 Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41

3.2 Systems Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

46

3.3 Systems with Three Unknowns . . . . . . . . . . . . . . . . . . . . . . . . . .

47

3.4 Systems of Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

50

4 Matrices

55

4.1 Matrices and Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . .

55

4.2 Determinants and Gaussian Elimination . . . . . . . . . . . . . . . . . . . .

63

iii

4.3 Matrix Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5 Quadratic Functions

72

79

5.1 Factoring Quadratic Expressions . . . . . . . . . . . . . . . . . . . . . . . . .

79

5.2 A Development of the Quadratic Function . . . . . . . . . . . . . . . . . . .

84

5.3 Transformations of the Quadratic . . . . . . . . . . . . . . . . . . . . . . . .

89

5.4 Applications of the Quadratic Function . . . . . . . . . . . . . . . . . . . . .

94

6 Complex Numbers 6.1 The Imaginary Number

99 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

99

6.2 The Geometry of Complex Numbers . . . . . . . . . . . . . . . . . . . . . . .

99

6.3 Complex Number Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 6.4 Complex Numbers and the Quadratic . . . . . . . . . . . . . . . . . . . . . . 104

7 Polynomial Functions

105

7.1 Factoring Polynomials of Degree 3 . . . . . . . . . . . . . . . . . . . . . . . . 105 7.2 Polynomial and Power Functions . . . . . . . . . . . . . . . . . . . . . . . . . 106 7.3 Division and Rational Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 7.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

8 Modeling and Data Analysis

123

8.1 Direct and Inverse Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 8.2 Piecewise Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 8.3

Regression and Function Fitting . . . . . . . . . . . . . . . . . . . . . . . . . 129

8.4 Applications and Regression Modeling . . . . . . . . . . . . . . . . . . . . . 131 iv

9 Rational and Radical Functions

135

9.1 Rational Expressions and Rational Functions . . . . . . . . . . . . . . . . . . 135 9.2 Applications of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . 147 9.3 Radical Expressions, Equations, and Functions (With Applications) . . . . . 147 9.4 Advanced Literal Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

10 Exponents, Logs, and e

159

10.1 e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 10.2 Exponential and Logarithmic Properties and Functions . . . . . . . . . . . . 160 10.3 Scientific Notation

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

10.4 Fermi Questions (and Orders of Magnitude) . . . . . . . . . . . . . . . . . . 175 10.5 Applications of Exponents and Logarithms . . . . . . . . . . . . . . . . . . . 177 10.6 Function Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

11 Sequences and Series

187

11.1 Arithmetic and Geometric Sequences . . . . . . . . . . . . . . . . . . . . . . 187 11.2 Defining a Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 11.3 Probability and Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . 194 11.4 The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

v

Unit 1 - Essential Prerequisites

1.1

Prerequisites

We start this Unit with an introduction to some of the most important notation and symbols that you may or may not have been introduced to previously. Knowledge of these items is generally assumed as you progress to higher mathematics courses. A definition of the concept of a set is also given. An understanding of what a set is will help you understand the concept function, which we will also visit as part of this module.

Sets Definition. Set. A set is a well-defined collection of elements.

By well-defined collection, we mean that there is no question as to what the elements of the set are representing. For example, suppose I defined my set to be the set of all small apples in a barrel. This set would not be well-defined, since one might ask, “What is your criteria for what constitutes a ‘small’ apple?” Whereas, if my set is “the positive integers,” there is no question as to which numbers belong to my set.

One names a set using {...} notation. For example, the set, S, of digits from 0-9 is written

S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

Set S is an example of a finite set, since the set contains a finite number of elements.

An example of an infinite set would be the set of all positive integers. We could name this set P , and it could be symbolized as

P = {1, 2, 3...}.

Last, it must be noted that a set may contain no elements at all. This set is called the empty set. It is symbolized ∅. Thus, if E is the empty set, we can write

E = ∅.

1

Notation and Symbols Here is a list of common symbols for number systems you have studied or will likely soon study.

N = {1, 2, 3, ...}

The Natural Numbers

Z = {..., −3, −2, −1, 0, 1, 2, 3, ...}

The Integers

Q = The set of all Rational Numbers a These are numbers of the form where a and b are elements of Z and b 6= 0. b Qc = The set of all Irrational Numbers Numbers that are not rational numbers are called irrational numbers.

R = The set of Real Numbers The set of real numbers consist of all of the rational and irrational numbers. Commonly, the real numbers are described as all numbers that have points on the number line.

C = The set of all Complex Numbers You have likely heard of complex numbers. Complex numbers are of the form x + yi √ where x and y are elements of the real numbers and i = −1. We will discuss this number system in detail later in the text.

The next few symbols of interest to us are commonly used as a shorthand way of writing some common mathematical statements. These are presented for your use.

Symbol

Meaning

∈

is an element of or belongs to, as in a ∈ A

∃

there exists

∀

for all or for every

: or |

such that

→

an implication or mapping (when used with functions). We can say that f : A → B is a function that maps or relates elements of set A to those in set B.

2

Exercise 1.1.1: Sets and Notation 1. Which of the following satisfy the requirements of a set? Explain why the requirements are or are not met. (a) All the current U.S. Senators from Texas (b) All the prime divisors of 1987 (c) All the tall people in Mexico (d) All the prime numbers between 8 and 10 2. Rewrite each statement using ∃, ∀, ∈ and | as appropriate. (a) There exists a positive number x belonging to the set R such that x2 = 5. (b) For every positive number M there is a positive number N such that 1 N< .

M (c) There exists m which belongs to the set M .

3. List the numbers in set M that are rational.

√ π 8 √ M = 0, 11, −0.06, 0.75, 0.7, 36, −49, , √ , −25 6 2 4. Why is π not a rational number?

5. Explain why 0.3 is a rational number. √ 6. If i = −1, what is i2 equal to? √ √ √ 7. If x ≥ 0, then −x = i x. Thus, −5 =?

3

Interval Notation In higher mathematics courses interval notation is often used to describe relevant sets of real numbers on the number line. The reader should become familiar with this notation. These sets can be closed intervals, open intervals, or half-open intervals. They are described below.

For a, b, and x elements of the real numbers: Interval [a, b] (a, b) (a, b] [a, b)

Meaning a≤x≤b a<x<b a<x≤b a≤x<b

Name closed interval open interval half-open interval half-open interval

We will also note that intervals (or any sets for that matter) can be strung together with a union symbol, ∪, which strictly implies a mathematical “or statement.” This means that all the elements are considered in any of the listed sets. For example,

[a, b] ∪ [c, d] means all elements to be considered belong to the interval [a, b] or the interval [c, d].

The Real Numbers As previously mentioned, the real numbers consist of all of the rational and irrational numbers. The real numbers are defined as all numbers that have points on the number line. In this sense the real numbers are complete in that they completely fill in the number line. Another useful property of the real numbers is that they constitute a field. The field of real numbers adheres to 11 rules which are also known as axioms.

There are five axioms for addition and five for multiplication, as well as one axiom that connects addition and multiplication.

4

The 11 field axioms for the real numbers: For all x, y, and z ∈ R: Addition Axioms

Multiplication Axioms

Closure

x+y ∈R

x·y ∈R

Commutative Rules

x+y =y+x

x·y =y·x

Associative Rules

(x + y) + z = x + (y + z)

(xy) z = x (yz)

Additive Identity

Multiplicative Identity

x+0=x

x·1=x

Identity Elements

Note that the additive identity, 0, is unique in that it is the only real number that has this property under addition.

Note that the multiplicative identity, 1, is unique in that it is the only real number that has this property under multiplication.

Inverse Rules

x + (−x) = 0

x·

1 = 1, for x 6= 0 x

Axiom Connecting Addition and Multiplication Distributive Rule x (y + z) = xy + xz

5

Exercise 1.1.2: The Field of Real Numbers 1. Complete the following table of the 11 field axioms for the real numbers.

Do this without referring back to the previous section.

For all x, y, and z ∈ R:

Addition

Multiplication

Closure x·y =y·x

Commutative Rules Associative Rules Identity Elements Inverse Rules

x + (−x) = 0

Distributive Rule

2. Why is it important that the real numbers have the property of being complete?

3. Give an example to show that the set of irrational numbers is not closed under the operation of multiplication.

4. Give an example to show that subtraction does not follow the associative rule.

6

1.2

Thinking Like a Mathematician

The Exercises in this section are meant to “get you started” on the right track regarding the expectations of the type of thinking expected in this course.

Exercise 1.2.1: A Tax(i)ing Problem Three people, Alice, Ben, and Calvin, are conversing at a taxi stand since taxis are the only ride service in this town. Although they haven’t met before, they realize that all are going along the same route to get to each one’s desired destination.

Alice’s destination is 20 miles away, Ben’s destination is 30 miles away, and Calvin’s destination is 40 miles away. The taxi costs 2 dollars per mile with tip included regardless of the number of passengers. Provide an argument as to how much each person should pay if the three share a cab to their respective destinations.

Exercise 1.2.2: Let’s Start with Integers 1. Can you devise an argument to show that the sum of any two odd numbers is an even number?

2. Can you prove that the product of any four consecutive positive integers is divisible by four?

1.3

Prerequisite Topics Review

The Exercises in this section are provided to illustrate the expectation of what mathematics the reader/learner should already be familiar with in order to continue in the course. Try to complete the following problems and make a note of any questions you have as you work.

7

Exercise 1.3.1: Prerequisite Knowledge Problems Note that some of these topics will be revisited within this course.

1. Solve the equation for x: 3x − (5x − 9) = 7 − 4 (x + 1).

2. Sketch the graph: 4y − 3x = 12.

3. Find the slope and y-intercept of the line given by 5y + 2x = −10.

4. Find the slope of the line containing the points (3, −2) and (−1, 6).

5. Draw the line with slope -2, which contains the point (3, 1).

( x + 3y = −5 6. Solve the system: 5x − 2y = 9

7. Simplify the expression:

−3ab2 4bc4

.

3 .

8. Perform the subtraction: 4x2 − 8xy + 3y 2 − 7x2 − xy − 3y .

9. Perform the distribution: (x − 7) (2x − 4).

10. Factor the expression: 8x2 y 4 − 12x5 y 3 + 4x2 y 3 .

8

11. Factor the expression: 4x2 − 12xy − x + 3y.

12. Solve for x in the equation: x2 − x = 20.

13. Jerry bought a coat on sale at 30% off. If the sale price for the coat was $63, what is the regular price for this coat?

14. If you can travel 25 miles in 35 minutes, how long should it take you to travel 12 miles?

15. Write in scientific notation: (a) 230,000,000,000 (b) 0.000076 16. Write in standard notation: (a) 3.3 × 10−7 (b) 3.1 × 1011 17. Multiply and give the result in scientific notation: 5 × 10−3

4 × 105 .

18. Hank invested a total of $20,000, part at 7% and part at 10%. How much did he invest at each rate if the total interest earned in one year was $1640?

19. Draw a best-fit line for the data in each plot and justify your drawing. (a) (b) (c)

9

Unit 1 Review

1. A rational number is any number that can be written as

.

2. List the numbers in set S below that are rational.

√ π 2 8 √ S = 0, 15, −0.06, 0.77, 0.7, 36, −49, , , √ −100 4 7 2 3. Why is π not a rational number?

4. Explain why 0.4 is a rational number.

5. List the numbers in set S above that are integers.

6. Explain how the set of non-positive integers differs from the set of negative integers.

7. Give an example to show that the set of irrational numbers is not closed under the operation of multiplication.

8. Which two basic arithmetic operations are commutative?

9. Give an example to show that subtraction is not an associative operation.

10. The additive identity element is

.

11. The multiplicative inverse of x (if x 6= 0) is

.

12. Explain what mistake is made in the following false statement.

3 (x + 4) (x + 7) = (3x + 12) (3x + 21)

10

Which field property is illustrated in each of the following? Include the appropriate operation.

13. a · (b + c) = a · (c + b)

14. a · (b + c) = (b + c) · a

15. a · (b + c) = ab + ac

16. a · (b + c) = a · (b + c) · 1

17. Is it ever possible that | x |= −x? Explain your answer.

Evaluate each expression for x = −1.

18. | 5 − 7x | −8

√

19. 4x2 − 5x − 11

20.

22. 4x − [6x + (x − 4)]

1 23. 6 x − (x − 1) 2

x2

Simplify each expression completely.

21. 7 [3 + 2 (x − 4) − 6 (x − 4)]

Multiply and write the answer in simplest form.

24. (5x + 6) (x + 4)

25. (3x − 10)2

11

26. (2x + 7) (2x − 7)

12

Unit 2 - Functions

2.1

Functions

In this section, we will define what a function is in mathematics, look at function properties, and list seven basic functions.

We start with a less rigorous definition of function.

Definition 1. Function. A function is a rule or relation in which there is exactly one value of the dependent variable for each value of the independent variable. The independent value is a member of the domain of the function and the dependent value is a member of the range of the function.

While this definition is a good definition of what a function is, we want to work toward a more rigorous understanding of function. In this regard, the next definition is a stronger and more general definition of function.

Definition 2. Function. Given two sets A and B, a function f from A to B is a rule or relation between A and B that assigns each element a ∈ A to exactly one element b ∈ B. This functional cross between sets A and B is often referred to as a Cartesian cross. This is symbolized A × B . The set A × B consist of all ordered pairs (a, b) where a ∈ A and b∈B. EXAMPLE: You are used to seeing functions written such that y is a function of x. That is, a unique value of y can be found for each value of x.

Given 2x − 3y = 15, we solve for y, which yields 2 y = x − 5.

3 Notice that in our example, y is a function of x. Thus, we can state in symbols that y = f (x).

13

Keep in mind, however, that functions can relate any two variables, not just y and x.

Other function (which of course includes linear functions) terminology that may need to be refreshed for you are: 1. x-intercept: The value of x when y = 0.

2. y-intercept: The value of y when x = 0.

Exercise 2.1.1: Functions 1. Let A = {a, b, c}, B = {4, 5, 6}, and f = Is f a function from A to B? Explain.

(a, 6) , (b, 4) , (c, 6) .

2. For the given function and domain, specify the associated range.

Function: Domain: Range:

1 {1, 2, 3, 4, 5, 6} y=5 x d = 2t3

t∈R

s =| x + 4 |

−6 ≤ x ≤ 6

Exercise 2.1.2: Function Investigation For each relation in problems 1-8 below, state the domain and range.

1. f (x) = 3x − 5

3. f (x) = 7 − 2x2

2. g(x) = 6

4. To the right is a piecewise function. Can you explain why?

14

5. The set of ordered pairs (t, m) where m is the remaining mass of a 500-gram sample of a radioactive substance that has been decaying for t hours.

6. The set of ordered pairs (m, T ) where T is the temperature of a cake that is removed from a 400◦ oven and cools in a 72◦ kitchen for m minutes.

7. The set of ordered pairs (t, d) where d is the distance in Astronomical units (AU) from the earth to the sun on day t, where t = 1 is January 1. This distance has an average value of 1 AU and varies from this average value a maximum of 0.017 AU. 8. The set of ordered pairs (n, V ) where V is the value in cents of n dimes.

For problems 9-17, decide which of the relations listed below are functions.

Justify each of your answers.

9.

11.

10.

x y

2 4

-6 4

13 13

4 -6

12. y = x2

13. The set of ordered pairs (h, m) where m is the remaining mass of radioactive compound that has been decaying for h hours.

14. The set of ordered pairs (M, D) where M is the mother of daughter D. 15. y =

2x − 1 x+5

16. The graph below

15

17. The line y = mx + b

Parent Functions We now provide the graphs of seven basic functions that you should become familiar with. These are common parent functions that generate a family of functions based on some manipulation of each.

Definition. Parent Function. A parent function is the simplest function for a family of functions of the same type. For example, the function f (x) = x is the parent function for the family of linear functions of the form f (x) = mx + b.

Here are the graphs of the important parent functions that you should recognize currently or become familiar with.

The Linear Function

The Quadratic Function

The Absolute Value Function

16

The Cubic Function

The Square Root Function

The Rational Function

The Exponential Function

Recall that e is an important constant that shows up in many branches of mathematics.

It is an irrational number. Does the behavior of f (x) = ex look any different than f (x) = bx , where b is a positive integer greater than 1?

GOOD ADVICE. You should start to become familiar with an important constant in mathematics, named e. This irrational number is approximately equal to 2.71828. Like the irrational number π , the number e shows up in many branches of mathematics. A development of the number e is usually provided in a Precalculus or Calculus course.

Exercise 2.1.3: Seven Parent Functions State the domain and range for each of the seven parent functions of this section.

Function: Increasing and Decreasing Definition. y = f (x) is defined on an Increasing and Decreasing. If a function

interval a, b , then f (x) is increasing on a, b if whenever x1 < x2 , then

f (x1 ) < f (x2 ). The function f (x) is decreasing on a, b if whenever x1 < x2 , then f (x1 ) > f (x2 ).

17

Exercise 2.1.4: Function Increase and Decrease 1. For each of the seven parent functions of this section, state the domain interval(s) on which each function is increasing or decreasing. You may use interval or inequality notation.

For problems 2-3, consider the graph of f (x) = − (x − 3) (x + 3).

2. On what interval(s) is this graph increasing?

3. On what interval(s) is this graph decreasing?

4. Do you think it makes sense to say, “This graph is increasing or decreasing at x = 0?” Why or why not?

5. For each table below, describe whether the table represents a function that is increasing or decreasing.

(a)

2.2

x 1 2 3 4 5

f (x) 3 6 12 24 48

(b)

x 1 2 3 4 5

g(x) 90 80 75 72 70

(c)

x 1 2 3 4 5

h(x) 200 190 170 140 100

(d)

x 1 2 3 4 5

z(x) 0 15 25 33 37

Linear and Absolute Value Functions

Linear Functions Although you have likely studied linear functions previously, we present a brief summary of the important properties of these functions. Don’t worry, there is much new material to come; we do want to start on a firm foundation.

A linear function is a function whose graph produces a line. Linear functions are often written in the standard slope-intercept form y = mx + b, where m is the slope of the function, also known as the constant rate of change of the function. The slope determines if the linear function is an increasing function or a decreasing function. Recall that slope =

y2 − y1 ∆y rise = = = m. run x2 − x1 ∆x 18

In function notation, if y1 = f (x1 ) and y2 = f (x2 ), slope is written equivalently as m=

f (x2 ) − f (x1 ) . x2 − x1

Also note that: f (x) = mx + b is an increasing function if m > 0 and f (x) = mx + b is an decreasing function if m < 0.

If m = 0, the rate of change of the function is zero, and the function is f (x) = b, a horizontal line passing through the point (0, b).

Properties of Linear Function Graphs Linear functions display a pattern that extends from the domain to the range of the function. This pattern is stated as: Addition—Addition Property for Linear Functions: For a linear function f , adding a constant c to a given domain value results in adding a constant k to the corresponding range value.

EXAMPLE 1: The following two tables display data that follow the addition – addition property, which is characteristic of linear functions. As an exercise, identify the constant added in the domain and the resulting constant that is added in the range to each of the data points.

Table 2.1 x f (x) 2 13 4 23 8 43 10 53

Table 2.2 x h(x) 0 5 5 -10 10 -25 15 -40

Some other important properties of linear functions that one should always be aware of are stated on the next page.

19

Given two linear equations f (x) = m1 x + b and g(x) = m2 x + b, 1. The lines will be perpendicular if one slope is the negative reciprocal of −1 the other. That is, if m1 m2 = −1, and thus m2 = , the lines are m1 perpendicular.

2. The lines are parallel if the slopes are equal. That is, the two lines will be parallel if m1 = m2 .

3. A vertical line (which is not a function) is written in the form x = k , where k is a constant.

4. A horizontal line is of the form y = c , where c is a constant.

Different Forms of the Equation of a Line

1. y = mx + b

Slope-intercept form, which can be constructed if one knows or can find the slope m and the y-intercept of a line.

2. y − y1 = m (x − x1 )

Point-slope form, which can be constructed if one knows or can find the slope m and one data point (x1 , y1 ) of the line. This form of a linear equation can be transformed into slope-intercept form through algebraic manipulation.

3. Ax + By = C

Standard form, where A, B, and C are integers.

This form of a linear equation can be transformed into slope-intercept form through algebraic manipulation.

Although all three of these forms of the equation of a line are quite useful to know, the slope-intercept form is the most commonly used in this course.

CHALLENGE: Solve for y in Ax + By = C. GOOD ADVICE. It is important that one can use two or more known points from linear data to construct the slope and equation of a line that passes through those points.

You will be asked to do so in the following exercise.

20

Exercise 2.2.1: Linear Functions and Properties For problems 1-3, plot the graph of each linear function.

2 1. y = − x + 5 2. (y − 3) = 2 (x + 5) 3 4. What are the coordinates of the y-intercept of the line whose equation is 5x − 2y = −6?

3. x − 4y = 12

5. What are the coordinates of the x-intercept of the line whose equation is 3x + 5y = 20?

7. Give the coordinates of one point 6. What is the slope of the line that on the line whose equation is passes through the points (−3, −8) 4 and (5, −2)? y = (x − 2).

3 1 8. Which of the following points lie on the line y − 5 = (x + 2)?

2 (−5, 2)

(12, 12)

(2, −5)

(−2, 5)

(−8, 2)

(0, 6)

For problems 9-16, write the equation of the line that is described.

10. Contains (4, −10) and has a slope of -3.

9. Contains a y-intercept of -3 and has a slope of 2.

11. Contains (5, 8) and is parallel to the line y = 7x − 6.

12. Contains (7, 2) and is parallel to the line 5x − 3y = 6.

13. Contains (3, −5) and is perpendicular to the line y = −8x + 5.

14. Contains a y-intercept of -9 and is perpendicular to the line 4x + 3y = −6.

15. Is horizontal and contains (−8, 9) .

16. Is vertical and contains (6, −2).

17. Do the points (15, −12), (28, 14), and (32, 21) all lie on the same line?

Explain why or why not.

21

For problems 18-23, determine if each function is increasing or decreasing.

18. f (x) = 4x + 3

19. g(x) = 5x + 6

20. a(x) = 5 − 2x

21. h(x) = −2x + 4

1 22. p(x) = x − 5 4

1 23. n(x) = − x − 2 3

For problems 24-25, find the slope of the line that passes through the two given points.

24. (2, 4) and (4, 10)

25. (−1, 5) and (4, 11)

For problems 26-27, find the linear equation for the supplied graphs.

26.

27.

For problems 28-30, you are given descriptions of two lines. Find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular or neither?

28. Line 1: Passes through (2, 3) and (4, −1) Line 2: Passes through (6, 3) and (8, 5) 29. Line 1: Passes through (1, 7) and (5, 5) Line 2: Passes through (−1, −3) and (1, 1) 30. Line 1: Passes through (0, 5) and (3, 3) Line 2: Passes through (1, −5) and (3, −2) 31. Write an equation for a line parallel to f (x) = −5x − 3 and passing through the point (2, −12).

32. Write an equation for a line parallel to g(x) = 3x − 1 and passing through the point (4, 9).

33. Start with the point slope formula y − y1 = m (x − x1 ) and divide both sides of the equation by the quantity (x − x1 ) . Does the result look familiar?

22

Linear Modeling

Exercise 2.2.2: Linear Modeling In this Exercise, use what you’ve learned about linear functions and their properties to model real-world situations.

1. In science class, you learned about converting temperatures between degrees Celsius (◦ C) and Fahrenheit (◦ F ), but recently you could not find a reference with the correct formulas. You then remembered that the relationship between ◦ F and ◦ C is linear. (a) Using this and the knowledge that 32◦ F = 0◦ C and 212◦ F = 100◦ C, find an equation that computes Celsius temperature in terms of Fahrenheit; i.e. an equation of the form C = “an expression involving only the variable F .” (b) Analogously, find an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form F = “an expression involving only the variable C.” 2. Physicist Jacques Charles (1746-1823) discovered that the volume of a gas at a constant pressure is linearly dependent on the temperature of the gas.

The table below illustrates this relationship. In the table hydrogen gas is held at a constant pressure of one atmosphere. The volume V is measured in liters and the temperature T is measured in degrees Celsius.

T -40 -20 0 20 40 60 80 V 19.15 20.79 22.43 24.08 25.72 27.36 29.00 (a) Consider V as a function of T and use data points from the table to find a linear function to model the data. (b) What is the x-intercept of your line and what does this intercept stand for in terms of temperature and volume?

3. It is believed that the world’s current oil reserves total approximately 2000 billion barrels. If, on average, the total is decreasing by about 25 billion barrels of oil each year: (a) Find a linear equation for the remaining oil reserves, R, in terms of t, the number of years beyond the present. (b) Ten years from now, how much oil will be in the reserves? (c) If the rate of depletion remains constant, when will the world’s oil reserves be depleted?

23

Absolute Value Function Definition. Absolute Value. For a real number x, we define the absolute value of x to be x if either x = 0 or if x > 0, and to be −x if x < 0. The absolute value of x is denoted by |x|. Thus, the absolute value function can be defined as

( x f (x) = |x| = −x

if x ≥ 0 . if x < 0

Notice the absolute value function is defined in a “piecewise” manner depending on whether the domain values are positive, negative, or zero.

Due to this piecewise nature, it should make sense that the parent absolute value function graph looks like this:

The absolute value function is commonly used to determine the distance between two numbers on the number line. Given two real numbers a and b, |a − b| yields the distance, a positive quantity, between these values, regardless of which value is larger.

One consequence of the definition of absolute value is this important theorem statement: If x and a are real numbers with a > 0, then |x| ≤ a iff −a ≤ x ≤ a.

The “iff ” used in this statement stands for “if and only if.” An “if and only if” statement is one that is true when read in both a forward or a backward manner. Thus, this one statement implies that both of these statements are true: 1. If x and a are real numbers with a > 0 and |x| ≤ a, then −a ≤ x ≤ a, and 2. If x and a are real numbers with a > 0 and −a ≤ x ≤ a, then |x| ≤ a.

24

The “if and only if” statement gives one insight as to how to interpret absolute value expressions and the absolute value function. An understanding of absolute value is essential to understanding more abstract statements in mathematics related to the distance between numbers on a number line or within a function.

EXAMPLE 1: First, we start with the task of solving an absolute value equation: |x − 6| = 10.

Considering distance on a number line, this equation can be interpreted as meaning, “The distance from a number to 6 is equal to 10.” Keep in mind that distance is always considered a positive quantity regardless of direction. For example, one never says, “I live negative 10 miles from here.” Thus, the quantity inside the absolute value must be equal to 10 or - 10. This leads to constructing two equations used to solve the original absolute value equation. x − 6 = −10 or x − 6 = 10.

Thus, the solutions are x = −4 or x = 16.

EXAMPLE 2: We now explore an absolute value inequality. Consider: |x − 3| < 5.

One way to interpret this statement is to notice that it can be read as, “The distance from x to 3 is less than 5 units.” Another way to think of this, according to our important absolute value theorem, is to notice that the quantity inside the absolute value must be less than 5 or greater than -5. This leads to constructing two inequalities used to solve the original absolute value inequality: −5 < x − 3 and x − 3 < 5 −2 < x and x < 8.

Thus, the solutions to this compound statement can be succinctly written as −2 < x < 8.

25

EXAMPLE 3: 1 Given the function f (x) = − |4x − 5| + 3, what are the values of x in the domain of f (x) 2 for which the function values are negative?

To answer this question, we introduce an inequality to determine where f (x) < 0: 1 f (x) = − |4x − 5| + 3 < 0.

2 To solve this inequality, begin by isolating the absolute value: 1 − |4x − 5| < −3.

2 Recall, when we multiply both sides by -2, it reverses the sense of the inequality: |4x − 5| > 6.

Next, solve the inequality |4x − 5| > 6.

This means the quantity inside of the absolute value has to be greater than 6 or less than -6 (recall our theorem statement).

4x − 5 < −6

4x − 5 > 6 4x > 11 11 x> 4

or

4x < −1 −1 x< 4

1 11 In interval notation, this is written −∞, − ∪ , ∞ . In terms of our original 4 4 1 function, f (x) = − |4x − 5| + 3, this means that for domain values in the intervals 2 stated above, the function values will be negative.

1 CHALLENGE: Graph the function f (x) = − |4x − 5| + 3 to see that your answer makes 2 sense. In a subsequent section we will learn how to graph f (x) quickly using function transformations on the parent absolute value function.

26

Exercise 2.2.3: Working with the Absolute Value Function 1. Use what you have learned about absolute value and x and y-intercepts to find the x and y-intercepts of the graph of f (x) = |3x + 1| − 5.

2. Graph f (x) = |x − 4|. (a) How could you use the graph to solve the inequality |x − 4| ≤ 4? (b) Solve the equation in (a) by algebraically and by using the graph.

(Hopefully your answers are the same).

3. Interpret the meaning of and solve the inequality 2|z − 4| ≤ 8 for z.

For problems 4-5, graph each function.

5. f (x) = 3|x − 2| − 3

4. f (x) = 2|x + 3| + 1

For problems 6-9, solve each equation.

6. 2|4 − x| = 7

7. 3|5 − x| = 5

8. 3|x + 1| − 4 = −2

9. 5|x − 4| − 7 = 2

For problems 10-13, solve each inequality.

10. |x + 5| < 6

11. |x − 3| < 7

12. |x − 2| ≥ 3

13. |x + 4| ≥ 2

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2.3

Composition of Functions

A composition of functions involves taking the output values from one function and using those as the input values for a second function. Visually, it looks like this:

Using a more general approach, the composition of functions can be thought of as a situation where one or more functions are nested inside another. An example of this would be:

h(x) =

p x3 + 2.

√ We think of this function as being composed of the function f (x) = x with the function g(x) = x3 + 2 nested inside of it. Thus, h(x) can be thought of as h(x) = f (g (x)), which can also be written as h(x) = (f ◦ g)(x) and stated “h of x is equal to f in composition with g.”

GOOD ADVICE. Note that when evaluating h(6) = f (g(6)), for example, one finds g(6) first and then applies the function f to that result.

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Exercise 2.3.1: Composition of Functions 1. Suppose c(p) gives the number of calories burned doing p push-ups, and p(t) gives the number of push-ups a person can do in t seconds. Interpret the meaning of the composite function c(p(45)).

2. For the functions g(t) = t2 − t and h(t) = 3t + 2: (a) Evaluate g(h(5)). (b) Write the equation for g(h(t)).

3. If f (x) = x2 − 3 and g(x) =

√

x + 3, find

(a) f (g(x)) (b) g(f (x))

For problems 4-7, use the table of values to evaluate each expression. x f (x) 4. f (g(3)) 0 7 1 6 2 5 5. f (g(5)) 3 2 4 4 6. g(f (6)) 5 6 6 7 7. g(f (3)) 7 4 8 2 9 1

g(x) 10 7 6 2 1 8 9 11 12 15

8. Let f (x) = 3x − 2, g(x) = 6 − 2x, and h(x) = x2 − 4. (a) Find f ◦ (g ◦ h) and (f ◦ g) ◦ h. (b) Based on your answer to part (a), state a hypothesis or conjecture that serves as a general statement for this situation.

29

2.4

A Function Recursive Process

We will define what is meant by function iteration. Iteration means to repeat a process over and over again. In mathematics this process is most often applied to a mathematical function. A discrete dynamical system can be characterized as a function under iteration, a function that is composed with itself over and over again.

As an example, we consider the simplest nonlinear function imaginable, namely F (x) = x2 . A capital ”F ” is commonly used to denote a function considered under iteration. To iterate x2 , begin with a seed for the iteration. This is a (real or complex) number, which is denoted by x0 . Applying the function x2 to x0 yields the new number x1 = (x0 )2 .

Now we iterate, using the result of the previous computation as the input for the next: x2 = (x1 )2 = (x0 )4

2 x3 = (x2 )2 = (x0 )4 = (x0 )8

and so forth. The list of numbers x0 , x1 , x2 , ... generated by this iteration has a name. It is called the orbit of x0 under iteration of F (x) = x2 . The orbit of a seed the sequence of outputs of the function under iteration. One of the principal questions in this area of mathematics is: What is the fate of typical orbits? Do they converge or diverge? Do they cycle or behave erratically?

Another interesting question to ask about functions under iteration is: what are the fixed points of the function? Fixed points are points such that F (a) = a. For example, the fixed points of F (x) = x2 occur at x = 0 and x = 1, since F (0) = 0 and F (1) = 1.

Using standard functional notation for this process, we define F n (x) to be the nth composition of F (x) with itself, or in other words, the nth iteration of F (x).

How does this relate to composite functions? Notice, for example, that F 2 (x) = F (F (x)) and F 3 (x) = F (F (F (x))) and so forth.

30

Exercise 2.4.1: Iteration √ 1. What happens to x0 under iteration of F (x) = x for 0 < x0 < 1? For x0 ≥ 1?

Also, name any fixed points for this function. Lastly, what happens to seed values under iteration for seed values in the neighborhood of the fixed point(s)?

2. What happens to x0 under iteration of F (x) = 3x ? For F (x) = (−1)x ? Try various seed values for x0 . Also, name any fixed points for each of these functions, if they exist.

3. What happens to x0 under iteration of F (x) = x2 − 2? Try various seed values for x0 . Also, name any fixed points for this function, if they exist.

4. Based on what you’ve learned about function iteration and fixed points, how could one algebraically find fixed points, if they exist, for any function under iteration?

5. Based on what you’ve learned about function iteration and fixed points, how could one graphically find fixed points, if they exist, for any function under iteration?

2.5

Function Transformations

A transformation of a function in the coordinate plane involves a horizontal or vertical slide or shift of the function, a reflection of the function about the x or y-axis, or a stretch or compression of the function. Keep in mind that multiple transformations can be applied to a function at the same time.

Horizontal Shift For a function f (x), define a new function g(x) as g(x) = f (x + k), where k is a constant.

It is then true that: (a) g(x) is a horizontal shift of the function f (x). (b) If the constant k is positive, then the graph will shift left. (c) If the constant k is negative, then the graph will shift right.

31

Vertical Shift For a function f (x), define a new function g(x) as g(x) = f (x) + k, where k is a constant.

It is then true that:

(a) g(x) is a vertical shift of the function f (x), where each output of the range has been increased by k. (b) If k is positive, then the graph will shift up by k units. (c) If k is negative, then the graph will shift down by k units.

Reflection about the y-axis For a function f (x), define a new function g(x) as g(x) = f (−x). It is then true that g(x) is a reflection about the y-axis of the function f (x).

Reflection about the x-axis For a function f (x), define a new function g(x) as g(x) = −f (x). It is then true that g(x) is a reflection about the x-axis of the function f (x).

Vertical stretch or compression For a function f (x), define a new function g(x) as g(x) = kf (x), where k is a constant. It is then true that g(x) is a vertical stretch or compression of the function f (x) such that: (a) If k > 1, then the graph will be stretched (i.e. the graph will move away from the x-axis at a faster rate). (b) If 0 < k < 1, then the graph will be compressed (i.e. the graph will move away from the x-axis at a slower rate). (c) If k < 0, then there will be combination of a vertical stretch or compression along with a vertical reflection.

Exercise 2.5.1: Function Transformation 1. Use the function f (x) = x2 to graph g(x) = −f (x) and h(x) = f (−x).

2. Given the following table for f (x), create a new table for the function g(x) = −f (x). x f (x)

2 1

4 3

32

6 7

8 11

For problems 3 through 12, describe how each of the provided functions is a transformation of the original function f (x).

3. f (x + 4)

4. f (x − 31)

5. f (x + 3)

6. f (x − 4)

7. f (x) + 5

8. f (x) + 8

9. f (x) − 2

10. f (x) − 7

11. f (x − 2) + 3

12. CHALLENGE: −f (x + 4) − 1 13. Write a formula for f (x) =

√

x shifted down 4 units and left 2 units.

14. Write a formula for f (x) = |x| shifted right 3 units and reflected about the x-axis.

15. Write a formula for f (x) =

1 shifted down 3 units and right 2 units. x

16. Write a formula for f (x) =

1 shifted up 6 units and left 5 units. x2

17. For the graphs in (a) and (b), identify the parent function and transformation function for each. (a) (b)

For problems 18-19, state the transformations used on the parent function g(x) = |x| to create f (x) in each case. Then graph each function based on the transformations.

19. f (x) = 3|x − 2| − 3

18. f (x) = 2|x + 3| + 1

33

Exercise 2.5.2: Transformations of Quadratic Functions Describe how each graph compares to the parent function y = x2 .

Function

Description:

1. y = −x2 1 2. y = x2 3 1 3. y = − x2 2 4. y = 5x2 5. y = (x + 4)2 6. y = (x − 2)2 7. y = (x − 4)2 8. y = x2 + 7 9. y = −x2 + 4 10. y = x2 − 3 11. y = (x − 2)2 − 2 12. y = − (x + 5)2 + 3 13. y =

1 (x − 1)2 − 4 4

34

Write the equation for each parabola (based upon the parent function y = x2 ).

2.6

14.

15.

16.

17.

Inverse of a Function

Interchanging the roles of the dependent and independent variables forms the inverse of a relation. Informally, this means that one switches the domain and range of a given function to find its inverse function (if it exists). Due to this switching of domain and range, one can think of a function and its inverse as “undoing each other.”

Thus, the inverse of a relation is formed by interchanging the roles of the dependent and independent variables. If the inverse of a function is itself a function, then the function is said to be invertible.

Definition. Inverse. If y = f (x), then the inverse of function f (if it exists) has the equation x = f (y). In symbols, we describe the inverse as f −1 (x).

GOOD ADVICE. The superscript -1 used in the notation for inverse functions is simply a notation and does not stand for an exponent or power of -1.

35

Definition. Invertible. If f −1 (x) is a function, then f is said to be invertible.

For a function f to be invertible, it must be injective (or one-to-one).

Definition. Injective. A function f : A → B is called injective (or one-to-one) if, for all a and a0 in A, f (a) = f (a0 ) implies that a = a0 .

GOOD ADVICE. In simpler terms, there can be no distinct y value that is associated with two different x values in an injective function. Graphically, this means that a horizontal line can pass through the graph of a function at most one time for the function to be one-to-one or invertible.

CHALLENGE: Is f (x) = x2 a one-to-one function? If you don’t think it is, could you restrict the domain of f (x) = x2 to make it one-to-one?

If f is invertible, then f −1 (f (x)) = x and f f −1 (x) = x. [TRY IT: Can you verify this property?] Thus, as previously mentioned, one can think of a function and its inverse as “undoing each other” when either one is applied to the other.

EXAMPLE 1: Let’s explore a classic example of the use of inverse functions. The formula

5 C = (F − 32) 9 where C stands for Celsius and F stands for Fahrenheit temperature, is used to convert Fahrenheit temperature into the Celsius temperature scale. Using this formula, we find that 75◦ F ≈ 24◦ C . You should verify this result. Notice that this function has an inverse since it is one-to-one. Since the inverse function for C switches domain and range, it should make sense that C −1 is the function that converts Celsius to Fahrenheit. The question, then is, “How do we find C −1 ?”

We start with the function for finding Celsius when Fahrenheit is given and we solve the equation for Fahrenheit.

5 C = (F − 32) 9 9 C = F − 32 5 9 F = C + 32 5 9 Thus, C −1 = F = C + 32 5 This new function F takes Celsius values of temperature and converts them to Fahrenheit.

36

GOOD ADVICE. To find the inverse of any one-to-one function, solve the equation for the domain variable. Thus, the original range variable now becomes the domain variable and the original domain variable now becomes the range variable.

EXAMPLE 2: Suppose we have the function y=

x − 1.

3

Solve for x to yield x as a function of y 3y + 3 = x, or x = 3y + 3.

Note that we have switched domain and range, so by convention we write f −1 (x) = 3x + 3 as the inverse of the original function.

EXAMPLE 3: It is also important to notice that, due to the switch of domain and range between a function and its inverse, the graph of the function f and its inverse f −1 will display symmetry about the line y = x. This is illustrated in Figure 2.1.

Figure 2.1

Exercise 2.6.1: Inverse of a function For problems 1-3, use C =

5 (F − 32) to: 9

1. Find the Celsius temperature when F = 0◦ .

37

2. Find the Celsius temperature when F = 32◦ 3. Find the Celsius temperature when F = 212◦ .

9 For problems 4-6, use F = C + 32 to: 5 4. Find the Fahrenheit temperature when C = 0◦ .

5. Find the Fahrenheit temperature when C = 21◦ .

6. Find the Fahrenheit temperature when C = 100◦ .

7. Solve for x in y = mx + b.

8. Solve for t in A = P + P rt. Then, write the equation in the form A−1 =.

9. Solve for w in P = 2w + 2l. Then, write the equation in the form P −1 =.

10. Find the conditions on m and b such that the linear function y = mx + b is its own inverse. Avoid guessing and checking. Prove this using algebraic techniques only.

11. A function f (t) is given as a table below, showing distance in kilometers that a car has traveled in t minutes. Find and interpret the meaning of f −1 (50). t (minutes) f (t) (km)

20 30

40 50

12. Using the graph shown to the right: (a) Find g(−1) (b) Solve g(x) = −3 (c) Find g −1 (−3) (d) Solve g −1 (x) = −1

38

60 70

80 90

For problems 13-16, find f −1 (x). You may have to restrict the domain of the given function to make it invertible. If so, state the restricted domain.

13. f (x) = 12x − 7 14. f (x) = x2 + 1 15. f (x) = 20 + 10x 3 16. f (x) = x + 2 4

17. Based on what you have learned about inverses, predict what the graph of x = y 2 would look like. Make a table if you need to. Note, this graph is not a function, but it is a relation.

18. What would the graph of x = −y 2 look like?

39

40

Unit 3 - Systems

3.1

Systems of Equations

We start and end this Unit with puzzling problems related to the topic. Indeed, these problems relate to our study of systems and matrices, but it is left to you to decide in what way.

Exercise 3.1.1: Shapes Puzzle – Warm-Up Each of the four shapes in the provided figure stands for a distinct positive whole number. The sum of the shapes in each row or column is shown at the end of each row or column respectively. Find the numerical value that each shape represents.

Using Substitution to Solve a System Chances are you have some previous experience with solving systems of linear equations and inequalities in two variables. To refresh your memory, we start with an example.

41

EXAMPLE 1: Consider this system of two linear equations. ( 3x − 3y = 9 (1) 2x + y = 6 (2) There are several ways to solve this system. One way is to solve one equation for a single variable and then substitute that equation into the other for the solved variable.

This is known as the substitution method. For our example, notice that equation (2) can be rewritten as y = −2x + 6. (3) Next, substitute equation (3) into equation (1) for y to yield 3x − 3(−2x + 6) = 9.

Distribute the -3, gather like terms, and solve for x.

3x + 6x − 18 = 9 9x = 27 x=3 Substituting x = 3 into either equation (1) or (2) will allow you to find that y = 0 when x = 3. Thus, the system has one solution, (3, 0). If you graph the linear equations of (1) and (2), you will find that the point (3, 0) is where the two graphs intersect, as displayed in Figure 3.1.

Figure 3.1: The solution to the system of equations.

Since this system has at least one solution, the system is said to be consistent. Another example of a system that is consistent is one in which there are infinitely many solutions.

42

EXAMPLE 2: Consider this system: ( 2x − 3y = −7 (1) . −4x + 6y = 14 (2) Solve equation (1) for y and use it to replace y in equation (2). Next, try to solve the equation for x. What happens?

You should have ended up with the simplified equation 0 = 0.

Since this last equation is true for any value of x, there are infinitely many solutions to this system. You might ask, how can this be?

CHALLENGE: Graph equations (1) and (2) of Example 2 and explain why this system has infinitely many solutions.

EXAMPLE 3: A system may have no solutions. This type of system is said to be inconsistent. Consider the system: ( x + 2y = 5 .

2x + 4y = −4 CHALLENGE: Solve this system. What does the result imply? Graph the system to verify your thoughts.

Using Gaussian Elimination to Solve a System A given system of equations is said to be equivalent to another if the two systems have the same solution. We make use of this fact when we employ the method of Gaussian Elimination to solve a system. For a given system, we can arrive at an equivalent system as long as we follow these three rules when performing operations on a given system.

We can 1. Interchange equations (especially useful when working on systems of three or more equations).

2. Multiply an equation by a nonzero constant.

3. Add a constant multiple of one equation to another in the system.

43

EXAMPLE 4: Consider the system ( 3x − 2y = 7 6x + 5y = 9

(1) .

(2)

Using rule 3, we can multiply equation (1) by -2 and add the result to equation (2). We do this to eliminate the x term from the system in order solve for y:

−6x + 4y = −14 +6x + 5y = 9 9y = −5 5 y=− 9 53 5 Next, substitute y = − into either original equation to find that x = . Thus, the 9 27 53 5 solution is the ordered pair , − . Graphically, what does this result mean?

27 9 GOOD ADVICE. You have learned two ways to solve linear systems of two equations.

You can use either method at any time based on which method you think will be easiest to use.

44

Exercise 3.1.2: Substitution and Elimination Methods 1. Return to EXAMPLES 1-3 in this section and redo them using the elimination method.

2. Return to EXAMPLE 4 in this section and redo it using the substitution method.

For problems 3-10, solve the following systems using either substitution or elimination. ( ( 11a − 7b = −14 −8r + 3s = 10 3.

4. a − 2b = −4 10r − 2s = −2  6 7 (   − =8 13x + 20y = −1 x 5.

6. 15 y14  19x − 15y = 87  − = 21 x y ( ( 5x + 2y = 10 5x + 2y = 11 7.

8.

5x + 2y = 20 x+y =4 ( ( 4x − 2y = 16 3x + 2y = 8 9.

10. y = 2x − 8 2x − y = 10 Graph the systems from problems 3-10 to see if your algebraic solutions make sense.

CHALLENGE: Based on what you have learned, use substitution to solve these systems algebraically involving a linear and a quadratic equation. ( ( y − x2 = 0 −x2 + y = −4 11.

12. y − 3x = 4 −2x + y = −1 13. Graph problems 11 and 12 to check your answers.

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3.2

Systems Applications

As an example, we will look at an application of using systems of equations to solve problems in real world situations.

EXAMPLE: Suppose one invested a total principal amount of $2,000 in two different savings accounts. Part of the $2,000 is in an account that pays 2% annual interest and the rest is in an account that pays 4% annual interest. If the accounts earn a total of $60 in the first year, how much was initially invested in each account?

To solve this problem, we’ll let x = amount of principal invested at 2%, and y = amount of principal invested at 4%. Thus, ( x + y = 2000 0.02x + 0.04y = 60

.

CHALLENGE: Use either substitution or elimination to solve for x and y.

Exercise 3.2.1: Using Systems of Two Equations to Solve Application Problems 1. Container R holds a solution that is 20% alcohol and Container S holds a solution that is 60% alcohol. How many liters of solution from each container should be used to create 2 liters of solution that is 50% alcohol?

2. Professional basketball player Wilt Chamberlain once scored 100 points in a single game, from a total of 64 shots made. Some of these shots were field goals worth 2 pts each and some were free throws worth 1 pt each (sorry, no 3 point shots in Wilt’s time). Set up and solve a system of equations to find out how many field goals and how many free throws Wilt made.

3. At a restaurant, 6 pretzels and 2 cups of lemonade costs $11.00. Two pretzels and 1 cup of lemonade costs $4.00. Find the cost of a pretzel and the cost of a cup of lemonade.

4. You plan to take a 7-day trip to California. You estimate that it will cost $275 per day in Los Angeles and $400 per day in San Francisco. Your total budget for the 7 days is $2,300. How many days should you spend in each location?

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3.3

Systems with Three Unknowns

Notice that when we worked with equations involving two unknowns, we needed two equations to solve our system. Analogously, to solve equations with three unknowns, we need a system of three equations. For that matter, when solving equations with n unknowns, we generally need n equations in our system. As you can imagine, this can become quite a tedious process as the number of unknowns gets large. In this section, however, we will learn how to deal with systems of equations with three unknowns.

As with systems of equations in two variables, we can imagine that there are many methods for solving systems in three variables. This will usually involve using a combination of the elimination and substitution methods. Outlined here is one method for how one might solve a general system of three linear equations in three variables.

Plan of action:

1. Select two of the three equations under investigation and eliminate one variable using the rules you learned for equations with two variables.

2. Select a different pair of equations from your original three and eliminate the same variable.

3. Using the resulting equations, which now only have two variables each, solve that system for both of the variables using the techniques you learned in the last section.

4. Substitute the two values you find into one of the original equations to find the remaining variable’s value.

5. It is wise to check and see that the three values you have solved for work in each of the original equations.

Just as with a system of two equations, if you ever get a false statement, there will be no solution. If you get an identity statement such as 0 = 0, there will be infinitely many solutions. One way to picture why this is so is to recall that with systems of two linear equations, the type of solution you found depended on whether the two lines intersected, were parallel, or were the same. To extend this idea, we note that linear systems of three variables can be thought of as finding solutions related to how three planes intersect in space. The planes can all intersect in a variety of ways, including intersecting along a single line, or they can be parallel.

47

Solving a Linear System of Three Variables EXAMPLE 1: Here is an outline of a method for solving a linear system of three variables according to our “plan of action” from earlier in the text. Let the system be

  3x + 4y + 2z = 11 2x + 3y − z = 4   5x + 5y − 3z = −1

.

1. Choose the first and second equations and eliminate the z variable by multiplying the second equation by 2 and adding it to the first.

3x + 4y + 2z = 11 4x + 6y − 2z = 8 7x + 10y = 19 2. Next, choose the second and third equations and eliminate the z variable by multiplying the second equation by -3 and adding it to third. −6x − 9y + 3z = −12 5x + 5y − 3z = −1 −x − 4y = −13 3. Now we are down to the familiar process of solving two linear equations with two unknowns. ( 7x + 10y = 19 . −x − 4y = −13 You can use any method previously learned to solve this system.

CHALLENGE: Solve this system.

4. You should have found that x = −3 and y = 4. Now substitute these values into any of the original equations to find that z = 2. The solution to this consistent system is the ordered tuplet (−3, 4, 2).

5. Always check your solutions in all of the original equations to verify that the solution works for all and that you have not made a mistake during the process.

48

Exercise 3.3.1: Solving Linear Systems of Three Variables Solve for x, y and z.

  2x + 3y − z = −1 1. −x + 5y + 3z = −10   3x − y − 6z = 5

  2x + y = 1 2. −2y + 5z = −5   3x + 4z = −2

  x + y + z = 6 4. 2x − y + z = 3   3x − z = 0

  3x − 3y + 6z = 6 5. 5x − 8y + 13z = 7   x + 2y − z = 5

  2x − 3y = 5 3. 4y + 2z = −6   5x + 7z = −15

6. Based on what you learned about 2 × 2 systems, show that this system has infinitely many solutions.   x + y + z = 2 x+y−z =2   2x + 2y + z = 4 7. Based on what you learned about 2 × 2 systems, show that this system has no solutions.   x − 2y + z = 4 x+y+z =2   3x + 3y + 3z = 14

CHALLENGE: Linear and Quadratic Equality. See if you can solve systems of a linear and a quadratic equation. [HINT: Either use elimination or set the two equations equal to one another and solve the quadratic.] ( ( y − x2 = −1 y − x2 = 1 8.

9. y − 2x = 2 y + 2x = 4 10. Use technology to check your answers for problems 8 and 9.

In future sections, we will learn how to use matrices to solve such systems as those of Exercise 3.3.1. First, however, we will briefly explore systems of inequalities.

49

3.4

Systems of Inequalities

Graphing the equations of a system of inequalities such as   y − 2x + 1 < 3 , y − x ≥ −1   y + 2x < 3

and looking for an “area of overlap” is the easiest way to find the solutions for a system of inequalities. For our purposes, we will restrict our investigation of systems of inequalities to linear equations of two variables. This is best explained through example.

First, the rules for working with a system of linear inequalities are as follows:

1. Solve each inequality for y and graph each as if the inequality was an equal sign.

2. If a provided inequality contains the symbol < or >, draw your graph as a dotted line. This signifies that points on the line are NOT part of the solution set. If a provided inequality in the system contains the symbol ≤ or ≥, draw your graph as a solid line. This signifies that the points on the line are a part of the solution set.

3. After solving for y, if a provided inequality contains the symbol y < ..., or y ≤ ..., shade all the points below the line. After solving for y, if a provided inequality contains the symbol y > ... or y ≥ ..., shade all the points above the line.

4. The set of points in the “area of overlap” in common to all inequalities is the solution set. This region is called the feasible region.

50

EXAMPLE: The graph corresponding to the system of inequalities

  y − 2x + 1 < 3 Line A y − x ≥ −1 Line B   y + 2x < 3 Line C is displayed in Figure 3.2.

Figure 3.2: The Feasible Region.

Note that the lines are solid or dashed according to the inequality rules stated previously. The shading in the graph represents the area of overlap of the shadings for each line. All points in this area of overlap represent the solutions to the system.

51

Exercise 3.4.1: Systems of Inequalities For problems 1-6, use the rules provided in this section to graphically display the solution sets for these systems of inequalities.   ( y < 3x − 4   y > 3x − 3 1.

2. 2x − 3y > 6  y ≤ −x + 5  y − 2 ≤ − 3 (x − 4) 5    x > 2 (  y ≤ 3 x+y ≥4 3.

4.  x + y ≥ −4 y ≤ 2x − 3    x + y < −1   x > −3   (   5x − 2y < 10 y≤2 5.

6. x − y ≥ −4 x+y <4    x − y < −1 7. A dietitian is creating a mix of a ounces of almonds and g ounces of granola.

She wants the mix to have less than 70 grams of fat and weigh less than 8 ounces. An ounce of almonds has 14 grams of fat, and an ounce of granola has 7 grams of fat. Write and graph a system of inequalities that models the situation.

52

8. On a trip, you are fishing for catfish and rock-fish, which are two species of bottom fish. Local laws allow you to catch no more than 14 catfish per day, no more than 8 rock-fish per day, and no more than 18 total bottom fish per day. Write and graph a system of inequalities that models the situation.

CHALLENGE: Linear and Quadratic Inequalities. For problems 9-10, solve these systems graphically, showing the shaded area that is the solution set. ( ( y − x2 ≥ −1 y − x2 > −4 9.

10. y − 2x < 2 y + 2x ≤ 4

11. There is a theorem called The Linear Programming Theorem, which states that the feasible region of a system of equations composes the feasible set S of solutions. Moreover, if the system attains a maximum value for (x, y) in S, then that maximum or minimum value is attained at a vertex of S. Research this topic on the internet and report on how Linear Programming is used in application.

53

54

Unit 4 - Matrices

4.1

Matrices and Matrix Operations

A matrix is a rectangular array of real numbers. This array may stand for many different things, such as coefficients in a linear system of equations or representative points from geometric objects in the plane, among other things. This module will strictly focus our attention on learning the basic properties of matrices and the associated operations of matrix addition and multiplication. Eventually, we will use matrices to solve linear equations, which are easily written in matrix form. In fact, matrices can also be used to find the coefficients in a general equation model and much more!

Each number in a matrix array is called an element. We will represent matrices using a capital letter. The dimensions of a matrix are the number of rows and columns in the matrix. A matrix with m rows and n columns is called an m × n matrix, where m and n are elements of the positive integers. As an example, if we want to represent the coefficients of the following three equations in this system

  2x + 3y − z = 0 −x + 5y + 3z = 0   3x − y − 6z = 0

(1)

as a matrix A, we would write  2 3 −1 3 .

A = −1 5 3 −1 −6 

Our matrix A would have dimensions 3 rows by 3 columns. Since the number of rows and columns are the same in A, we call this a square matrix. Notice that this A matrix symbolizes a shorthand way to express the three equations of (1).

Definition. Matrix Addition. For the two matrices A and B, which have the same dimensions, their sum, A + B, is a matrix of the same dimensions whose elements represent the sums of the corresponding elements of A + B.

55

ADDITION (Remember, addition requires that the matrices are of the same dimensions.)

a11 a12 b11 b12 a11 + b11 a12 + b12 A= B= A+B = a21 a22 b21 b22 a21 + b21 a22 + b22 EXAMPLE:

−3 −2 5 6

+

4 8 −7 −3

=

1 6 −2 3

In matrix algebra, a real number is called a scalar. When multiplying a matrix by a scalar, one simply multiplies each element in the matrix by the scalar.

SCALAR MULTIPLICATION (where k is a real number)     a11 a12 a13 k · a11 k · a12 k · a13 A = a21 a22 a23  kA = k · a21 k · a22 k · a23  a31 a32 a33 k · a31 k · a32 k · a33 EXAMPLE:   −1 2 −3 6 0 −3 ·  4 5 −2 4

 =

 3 −6 9 −12 −18 0  −15 6 −12

Also notice that subtraction of two matrices, A and B, can be accomplished by adding matrix A to the negative of matrix B. The matrix B can be made negative by multiplying the scalar,−1 , by matrix B. The matrix operations of addition and multiplication allow one to think of matrix operations like that of a separate number system. We point out that, under addition, matrices follow the rules associated with that of a field of numbers. Be cautioned, however, that matrix multiplication DOES NOT necessarily follow all of the field properties for multiplication.

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Definition. Matrix Multiplication. The product of two matrices Am×n and Bn×r is the matrix Cm×r . Any element in the ith row and j th column of the product matrix, C, is the sum of the products of the corresponding elements of the ith row of A and the j th column of B. This is best shown by example.

MATRIX MULTIPLICATION (By definition, the number of columns in matrix A must equal the number of rows in matrix B in order to perform matrix multiplication.)

Am×n · Bn×r = Cm×r For

a11 a12 a13 A= a21 a22 a23

  b11 b12 B = b21 b22  b31 b32

a11 b11 + a12 b21 + a13 b31 a11 b12 + a12 b22 + a13 b32 A·B = a21 b11 + a22 b21 + a23 b31 a21 b12 + a22 b22 + a23 b32 EXAMPLE:

  −6 4 B = −1 3  7 −4

−38 26 A·B = 11 −7

−1 2 −6 A= 3 −1 4

GOOD ADVICE. One way to help you think about how matrix multiplication works is to have you place three fingers of your right hand along the elements of row 1 of matrix A. Then imagine picking those fingers up with the row elements attached. Next, turn those same fingers sideways (a 90◦ rotation) and place them just to the left of column 1 of matrix B. Multiply across and add the products to get the first element in the first row of your product matrix C. Continue in this fashion to get the rest of the elements.

A word of caution, matrix multiplication does not follow all of the field properties of multiplication like the real numbers do. Matrix multiplication is NOT necessarily commutative! For matrices, the product A × B is not necessarily equal to the product B × A.

57

Exercise 4.1.1: Matrices and Matrix Operations Given:

−3 2 A= −1 4

  −2 −3 5 −3 −1 8 B= C= D =  4 −6 3 −6 4 −2 −5 1    

−3 −1 −1 0 1 1 4 −1 3 E= F = 5 G =  0 1 −1 −2 5 −4 2 4 −1 0 1

For problems 1-12, compute, if possible. If the operation requested is not possible, state why. Perform these operations by hand.

1. A + B

2. B − C

3. D + E

4. A + B + C

5. 3C

6. −1F

7. BC

8. CB

9. F E

10. F G

11. GF

12. 2A + EF

13. Looking at your answers for problems 7 and 8 above, what conclusion can you make about matrix multiplication?

Next, we define what is meant by the identity matrix.

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Definition. Identity Matrix. The identity matrix is a square matrix, I, of dimensions n × n. The main diagonal of elements of I (from the upper left corner to lower right corner) contains only 1s. All other elements in I are 0s. The product of I and any other matrix A is equal to A. In fact, A · I = A = I · A. For example, if     1 0 0 −1 2 −3 6 0 , I = 0 1 0 , and A =  4 0 0 1 5 −2 4 then           1 0 0 −1 2 −3 −1 2 −3 −1 2 −3 1 0 0 0 1 0 ·  4 6 0  · 0 1 0 .

6 0 = 4 6 0 = 4 0 0 1 5 −2 4 5 −2 4 5 −2 4 0 0 1 Last, for completeness, we inform the reader that there also exist square matrices of equal dimensions, A and B, that have the property A · B = B · A = I, where I is the identity matrix. In this case, we say that B = A−1 , which means matrix B is equal to the multiplicative inverse of the A matrix, and A = B −1 , which means matrix A is equal to the multiplicative inverse of the B matrix. Thus, A · A−1 = B · B −1 = I.

59

Using Matrices to Solve Systems Next, we explore a different use of matrices to solve systems. The previously stated multiplicative inverse property of matrices is quite useful to us in practice. For example, suppose we have a 3 × 3 matrix A that represents a system of equations along with a 3 × 1 solution matrix B . Note that, in this situation, one is not trying to solve for x and y. Rather, we are trying to solve for three unknown coefficients a, b, and c in the following equation: y = ax2 + bx + c.

Since there are three unknowns a, b, and c in the equation, we would need three data points, (x1 , y1 ), (x2 , y2 ), and (x3 , y3 ), to fill in our A and B matrices in order to solve for the C matrix in [A] · [C] = [B].

For clarity, we have bracketed the matrix symbols as well in this example. For our example, this will look like:

      (x1 )2 x1 1 a y1 (x2 )2 x2 1 ·  b  = y2  . (x3 )2 x3 1 c y3

As an exercise, multiply the matrices on the left of the equation above to see the actual system that you are solving.

Keep in mind that our goal is to find the values of the coefficients of the matrix C. Based on our knowledge of the inverse of a matrix, the way to accomplish this task is to multiply both sides of the equation by the inverse of the matrix A (provided that it exists) to yield [A−1 ] · [A] · [C] = [A−1 ] · [B], where [A−1 ] · [A] is equal to the identity matrix. This simplifies to [C] = [A−1 ] · [B].

60

Of course, one should ask, “How do I find the inverse of a square matrix, assuming it exists?” The next example illustrates how this can be done by hand. Once this process is mastered by hand, one could also consider using graphing software that performs this process. The next example illustrates how this is done.

EXAMPLE: How to Find the Inverse of a 3 × 3 Matrix The goal is to find A−1 of 

 1 0 1 A =  2 1 3 . −1 1 1 Begin by forming the following 3 × 6 augmented matrix. By augmented matrix, we mean we adjoin the identity matrix with the given matrix separated by a vertical line (by convention), as pictured below.

Next, we perform matrix row operations to obtain the identity matrix on the left side and perform the same operations on the right side of this matrix. These types of row operations are discussed at length in the next section which details Gaussian elimination, but we summarize them here.

1. Switch any two rows in the matrix.

2. Multiply an row in the matrix by a nonzero constant.

3. Add a constant multiple of one row to another row.

 1 0 1 1 0 0  2 1 3 0 1 0  −1 1 1 0 0 1  1 0 1 1 0  R2 − 2R1 → 0 1 1 −2 1 R3 + R1 → 0 1 2 1 0  1 0 1 1 0  0 1 1 −2 1 R3 − R2 → 0 0 1 3 −1 

 0 0  1  0 0  1

  R1 − R3 → 1 0 0 −2 1 −1 R2 − R3 →  0 1 0 −5 2 −1  0 0 1 3 −1 1

61

The right side is equal to A−1 . That is,   −2 1 −1 A−1 = −5 2 −1 .

3 −1 1 It can be verified that A−1 A = I3 = AA−1 .

CHALLENGE: Based on this example, can you provide some general informal justification for why this process works?

NOTE: If it is not possible to obtain the identity matrix on the left side of the augmented matrix by using matrix row operations, then A−1 does not exist.

Exercise 4.1.2: The inverse of a Matrix A calculator should not be used on problems 1 and 2.

2 7 −1 .

1. Find A if A = 1 5 2. Find the inverse of matrix A, where:   0 1 2 A = 1 0 3 .

4 −3 8 Show how you can check your answer to problem 2.

3. Use the matrix method of this section to write the particular equation of a quadratic function in the form y = ax2 + bx + c. Use a calculator to find the inverse matrix when solving these systems of equations. (a) containing (−2, −11), (4, 13), and (6, 29). (b) containing (0, 5), (2, 13), and (3, 26). (c) containing (2, −1), (−3, −11), and (7, 9). (d) containing (1, 2), (2, −1), and (3, −2).

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4.2

Determinants and Gaussian Elimination

There is another way to find the inverse of a matrix or to solve matrices. This method involves first finding the determinant of a matrix. With each square matrix (i.e. a matrix with the same number of rows and columns), we can assign a representative number called the determinant of A. This determinant is denoted by the name det A and symbolized as |A|.

Definition. Determinant. If a 2 × 2 matrix A = determinant of A, symbolized as

a b , then the c d

a b , is written and evaluated as c d

det A = ad − bc.

Do you notice the pattern of multiplication and subtraction used in the general formula for the determinant of a 2 × 2 matrix?

Things get more involved quickly when we look at the determinant of a 3 × 3 matrix.

  a1 b1 c1 If a 3 × 3 matrix A = a2 b2 c2 , then the determinant is evaluated a3 b3 c3 as det A = a1 b2 c3 − a1 b3 c2 + a2 b3 c1 − a2 b1 c3 + a3 b1 c2 − a3 b2 c1 .

Do you notice the pattern of multiplication and subtraction used in the general formula for the determinant of a 3 × 3 matrix? This pattern is a bit harder to discern.

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Exercise 4.2.1: Determinants For problems 1-5, use the following matrices.

−1 5 10 10 A= B= 4 20 5 −7

7 −7 2 0 2x −5 C= D= E= −1 4 −1 −9 x 10

1. |A|

2. |B|

3. |C|

4. |D|

5. |E|

CHALLENGE: For problems 6-7, evaluate the determinants of the given 3 × 3 matrices.

6.

4 0 3 5 1 2 −2 1 −1

1 0 2 7. 3 −1 4 5 −3 −2

Using the Determinant to Find the Inverse of a Matrix Determinants can also be used to find the inverse of a matrix, if it exists. Note that we will restrict our focus to 2 × 2 matrices although determinants can be used to find inverses for higher dimension square matrices as well.

a b , and det A 6= 0, then the inverse of A is If A = c d

−1

1 1 a b d −b d −b = = . c d detA −c a ad − bc −c a

It follows from the result that if det A = 0, then A−1 is undefined and A is not invertible.

64

If we assume that A is an invertible matrix, the proof of this statement can be verified by creating a matrix

w x B= y z such that AB = I, where I is the identity matrix

1 0 I= .

0 1 This would mean that B is the inverse of matrix A. Thus, we solve for w, x, y, z in

a b w x 1 0 = c d y z 0 1 to find B = A−1 . We note that

aw + by ax + bz a b w x . = cw + dy cx + dz c d y z Thus,

1 0 aw + by ax + bz . = 0 1 cw + dy cx + dz

This means that

aw + by cw + dy

= =

1 0

(1)

ax + bz cx + dz

and

= =

Solving system (1) for y and then for w yields

y=

−c d and w = . ad − bc ad − bc

Likewise, solving system (2) for x and then for z yields

x=

−b a and z = . ad − bc ad − bc

65

0 1

(2).

Therefore, B=A

−1

−1

1 1 a b d −b d −b = = = . c d det A −c a ad − bc −c a

CHALLENGE: Show that all of the steps of this verification are correct.

Exercise 4.2.2: Using the Determinant to Find the Inverse of a 2 × 2 Matrix For problems 1-5, use determinants to find the inverse of these matrices (if they exist) and verify your answers by showing that your answer times the original matrix yields the identity matrix.

1.

4.

3 2 2 −1

0 1 1 0

2.

5.

6 −2 4 −1

7 4 3 2

3.

2 1 4 3

CHALLENGE: For problems 6-8, solve for matrix B by using determinants and inverses. That is, find the inverse of the matrix on the left side of the equation and multiply both sides of the equation by the inverse, as in A−1 AB = A−1 C, to yield B = A−1 C.

1 2 2 1 B= 1 4 4 3

3 2 5 0 B= 2 2 −1 −1

6.

8.

3 −1 2 1 B= 7.

1 1 2 2

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As mentioned previously, determinants can be used for many purposes. We will now explore a method that uses determinants to solve linear systems. This method is known as Cramer’s Rule. Again, we will restrict our focus to solving systems of two equations, although Cramer’s rule can be used for higher dimension square matrices as well.

Cramer’s Rule To illustrate the use of Cramer’s Rule, we start with the general system of equations ( ax + by = e . cx + dy = f

In matrix form, this system is written

e a b x . = f c d y

You should verify that this is true.

We use the fact that −1

−1

e a b a b x a b = f c d c d y c d to yield our desired result

1 x d −b e = . y f det A −c a Recall that the inverse matrix in our example can be found using determinants.

Assuming det A does not equal zero, we can write

1 1 x d −b e d −b e = = . y f f det A −c a ad − bc −c a

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Therefore (by performing the matrix multiplication) we find that the general solution to our system is

x=

de − bf af − ce ,y = det A det A

x=

de − bf af − ce ,y = . ad − bc ad − bc

or

Moreover, this can also be written in determinant form as

x=

e b f d a b c d

,y =

a e c f a b c d

.

Finally, Cramer’s Rule is written succinctly as x=

Dy Dx ,y = D D

where D is the determinant of the original coefficient matrix, and Dx and Dy are the respective numerators de − bf and af − ce in the fractions for x and y above.

NOTE: 1. If det A, de − bf , and af − ce all equal zero, the system is dependent and there are infinitely many solutions.

2. If det A equals zero and de − bf or af − ce does not equal zero, the system is inconsistent and has no solution.

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EXAMPLE Use Cramer’s rule to solve the system ( −2x + y = 5 x + 3y = 7

.

This means D = (−2 · 3) − (1 · 1) = −7 Dx = (5 · 3) − (7 · 1) = 8 Dy = (−2 · 7) − (1 · 5) = −19.

Thus the solution is 8 19 x = − ,y = .

7 7 CHALLENGE: Use one of the other methods you have learned to solve systems to verify this result.

Exercise 4.2.3: Using Cramer’s Rule Use Cramer’s Rule to solve these systems. Then employ a different method of solving systems to check your answer. ( 3x + 4y = 2 1.

5x − 7y = 17

( 5x − 3y = 22 2.

6x − 7y = 41

( x − y = −11 3.

7x + 4y = −22

( 2x + 3y = 9 4.

3x + 6y = −3

( 4x − 3y = 11 5.

5x − 6y = 9

( 7x − 2y = 10 6.

9x + 3y = 24

( x − 2y = −6 7. −2x + 4y = 6

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Gaussian Elimination In this section, we will use a combination of the elimination and substitution methods to solve linear equations. This combined method is referred to as Gaussian elimination.

This method depends on the fact that we create an equivalent system to one that is given in order to make it easier to solve the system. The rules of Gaussian elimination allow us to: 1. Interchange any two rows in the system.

2. Multiply an equation in the system by a nonzero constant.

3. Add a constant multiple of one equation to another.

The goal is to try to eliminate one variable in order to solve for another. Then, substitution is also usually employed to solve for the other variable.

We apply this method to matrices to create a row echelon form, where we try to do operations that result in placing zeros in strategic locations in a matrix in order to solve a system. Keep in mind that we are simply performing the operations we have done on systems written in the traditional way. In fact, Gaussian elimination was employed in the example of the previous section used to illustrate how to find the inverse of a matrix A. Along the way, we create an augmented matrix. This will be detailed in the following Exercise.

The next Exercise will guide you through the use of Gaussian elimination.

Exercise 4.2.4: Gaussian Elimination and Row Echelon Form In this Exercise we will further the understanding of the connections between solving systems with Gaussian Elimination and using reduced row echelon form of a matrix.

1. First , for the system ( 4x − 5 = −5y 3y − 16 = −5x

,

algebraically move all x and y terms to the left of the equal sign and move all constants to the right side of the equation.

2. Create an augmented matrix from your work in problem 1 that looks like this, where the constants are on the right side of the vertical line within the matrix:

# # # . # # #

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3. Recall, there are three rules in using Gaussian Elimination.

• Interchange any two rows in the system.

• Multiply an equation in the system by a nonzero constant.

• Add a constant multiple of one equation to another.

Use the three rules above to manipulate your augmented matrix from # # # . problem 2 into the form 0 # # 4. Use the row echelon form from problem 3 to solve your system.

5. Using the same process, solve the systems below. ( ( x − 3y = −1 2x − y = −10 (a) (b) 3x + 2y = 8 7x + 5y = −1

CHALLENGE: For the next two problems, your goal for the augmented matrix is to get it to look like   # # # #  0 # # # .

0 0 # #

HINT: Start by doing operations that will yield the two zeros in the left most column first, then move on to getting the middle column zero that is needed in this space.

6. Set up an augmented matrix, use Gaussian elimination to create echelon form, and solve.   2x + y − z = 8 −3x − y + 2z = −11   −2x + y + 2z = −3 7. Set up an augmented matrix, use Gaussian elimination to create echelon form, and solve.   3x + y − 2z = −7 2x + 2y + z = 9   −x − y + 3z = 6

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4.3

Matrix Applications

An Interesting Postage Problem

Exercise 4.3.1: The Pay For Postage Problem The post office has only 5 cent and 7 cent stamps. By combining different numbers of 5 and 7 cent stamps, customers can usually get the amount of postage they need. For example, two 5 cent and one 7 cent stamps makes 17 cents in postage, two 7 cent stamps makes 14 cents in postage, etc.

1. Describe how to make each of the following postage amounts by combining 5 and 7 cent stamps. (a) 30¢ (b) 42¢ (c) 24¢ (d) 51¢

2. Which amounts of postage are impossible to make using only 5¢ and 7¢ stamps? To make the problem more manageable, consider only postage amounts up to $1.08. (Make sure you can explain how each of the amounts you found are impossible really are impossible, and how each of the amounts you found possible really are possible.)

Hint: If you are struggling with this problem, perhaps it would help to find a way to organize and display your thoughts in a chart or array.

Working With an Array By this point, you should have realized how to create a number chart array to represent this problem. For problems 3 and 4 in the next page, assuming that the values in the charts below were determined in the same way, can you figure out what goes in the “?” spot?

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3.

4.

5. Consider the following system of equations: ( x + 2y = 35 .

3x + y = 55 Solve this system by elimination and then try solving this system on the number grid.

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6. Solve this system by elimination. ( 2x + 3y = 35 4x + 5y = 61

Now solve this system using only the grid.

CHALLENGE: The Pay Postage Problem Generalized 7. Suppose now that there are only 6 cent and 9 cent stamps. Which amounts of postage are impossible to make using only 6 cent and 9 cent stamps?

(Make sure you can explain how each of the amounts you found are impossible really are impossible, and how each of the amounts you found possible really are possible.)

8. Play around with different stamp values. Under what conditions on m and n, with m cent and n cent stamps, will there be a largest postage amount that can’t be made? What will that next value be?

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Using Matrices to Affect Transformations in the Plane We start this section by exploring transformations on a Point in the Plane.

1 0 Start with the Identity (2 × 2) matrix I = and the ordered pair (x, y) written as 0 1

x matrix point p = . Verify the identity equation: y I · p = p.

Exercise 4.3.2: Transformations on a Point in the Plane

x 1 0 Using the point p = and the matrix I = as a template, develop a y 0 1 matrix that will:

1. reflect p about the x-axis.

2. reflect p about the y-axis.

3. reflect p about the y = x.

4. reflect p about the y = −x.

Let’s now extend the idea of transforming points in the plane to transforming objects in the plane using matrices. This concept will also be combined with the topic of function iteration that we have previously discussed.

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Exercise 4.3.3: An Iterated Function and Matrices Recall: Definition. Fixed Point. If a is an element of the domain of a function f and f (a) = a, then we say that a is a fixed point of f .

We will use a matrix notation to denote points in the plane. Let

x p0 = 0 . y0

1. Let p0 be a point in the plane and suppose the real number k is such that 0 < k < 1. We define a transformation function that works on matrices:

x − x0 x x =k T + 0 . y y − y0 y0 What is the fixed point for this function?

1 9 and k = . Perform five iterations of T 2. Use the function T with p0 = 2 3

−2 and describe the orbit (or path) of p under iteration. starting with p = 9 Including the initial value (−2, 9), graph the sequence of outputs of T −2 , and explain why the points line up generated by the seed value p = 9 like they do.

3. The function T is an example of a linear contraction of the plane. Why do you think the function has this name?

4. If you consider the order of the outputs of T that were generated, what point do these outputs seem to be approaching?

1 9 5. Now use the same equation for T provided above, with p0 = and k = .

2 4

1 Perform six iterations of T starting with p = and describe the orbit (or 8 path) of p under iteration.

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Using Matrices to Balance Chemical Equations A fundamental skill used in chemistry is the ability to balance equations. Balancing equations means writing chemical equations such that the amount of substance you start with in the reaction equals the amount of substance you end up with as a product of a reaction.

By way of example, one can form water by combining hydrogen gas, H2 , and oxygen, O2 , in the presence of electricity. The reaction proceeds like this: H2 + O2 → H2 O. One might notice that there are two oxygen atoms on the left and only one on the right! We need to get things in the correct proportions for this reaction to be balanced.

The balanced reaction looks like this: 2H2 + O2 → 2H2 O. While this process can be done by inspection, the more complicated the reaction, the harder it is to balance such equations. Therefore, it is of interest to explore a way to balance equations using matrices.

Thus, we will now discuss balancing chemical equations using matrices.

EXAMPLE: Consider this simple reaction: a · Cr + b · O2 → c · Cr2 O3 We have two different elements, Cr and O, so we will need two different equations. We are trying to calculate the values of a and b, the coefficients on the left side of the reaction. We will initially assume the value c is equal to 1. However, this may change if our a and b coefficients are fractions. The two equations look like this: ( Cr: 1a + 0b = 2c .

O: 0a + 2b = 3c Thus, we create A, a 2 × 2 matrix system for a and b,

1 0 A= , 0 2 and B, a 2 × 1 matrix for c.

2 B= .

3

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To obtain the values of a and b, perform the operation [C] = [A]−1 [B] .

This yields the C matrix, which solves for a and b. Use this answer to balance the chemical equation. We will assume the coefficient c is 1 unless the solutions for a and b are not integers. In fact, the C matrix for our example is " # 2 C= 3 .

2 Our initial result is 3 2Cr + O2 → 1Cr2 O3 .

2 Thus, we multiply by a scale factor of 2 to rid the result of fractions. This yields the final balanced chemical equation: 4Cr + 3O2 → 2Cr2 O3 .

GOOD ADVICE. The matrix solution may not always be expressed as integers as in the previous example. If the solution matrix is fractional (rational), simply multiply both sides of the reaction by a least common multiple (LCM) integer to rid all fraction denominators, leaving only integers.

In fact, the determinant of the A matrix, with few exceptions, can be used to rid your solution of fractions. Simply multiply both sides of the chemical equation matrix solution by det A.

Exercise 4.3.4: Balancing Chemical Equations Use matrices to balance these chemical equations. Note: You will need a 3 × 3 matrix to solve problems 3 and 4, but the process is the same. If more than one molecule or element is on the right side of the chemical equation, subtract the more complicated molecule from each side first.

1. H2 + O2 → H2 O 2. Fe + O2 → Fe2 O3 3. MgO + Fe → Fe2 O3 + Mg 4. Cu2 S + O2 → Cu + SO2

GOOD ADVICE. In this Unit, you have learned multiple methods for solving systems of equations, and you have learned much about using matrices to aid in solving systems and to find function models to represent data. Ultimately, you should choose and use the method or methods that you feel most comfortable with to solve systems or model data.

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Unit 5 - Quadratic Functions

5.1

Factoring Quadratic Expressions

Partly for review purposes and partly as a precursor for factoring polynomials of degree higher than 2, we provide you with some quadratic factoring exercises to start this section. Recall that, from a functions point of view, factoring of polynomial functions such as the quadratic function allows one to find the roots of the function.

Exercise 5.1.1: Factoring Review 1. Factor these quadratic expressions into the form (x + a)2 . (a) x2 + 2x + 1

(b) x2 + 4x + 4

(d) x2 + 8x + 16

(e) x2 + 10x + 25

(c) x2 + 6x + 9

(f) Do you notice any pattern that alerts you to the fact that problems (a) (e) factor into the form (x + a)2 ?

2. Factor these quadratic expressions into the form (x − a)2 . (a) x2 − 2x + 1

(b) x2 − 4x + 4

(d) x2 − 8x + 16

(e) x2 − 10x + 25

(c) x2 − 6x + 9

(f) Do you notice any pattern that alerts you to the fact that problems (a) (e) factor into the form (x − a)2 ?

3. Factor these quadratic expressions into the form (x − a) (x + a). This factorization is known as a difference of squares. (a) x2 − 1

(b) x2 − 4

(c) x2 − 9

(e) x2 − 5

(f) x2 − 13

(g) x2 − a

(d) x2 − 16

4. Factor these quadratic expressions into the form (x + a) (x + b). (a) x2 + 3x + 2

(b) x2 + 9x + 18

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(c) x2 + 11x + 30

(d) x2 + 14x + 45

5. Factor these quadratic expressions into the form (x − a) (x − b). (a) x2 − 3x + 2

(b) x2 − 10x + 21

(c) x2 − 16x + 60

(d) x2 − 19x + 90

6. Factor these quadratic expressions into the form (x + a) (x − b). (a) x2 − 3x − 4

(b) x2 − 6x − 27

(c) x2 + 4x − 60

(d) x2 + 7x − 30

7. Now factor these. (a) x2 + 10x + 9 (d) x2 − 11x + 30

(b) x2 − 81

(c) x2 + 9x + 14

(e) x2 − x − 6

CHALLENGE: Factor by Grouping Suppose one wants to factor the expression 9x2 − 56x + 12. One method to try is to multiply the coefficient of the first and last term. In this case, this equals 108. Then ask, ”What factors of 108 added together sum to the number -56?” A little thought brings one to the conclusion that -2 and -54 are factors of 108 that add to -56.

Thus we rewrite the original expression as 9x2 − 54x − 2x + 12.

Then we ”factor by grouping” two terms at a time in the expression to yield 9x (x − 6) − 2 (x − 6) .

Notice that both terms have (x − 6) in common, so we can regroup the expression and write it as (9x − 2) (x − 6) .

Now try this method with these equations.

8. 6x2 − 26x − 20

9. 12x2 + 25x + 2

10. 20x2 + 39x + 18

GOOD ADVICE. One main reason you learn factoring of quadratic expressions is to allow you to find the roots of quadratic functions. While you have learned many patterns and rules to help you accomplish the factoring, don’t forget that the quadratic formula is always at your disposal to find roots (real or complex) of quadratic functions when all else fails.

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Exercise 5.1.2: A Related Area Investigation You are all probably familiar with area models representing algebraic distribution or factoring of quadratics. For example, what algebraic expression is represented by the following image?

Now, let’s take this idea to another level of abstraction.

Squaring Numbers and Grids

1. Start with a 5 × 5 grid as pictured below. What is the area of the entire grid?

2. Next, take the 4 darkly shaded cells from the top of the grid and place them on the side of the grid as shown. Find the total area of the new 4 × 6 grid that you have created.

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3. Compare your answers for problems 1 and 2. What do you notice about the difference in the areas of the grids? Can you write a number sentence that represents this situation?

4. Now, start with a 6 × 6 grid as pictured below. What is the area of the entire grid?

5. Next, take the 5 darkly shaded cells from the top of the grid and place them on the side of the grid as shown. Now find the total area of the new 5 × 7 grid that you have created.

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6. Compare your answers for problems 4 and 5. What do you notice about the differences in the areas of the grids? Can you write a number sentence that represents this situation?

7. Start with other “square grids” and repeat the procedure that you did above to find the difference between the original grid areas and the adjusted grid areas that you created by moving the boxes.

8. Do you notice a pattern in the difference of areas? Explain your answer to your group or class.

CHALLENGE: 9. Will the pattern you noticed work for all square grids when you repeat the procedure that you did for this exercise? How do you know?

10. Go back to problem 1. and start by removing 2 rows of 3 units (that is, 2 less than the given 5 units) and placing them on the side of the figure to make a rectangle. Then proceed in this manner through problems 2-8. Can you come up with a more general statement than in problem 9?

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5.2

A Development of the Quadratic Function

The quadratic function f (x) = ax2 + bx + c, where a 6= 0, can certainly be said to be everywhere present in many levels of mathematics curriculum. The graph of a quadratic function is a U-shaped curve called a parabola. A brief summary of the properties of the quadratic function is now presented.

a. If a > 0 , the parabola opens upward and the y-coordinate of the vertex is the minimum value of the function.

b. If a < 0, the parabola opens downward and the y-coordinate of the vertex is the maximum value of the function.

c. If f (c) = 0, then the value c is a zero of f . Moreover, the roots of the equation f (x) = 0 are the zeros of f . For √ quadratic functions, these are given by the −b ± b2 − 4ac quadratic formula, x = . If these zeros are real numbers, then they 2a represent the x-intercepts of the parabola.

d. If the discriminant b2 − 4ac is equal to zero in the quadratic formula, then the parabola touches the x-axis at only one point. This will also mean that the quadratic is a perfect square with one real root having multiplicity 2 (i.e. the root for this quadratic is the same value twice.) CHALLENGE: Generally, what is the x value for the point of tangency in this case?

e. The axis of symmetry is an imaginary vertical line that divides the parabola into two symmetric halves. Since the quadratic formula can be written √ 2 −b b − 4ac x= ± , we can see that the axis of symmetry occurs at 2a 2a x=

−b , 2a

which is halfway between each root (if we assume the discriminant quantity, b2 − 4ac, is not equal to zero).

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f. If the discriminant, b2 − 4ac , in the quadratic formula is negative, then the roots are complex numbers and the parabola does not intercept the x-axis.

g. The parameter a influences the appearance of a parabola. The value of |a| determines the rate at which the points of the parabola move away from the y-axis, thus making the parabola appear skinnier or wider.

h. Last, the vertex form of the quadratic function can be found by completing the square on the quadratic equation and can be written f (x) = a (x − h)2 + k, where (h, k) is the vertex of the associated parabola and x = h is the equation of −b the axis of symmetry. Recall that the axis of symmetry is x = . So, the vertex 2a (h, k) can be written

b b . − ,f − 2a 2a We will soon discuss what is meant by the phrase “completing the square”.

However, one might ask how and why was the quadratic developed and studied? Many students don’t realize the fact that algebra can be viewed as an abstraction of geometry. Algebra is a language with equations as sentences. Algebra is developed based upon a logical expansion of definitions and axioms, or first principles. Thanks to Rene Descartes, the abstraction of algebra can be developed directly from the geometry of the plane when geometric figures are viewed as collections of identifiable points in the Cartesian, or Rectangular, coordinate system.

In this section, we turn our attention to one type of conic section. A conic section can be defined as the curve of intersection of a plane with a right circular cone. The curves created by these intersections are the circle, the ellipse, the hyperbola, and the parabola. There are also degenerate cases of these curves created, such as the point, the lines, or two intersecting lines. Associated with each of these conics is an analytic definition. For this course, we will restrict our investigation to the analytic definition of the conic section called the parabola.

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Figure 5.1: The Parabola as a Conic Section.

Definition. Parabola. A parabola is the set of all points in the plane that are equidistant from a fixed point and a fixed line. The fixed point is called the focus of the parabola and the fixed line is called the directrix of the parabola. A special point on the parabola is the vertex. The vertex is the midpoint of the perpendicular segment connecting the focus and the directrix.

Of course, the question is, “How does this definition logically lead to the quadratic equation for a parabola that we are so accustomed to seeing, f (x) = ax2 + bx + c ?”

This is shown as an Exercise.

Exercise 5.2.1: From a Definition to an Equation Notice that the analytic definition of the parabola states that any point on the parabola is equidistant from a fixed focus and directrix. As a result, we will make use of the distance formula in our derivation.

PART 1 - The Distance Formula Definition. The Distance Formula. For any two points in the plane, (x1 , y1 ) and (x2 , y2 ) , the distance, d, between the two points is found by d=

q (x1 − x2 )2 + (y1 − y2 )2 .

1. Find the distance between the points (3, −5) and (−6, −1) .

2. Find the distance between the points (5, 1) and (−2, 7) .

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PART 2 - The Derivation Without a loss of generality, we place an up-oriented parabola with vertex at the origin of a coordinate system.

Given: A fixed point F and a fixed line l. The parabola consists of all points P such that F P is equal to P L. The distance from F to P is the same as the distance from P to the line l ; in symbols F P = P L .

3. Without loss of generality, orient the parabola so that the point F (called the focus of the parabola) is on the y-axis and the line l (called the directrix of the parabola) is parallel to the x-axis.

4. From the figure above, let P be a general point on the parabola; its coordinates are ( , ).

5. Use the distance formula to find F P .

6. Use the distance formula to find P L. 7. Substitute the expressions in problems 5 & 6 in the equation F P = P L , and simplify using algebraic techniques in order to derive a general equation that models a parabola centered at the origin.

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From Exercise 5.2.1, you should have found that the analytic definition of the parabola leads to the derivation of the quadratic equation y=

1 2 x 4c

for a parabola centered at the origin. We can let a =

1 . This yields the equation 4c

y = ax2 for a parabola centered at the origin. Combining this result with what you’ve previously learned about function translations produces a more general equation of the parabola that is not necessarily centered at the origin: y = a (x − h)2 + k.

This is an equation of the parent function y = x2 that has been dilated by a factor of a, moved h units horizontally and k units vertically.

Exercise 5.2.2: The Familiar Equation for the Quadratic 1. Start with the equation y = a (x − h)2 + k, perform the squaring of the term in parentheses and distribution of the coefficient a. Gather like terms to show that this equation can be written in the general form f (x) = y = ax2 + bx + c.

2. If the coefficient a in the general quadratic equation is equal to 4, what is the focal length for the parabola, c0 , equal to?

3. If the coefficient a in the general quadratic equation is equal to 3, what is the focal length for the parabola, c0 , equal to?

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5.3

Transformations of the Quadratic

You may recall, when you worked with transformations of functions in Unit 2, you were asked to describe the transformations of some quadratic functions, such as y = − (x + 5)2 + 3 and y =

1 (x − 4)2 − 4.

4

Based on the function transformation rules you learned and on the previous section, you may have noticed a generalized way to view quadratics of this form. We proceed with a definition.

Definition. Vertex Form of a Quadratic. The vertex form of a quadratic is written y = a (x − h)2 + k, where (h, k) is the vertex of the associated parabola and x = h is the equation of the axis of symmetry. Moreover, the constant a determines how fast the y values move away from the x-axis, and whether the parabola is up-oriented (a is positive) or down-oriented (a is negative).

Figure 5.2: The Parabola.

To convert the function y = ax2 + bx + c to the vertex form, y = a (x − h)2 + k, one can complete the square.

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Completing the Square For the function y = ax2 + bx + c , if the coefficient a is equal to 1, as with x2 + bx, one finds the number that completes the square by using half of the coefficient b and squaring it. Thus, the result is then 2 b x + bx + , 2 2

which can be factored into the perfect square

b x+ 2

2 .

If the coefficient a is NOT equal to 1, start by dividing each term by a.

EXAMPLE 1: An example of completing the square to create a vertex form is y = 3x2 + 18x + 5.

Thus, y − 5 = 3 x2 + 6x

y − 5 + 27 = 3 x2 + 6x + 9 .

Notice that we divided the coefficient b (equal to 6) by 2 and squared the result to add 9 to the quantity inside the parentheses. Since the terms inside the parentheses are being multiplied by 3, this means we have added 3 times 9, or 27, to the right side of the equation. Thus, we must also add 27 to the left side of the equation. This simplifies to y = 3 (x + 3)2 − 22, which is now in vertex form. Note that this equation can be thought of as y = 3 (x − (−3))2 − 22.

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EXAMPLE 2: An alternate way to find the vertex form of the quadratic equation is by using algebraic methods applied to the general form y = ax2 + bx + c.

Suppose we, once again, start with the quadratic y = 3x2 + 18x + 5.

First, find the vertex of this quadratic using the fact that (h, k) can be written

b b − ,f − .

2a 2a Thus, for the particular equation y = 3x2 + 18x + 5, (h, k) is equal to (−3, −22) which yields y = a(x + 3)2 − 22. (1) Next, to solve for a, use a convenient point from the given equation

y = 3x2 + 18x + 5.

Notice the ordered pair (1, 26) satisfies the given equation. Use this fact to solve for a in equation (1). Therefore

26 = a (1 + 3)2 − 22, which mean a = 3. This means the final vertex form of the equation is y = 3 (x + 3)2 − 22.

Notice this is the same result as obtained by completing the square on the original equation, as we did previously. This displays the power of completing the square generally to produce the vertex form of a quadratic, since the algebraic method of EXAMPLE 2 relied on having the general form to start with.

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Exercise 5.3.1: Completing the Square and Vertex Form of a Quadratic 1. Complete the square: (a) x2 + 20x +

(b) x2 − 12x +

(c) x2 + 9x +

(d) x2 − 5x +

2. Solve: ax2 + bx + c = 0 for x by completing the square.

3. Change y = x2 − 6x + 2 to the form y = a (x − h)2 + k and sketch the graph without using technology. Also find the zeros of the function.

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4. Change y = x2 + 8x + 21 to the form y = a (x − h)2 + k and sketch the graph without using technology. Also find the zeros of the function.

5. Change y = −3x2 − 24x + 11 to the form y = a (x − h)2 + k and sketch the graph without using technology. Also find the zeros of the function.

6. CHALLENGE: Use the vertex form of the parabola to find the equation of the parabola with vertex at (2, −5) and containing the point (3, 1). Then find the zeros of the function.

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Exercise 5.3.2: A Geometric View of Completing the Square Use the geometric illustration below to develop and describe the process of Completing the Square.

5.4

Applications of the Quadratic Function

The “Live” Cannonball activity is an application Exercise that will allow you to use and connect a few different concepts that have been covered thus far in the course.

Exercise 5.4.1: The “Live” Cannonball and Other Applications

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1. Earlier this week the famous ”Live Cannonball” - Gunter Money was shot from a cannon at the Austin Expo. Gunter flew to a height of 70 feet and traveled a distance of 160 feet only to land perfectly in a safety net. Assume ( 0,0 ) was the starting point for Gunter’s flight and that he landed in the net at ( 160,0 ) . (a) Use matrices to find the equation for Gunter Money’s path.

(b) Write the equation for Gunter Money’s path in vertex form.

(c) CHALLENGE: Devise a method to estimate the angle at which Gunter Money takes off.

2. Texas University quarterback, Michael Daniels, is standing on his own 10-yard line. He throws a pass toward the opposite goal line. The football is 2 yards above the ground when the quarterback lets it go. It follows a parabolic path, reaching its highest point, 30 yards above the ground. It is caught 50 yards downfield at a point 2 yards above the ground. Let x be the number of yards the football travels horizontally, and let y be the number of yards the ball is above the ground. (a) Write the general equation of the parabola.

(b) What are the roots of the equation?

(c) How high is the ball when it crosses the 50 yard line?

(d) Write the equation of the football’s path in vertex form.

(e) Draw the graph of the function.

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3. A rectangular pool 6 meters by 8 meters is to be surrounded by a walkway of width x meters. This problem has to do with the rectangular region taken up by the pool and walkway. (a) Write the length and width of the region as expressions of x. (b) Write the area of the region as a function of x. (c) What is the area of the full rectangular region when x = 3? (d) Write this function in vertex form. (e) Sketch the graph in a reasonable domain. (f) Find the value of x which makes the walkway have an area equal to the area of the pool.

4. The table below shows data collected for average braking distance, f , in feet of a car on icy pavement traveling at m miles per hour. It is known that the relationship between feet as a function of miles-per-hour in this situation is a quadratic function. m (mph) 10 15 20 25 30 35 40 45 50 55

f (f t) 11.21 25 45.40 69.44 100 137.11 177.78 225 278.28 337.05

(a) Use matrices to find a quadratic function of feet as a function of mph that models this situation. (b) Write your equation in vertex form. (c) Graph the equation in a domain reasonable for the data.

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5. The distance s of a projectile, such as a rocket, above the Earth’s surface (under the influence of gravity) after t seconds is modeled by the quadratic function 1 s(t) = gt2 + v0 t + s0 , 2 where t ≥ 0, g (force of gravity acting downward and thus negative) is approximately equal to −32 ft/sec2 when s is measured in feet, v0 is initial velocity (when t equals zero), and s0 is the initial height of the rocket above the ground (when t equals zero).

A rocket is fired straight upward from the top of an 80 ft tower with an initial velocity of 64 ft/sec. Use what you have learned in this unit to find the rocket’s maximum altitude and the time at which the rocket hits the ground.

Then draw a graph of the situation and write the quadratic function in vertex form.

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98

Unit 6 - Complex Numbers

6.1

The Imaginary Number

You may have encountered complex numbers in a previous course when studying quadratics. In this section, this number system is explored in detail. The complex number system is the set generated by the real number system and the imaginary number i . The symbol i is introduced to serve as the solution to the equation

x2 + 1 = 0, which can be written as x2 = −1.

Of course, this means that the solution, x = i , cannot be a real number. Therefore, we consider i as part of a number system called the Complex Numbers. A complex number z can also be written in the form z = x + yi, where x and y are real numbers and i is the imaginary number. Note that x or y could be 0. Thus, the real numbers are also included in C, where C is the set of complex numbers.

6.2

The Geometry of Complex Numbers

The complex number z = x + yi can be associated with a point in the complex plane whose coordinates are x and y. We are now moving beyond naming points on the real line and moving into two-dimensional space, or R2 , where points are located in the plane. The x-axis in this case is called the real axis and the y-axis is named the imaginary axis. This is known as the complex plane. For example, the number z = 1 + 3i can be represented by the point (1, 3) in the complex plane as shown in Figure 6.1.

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Figure 6.1: A Number in the Complex Plane

The Modulus of a Complex Number The modulus, or absolute value, of z = x + yi is represented by the symbol |z| and is defined as

|z| =

p

x2 + y 2 .

Note that this modulus is really just a result of the Pythagorean Theorem.

The geometric interpretation of the modulus of z as defined above is that |z| is the distance from the origin to the point (x, y) . This distance can be represented by a line segment from the origin to the point (x, y) as seen in Figure 6.2.

Figure 6.2: The Modulus of z.

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6.3

Complex Number Operations

Like the real numbers, the complex numbers are a field. This means that the complex numbers adhere to the 11 axioms that constitute a field. For our purposes, we will only explore addition and multiplication of complex numbers.

ADDITION [Which is both closed in the field and commutative.]

Given the two complex numbers z1 = a + bi and z2 = c + di, z1 + z 2

=

(a+bi)+(c+di)

=

(a+c)+(bi+di)

=

(a + c) + (b + d)i

This method of addition of complex numbers can be verified by exploring the geometry of the addition of two complex numbers in the plane. Thus, a general rule for addition of two or more complex numbers is to “add the real terms and add the imaginary terms.”

MULTIPLICATION [Which is both closed in the field and commutative.]

Given the two complex numbers z1 = a + bi and z2 = c + di, z1 · z 2

=

(a + bi)(c + di)

=

ac+adi+bci+bdi2

Since i2 = −1, the term bdi2 is equal to −bd , thus z1 · z2

=

(a + bi)(c + di)

=

ac + adi + bci + bdi2

=

(ac − bd) + (ad + bc)i

Note that there is a geometric way to interpret the multiplication of complex numbers, but this is beyond the scope of our course. Also observe that this multiplication result reveals that we can use the distribution rule with complex numbers in the same way we do with real numbers.

We can also define quotients of complex numbers such as a + bi . c + di 101

While it may not be intuitive or obvious, it can be shown that division of complex numbers is accomplished by multiplying the numerator and denominator of the quotient by the complex conjugate of the denominator.

Definition. Complex Conjugate. The complex conjugate of a complex number a + bi is a − bi.

Thus, a + bi c + di

=

a + bi c − di · c + di c − di

=

ac + bd bc − ad + 2 i c2 + d2 c + d2

Notice the result is written in a + bi form, as is convention.

CHALLENGE: Do the conjugate multiplication to verify this result.

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Exercise 6.3.1: Complex Numbers 1. If i =

√

−1, what is i2 equal to?

2. if x ≥ 0, then

√

√ √ −x = i x. Thus, −5 =

.

3. Given two complex numbers z1 = a + bi and z2 = c + di, construct a picture in the coordinate plane that justifies the fact that z1 + z2

= = =

(a+bi)+(c+di) (a+c)+(bi+di) (a + c) + (b + d)i

4. Find the modulus of the following complex numbers. (a) |7 + 2i| (b) | − 2 + 4i| 5. If w = 1 − 4i and z = 3 − i, which point is closer to the origin?

6. Describe the set of all complex numbers that are at a distance of 2 units from the origin.

7. Find the distance between w = 1 + 2i and z = 3 − i.

8. Sketch a graph of the set of all complex numbers that satisfy the equation |z − 2 − 2i| = 3. [Note: this equation can be thought of as |z − (2 + 2i)| = 3 ] For problems 9-11, you are given z1 = 7 − 8i and z2 = 2 + 11i . Find: 10. z1 − z2

9. z1 + z2

11. z1 · z2

12. Multiply: (4 + 6i)(4 − 6i).

13. Find, generally, the product of complex conjugates (a + bi)(a − bi).

14. Perform the division and write the result in a + bi form. (a)

4 + 3i 5 − 2i

(b)

1 1+i

(f)

7 2 − 2i

(g)

−2 + i 5i

(c)

5 i

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(d)

6+i 3 + 2i

(e)

−4i 3−i

6.4

Complex Numbers and the Quadratic

The Quadratic and Complex Roots of a Quadratic For a quadratic function f (x) = y = ax2 + bx + c, the zeros can be found by using the quadratic formula, √ −b ± b2 − 4ac x= .

2a If these zeros are real numbers, then they represent the x-intercepts, or roots, of the parabola.

However, if the discriminant, b2 − 4ac , in the quadratic formula is negative, then the roots are complex numbers and the parabola does not intercept the x-axis.

Exercise 6.4.1: Quadratic Functions and Complex Roots 1. Find the roots of x2 − 2x + 9 = 0. Also, what does the discriminant tell you about the solution?

2. If the discriminant of −2x2 + 3x − 5 = 0 is −31, what does this tell you about the graph of the parabola?

3. If a quadratic function has complex roots, they will be complex conjugates of the form (m + ni) and (m − ni). Justify this statement using the quadratic formula.

4. Find the complex conjugate roots for the following quadratics. (a) f (x) = x2 + 14x + 58

(b) f (x) = x2 − 10x + 34

(c) f (x) = x2 − 4x + 29

(d) f (x) = x2 + 3x + 19

(e) f (x) = x2 − 6x + 73

(f) f (x) = x2 − 10x + 26

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Unit 7 - Polynomial Functions

7.1

Factoring Polynomials of Degree 3

At this point, you have explored multiple ways to factor quadratic (degree 2 polynomial) functions. In fact, you have found that the quadratic formula will always allow one to factor a quadratic equation. We perform this factoring, usually, to find the roots of polynomial functions. Unfortunately, polynomial functions of higher degree are often harder to factor. Recall that a polynomial is a monomial or sum of monomials.

Moreover, a monomial is a term, which is either a constant, a variable, or a product of a constant and one or more variables. Mathematicians such as Cardano and Ferrari have developed methods for factoring some cubics (degree 3) and quartics (degree 4), but these methods are cumbersome, limited, and are not widely used.

In this section, however, we offer some standard factorization templates for factoring cubics of specific forms. These are: Products and Factors of Special Cubics 1. a3 + 3a2 b + 3ab2 + b3 = (a + b)3 2. a3 − 3a2 b + 3ab2 − b3 = (a − b)3 3. a3 + b3 = (a + b)(a2 − ab + b2 ) 4. a3 − b3 = (a − b)(a2 + ab + b2 ) One more general rule is for cubics and higher degree polynomials:

5. Always try factoring out greatest common factors of groups of the given terms to see if this leads to a factored form. For example: 10x3 − 15x2 = 5x2 (2x − 3).

Try out these rules in the next Exercise.

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Exercise 7.1.1: Factoring Cubics For problems 1-3, use factor by grouping with the following polynomials.

1. x3 + 6x2 − 5x − 30

2. 12x3 + 2x2 − 30x − 5

3. 8x3 − 64x2 + x − 8

For problems 4-13, use any rules learned in this section to factor the following cubics.

4. x3 − 8y 3

5. x3 + 8

6. x3 + 1

7. x3 + 27

8. x3 − 64

9. x3 − 6x2 + 9x

10. 2x3 − 128

11. 64x3 + 8

12. x3 − x2

13. 6x3 − x2

7.2

Polynomial and Power Functions

First, we define the term polynomial function.

Definition. Polynomial Function. A polynomial Function is one that can be written in the form f (x) = an xn + · · · + a2 x2 + a1 x + a0 .

A polynomial function can consist of one term or many terms. By term, we mean any one ai xi of the polynomial. Each of the ai constants are called coefficients.

The leading term of the polynomial is the one with the highest degree, or power, of the variable.

Also note that power functions are related functions that can be written in the form f (x) = axk , where k can be any number.

When considering the graphs of polynomial and power functions, the end behavior of a given function is determined by the behavior of the leading term of the function. By “end behavior” of a function, we mean what happens to f (x) as x → −∞ or x → ∞.

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For example, for the power functions graphed in Figure 7.1, the end behavior of these even powered functions is that as x approaches positive or negative infinity f (x) approaches positive infinity.

Figure 7.1: Even Powered Power Functions

CHALLENGE: How would the graphs and end behavior of the functions pictured in Figure 7.1 be changed if each function had a “-1” as a leading coefficient?

Moreover, for the power functions pictured in Figure 7.2, the end behavior of these odd powered functions is that as x approaches positive infinity, f (x) approaches positive infinity and as x approaches negative infinity, f (x) approaches negative infinity.

Figure 7.2: Odd Powered Power Functions

CHALLENGE: How would the graphs and end behavior of the functions pictured in Figure 7.2 be changed if each function had a “-1” as a leading coefficient?

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The situation of “end behavior” of general polynomial functions is left for you to explore in the next couple of Exercises.

Exercise 7.2.1: Introduction to End Behavior

1. Without the use of graphing software, fill in the tables below and sketch a graph for each function (scale the y-axis as needed). (a)

(b) f (x) = x2

g(x) = x3

-3 -2 -1 0 1 2 3

-3 -2 -1 0 1 2 3

(c)

(d) h(x) = −x4

k(x) = −x5

-3 -2 -1 0 1 2 3

-3 -2 -1 0 1 2 3

2. Describe geometrical symmetry in the plane for each graph from problem 1.

3. Describe the end behavior for each function (as x increases positively and negatively, describe the effects of the y value).

4. Make a conjecture for the end behavior of odd degree and even degree functions generally. Be sure to discuss the situation of positive or negative leading coefficients for the polynomials as well.

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5. Based on your conjecture, without graphing, describe the end behavior for each of the following functions: (a) f (x) = −x3 (b) f (x) = x3 − 2x + 3 (c) f (x) = 2x4 − x3 (d) f (x) = 4x − 2x2 6. At this point you may use graphing software to view the graphs of the functions of problem 5 and verify your conjecture about end behavior. If your conjecture was correct, restate it here. If your conjecture was incorrect, create a correct rule that would describe the end behavior for all polynomials.

CHALLENGE: Evaluate each function in problem 1 with f (−x) and describe the similarities and differences between f (x) and f (−x). Note: An even function is defined as f (−x) = f (x) , and an odd function is defined as f (−x) = −f (x).

Exercise 7.2.2: End Behavior and Other Characteristics of Polynomial Functions For problems 1-9, use technology to graph these polynomials. Then identify the degree, the leading term, leading coefficient, and end behavior of each.

1. f (x) = 3 + 2x2 − 4x3

2. g(t) = 5t5 − 2t3 + 7t

3. h(p) = 6p − p3 − 2

4. f (x) = −2x4 − 3x2 + x − 1

5. f (s) = 6s5 − 2s4 + s2 + 3

6. h(x) = −3x3 + x2 − x + 3

7. f (t) = 2(t − 1)(t + 2)(t − 3)

8. f (x) = 3(x + 1)(x − 4)(x + 5)

9. g(n) = −(3n − 1)(2n + 1)

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10. A polynomial of degree n will cross the x-axis, at most, n times. It will also have at most n − 1 places where the graph changes direction. For problems 1-9 above, identify how many x-intercepts (i.e. zeros) and how many direction changes are present in each function.

For tproblems 11-13, identify what type of polynomial each grapht is likely to be based on the graph’s characteristics.

11.

12.

13.

CHALLENGE: For the graph of problem 11, it is clear to see that it has only two real roots. What must be true of the other roots? Along these lines, what can be said about the roots of problem 13?

The Behavior of x-intercepts of Polynomials A quadratic is a polynomial of degree 2. As was mentioned previously, a quadratic such as

f (x) = (x − 2)2 = (x − 2) (x − 2) has a linear factor (x − 2) repeated twice, so we call this a double zero. Thus, we also say the zero has multiplicity 2. Since polynomials of higher degree can have various combinations of factors with multiplicity, we state a few observations.

For a polynomial containing a factor of the form (x − h)k , the behavior near the x-intercept is determined by k.

k=1

k=2

110

k=3

Generally, for higher even powers of multiplicity, the graph will be tangent to the x-axis but will “flatten more” around the root with increasing powers. For higher odd powers of multiplicity, the graph will be “flatter” as it passes through the x-axis around the root with increasing powers.

Exercise 7.2.3: Multiplicity For problems 1-4, use graphing software to sketch each graph and note the behavior of the graph around each root. Then state the degree of each polynomial along with the multiplicity of each root in the polynomial.

1. h (t) = 3(t − 5)3 (t − 3)3 (t − 2)

2. k (x) = 2(x − 3)2 (x + 1)3 (x + 2)

3. h (x) = −2x(x − 1)(3 − x)2

4. q (t) = −4t(2 − t)(t + 1)3

For problems 5-9, write an equation for a polynomial function containing the required features.

5. Degree 3; zeros at x = −2 , x = 1 , and x = 3 ; y-intercept at (0, −6).

6. Degree 5; double zero at x = 1, and triple zero at x = 3; passes through the point (2, 8).

7. Degree 4; root of multiplicity 2 at x = 4, and roots of multiplicity 1 at x = 1 and x = −2 ; y-intercept at (0, −3).

8. Degree 3; zeros at x = −4, x = −2, and x = 1; y-intercept at (0, 9).

9. Degree 5; roots of multiplicity 2 at x = 3 and x = 1, and a root of multiplicity 1 at x = −3; y-intercept at (0, 7).

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7.3

Division and Rational Zeros

Long Division If a polynomial is reducible, we can write it as a product of factors, such as in f (x) = (x + 1)(x − 4)(x + 5), where each factor corresponds to an x-intercept. Formally, a polynomial is reducible if it can be written as a product of two other polynomials that are of positive degree. If a polynomial is not reducible it is said to be irreducible. A polynomial is completely factored when it has been written as a product of irreducible polynomials.

The hard part, however, is actually finding ways to factor higher degree polynomials.

However, if we know one root of a given polynomial, we can try to use long division to find others. If, for example, we know that x = 2 is a root of the polynomial function f (x) = x3 + 4x2 − 5x − 14, we might guess that the polynomial could be factored as x3 + 4x2 − 5x − 14 = (x − 2)(g(x)), where g(x) is another polynomial. To find g(x) we can use polynomial long division.

EXAMPLE 1: To divide x3 + 4x2 − 5x − 14 by (x − 2), we start by writing the problem out in long division form just as you did with numbers when you were younger.

Next, we notice x3 ÷ x = x2 , so, multiply that x2 by (x − 2) , write the result below the dividend, and subtract. Notice, we write the x2 term above the x2 term of the dividend by convention.

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Now subtract that expression from the dividend.

Now, divide the leading term of the remainder 6x2 − 5x − 14 by x. Thus, since 6x2 ÷ x = 6x, we multiply 6x by (x − 2) and subtract.

Repeat the process one last time, this time multiplying (x − 2) by 7 and subtracting.

This tells us x3 + 4x2 − 5x − 14 divided by x − 2 is x2 + 6x + 7 , with a remainder of zero.

Since the remainder is zero, this also means that we can factor x3 + 4x2 − 5x − 14 as (x − 2)(x2 + 6x + 7) . Thus, the number 2 is a root of the function and (x − 2) is a linear factor of x3 + 4x2 − 5x − 14.

Now that you have shown that the quadratic is also a factor of the original polynomial function, you could use the quadratic formula to find the other two roots of the original function.

CHALLENGE: Find these roots. This gives us one way to find the intercepts of this polynomial.

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Synthetic Division Analogous to matrices being used as a “shorthand” method of working with systems of equations, we can perform a “shorthand” version of long division of polynomials. This shorthand version is called synthetic division. Let’s look at the division that was done in Example 1, yet we’ll perform the process this time using synthetic division in Example 2.

EXAMPLE 2: For this method, we construct a table. To divide the polynomial x3 + 4x2 − 5x − 14 by the linear factor x − 2, we write 2 in the place of the divisor, which is placed in the top most left-hand cell. The reason we write 2 and not -2 is twofold. First, if (x − 2) is a factor of the polynomial, this means that 2 is the root. Second, in traditional long division we subtract the second line from the first; subtracting a negative leads to a positive. Thus, we will use positive 2 and then add our second line in synthetic division.

Only the coefficients of x3 + 4x2 − 5x − 14 are written in for the dividend along the top of the table. Then “bring down” the first coefficient of the dividend. In this case, the first coefficient is a 1.

Step 1.

Synthetic Division Table

2 1 4 −5 −14 ↓ 1

Step 2.

Next, using the 2 from the divisor, multiply the 2 by the 1 that was “brought down” to yield 2. Write this underneath the 4, then add to get 6 as shown in the next iteration of the table.

2 1 4 −5 −14 ↓ 2 1 6

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Step 3.

Now, using the 2 from the divisor, multiply by the 6 to yield 12. Add that to the -5 to get 7.

2 1 4 −5 −14 ↓ 2

12

1 6

7

Step 4.

Repeat the same process, this time with the 2 and the 7 to fill in the last column.

2 1 4 −5 −14 ↓ 2

12

14

1 6

7

0

Notice that the remainder is zero, displaying the fact that (x − 2) is a factor of the original polynomial function and thus f (2) = 0.

[NOTE: If our divisor in synthetic division is of the form x + 2, this can be thought of as x − (−2). Thus, we would write -2 in the top left-hand corner of our table and still use the “multiply-add” method to work through the table.]

Looking at either our long division or synthetic polynomial division methods, intuition tells us that, in general, if we divide a polynomial by a term of the form x − c, then the remainder will be zero or a constant. This observation can be proved and written as the Remainder Theorem.

The Remainder Theorem. If p (x) is a polynomial of degree 1 or greater and c is a real number, then when p (x) is divided by (x − c), the remainder is p (c).

This implies that if x − c is a factor of the polynomial p, then p (x) = (x − c)(q(x)) for some polynomial q. Thus, p (c) = (c − c)(q(c)) = 0, meaning c is a zero of the polynomial.

More Related Theorems Except for the Rational Zeros Theorem, the following theorems will not be proven, yet they provide important and useful information about polynomial functions. You will be asked to prove the Rational Zeros Theorem as an Exercise.

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The Factor Theorem. A number c is a solution of the polynomial equation p (x) = 0 if and only if x − c is a factor of p (x).

We will take a moment to express a few more related theorems. The next theorem applies to polynomials with complex number coefficients, but keep in mind that this theorem applies to integral, rational, and real coefficients as well. This is because any real, rational, or integer number a can be written as a complex number a + 0i.

The Fundamental Theorem of Algebra. Every polynomial function of degree ≥ 1, with complex coefficients has at least one complex zero.

A corollary that logically follows from this theorem is that each polynomial of degree n ≥ 1 has exactly n complex zeros and at most n − 1 turning points. Turning points occur when the graph of a function changes direction at a point. That is, the function goes from increasing to decreasing or decreasing to increasing.

The Complex Conjugate Theorem. If a complex number a + bi is a zero of a polynomial function with real coefficients, then the conjugate a − bi is also a zero of the polynomial.

Rational Zeros Theorem. Assume the coefficients of f (x) = an xn + · · · + a2 x2 + a1 x + a0 p are integers. If a rational number , where p and q are relatively prime (i.e. p and q have q no factors in common), is a zero of the function, then p is a factor of a0 and q is a factor of an .

GOOD ADVICE. Don’t forget that the factors discussed in the Rational Zeros Theorem can each be either positive or negative.

CHALLENGE: After reviewing these theorems, return to Exercise 7.2.2 problems 11-13.

Explain what type (real or complex) and how many zeros there are for each of the polynomial function graphs based on your previous answers to the problems.

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Exercise 7.3.1: Division and Finding the Zeros of Polynomial Functions Part 1: Choose either synthetic division or long division to solve the following problems.

1. Divide x3 + 8 by (x + 2). (Note: You must enter 0 for the coefficients of x2 and x in the dividend).

2. Divide 2x3 − 3x2 + 4 by (x + 2). (Note: You will have a remainder).

3. Divide 5x3 − 2x2 + 1 by (x − 3).

4. Determine whether (x − 5) is a factor of f (x) = x4 − 3x3 + 7x2 − 60x − 125.

You are given that (x + 1) is a factor of P (x) = x3 + 4x2 − 2x − 5. Thus, 5. P (−1) =

6. One zero of x3 + 4x2 − 2x − 5 is

.

7. Another root of x3 + 4x2 − 2x − 5 = 0 is 8.

.

.

(a) Find all the roots of x4 − 8x2 − 8x + 15 = 0. (b) Thus, the zeros of f (x) = x4 − 8x2 − 8x + 15 are what?

You are given that (x + 3) is a factor of P (x) = x3 − 7x + 6. Thus, 9. P (−3) =

.

10. One zero of x3 − 7x + 6 is

11. Another root of x3 − 7x + 6 = 0 is

.

.

12. What does the Fundamental Theorem of Algebra and its corollary say about the zeros of P (x) = x7 + 11x3 − 4x + 20?

13. Find the quotient of

a3 + b3 . (HINT: a3 + b3 = a3 + 0a2 b + 0ab2 + b3 ) a+b

14. Find the quotient of

a3 − b3 . a−b

CHALLENGE: Find the quotient of

ax2 + bx + c by using both long and synthetic dx + e

division.

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Part 2: Use The Rational Zeros Theorem to make a list of possible rational zeros for each of these polynomial functions.

15. f (x) = 2x4 + x3 − 7x2 − 3x + 3

16. f (x) = x3 − 2x2 − 5x + 6

17. f (x) = x4 + 2x3 − 12x2 − 40x − 32

18. f (x) = x3 + 4x2 − 11x + 6

19. f (x) = x3 − 7x2 + x − 7

20. f (x) = 3x3 + 3x2 − 11x − 10

21. f (x) = −17x3 + 5x2 + 34x − 10

22. f (x) = x4 − 9x2 − 4x + 12

Find the real zeros of each polynomial.

23. f (x) = x4 + 2x2 − 15

24. f (x) = x4 − 9x2 + 14

25. f (x) = x3 − 7x2 + x − 7

26. f (x) = −17x3 + 5x2 + 34x − 10

27. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1

28. f (x) = 3x3 + 3x2 − 11x − 10

Find all of the zeros of the polynomial, then completely factor it over the real and complex numbers. Remember, you can use long or synthetic division as part of the process. You should graph these problems to check your answers.

29. f (x) = x3 + 6x2 + 6x + 5

30. f (x) = 3x3 − 13x2 + 43x − 13

31. f (x) = x2 − 4x + 13

32. f (x) = x2 − 2x + 5

33. f (x) = 3x2 + 2x + 10

34. f (x) = x3 + 7x2 + 9x − 2

35. f (x) = x3 + 3x2 + 4x + 12

36. f (x) = x4 + 5x2 − 24

37. f (x) = −3x4 − 8x3 − 12x2 − 12x − 5

38. f (x) = 8x4 + 50x3 + 43x2 + 2x − 4

39. f (x) = 9x3 + 2x + 1

40. f (x) = 2x4 − 7x3 + 14x2 − 15x + 6

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Exercise 7.3.2: Justifying the Rational Zeros Theorem End behavior and y-intercepts are useful to know and can be quickly obtained.

Zeros of the function are also useful to know but may not be easily attainable. A general polynomial function with integer coefficients can be written as f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 Our goal is to find the roots of the function, the x-values that satisfy f (x) = 0, and to understand the method.

Recall that the Rational Zero Theorem states: Assume the coefficients of f (x) = an xn + an−1 xn−1 + · · · + a2 x2 + a1 x + a0 p are integers. If a rational number , where p and q are relatively prime (i.e. p and q q have no factors in common), is a zero of the function, then p is a factor of a0 and q is a factor of an .

Justifying the Rational Zero Theorem p 1. Suppose that , where p and q are relatively prime, is a root of f (x). q p Substitute into f (x). Make sure you set the function equal to zero. q 2. The next steps are going to be procedural. (a) Subtract the term a0 from both sides of the equation. (b) Multiply both sides by q n to rid the equation of fractions. (c) Now, factor out the greatest common factor. (d) Since both sides of the equation must be integers, what does this mean regarding the possibility of p being a factor of a0 ? Keep in mind that p and q have no factors in common.

3. The next steps, once again, are going to be procedural. n p (a) Return to the equation you made in step 1. Move the an term to q the other side of the equation by subtracting this term from both sides. (b) Multiply both sides by q n . (c) Now factor out the greatest common factor. (d) Since both sides of the equation must be integers, what does this mean regarding the possibility of q being a factor of an ? Remember, p and q have no common factors.

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4. Now create a statement about possible zeros for a polynomial function based on what you have learned so far. (The statement you create should be what is known as the Rational Zero Theorem.)

5. Can you think of a less formal way of explaining the Rational Zero Theorem?

HINT: Try to create a product of factors for a general polynomial function.

7.4

Applications

This section is done purely as an Exercise. Use what you have learned to solve these polynomial application problems.

Exercise 7.4.1: Applications of Polynomial Models Linear Model 1. A new linear road is planned to be constructed perpendicular to an existing one. On a representative grid, a surveyor determines that the existing linear road passes through the points (4, 7) and (2, 15). The roads must intersect at the point (4, 7). (a) Determine the slope of the new road. (b) Determine the equation of the new road.

Quadratic Model 2. A certain 12-volt car battery has a resistance of 0.04 ohms. Due to this fact the available battery power to start the car is modeled by the function P (I) = 16I − 0.04I 2 , where the power, P , is measured in watts and the electric current I is measured in amperes. (a) What are the zeros of the function? (b) What current is needed to create 1599 watts of power to start the car? (c) Find the line of symmetry of this function.

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Cubic Model

3. Suppose a company is testing the flexibility of a new type of fiberglass. The company decides to test this flexibility by creating a diving board out of the fiberglass material. The bend deflection from the horizontal of a diving board when someone stands on it is a cubic function of the form y = ax3 + bx2 + cx + d where y is the deflection of the board (in 1000ths of an inch) from the horizontal rest position, and x is the distance from the secured end of the board. Below is a table of distance and deflection measurements. x feet 0 1 2 3

y deflection 0 116 448 972

(a) Use a coefficient matrix and the supplied data to find the coefficients a, b, and c in the equation y = ax3 + bx2 + cx + d, and write the particular function that models this data. [NOTE: You only need a 3 × 3 matrix to find a, b, and c since the first data point from the table, (0, 0) tells you that d must be equal to zero. Why is this true?] (b) What is the multiplicity of 0 as a zero of the function? (c) What is the deflection of this board at the tip, if it is 8 feet long? (d) Some functions used to model data only work in limited intervals of the domain. Explore the graph of your particular function and identify an interval where the model breaks down as a reasonable model for describing the diving board’s deflection.

121

Quartic Model

4. A dye is used in the blood to measure a patient’s cardiac output when there is potential for a heart attack. In a normal heart, the concentration of the dye is modeled by the function C (t) = −0.005t4 + 0.139t3 − 0.051t2 + 1.800t where C is concentration and t is time, measured in seconds. (a) Use graphing software to graph and find the real zeros of the function. (b) What must be true of two of the zeros of this function? (c) Explain why the end behavior of this function makes sense based on its leading coefficient and degree.

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Unit 8 - Modeling and Data Analysis This Unit deals specifically with applying models to data or application situations using the concepts of variation and function regression. Thus far, throughout the text, you have learned a few methods of fitting models (representative functions) to data and general real-world situations using matrices and polynomials. In this Unit, we extend the concept of modeling by presenting more techniques that can be considered when trying to apply a function to real-world phenomena.

8.1

Direct and Inverse Variation Definition. Direct Variation. A function of the form y = k · xn , where k is a constant belonging to the real numbers and n is a positive integer is known as a direct variation. We say that y is directly proportional to xn . The constant k is known as the constant of proportionality.

Definition. Inverse Variation. A function of the form y=

k , xn

where k is a constant belonging to the real numbers and n is a positive integer, is known as an inverse variation. We say that y is inversely proportional to xn . The constant k is known as the constant of proportionality.

As might be imagined, there are situations where the two types of variations defined are combined in one way or another. For example, the variation equation

T =

kr2 , s3

(1)

where r and s are variables and k is a constant, is a combined variation.

CHALLENGE: Describe the variations that exist in equation (1) between T , r and s.

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One famous variation function is given by the statement of Sir Isaac Newton’s Second Law of Motion. This Law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

The first part of Newton’s Second Law is conventionally written as

F ∝ a.

The “∝” symbol stands for the statement “is directly proportional to.” This symbol can be replaced by an equal sign, once a constant of proportionality is found and inserted into the situation. Newton found that the constant of proportionality needed for his Second Law is the mass of the object the net force is acting upon. Thus, as an equation, the Second Law is written as F = ma.

Notice that the statement of Newton’s Second Law also describes the inversely proportional relationship between the acceleration and mass of an object when subjected to a net force. This is written as

a=

F . m

In application, these variation models are used quite often. In fact, data is frequently used in these situations to find the constant of proportionality, k, that fits a particular model. You will get practice solving for k in the next Exercise.

Exercise 8.1.1: Variation Functions and Applications For problems 1-4, assume y varies directly with x. Find the constant of proportionality and a representative function for each problem.

1 4

1. y = 2.8 when x = 2

2. y = 2 when x =

3. y = 0.6 when x = 0.4

4. y = 5 when x = 0.5

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5. Graph the functions found for problems 1-4 together on the provided graph.

For problems 6-9, assume y varies inversely with x. Find the constant of proportionality and a representative function for each problem.

1 3

6. y = 4 when x = 3

7. y = 24 when x =

8. y = 9 when x = 5

9. y = 15 when x = 3

10. Graph the functions found for problems 6-9 together on the provided graph.

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For problems 11 and 12, write the variation function that is described in the given statement.

11. y varies inversely with the square of x.

12. y is directly proportional to the cube of x.

Some Models Involving Variation Functions.

13. It is known that the distance d that a spring is stretched by a hanging mass varies directly with the mass m. This is known as Hooke’s Law. If the distance a mass stretches a spring is 0.20 meters when the mass is 4 kilograms, what is the proportionality constant k for the spring? What distance would this spring be stretched by a 7 kg mass?

14. A person’s weight w on Mars varies directly with that of her weight on Earth E. A person who weighs 140 lbs on earth weighs 56 lbs on Mars. How much would a person who weighs 185 lbs on Earth weigh on Mars? Also, what is the proportionality constant for this situation?

15. The braking distance d of a car after the brakes have been applied varies directly as the square of the car’s speed s. (a) Write a variation function that models this relationship. (b) A certain car going 60 mph can stop in 180 ft. At what speed is the car traveling if it can stop in 100 ft?

16. The weight of an object on Earth varies inversely as the square of its distance d from the center of the Earth. If a pilot weighs 175 lbs at sea level on Earth (approximately 6400 kilometers from the center of the Earth), how far above the Earth’s surface must the pilot fly in order to weigh 150 lbs?

17. According to Boyle’s Law, the volume V of a gas varies inversely as the amount of pressure placed on it. If the volume of a gas is 200 cm3 as a result of a pressure of 32 kg/cm2 , what volume will the gas have under a pressure of 44 kg/cm2 ?

18. Describe the variation given by the equation F =

km1 m2 , d2

where F is a force, k is a constant of proportionality, m1 and m2 are each masses, and d is the distance between masses.

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19. The amount of force one exerts on a wrench handle to loosen a bolt is inversely proportional to the length of the wrench handle. With a wrench 12 inches long it takes 96 pounds of force to loosen the bolt. (a) Write the variation function for the force in this situation. (b) If a person can only exert 60 pounds of force on a wrench, what length wrench must be used in this situation?

20. The intensity of light reaching a person from a light source varies inversely as the square of the distance from the source to the person. Assume that at 3 meters the intensity is 120 units, and at 6 meters the intensity is 30 units. (a) Write the variation function for the intensity of light in this situation. (b) Find the light intensity at 9 meters.

8.2

Piecewise Functions

Many physical situations can be modeled using piecewise functions. This implies that the phenomena under investigation changes behavior in such a way that different types of functions must model different intervals of the data related to the phenomena. We have mentioned piecewise functions previously in the text, but we take this chance to formally define the term.

Definition. Piecewise Function. A piecewise function f is a function in which different rules are applied over different “pieces” (or intervals) of the domain. We symbolize this as:

  expression 1, f (x) = expression 2,   expression 3,

if domain for expression 1 if domain for expression 2 . if domain for expression 3

An example of a piecewise function is: ( 30 if 0 < x < 3 f (x) = 30 + 8(x − 3) if x ≥ 3 CHALLENGE: Graph this piecewise function.

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An example of the graph of a piecewise function is

CHALLENGE: Write the piecewise function that generated this graph.

Exercise 8.2.1: A Brief Investigation of Piecewise Functions For problems 1-4, sketch a graph of these functions.  2 (  x 3x if 0 < x < 10 1. f (x) = 2. f (x) = 3  40 if x ≥ 10  x ( x2 − 2 if x < 2 3. f (x) = 4 + |x − 5| if x ≥ 2

if x ≤ 1 if 1 < x ≤ 2 if x > 2

( 5t 4. H (t) = 2t + 6

if 0 < t < 5 if t ≥ 5

For problems 5-6, write a formula for the piecewise functions graphed below.

5.

6.

A Piecewise Function as a Model 7. An amusement park charges $10 per person for a group of 1 to 10 people, or a flat $100 fee for 10 or more people. Write a piecewise function to represent the fee charged as a function of the number of people attending.

Also, graph the situation.

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8.3

Regression and Function Fitting

Using Statistical Regression to Fit a Function to Data A regression method called the method of least squares is used in this section to fit a best-fit linear function to data that is conjectured to be linear. A simple linear regression is accomplished by finding a line that minimizes the distance between actual data points and the predicted values, ŷ, on the best-fit line. The difference between the predicted value at the ith data point, ŷi , and the observed value, yi , is called a residual, symbolized ei , such that ei = yi − ŷi .

The method of least squares used to find the predicted regression line employs an optimization technique from calculus to minimize the sums of the squares of the deviations yi − ŷi such that an estimated or fitted regression line of the form ŷ = bx + a can be produced.

An example of a scatter plot of data points and a best-fit regression line is displayed in Figure 8.1.

Figure 8.1: A Scatter Plot and “Best-Fit” Line

One can use technology to avoid some of the tedious mathematics involved in producing the regression equation of the best-fit line. Graphing calculators, statistical software, or spreadsheet programs such as Microsoft Excel will calculate a linear regression equation to fit data.

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A tool available that helps one decide which type of regression to use to fit a model or function to data is the correlation coefficient, associated with linear regression. The correlation coefficient, r, measures and describes a relationship or trend between two variables. Note that a correlation between two variables does not imply a causal relationship between the two variables. The relationship can, however, be characterized as having a positive correlation or a negative correlation depending on whether the linear model associated with the data is increasing or decreasing respectively. The correlation coefficient r can take on the values −1 ≤ r ≤ 1.

The actual derivation of the formula for r is beyond the scope of this course. If the value of r is close to negative one, this indicates a strong linear correlation between variables from data with a negative trend. If r is close to positive one, a strong linear correlation between variables from data with a positive trend is indicated. A value of r close to zero indicates little linear correlation between the variables under investigation. This is summarized as:

The Correlation Coefficient r

• r > 0 indicates a positive (increasing) trend • r < 0 indicates a negative (decreasing) trend • The closer the r value is to 0, the data is more scattered • The closer the r value is to 1 or -1, the data is less scattered

Exercise 8.3.1: Data Trends State whether the data in the scatterplot shown has a strong positive trend, a strong negative trend, or neither. Then draw an estimate of a best-fit linear regression function for each graph that has an identified trend.

1.

2.

3.

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8.4

Applications and Regression Modeling

Exercise 8.4.1: Examples for Linear Regression Practice NOTE: This Exercise should be done as a full-class activity as a way for your instructor to teach you how to perform a linear regression investigation on relevant technology.

For problems 1-3, consider the following scenario.

Various pounds of force are applied to a spring, which consequently stretches the spring y inches away from a frame of reference at (0, 0). The data for this situation is located in the following table.

x (pounds) 0 20 30 40 50 80

y (inches) 0 5.9 9.2 12.1 15.3 24.3

1. Make a scatter plot of the data and use linear regression to write a particular function that models this data.

2. What is the spring constant, k, for this direct variation?

3. What is the correlation coefficient for this linear regression, and what does it say about the strength of linear fit for this situation?

NOTE: For the next data situation, use relevant technology and work on your own to perform the linear regression.

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For problems 4-7, consider the following scenario.

Taken from a classic experiment, the following table displays the number of cricket chirps C recorded in 15 seconds at various air temperatures T (in degrees Fahrenheit). It is conjectured that this data is linearly related.

Chirps T (◦ F )

44 80.5

35 70.5

20.4 57

33 66

31 68

35 72

18.5 52

37 73.5

26 53

4. Make a scatterplot of the data.

5. Does the data appear to display a trend, and does the data appear to be linear?

6. Perform a linear regression on the data and comment on the appropriateness of the result and write the regression equation. Do you think there is evidence that this data truly is linear?

7. What is the correlation coefficient for this linear regression, and what does it say about the strength of linear fit for this situation?

NOTE: As a result of similar methods to that of the Method of Least Squares, regression equations to fit quadratic, power, exponential, logarithmic models or other functions to supplied data have also been devised. At this point, we will also use statistical regression methods to find functions that model certain phenomena governed by other types of functions that are taught in College Algebra. Microsoft Excel, other statistical packages, and graphing calculators all have the capability of using regression to fit data to various types of functions.

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Exercise 8.4.2: Other Types of Function Models Found by Statistical Regression Part 1: This data represents the average height of a sample of children in inches as a function of their corresponding ages.

t (years) H (inches)

5 30

6 33

7 39

8 41

9 47

10 54

It is conjectured that this data is exponential of the form H (t) = abt , where H, the average height of a child, is a function of t, a child’s age in years.

1. Make a scatterplot of the data.

2. Does the data appear to display a trend, and does the data appear to be exponential?

3. Use technology to perform an exponential regression on the data. Record your regression equation and comment on the appropriateness of the result. Is there evidence that this data truly is exponential?

4. What is the correlation coefficient for this exponential regression, and what does it say about the strength of fit for this situation?

[NOTE: Even though this is not a linear model, an r value may be reported by your technology program. This is because an exponential function can be statistically linearized. A discussion of this linearization method is beyond the scope of this course, but is discussed in the OnRamps Precalculus course.]

5. At age 18 years, how tall will this child be according to the model? What does this tell you about the limitations of using a limited interval of data to find a general regression model?

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Part 2: A ball is dropped from the top of a building. The object’s height in feet above ground after t seconds is given by the function h (t) = 145 − 16t2 for 0 ≤ t ≤ 3.

t 0 1 2 3

h (t) 145.1 127.2 81.1 0

From physics, we know that this data is quadratic.

6. Make a scatterplot of the data.

7. Does the data appear to be quadratic?

8. Use technology to perform a quadratic regression on the data. Record your regression equation and comment on the appropriateness of the result.

9. Why is it not useful to investigate your model for t values less than 0 seconds or greater than 3 seconds?

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Unit 9 - Rational and Radical Functions

9.1

Rational Expressions and Rational Functions

Rational Expressions Before we explore rational functions at length, we will investigate the algebra of adding, subtracting, multiplying, and dividing rational expressions. Although you have probably simplified algebraic expressions in a previous algebra class, properly performing algebraic manipulations on rational functions is an essential skill to carry forward into future mathematics and science courses. Thus, we review some useful rules.

GOOD ADVICE. Notice that rational expressions represent fractions. Therefore, we apply fraction rules to them with regard to addition, subtraction, multiplication, and division.

Addition or Subtraction When adding or subtracting fractions, one must first find a common denominator. An example is:

x y b x a y bx + ay + = + = . a b b a a b ab Apply this rule to: x+3 x−3 + . x−3 x+3 Verify that finding a common denominator yields the expression (x + 3)2 + (x − 3)2 . (x − 3)(x + 3) Notice that the terms in the numerator can be expanded to yield the expression x2 + 6x + 9 + x2 − 6x + 9 . (x − 3)(x + 3) To simplify the overall expression, one can now combine the terms common in the numerator and distribute the denominator, which produces the expression 2x2 + 18 . x2 − 9 Lastly, the numerator can be rewritten to yield 2(x2 + 9) . x2 − 9 135

Multiplication When multiplying fractions, multiply the numerators of each fraction and the denominators of each fraction respectively, as in r t rt · = . a b ab Apply this rule to: x2 + 3x − 10 x2 + 2x − 3 · . x2 − 7x + 6 x2 + x − 6 GOOD ADVICE. ALWAYS try to factor and cancel first before performing multiplication of expressions. Note that the expression above, in factored form, is (x + 5)(x − 2) (x + 3)(x − 1) · . (x − 6)(x − 1) (x + 3)(x − 2) Since all terms are being multiplied, verify that one can cancel like terms in numerators and denominators to yield the simplified expression x+5 . x−6 This is as far as we can go for this expression.

Division When dividing fractions as in the provided example, recall that one multiplies the first fraction by the reciprocal of the second, as in r x r y ÷ = · . t y t x CHALLENGE: Can you explain why this is done? Thus, to simplify the expression x2 + 13x + 40 (x + 8) ÷ , (x − 5) 1 one first factors all relevant terms and then multiplies the first fraction by the reciprocal of the second to produce (x + 5)(x + 8) 1 · . (x − 5) (x + 8) Last, canceling like terms in the numerator and denominator yields the simplified answer (x + 5) . (x − 5)

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CHALLENGE: Devise a method for simplifying the following rational expression which involves a complicated fraction over a fraction: 5 x−3.

4 x+2+ x−3 x+3+

Exercise 9.1.1: Review of the Algebra of Rational Expressions Perform the indicated operations and simplify each result.

1.

x2 − 2x + 1 x2 − 1 x2 + 3x − 10 2x + x2 x2 − 5x + 6 x2 + 2x − 8 ÷ 2. · 3. ÷ 12 6 x2 − x − 6 x2 − 2x x2 − 3x − 4 x2 + 4x + 3

4.

x2 − 25 x2 − 2x · x2 − 10x + 25 6x − 12

x2 − 16 x+4 7. ÷ 3 2 x x − 4x2

1− 10.

1−

13.

1 x2

2 3 − 2 x x

6x x2 − 36

+

x−6 x+6

5.

x2 y x2 − 81 · x−9 xy 2

9 x 8. x 1+ 3 x−

6.

x2 + 5x + 6 x2 − 3x − 10 ÷ x2 + 5x − 24 x2 − 8x + 15

2 x+1 9.

2 2+ x+1 2−

11.

x − 5 3x − 1 − 4+x x

12.

x2 4 − x−4 4−x

14.

5 6 + x−1 x+1

15.

1 2 + x+4 x+3

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Rational Functions We are now ready to define the term rational function.

Definition. Rational Function. A rational function is a function of the form f (x) h (x) = , g (x) where f (x) and g (x) are polynomials.

One of the most interesting properties of rational functions concerns the possible presence of asymptotes, both vertical and horizontal, inherent to their graphs.

Definition. Asymptote. An asymptote is a line such that the distance between the graph of a point P on the curve and the line approaches zero as the distance from point P to the origin increases without bound.

We will confine our study, at this time, to that of vertical or horizontal asymptotes. For a vertical asymptote, the distance between a graph and a line of the form x = c, where c is a constant, approaches zero. For a horizontal asymptote, the distance between a graph and a line of the form y = k, where k is a constant, approaches zero. An example of a common function that you have seen, which contains a horizontal asymptote, is the exponential function f (x) = bx , where b is greater than 1. See the figure below.

Figure 9.1: Horizontal Asymptote at y = 0, for f (x) = bx Notice that there is a horizontal asymptote inherent to f (x) = bx . As x approaches negative infinity, the graph of f (x) = bx approaches the line y = 0, but never actually touches the line.

The next Exercise provides you with a chance to think about a few basic rational functions, some of which you have seen previously.

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Exercise 9.1.2: An Introduction to Rational Functions

1. Create a graph for f (x) = x

f (x) =

1 x

g (x) =

1 x2

1 1 and g (x) = 2 by filling in the tables below. x x

-5 -2 -1 -1/2 -1/4 0 1/4 1/2 1 2 5 x -5 -2 -1 -1/2 -1/4 0 1/4 1/2 1 2 5

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2. Explain the end behavior of the graph of each function.

3. Based on the graphical symmetry, categorize each function as even, odd, or neither.

4. Describe the behavior of each graph around x = 0.

5. Predict and draw a graph for the following functions without creating a table or using a calculator.

1 1 1 1 1 1 f (x) = 3 , f (x) = 4 , f (x) = − , f (x) = − 2 , f (x) = , f (x) = + 2 x x x x x−2 x

6. Write the domain and range for each function in problem 5 using interval or inequality notation.

In addition to horizontal and vertical asymptotes, rational functions may have “holes” in them called removable discontinuities. The next Exercise should lead you to discover this condition. Following the next Exercise, we will summarize some properties of rational functions.

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Exercise 9.1.3: Exploring Rational Functions

1. Generate a set of ordered pairs, plot the points, and sketch the graph for each function below. Sketch any vertical asymptotes you find as dashed lines. Then, write the equations for the vertical asymptotes. (a) h (x) =

3x − 2 x+4

Equation(s) of vertical asymptote(s):

(b) h (x) =

x+5 x2 − 16

Equation(s) of vertical asymptote(s):

(d) Your conjecture: The rational function h (x) = asymptotes when

(c) h (x) =

x2 + 8 x2

Equation(s) of vertical asymptote(s): f (x) will have vertical g (x) .

2. Look at the 3 graphs that you drew in problem 1, and for each function give the equations of any horizontal asymptotes that you detect or predict. You may use a calculator to type in the values of x that you want to substitute into the function. A good method would be to choose values of x that are powers of 10 (i.e., let x = 1, 10, 100, 1000, etc.). (a) As x increases, is the value of h (x) in each of 1(a), 1(b), and 1(c) above, getting close to the value that you predicted?

(b) Your conjecture: The rational function h (x) = asymptotes when

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f (x) will have horizontal g (x) .

(c) Use your conjecture to predict where the rational function x2 − 25 f (x) = will have a vertical asymptote. Graph the function on x−5 your calculator and see whether or not your conjecture predicted correctly. Specifically, find out what happens to the function when x = 5. This situation occurs when there is a removable discontinuity in the graph of the rational function. [Hint: Try factoring the numerator of this rational function to help see what is happening.] If necessary, modify your conjecture in light of this example.

3. Try to rigorously articulate your findings about horizontal asymptotes, vertical asymptotes, removable discontinuities, and about end behavior of rational functions. More importantly, see if you can explain some rules as to how and when these things occur.

4. Without the use of a graphing calculator, find and sketch the vertical and 4 − x2 horizontal asymptotes of the function f (x) = . [Hint: Do some 2 − x − x2 factoring first.]

For problems 5-10, find the x-intercept(s), the y-intercept, the vertical asymptotes, the horizontal asymptotes, and any removable discontinuities of each function. Also, discuss the end behavior of each function. Use that information to sketch a graph. Do this without a calculator.

5. q (x) =

x−5 3x − 1

6. s (x) =

4 (x − 2)2

7. r (x) =

5 (x + 1)2

8. g (x) =

x2 + 2x − 3 x2 − 1

9. b (x) =

x2 − x − 6 x2 − 4

10. z (x) =

3x2 + 4x − 4 x3 − 4x2

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For problems 11-12, write an equation for a rational function with the given characteristics.

11. Vertical asymptotes at x = 5 and x = −6; x-intercepts at (1, 0) and (−1, 0); y-intercept at (0, 4).

12. Vertical asymptotes at x = −4 and x = −2; x-intercepts at (1, 0) and (5, 0); y-intercept at (0, 9).

For problems 13-17, let f (x) =

x2 − 1 . x2 + 2x − 3

13. Write f (x) in factored form.

14. For what value(s) of x does f (x) have a removable discontinuity?

15. For what value(s) of x is f (x) = 0?

16. For what value(s) of x does f (x) have a vertical asymptote?

17. Sketch a graph of f (x) showing all the features described in problems 13-16 (use a dashed line for any vertical asymptotes).

CHALLENGE: Topics Connections Challenge ax + b where c 6= 0. Find the cx + d condition(s) on the parameters a,b,c, and d such that f = f −1 .

18. Consider the rational function f (x) =

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Rational Functions and Properties To summarize some of the properties of rational functions that you have probably noticed by now, we can state:

1. A removable discontinuity, or “hole in the graph”, may occur in the graph of a rational function if an input causes both the numerator and denominator to be 0 zero. That is, if f (c) = then there is a removable discontinuity or hole in the 0 graph at x = c. c 2. If c 6= 0 and f (c) = then there is a vertical asymptote that occurs in the graph at 0 x = c.

3. If c 6= 0 and f (c) =

0 then there is a root of the function at x = c. Thus f (c) = 0. c

4. If the numerator and denominator of the rational function have the same polynomial highest degree, then there will be a horizontal asymptote at the ratio of the coefficients of each of the highest degree terms.

These are just some of the observable properties of rational functions that you may have noticed. There are more that can be noted, some of which you may have noticed already.

Exercise 9.1.4: Rational Functions, Division, and Another Kind of Asymptote Part 1: 1. Consider the rational function f (x) =

x+2 . x2 − x − 6

(a) What happens at x = 3? Why? (b) What happens at x = −2? Why? (c) What happens as x gets large when x is positive; when it is negative?

Why? 2. Without graphing, explain what the graph of f (x) =

x2 − 2x − 15 will look x2 − 7x + 10

like.

3. Without graphing, explain what the graph of f (x) = will look like.

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(x + 3)(x − 1) (x − 1)(x − 2)(x + 3)

4. Is it possible for the graph of a rational function to intersect its own vertical asymptote? Why or why not? If you think that it is possible, see if you can find such a function.

5. Is it possible for the graph of a rational function to intersect its own horizontal asymptote? Why or why not? If you think that it is possible, see if you can find such a function.

For problems 6-7, write an equation for each function graphed.

6.

7.

Part 2: A Topics Connections Activity Consider the rational function f (x) =

2x2 − 5x + 8 for problems 8-11. x−3

8. Based on what you have learned, describe the end behavior, vertical and horizontal asymptotes (if any), and y-intercept. Then, graph the function using technology to check your description.

9. Next, divide the numerator of the function by the denominator using long division. Then, write the answer as a linear function including the remainder. Make sure to write the remainder as a fraction with the remainder in the numerator and the linear factor (x − 3) in the denominator.

10. As a review, redo the previous problem’s division using synthetic division.

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11. Now, graph the linear function WITHOUT the remainder along with the original rational function. Note that the linear function you have graphed serves as a slant or oblique asymptote. Based on this situation, define the term slant asymptote. x2 − 3x − 4 , what role does the remainder x−5 play in this slant asymptote situation?

12. For the rational function f (x) =

13. Based on what you have learned, describe the end behavior, vertical and horizontal asymptotes (if any), and y-intercept. Then, graph the function using technology to check your description.

14. Next, divide the numerator of the function by the denominator using long or synthetic division. Then, write the answer as a linear function, including the remainder. Make sure to write the remainder as a fraction with the remainder in the numerator and the linear factor (x − 5) in the denominator.

15. Draw a graph of the rational function from problem 12 along with the linear function that defines the slant asymptote.

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9.2

Applications of Rational Functions

In the next Exercise, we explore some applications of rational expressions and functions. Use what you’ve learned in this Unit to explore these applications.

Exercise 9.2.1: A Few Applications of Rational Expressions and Functions 1. The equation that relates the resistance R of two resistors, r1 and r2 , in a parallel electronic circuit is written 1 1 1 = + .

R r1 r2 (a) Solve this equation for r1 . (b) Find r1 when R = 4.50 ohms and r2 = 5 ohms.

2. The formula f=

sg s+v

represents a phenomenon called The Doppler Effect, where f represents the frequency of a sound that is moving at a velocity, v, towards a person who in turn hears the same sound at a frequency g. The speed the sound moves through a medium (such as air) is labeled s. Solve this rational equation for s.

9.3

Radical Expressions, Equations, and Functions (With Applications)

Radical Expressions Before we discuss radical expressions, equations, and functions, a brief review of some rules for working with radical expressions is presented.

Rules for working with radical expressions RULE √ n

√ √ n ab = n a b

r n

√ n a a = √ n b b

Example √ 3

27a6 =

r

√ 3

√ 3 27 a6

√ x8 x8 =√ 144 144 147

Note that the examples are not in simplified form. We shall consider radical expressions simplified when:

1. The radicand (expression under the radical nth root symbol) of an nth root contains no nth powers as factors, 2. The root index is as small a value as possible, and 3. There are no radicals in the denominator. Note that, for simplification purposes, this rule is often neglected in higher courses. (But not this course!)

GOOD ADVICE. Remember that, to clear a denominator of a radical, multiply the expression by a clever form of 1 involving the radical of the denominator.

Another rule related to radical expressions has to do with real numbers raised to rational number powers.

Definition. Rational Exponents. If a is an integer, b is a positive integer, and x is a real number, then √b a x b = xa .

A way to remember this is to think of x raised to a rational exponent to mean power x root .

Since nth roots can be written as exponents with rational number powers, the Laws of Exponents are summarized here along with the rules for rational exponents.

Laws of Exponents Let x and y be non-zero real numbers, where a, b and n are positive integers. xa 3. (xa )b = xab 2. b = xa−b 1. xa · xb = xa+b x a x xa 1 4. (xy)a = xa y a 5. = a 6. x−n = n y y x 7.

1 x−n

= xn

8. x0 = 1, if x 6= 0

√ 1 9. x n = n x

√ a 10. x b = ( b x)a CHALLENGE: Why does x0 = 1? Also, what happens in this case if x = 0?

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Exercise 9.3.1: Working with Radical Expressions For problems 1-15, simplify these expressions using the rules provided in this section.

1.

5.

9.

13.

√

√

288

2.

√ 6

450

6.

√ 4

(x2 )5 (x4 )3 √ 5

64

2 3. √ 5

1 4. √ 3 2

81

2 7. √ 2

5 8. √ 3 16

10. (7y)0

11. 7y 0

12.

1

m30 k 32

p x4 y 16

1

14. x 3

15. 9 2

2

(6x 7 y −4 z 0 )3 16. Simplify: . Express your answer with no variable in the 9x2 y 5 z −8 denominator.

For problems 17-24, simplify the following expressions.

17.

√ 3

√ √ 3 3 108 + 10 32 + 500

10 20. √ 3 25

23.

21. √

1 5−1

√ 19. (2 5 − 3)2

22. √

√ √ 24. (5 + 3 3)(5 − 3 3)

2 √ 5+ 3

25. CHALLENGE: Express

12 √ 18. √ + 6 6

p

√ √ √ 10 + 2 21 in the form x + y .

149

4 √ 7+ 3

For problems 26-31, simplify and express each value without using an exponent.

3

26. −(−5)

27. 64 2

1 29. 163 2

30.

2

28. 64− 3

(312 )3

4

31. 27− 3

(38 )4

For problems 32-36, simplify and express each expression with no variables in the denominator.

32.

35.

10p12 n7 5p3 n5

√ 3

6

m30 k 45

33. 7x−3 · 8x9

36.

34.

x−1 y −4 z −3 x−2 yz −3

32 w−5 x2 z −3 3−1 w−2 x−5 y 4

Radical Equations A radical equation has variables in one or more radicands. Examples of such equations are √ √ 3 7 = 5 + x + 2 and 8x + 2 = 10.

To work with such equations, we must use the Property of Powers, which states: If a = b, then an = bn .

GOOD ADVICE. Notice that the converse (that is, the statement read backwards) of the Property of Powers is NOT true. It is not the case that, if an = bn , then a = b. For example, 22 = (−2)2 but 2 does not equal -2.

Thus, when working with radical equations of even roots, one can generate extraneous solutions, or solutions that do not produce an equivalent relation. This can occur, as we shall see, when radical equations are raised to even powers in the solution-finding process. The following examples provide a process for working with radical equations.

150

EXAMPLE 1: Suppose one wants to solve the radical equation √ x − 5 = 3.

First, the radical is isolated:

√

x = 8.

Then both sides of the equation are squared, yielding x = 64.

Due to the possibility of extraneous solutions, a check of the answer shows that our solution works √ 64 − 5 = 3. True.

EXAMPLE 2: Now, suppose we are given this equation to solve: √ x − 5 = −7.

Isolating the radical and squaring both sides of the resulting equation yields x = 4.

However,

√

4 − 5 6= −7.

Therefore, the generated solution x = 4 is extraneous. In fact, there is no real number solution to this equation.

To summarize, when working with radical equations:

1. Isolate one of the radical terms.

2. Use the Property of Powers to “undo” the radical(s) in the equation. Note: Sometimes, there is more than one radical term in an equation, so you must iterate steps 1 and 2.

3. Check your solutions to eliminate any extraneous solutions generated.

Use the process detailed above to find the solutions to the supplied equations in the following Exercise.

151

Exercise 9.3.2: Radical Equations Solve these equations.

1.

4.

7.

9.

√

√

√

√

2.

2x + 3 = 5

x+

√

7=

√

x+7

x2 + 5x − 6 +

x+

√

11. 5 = 3 +

8= √

√

√

5.

√

√

x−1+x=3

x−7=

x2 + 3x − 3 = 1

√

3. x + 5 =

x−7

8.

√

6.

√

√

x+5+6

x2 − 9 = 4

x2 + 3x + 6 −

√

x2 + 3x − 1 = 1

√ √ 10. 2 x − 4x − 3 = 0

x+8

12. 0 = 3 +

x+2

√

x+2

CHALLENGE: q p √ 13. x = 12 + 12 + 12+...

Radical Functions and Applications Functions involving nth roots are often called radical functions. These functions often arise naturally when finding inverses of polynomial functions.

EXAMPLE: If one desires to find the inverse of f (x) = (x − 2)2 − 3, we first notice, using our knowledge of transformations of functions, that this is a parent quadratic that is transformed 2 units right and 3 units down. Remember, to find an inverse of a function, it must be a one-to-one function to start with.

152

Thus, we restrict the domain to x ≥ 2. Then, the inverse of y = (x − 2)2 − 3 is found by:

1. Adding 3 to both sides of the equation: y + 3 = (x − 2)2 . √ 2. Taking the square root of both sides: ± y + 3 = x − 2.

(Remember, this introduces a ± on the left side.)

3. Adding 2 to both sides: 2 ±

√

y + 3 = x.

We can write this as a radical equation: f −1 (x) = 2 ±

√ x + 3.

Use the next Exercise as a chance to explore radical functions and some of their uses.

Exercise 9.3.3: Radical Functions and Applications √ For problems 1-5, consider the function f (x) = 1 − 3 x + 4.

If x and y are real numbers, find 1. Domain:

2. Range:

3. x-intercept:

4. y-intercept:

5. Sketch a graph of f (x) without using technology.

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For problems 6-9, state the transformations of the parent square root or cube root function present. Then, graph each radical function using technology to check your answer.

6. f (x) = 9 +

√

4x − 4

√ 8. f (x) = 9 + 2 3 x

7. f (x) =

√

6x − 8 + 5

√ 9. f (x) = 3 − 3 x

10. Find the inverse of the function in problem 6.

11. Find the inverse of the function in problem 7.

12. Find the inverse of the function in problem 8.

13. Find the inverse of the function in problem 9.

For problems 14-17, find a domain on which the function is one-to-one and non-decreasing, then find an inverse of the function on this domain.

14. f (x) = (x − 4)2

15. f (x) = (x + 2)2

16. f (x) = 12 − x2

17. f (x) = 9 − x2

CHALLENGE: 18. Find the inverse of the function f (x) = 5x3 + 1, then graph it using technology.

Applications 19. The escape velocity, v, of an orbiting spaceship or satellite can be found using the formula ! r p h v = 2gr , r+h where g is the force of gravity, r is radius of the planet being orbited, and h is the height of the orbiting object above the surface of the planet. (a) Solve this formula for r. (b) Solve this formula for h. (You will use this type of process extensively in the next section.)

154

√ 20. Insurance investigators and police use the radical function v = 20l to estimate the speed of a car, v , in feet per second, based on the length, l, in feet, of its skid marks when abruptly braking on a dry asphalt road. (a) If a car’s skid marks are measured to be 180 feet long, approximately how fast was the car traveling? (b) At the scene of an accident, if a police officer measures a car’s skid marks to be 125 feet long, approximately how fast was the car traveling? (c) Find the inverse of this function. What information does this tell us?

21. A more detailed radical function of velocity, v, in feet per second, of a car that slams on its brakes can be determined based on the length of skid marks that the tires left on the ground is p v(d) = 2gf d.

In this function, g represents acceleration due to gravity (32 f t/sec2 ), d is the length of the skid marks in feet, and f is a constant representing the friction coefficient of the surface. (a) A car lost control on asphalt, with a friction coefficient of 0.5, leaving 200 feet skid marks. How fast was the car traveling when it lost control? (b) Find the inverse of the function v(d). What information does this tell us? s (x + 2)(x − 3) 22. Find the domain of the function f (x) = . (x − 1) [HINT: Since a square root is only defined when the quantity under the (x + 2)(x − 3) ≥ 0]. radical is nonnegative, you must determine where (x − 1) CHALLENGE: √ 23. Given f (x) = 1 − 3 x + 4, if x and y are real numbers, find the domain, range, x-intercept(s), and y-intercept. Then, sketch a graph of f (x) without using technology.

155

9.4

Advanced Literal Equations

You likely have had some previous experience with algebraically manipulating literal equations in a previous course. In fact, you were asked to do this as one of the problems in the last Exercise.

Definition. Literal Equation. A literal equation is one in which most or all constants and variables are represented by letters.

For example, xn + y n = z n or ax + by + cz = 0 are literal equations, whereas 4x + 5 = 19 is not a literal equation. This equation is an example of a numerical equation.

Fortunately, you practiced the skills needed for manipulating literal equations throughout the text. They are the same ones you have always used in order to solve numerical expressions or functions. One can add, subtract, divide, multiply, raise to a power, take an nth root, and factor as needed. Of course, all the real number equality and field properties that you explored in Unit 1 apply as well. The difference is that, instead of solving to find a numerical solution to a situation, the goal of manipulating a literal equation is to solve for one or more of the variables or constants in relation to the others in the equation. The skill to work with literal equations is extremely important to mathematics, the sciences, statistics, economics, and business to name a few disciplines.

Let’s simply dive in and get to work.

Exercise 9.4.1: Advanced Literal Equations Part 1: For problems 1-6, solve for the indicated variable or constant (some of these equations are true statements in math and science and some are contrived).

1. Solve B = P + P rz for z.

2. Solve B = P + P rz for P .

3. Solve Ax + By = z for y.

4. Solve V = pr2 h for h.

5. Solve V = pr2 h for r.

6. Solve A =

156

h(b + c) for h.

2

For problems 7-16, solve for the indicated variable or constant (some of these equations are true statements of mathematics and science and some are contrived). h(b + c) 7. Solve A = for b.

2 9. Solve

a−b 2a + 3b = for b.

7 2

8. Solve 5(bx − 2ab) = b(x − 7a) + 5ab for x.

10. Solve

2a + 3b a−b = for a.

7 2

11. Solve y − y0 = m(x − x0 ) for m. 12. Solve x = 13. Solve A = 2(L + W ) for L.

7 − 2k for k.

9

14. Solve −6(x + y) = 3 + 5x for y.

15. Solve −6(x + y) = 3 + 5x for x. 16. Solve v = Qs + Rt for t.

Part 2: For problems 17-26, solve for the indicated variable or constant. r

θ1 − θ2 17. Solve α = for t2 . t1 − t2

18. Solve T = 2π

19. Solve P V = nrT for V .

20. Solve F =

21. Solve F =

Gm1 m2 for G. d2

1 23. Solve s = at2 + v0 t for a.

2 1 25. Solve T = s

n1 n1 + n2

m for k. k

Gm1 m2 for d. d2

1 22. Solve s = at2 + v0 t for v0 .

2 24. Solve for y.

p p x2 + (y − c)2 = (y + c)2

for n2 . 26. Solve 0 = ax2 + bx + c for c.

157

158

Unit 10 - Exponents, Logs, and e

10.1

e

The Natural Exponential There is one very important number that arises in the development of exponential functions, and that is the “natural” exponential.

The number e is the “natural” exponential because it arises naturally in mathematics and the physical sciences (that is, in real life situations), just as the number π arises naturally in geometry. The real number e is an irrational number. Leonard Euler, who described the number and named it e, discovered this number. There is no evidence, however, that Euler was naming the number after himself.

What is e?

Let’s take a moment to investigate the number e. We do this by considering the situation of an initial amount of money, A0 , invested at an annual interest rate, r, compounded n times per year. It is not hard to show that the value A of the investment after t years is

r nt A(t) = A0 1 + . n n r 1 If x = , then we can replace with , and we can replace the n in the exponent with r n x xr. We also let A0 = 1.

Making these substitutions, we write the revised formula here:

rt 1 xrt 1 x A(t) = 1 + = 1+ . x x

1 x We will now investigate the expression 1 + as an Exercise. x

159

Exercise 10.1.1: In the Long Run

1. Without

the use of technology, predict what happens to the value of 1 x 1+ as x gets large. State your prediction and your justification for it. x 2. Using technology, fill in the following table.

1 x 1+ x

x 1 10 100 1,000 10,000 100,000 1,000,000 10,000,000 100,000,000

(a) Does the sequence of outputs on the right side of the table seem to be approaching a certain value? If so, approximate that value. (b) If you’ve seen this value before, what is it?

10.2

Exponential and Logarithmic Properties and Functions

GOOD ADVICE. While e is considered a special irrational number in mathematics, keep in mind that it is simply another real number that adheres to all the properties of real numbers.

Since e is a real number, it follows all the Laws of Exponents that were previously presented in the last section. The next Exercise provides more opportunity to work with exponents.

160

Exercise 10.2.1: More Exponents Including e For problems 1-12, simplify the following expressions.

1. e8 · e0 · e−6

2. (7x4 y 3 z)2

4. 8−2

5. 10, 000 4

6. 9 2

1 7. (e3 ) 3

8. 184 · 18−4

−2 3 9.

8

1 2

10. 64 ÷ 64

1

1 3

11.

√

2·

√ 3

3. (4xy 2 )0 3

12.

2

e15

5

(e8 )9

For problems 13-16, express each of the following with no variable in the denominator.

5 8 5 2 3 8x y 3 −3 −1 13. 4x −2x y 14.

4x3 y 7

15.

e2 a−7 b3 d−4 , where e is the natural exponential. e0 a−4 b−5 c6

5 7 8 10 x y 16. · 6 y x9

The Exponential Function In your mathematical experience, you will eventually extend your knowledge of exponents to understanding the exponential function. Although there are many kinds of exponential functions that describe a wide variety of real-life situations, the most basic of these is of the form f (x) = a · bx , where a and b are constants.

Generally, we place two restrictions on the value of the constant b. The first is that b > 0 and the second is that b 6= 1. NOTE: See if you can explain why each of these restrictions is necessary.

161

Below is an exponential function of the form f (x) = bx where b > 1.

Below is an exponential function of the form f (x) = b−x where b > 1.

The graph of the case where b is between 0 and 1 will be left to you as an Exercise.

Let’s observe some general properties of the basic exponential function f (x) = bx , where b > 1.

As you trace the graph from left to right, it always goes up. This is an example of an increasing function, which is defined as follows: For all x2 and x1 in the domain of f , if x2 > x1 , then f (x2 ) > f (x1 ). This is just a mathematical way of stating the observation that the graph always has an upward trend.

Similarly, the graph of f (x) = b−x , where b > 1 is an example of a decreasing function. Recall, a decreasing function is such that for all x2 and x1 in the domain of f , if x2 > x1 , then f (x2 ) < f (x1 ).

162

Exercise 10.2.2: The Exponential Function For problems 1-5, complete the following: (a) Graph the function. (b) State the domain of the function. (c) State the range of the function.

1. f (x) = bx , where b is greater than 1 2. f (x) = b−x , where b is greater than 1 3. f (x) = bx , where 0 < b < 1. Also, explain how and why this graph is related to the graph of problem 2 4. f (x) = ex 5. f (x) = e−x

Transformations 6. Use technology to sketch the graph of f (x) = e(x+a) for various values of a and discuss your findings. [Note: Using exponential properties, how else could one write f (x) = e(x+a) ?]

7. Use technology to sketch the graph of f (x) = e(x+a) + k for various values of a and various values of k, and discuss your findings.

163

Logarithms and Logarithmic Functions The logarithmic function is the inverse of the exponential function. That is, the logarithmic function switches the domain and the range of the exponential function in such a way that if baseexponent = answer, then logbase answer = exponent.

For example, if 102 = 100, then log10 100 = 2. Try this yourself: If log5 125 = x, what is x?

Therefore, as a general rule we write y = logb x if and only if by = x, for b > 0, b 6= 1.

NOTE: By convention, if no base is written for a log, it is assumed to be base 10. For example, log10 (x) is usually written simply as log(x).

As a historical note, it was the Scottish mathematician, John Napier, who developed the mathematics of the inverse of the exponential function. It is believed that Napier called this new function the logarithmic function, because to use it was to perform “logical arithmetic” or “logarithmic” for short.

Note that blogb x = x and logb bx = x

(1).

CHALLENGE: Why is this statement true?

Finally, we also note that when b = e, the natural exponential, (1) can be written eln x = x and ln ex = x, where ln(x) is the “natural log.” The symbol ln(x) means loge (x).

164

Logarithm Properties. For a > 0, b > 0, b 6= 1, M > 0, N > 0, and k ∈ R. Use the supplied rules to fill in the “Solution” column of the following table.

Property

Example

logb (M ) + logb (N ) = logb (M N )

M logb (M ) − logb (N ) = logb N

log5 (12)

5 logx 2

logb (M k ) = k logb (M )

2 loga 7

The Change of Base Formula

Write log2 (4) as a logarithm with base e

loga (x) =

logb (x) logb (a)

Solution

The Logarithm properties follow directly from the laws of exponents. Can you prove this?

Exercise 10.2.3: Proofs of the Properties of Logarithms and More For problems 1-5, prove these statements using the definitions of this section.

Hint: Your proof should start: Let x = logb M and y = logb N . Then bx = M and by = N .

For b > 0, b 6= 1, M > 0, N > 0, and k ∈ R: 1. logb (M ) + logb (N ) = logb (M N )

2. logb (M ) − logb (N ) = logb

M N

165

3. logb (M k ) = k logb (M )

4. Given that rD = n, show that D =

5. Show that

log n . log r

loga n logb n = for bases a and b. loga r logb r

[Hint: This can be accomplished by starting with the equation rlogr n = n, applying a logx to both sides of the equation for strategic values of x, and then using some log properties and algebra.]

6.

(a) Use the change of base formula to write log10 100 as a natural logarithm. (b) Use the change of base formula to write log5 125 as a natural logarithm.

For problems 7-13, write the given expressions as a single logarithm of a single argument and simplify.

7. 4 log2 3

8. 4 log8 3 − log8 6

9. log2 225 − log2 5 + log2 3

10. log3 6 + log3 7

11. ln e2 + ln e5

12. log5 54 − 3 log5 3

13. log7 64 − log7 4 + log7 5

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Exercise 10.2.4: More Logarithm Practice Part 1: Solve for x without using technology.

1. log3 x = −4

2. log2 x = 5

3. log 1 x = 8

4. ln e3 = x

5. log3 27 = x

6. log 1

7. ln e = x

8. logx 16 = −4

9. ln x = 1

10. ln x = 2

11. log2 x = −3

12. log5 25 = x

5 2

15. x = 7log7 10

13. ln

1 =x e

14. logx 32 =

2

2

Part 2: Evaluate the following without using technology.

16. log2 4 =

17. log3 27 =

18. log25 25 =

19. log3 9 =

20. log4 16 =

21. log10 1000 =

22. log10 10 =

23. log7 49 =

24. log4 4 =

25. log6 36 =

26. log1 1 =

27. log5 1 =

28. log2 16 =

167

1 =x 16

Exercise 10.2.5: Fun Log Challenge Given real numbers a, b, c > 1, show that the sum of the terms in the array   loga a loga b loga c  logb a logb b logb c  logc a logc b logc c is greater than or equal to 9.

The Logarithmic Function The graph of the logarithm function is the inverse of the exponential function.

Informally speaking, this means that any ordered pair (x, y) for the exponential function will become the ordered pair (y, x) for the logarithmic function. Due to this “inverse action,” the graph of the logarithmic function can be thought of as a reflection of the exponential function about the line y = x. Here is the graph of the basic logarithmic function f (x) = logb x, where b > 1.

168

Exercise 10.2.6: The Logarithmic Function 1. State the domain and range of f (x) = logb x, where b > 1.

2. State the domain and range of f (x) = ln x.

3. Graph the functions y = ln x, y = log10 x, and y = log3 x on the same grid and discuss the similarities and differences between the graphs.

4. Transformations (a) Use technology to sketch the graph of f (x) = log(ax) for various positive values of a on the graph below and discuss your findings.

[Note: Using log properties, how else could one write f (x) = log(ax)?]

169

(b) Use technology to sketch the graph of f (x) = log(x + c) for various positive values of c on the graph below and discuss your findings.

(c) Use technology to sketch the graph of f (x) = log(−x + c) for various positive values of c on the graph below and discuss your findings.

(d) Lastly, use technology to sketch the graph of f (x) = log(x + c) + k for various positive values of k on the graph below. Discuss your findings.

170

CHALLENGE: An Interesting Inverse 5. We know that the function f (x) = ex is a continuous increasing function whose domain is the set (−∞, ∞) and range (0, ∞).

Let L(x) be the inverse function for f (x). (a) What is the domain of L(x)? What is the range of L(x)? What is L(1), L(2), etc.? (b) In terms of composition of functions, what does it mean for L to be the inverse of f ? (c) Use (b) to find L−1 (x). (d) Sketch a graph for L(x).

Using Logarithms to Solve Exponential Equations of the Form f (x) = a · bx Recall these two previously mentioned important facts:

blogb x = x and logb bx = x, and eln x = x and ln ex = x, which are a consequence of the inverse relationship between exponents and logarithms. We can exploit these relationships, sometimes in conjunction with the Logarithm Properties, to devise procedures that allow for the use of logarithms to solve certain exponential equations.

171

EXAMPLE 1: We can use a logarithm to solve the equation

3x = 29.

First notice that an estimate of the answer to this equation should be a number between 3 and 4, which is closer to 3 than 4. Why? We use a logarithm to find the answer.

3x = 29 log10 3x = log10 29 x log10 3 = log10 29 (by log properties) log10 29 ≈ 3.01 using a calculator Thus x = log10 3

GOOD ADVICE. Notice that you could have performed the exact same procedure used to solve 3x = 29 by applying the natural logarithm, ln x, to both sides of the equation or even log3 x. Why might this logarithm be convenient?

NOTE: This text will primarily use base 10 logarithms or natural logarithms to solve exponential equations simply because most graphing calculators or computer algebra systems only have base 10 or natural logarithm calculation capabilities. Of course, the Change of Base Formula allows one to turn any base logarithm into another as well.

EXAMPLE 2: Use the natural logarithm to solve the equation

5(2x ) = 73.

One way to solve this equation is as follows: 5(2x ) = 73 73 2x = 5 73 ln 2x = ln 5 73 x ln 2 = ln 5 73 ln 5 ≈ 3.87. x= ln 2

172

Notice, in the last line of EXAMPLE 2 that Logarithm Properties could have been used to write

x=

ln 73 − ln 5 ≈ 3.87. ln 2

I hope you can see that the more one knows about logarithms and their properties, the more ways one will have to solve exponential equations. Can you think of a different logarithm expansion property that could have been used in EXAMPLE 2 to solve it in a different manner?

Exercise 10.2.7 Using Logarithms to Solve Exponential Equations of the Form f (x) = a · bx For problems 1-8, first estimate. Then, use a logarithm to solve each equation for the given variable. You may use a calculator.

1. 5x = 92

2. et = 17

3. 8s = 258

4. 6(4x ) = 258

5. 9(ex ) = 115

6. 5n = 78

7. 2en = 36

8. 10(6t ) = 3041

CHALLENGE: For problems 9-10, use a logarithm to solve each equation for the given variable. You may use a calculator.

9. 2(3x−2) = 32

10. 4(x−1) = 40

For problems 11-12, use the techniques of this section to solve these word problems. You may use a calculator.

11. Initially, there are 50 bacteria in a lab culture. Since the number of bacteria doubles each hour, the number of bacteria after t hours is given by the equation B = 50(2t ) where B is the number of bacteria present at time t, and t is measured in hours. How long will it take for the lab culture to reach 12,000 bacteria?

12. Starting with 250 grams of a radioactive substance, one can determine the amount S of this substance remaining after t year by using the equation S = 250(0.60t ).

After how many years will 30 grams of the substance remain?

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10.3

Scientific Notation

Scientific Notation You may recall from previous mathematics courses that unusually large or unusually small numbers, such as the ones below, are often written more efficiently in scientific notation.

93,000,000

186,000

0.00000001016

Definition. Scientific Notation. Numbers written in the form b × 10n , where 1 ≤ b < 10 are said to be written in scientific notation. Note that b is a real number and n is an integer.

EXAMPLES When using scientific notation, the digit term indicates the number of significant figures in the number. The exponential term only places the decimal point. As an example, 76, 600, 000 = 7.66 × 107 .

This number only has 3 significant figures. The zeros are not significant; they are only holding a place. As another example, consider: 0.00064 = 6.4 × 10−4 .

This number has 2 significant figures. The zeros are place holders in this case.

Addition/Subtraction example: (8.96 × 104 ) − (3.46 × 103 ) = (8.96 × 104 ) − (0.346 × 104 ) = 8.61 × 104

Multiplication example: (3.4 × 108 )(4.2 × 103 ) = (3.4)(4.2) × 10(8+3) = 14.28 × 1011 = 1.4 × 1012

Division example: (6.4 × 106 )/(8.9 × 102 ) = (6.4)/(8.9) × 10(6−2) = 0.719 × 104 = 7.2 × 103

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Exercise 10.3.1: Scientific Notation Practice 1. Write these numbers in standard form not using scientific notation. (b) 2.5 × 10−3

(a) 7.28 × 106

(c) 320 × 104

2. Why does problem 1(c) need some work? Fix it by putting it into proper scientific notation.

3. Based on what you know about multiplication properties and exponent properties, how would one perform the operations in problems (a) - (e)? (a) (3 × 1017 )(2 × 10−13 )

(b) (6 × 108 )(4 × 10−10 )

(c) (7 × 10−3 )(5 × 109 )

(d)

(e)

8 × 107 2 × 10−2

3 × 1011 6 × 1015

For problems 4-5, express each quantity as a single number in scientific notation.

4. 8, 000 × 600, 000

5.

9 × 10−7 4 × 10−2

6. Estimate and write your answer in scientific notation: 5, 870, 000, 000, 000 × 2000

10.4

Fermi Questions (and Orders of Magnitude)

There are many situations involving the application of scientific notation and estimation, which are quite useful in the sciences for providing a reasonable range of values within which an answer must lie. The order of magnitude associated with these reasonable answers refers to a power of 10 number that fits the value. These order of magnitude approximations can be quite enjoyable to perform. In fact, these types of questions are often used in job interviews to decipher the creative thinking ability of job seekers. These types of questions are often called Fermi questions, named after the famous physicist Enrico Fermi who was known for finding interesting ways of estimating answers to such questions.

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For example, one might ask. “How many ping pong balls fit in a typical airline travel bag?”

Answer - We’ll start by making the reasonable assumption that the typical airline bag is 2800 × 900 × 2400 . Thus, the volume of the bag is approximately equal to 6 × 103 in3 . Also assume that a ping pong ball is approximately 1” in diameter, or about 0.5” for radius.

Recall that the volume V of a sphere of radius r and is found by: 4 V = πr3 .

3 The quantity π is about 3. Thus, the volume of the ping pong ball is approximately 1 in3 (we round up pretty generously since the balls in the bag don’t fit perfectly together).

Now, the volume of our bag divided by the volume of a ping pong ball gives us the number of ping pong balls that should fit. Therefore, we have approximately 6 × 103 , which rounds up to about 104 ping pong balls.

These types of order of magnitude questions are a great way to work on your creative and critical thinking skills. Some knowledge of dimensional analysis is also helpful. Now you are ready to try some on your own or in groups.

Exercise 10.4.1: Estimating the “Size” of Quantities Using Orders of Magnitude Use what you’ve learned about scientific notation and orders of magnitude to estimate the answers to these questions.

1. How many hairs are on a human head?

2. How many restaurants are there in Houston? [Assume there are approximately 2 × 106 people in Houston.]

3. How many cells are there in the human body? [Assume the volume of a given cell is 10−15 m3 .]

4. The average flea is 1.5 mm in length and can jump 18 cm vertically into the air. If humans could proportionally jump as high as fleas can, how high could the average human jump?

5. Make up a “Fermi question” of your own and describe how you would find the answer.

CHALLENGE: Research Fermi questions on the internet and see if you agree with the proposed solutions.

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10.5

Applications of Exponents and Logarithms

In the following Exercise, you are asked to use what you have learned in this section to solve a few application problems using exponential and logarithmic functions. At times, you might have to apply a logarithm to each side of an exponential function or equation to “undo it” and solve for a specified value. Likewise, you might have to exponentiate each side of a logarithmic function or equation to “undo it” and solve for a specified value. By exponentiate, we mean treat both sides of a logarithmic equation as the exponent for a specified exponential base of your choice.

You might ask, “Why does this ‘undoing’ process work in solving for specific values in exponential or logarithmic situations?” Well, since the exponential and logarithmic functions are inverses of each other, we know that one undoes the other due to this fact: f −1 (f (x)) = x and f (f −1 (x)) = x.

Use this inverse relationship to simplify and solve equations in the next Exercise.

Exercise 10.5.1: Exponential and Logarithmic Function Applications

r nt For problems 1-3, use the Amortization Function, A(t) = A0 1 + , where an n initial amount A0 of money is invested at an annual interest rate r compounded n times per year over t years. NOTE: Be sure to write your interest rate r in decimal form, not as a percent.

1. If you invest $2000 at an annual rate of 4.3% interest compounded quarterly for 5 years, what amount will you end with?

2. If interest is compounded monthly, at what annual rate must you invest $5,000 to have $7,500 at the end of 5 years?

3. If you invest $10,000 at an annual rate of 5% compounded quarterly, how long will it take to double your money?

For problems 4-6, we notice that, due to our previous discussion of the value of e, for continuous compounding (that is, if n gets large) the Amortization Function becomes A(t) = A0 ert , where an initial amount of money A0 is invested at an annual interest rate r (written as a decimal). Use this fact to answer problems 4 and 5.

4. If you invest $20,000 at an annual rate of 10% compounded continuously, how long will it take to double your money?

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5. What amount of money must you invest at 5.5% compounded continuously, in order to have $15,000 after 10 years?

6. A population of bacteria is growing according to the equation P (t) = 1600e0.21t , with t measured in years. Estimate when the population will exceed 7569.

For problems 7-11, a person is driving along a straight, level highway at 75 miles per hour (mph) when her car runs out of gas. As she slows down, it is known that her speed decreases exponentially as a function of the number of seconds since she ran out of gas. Her speed decreases to 44 mph after 10 seconds.

7. Start with the general exponential function, f (t) = abt , where a and b are constants. Then write the particular exponential equation that expresses the driver’s speed as a function of time in seconds.

8. To check your answer for problem 7, perform an exponential regression on the limited data given.

9. Draw and label a graph of the situation.

10. What is the person’s speed after 15 seconds?

11. At what time will the person’s speed be approximately 10 mph?

For problems 12-14, a certain radioactive material decays at a rate that is governed by the function M = M0 e−kt , where M is measured in grams and t is measured in years.

12. If M0 = 500 grams, find k if M = 400 grams when t is equal to 1000 years.

13. Find M when t is 1500 years.

14. Draw and label a graph of the situation.

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For problems 15-17, consider the following scenario. In chemistry, the pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral, less than 7 is more acidic, and more than 7 is more basic. The pH of a solution is governed by the function pH = − log[H + ], where [H + ] is the hydrogen ion concentration in moles per liter.

15. The hydrogen ion concentration of milk is 2.0 × 10−7 . Find the pH of milk.

16. If the pH of acid rain is 4.52, find the hydrogen ion concentration of rain in moles per liter.

17. Using the logarithm properties, what are two other ways to write pH = − log[H + ]?

An Exponential Application Exploration A catenary is the curve that an idealized hanging chain or cable assumes under its own weight when supported only at its ends. An equation of the form x

−x a ea + e a f (x) = +k 2 generates the graph of the catenary. The values of the constants a and k in application is determined by the properties of the chain or cable under investigation.

The graph or shape of the catenary is often mistaken for a parabola, but the two are distinctly different. The next Exercise illustrates this fact.

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Exercise 10.5.2: Catenary Exploration Part 1: 1. Precisely graph the catenary function with a = 1 and k = 0.

2. On the same grid precisely graph the quadratic function f (x) = x2 + 1.

3. Describe similarities and differences between the graphs.

4. Use a graphing utility to graph the catenary using various values for a. How do your choices for a affect the graph?

5. Graph the catenary of problem 1 using a graphing utility. Choose 3 points from the graph of the catenary and use them to perform a quadratic regression on those points. Graph the quadratic regression equation and the catenary in the same window of your graphing utility. Do you notice distinct differences? Explain.

6. The famous landmark, The Gateway Arch, in St. Louis, Missouri is often mistaken for a parabola in shape. In fact, it is a catenary. The equation for the Arch is

−x 127.7 x y = 757.7 − e 127.7 + e 127.7 2 where x and y are measured in feet. Graph this equation on the interval [−315, 315] and determine the height of the Gateway Arch.

7. Based on your work in this Exercise thus far, make a conjecture as to how one could introduce vertical and horizontal translations into the general catenary equation. Test your conjecture with a few graphs and then explain your findings.

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8. Research the Gateway Arch and find actual pictures of the structure.

Part 2: A Catenary Activity 9. Hang a link chain or a metal necklace from its endpoints over a large piece of graph grid paper taped to a wall or vertical board as pictured.

10. Choose 3 to 5 points on the chain and use your grid to assign ordered pairs to those points.

11. Do a quadratic regression on your chosen points and generate a quadratic regression equation that fits the points. Carefully graph the quadratic on your coordinate grid along with the hanging chain or necklace. Can you see the distinct differences? Explain.

12. Try to find a value for a in the general catenary equation that will create a best fit catenary for your hanging chain.

13. The next time you are outside, look around for hanging cables and chains.

You will be looking at catenary situations.

CHALLENGE: Research this hyperbolic trigonometric equation and explain the graph generated by the equation x . y = acosh a

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10.6

Function Patterns

Functions Defined by Patterns We start this section with an Exercise. The purpose of this Exercise is to detect patterns in the domain of a given data set that result in patterns in the range, which we can use to identify the type of function that models the data.

Exercise 10.6.1: Functions Defined by Patterns The goal of this Exercise is to try to find a pattern in each domain and related range of a given function in order to identify what kind of model may be present (i.e., exponential, linear, quadratic, etc.). Some of these patterns are harder to detect than others. You might try taking differences (subtracting) between consecutive range values - sometimes more than once. This is called the triangular difference method.

For each table of data: 1. Plot the data points and make a conjecture as to what kind of function is present.

2. Find a pattern in the domain and the range for each function provided and try to understand how this pattern affects the shape of the graph that you have plotted (assuming that the graph is continuous).

3. Make a conjecture as to what kind of function can be used as a model for the given data. (a)

(b) x 2 4 6 8 10

f (x) 4 9 14 19 24

x 6 18 54 162

f (x) 1 2 3 4

(c) x 1 3 5 7 9

f (x) 15 5 19 57 119

(d)

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x 1 3 5 7

f (x) 15 135 1215 10935

Here are the domain - range patterns that you should have identified, which are related to some of the more common functions studied in mathematics courses:

a. Sum - Sum: Linear Function b. Sum - Constant Second Difference: Quadratic Function c. Sum - Product: Exponential Function d. Product - Sum: Logarithmic Function

Of course, at this point your pattern identifications and associated functions are simply conjectures. The general function pattern statements are stated below without verification. Verification will come in a later course (possibly OnRamps Precalculus).

1. Sum - Sum: A Linear Function For a linear function f (x) = mx + b, adding a constant c to a given domain value results in adding a constant to the corresponding range value.

2. Sum - Constant Second Difference: A Quadratic Function For a quadratic function of the form f (x) = ax2 + bx + c with domain values k units apart, the second differences between consecutive f (x) values are constant and equal to 2ak 2 .

3. Sum - Product: An Exponential Function For an exponential function f (x) = abx , adding a constant c to a given domain value results in multiplying the corresponding range value by a constant.

4. Product - Sum: A Logarithmic Function For a logarithmic function f (x) = a + b logn x, multiplying a given domain value by a constant c results in adding a constant to the corresponding range value.

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Exercise 10.6.2: Function Patterns

1. Write the general equation for a linear function.

2. Write the general equation for a quadratic function.

3. Write the general equation for an exponential function.

4. Write the general equation for a logarithm function.

5. Identify the function pattern and function for each of the following function tables. (a)

(b) x −2 −1 0 1 2

f (x) −4 −1 2 5 8

(c)

x −2 −1 0 1 2

(d) x 2 3 4 5

f (x) 11 16 21 26

f (x) 3/4 3/2 3 6 12

(e) x 1 3 5 7

f (x) 3 11 27 51

x 0 2 4

f (x) 5 45 405

6. Given f (2) = 5 and f (6) = 20, use function patterns only to find f (18) if f (x) is (a) A linear function.

(b) An exponential function.

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7. The table below shows the height h (measured in ft) for a model rocket under the influence of gravity only (after 2 seconds of thrust) at different times t (secs). t (sec.)

2 3 4 5 6 7

h (ft.)

166.2 215.7 234.5 219.6 174.0 97.8

(a) Make a scatter plot of the height of the rocket as a function of time.

Record the graph and make a conjecture as to what type of function this data might represent.

(b) Use your “function patterns” in order to justify or question your conjecture above. Note: this is real data, so there is measurement error inherent. Thus this task is a little tougher than previous problems you have encountered using function patterns.

(c) Find a regression equation based on your findings. What is the function equation that you found?

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Unit 11 - Sequences and Series

11.1

Arithmetic and Geometric Sequences

Defining a Sequence We will start this Unit by recalling, from your work in Unit 1, that a sequence is formally defined as a function with a domain of the positive integers, sometimes in union with 0.

The domain values are the term numbers of the sequence, as in “first term,” “second term,” etc. of the sequence. Notice that this implies that sequences are discrete functions with distinct range values as opposed to the continuous functions that you have generally worked with in the past. Sequences can contain a finite or an infinite list of terms. Here is an example of a sequence of numbers that you may be familiar with: 1, 2, 4, 8, 16, 32, . . .

A sequence can be “self-explained” (i.e. explicitly explained) as a function of n, or it can be defined recursively where any next term depends upon a previous term or terms. An example of an explicitly defined sequence is an = 10n + 3, where it is assumed that the domain is the positive integers. On the other hand, an example of a recursive function is a1 = 8 and an = 2an−1 − 3 where, to find the next term of this sequence, one must substitute the given first term into the sequence.

By convention, a sequence can be symbolized in many ways, such as an , a(n), f (n), {an }, or {an }∞ n=1 .

Sequences can be finite or infinite. Lastly, sequences can be characterized as being arithmetic, geometric, or neither.

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Arithmetic Sequences Definition. Arithmetic Sequence. An arithmetic sequence is a sequence in which the difference between any two consecutive terms is constant. This constant is called the common difference and is usually symbolized as d.

Exercise 11.1.1: Working with Arithmetic Sequences

1. If the first term of an arithmetic sequence is 4 and the difference between consecutive terms is 5, what is the second term? The third term? The twelfth? The eightieth? The nth?

2. An explicit, or self-explained, sequence is given by an = 33 + 2n. Is this sequence arithmetic? How can you tell? Graph this explicit sequence using a domain of positive integers, n, from 1 to 10 inclusive.

3. How much information does one need to determine an arithmetic sequence? That is, if a person tells you that she is thinking of an arithmetic sequence, how many, or which, terms must be revealed such that you will be able to create the function that generates the terms of the sequence?

4. Write a recursive formula for the sequence: 7, 17, 27, 37, 47, . . .

5. Write an explicit formula for the sequence: 1, 7, 13, 19, 25, 31, . . .

In order to find a specific term value of an arithmetic sequence, one can use the formula an = a1 + (n − 1)d.

6. Find the 6th term of the arithmetic sequence that has a1 = 12 and a2 = 20.

Find the 50th term of the above sequence.

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Geometric Sequences Definition. Geometric Sequence. . A geometric sequence is a sequence in which the ratio between any two consecutive terms is the same. This common ratio is usually symbolized as r.

An example of a geometric sequence is 2, 6, 18, 54, . . .

What is the common ratio for this sequence?

To find any term of a geometric sequence, one can use the formula an = a1 · r(n−1) .

Exercise 11.1.2: Working With Geometric Sequences

1. If the first term of a geometric sequence is 3 and the common ratio between terms is 2, what is the 2nd term? The 3rd term? The 8th? The 12th? The nth?

2. How much information does one need to determine a geometric sequence?

That is, if a person tells you that she is thinking of a geometric sequence, how many, or which, terms must be revealed such that you will be able to create the function that generates the terms of the sequence?

3. Write an explicit formula for the sequence 1, 2, 4, 8, 16, . . .

4. Find the 7th term of the geometric sequence which has a1 = 2 and a2 = 8.

Find the 20th term of this sequence.

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Exercise 11.1.3: Arithmetic, Geometric, or Neither?

Fill in the table.

Sequence

Arithmetic, Geometric, or Neither?

Find d or r

Find the 20th term

nth Term Formula

−7, −3, 1, . . .

1, 4, 9, . . .

81, 54, 36, . . .

25, 50, 100, . . .

1, 2, 6, 24, . . .

As stated previously, not all sequences are arithmetic or geometric. A famous sequence that is neither arithmetic nor geometric is the Fibonacci Sequence, which can be written as {1, 1, 2, 3, 5, 8, 13, 21, 34, . . . }. You might wish to do some research on the Fibonacci Sequence, for it has a rich history and many connections to other topics in mathematics. For example, most of you have probably heard of Pascal’s Triangle. Can you find the Fibonacci Sequence within Pascal’s Triangle?

Quick Exercise:

1. How is the Fibonacci Sequence generated?

2. Can you create three more sequences that are neither arithmetic nor geometric?

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11.2

Defining a Series

A series is simply the sum of the terms of a sequence. Thus, one can work with series that are arithmetic, geometric, or neither. Our focus is on geometric series. We will start this section by exploring geometric series properties.

Geometric Series - Some Background A geometric series is simply the sum of the terms of a geometric sequence. It looks like: Sn = a1 + a1 · r + a1 · r2 + a1 · r3 + · · · + a1 · rn−1 , where r is the common ratio of the geometric series and a1 is a coefficient. The sum of a finite geometric series can be found by: Sn =

a1 (1 − rn ) .

1−r

Exercise 11.2.1: Derive the Formula for the Sum of a Finite Geometric Series The sum of a finite geometric series can be found by Sn =

a1 (1 − rn ) .

1−r

1. Derive the formula for the sum of a finite geometric series: (a) Start with the general geometric series.

Sn = a1 + a1 · r + a1 · r2 + a1 · r3 + · · · + a1 · rn−1 (b) Multiply both sides of this equation by r. (c) Subtract the result of (b) from (a) and then do some algebraic rearranging.

2. Find S10 for the geometric series with a1 = 5 and r = 3.

At this point, you should be ready for some application problems.

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Exercise 11.2.2: Applications of Geometric Series

1. For a certain drug to have the desired healing effect and be safe, the amount in a person’s system should be between 60 and 70 micrograms.

After 6 hours, the amount of a dosage of the drug left in the body is 70% of the original dosage. Each dosage of the drug delivers 20 micrograms to the body. (a) Assume a person receives one dosage of the drug every six hours.

Write a geometric series representing the amount of the drug in a person’s system after n dosages. (b) When (if ever) would the amount of the drug in a person’s system be greater than 60 micrograms?

CHALLENGE: Will the amount of the drug in a person’s system ever be greater than 70 micrograms? Explain your reasoning.

2. A ball is dropped from a height of 30 feet. After hitting the ground, it rebounds to a height of 75% of the previous distance it fell. (a) On the axes below, create a graph that shows the path of the ball where h the height of the ball is a function of the bounce number of the ball starting with bounce 0 (the height from which the ball was dropped). Mark the points that correspond to the height of the ball at each of the first 10 bounces after the initial drop height.

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(b) Fill in the table below corresponding to the height of the ball as a function of bounce number.

Bounce 0 1 2 3 4 5 6 7 8 9 10

Height

(c) Use function patterns to decide what type of function this is. Justify your claim. (d) Now find the particular function that represents your data. Although your data is discrete (i.e. not continuous), we will treat our function as a continuous function. (e) Now check your answer to problem 2(d) by conducting a regression on the data based on your function pattern decision. (f) Lastly, use your sequence of bounce heights to create a series that determines the total distance the ball travels UP AND DOWN from the initial drop to the end of the 10th bounce. Assume the ball actually bounces exactly straight up and down with each bounce. What type of series is this?

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11.3

Probability and Combinatorics

A Momentary Diversion Toward the Binomial Theorem In this lesson, we will allow ourselves a momentary diversion into the area of probability and, specifically, the field of combinatorics in mathematics. We introduce useful counting techniques that are pertinent to probability theory. Our ultimate goal, however, is to develop the theory needed to explore our ultimate topic of interest, the binomial theorem. The binomial theorem will be explored at length in the next lesson.

At this point in your mathematical careers, it is quite likely that you have had some experience with calculating the probability that certain equiprobable outcomes occur within finite sample spaces. The probability of such an event occurring consists of the number of events constituting the event divided by the total number of outcomes in the sample space. While this counting method may sound fairly straightforward, it is often the case that actually calculating the way certain outcomes can occur can be quite an unruly process.

In order to perform these types of calculations in probability, we appeal to the field of combinatorics.

Definition. Combinatorics. A study of different possibilities for arrangements of objects is known as combinatorics.

Essential to such a study is the rule known as the counting principle.

Definition. The Counting Principle. The counting principle states that if one event can be chosen in m different ways, and another event can be chosen in n different ways, then the two events can be chosen in m · n different ways.

The counting principle can be extended to any number of independent events.

We focus on an extension of the counting principle known as a combination where n objects are chosen without repetition.

Definition. Combination. A combination is an arrangement of objects without repetition where order is not important. The number of combinations of n objects chosen r at a time is symbolized as:

C(n, r) =

n! . (n − r)! r !

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Note: The symbol n! = n(n − 1)(n − 2) . . . (2)(1). This is known as a factorial. Also note that 0! = 1 and 1! = 1. Lastly, we point out that the expression C(n, r) can also be denoted:

n C(n, r) = . r

Exercise 11.3.1: Counting Part 1: 1. Suppose one conducts an experiment where a die is rolled and then a coin is flipped. Each outcome pair is recorded. For example, one possible outcome pair is 2H where 2 is the number that is displayed after rolling a die and H stands for the coin flip that resulted in a “heads”. Use the counting principle to find the number of outcome pairs that are possible for this experiment. A tree diagram may be used to help visualize your result.

2. A pizza shop offers 3 different kinds of meat and 5 different kinds of vegetables as possible cheese pizza toppings. How many different choices does a customer have if she wants 1 meat topping and 1 vegetable topping on her cheese pizza? A tree diagram may be used to help visualize your result.

3. An automobile license plate usually has 3 letters of the alphabet on it followed by 4 numbers, each number belongs to the set {0, 1, 2, 3, . . . , 9}.

These letters or numbers can be repeated on the plate. Use this information to write an equation that will find N , the total number of different license plates possible.

Part 2: Use what you’ve learned about combinations to answer the following questions.

4. From a list of 10 different songs, how many 6-song playlists can be created, assuming song order doesn’t matter?

5. How many different groups of 5 basketball players can be selected from a team of 10 players?

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Definition. Permutation. A permutation consists of different arrangements of objects where the order of the objects is important.

For example, if one wanted to list the number of possibilities for choosing a president, vice president, and secretary from a group of individuals, one would use permutations.

The number of permutations of n objects taken n at a time is P (n, n) = n!.

The number of permutations of n objects taken r at a time is P (n, r) =

n! . (n − r)!

Exercise 11.3.2: Counting Continued

1. A Track and Field relay team consists of 4 team members lined up in order around a track in the 4 × 100 meter relay. How many different line-ups are possible if the 4-man relay team is chosen from among 6 possible team members?

2. Show that C(n, r) =

P (n, r) . r!

3. How many permutations of the 3 letters in MAD can be made?

4. How many combinations of the 3 letters in MAD can be made?

5. How many 3-letter permutations can be made from the 4 letters ABCD ?

6. How many 3-letter combinations can be made from the 4 letters ABCD ?

7. In how many different ways can you form a committee of 5 people from a group of 9 people?

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11.4

The Binomial Theorem

Consider the expansion of the binomial powers, (x + y)n , as listed in the figure.

Figure 11.1: A Sample of Binomial Expansions for n = 0, 1, 2, 3, 4, 5.

Based on previous experience, you might have already noticed that the binomial coefficients of the expansions above are the same as the array known as Pascal’s Triangle. Pascal’s triangular array contains many interesting properties and patterns, some of which will be explored in this lesson. The first few rows of Pascal’s Triangle are listed in the next figure.

Figure 11.2: Pascal’s Triangle for Rows 0, 1, 2, 3, 4, 5.

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In fact, the binomial coefficients are simply an array of combinations of the form

n . k This is also true of the Pascal’s Triangle array as displayed in the figure:

Figure 11.3: Pascal’s Triangle - An Array of Combinations

We define the Binomial Theorem: Definition. The Binomial Theorem. For all positive integers n and x, y belonging to the real numbers:

n n 0 n n−1 1 n n−2 2 n n 0 n 1 n−1 (x + y) = x y + x y + x y + ··· + x y + x y 0 1 2 n−1 n n X n n−k k = x y . k n

k=0

The last line of the series formula above was written in summation notation. A series Sn can be symbolized using summation notation: Sn =

n X

an .

k=0

This is read: “The series Sn that is the sum of the first n terms of the sequence an .”

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The questions is, “Why does the binomial expansion of (x + y)n for various powers of n lead to an array of combinations?” An informal justification of this fact starts with noticing that (x + y)n = (x + y)(x + y)(x + y) . . . (x + y) . | {z } n times

It is also worth noting that each term of the distributed expression of (x + y)n has the form xr y s where r and s are integers greater than or equal to zero, such that r + s = n.

One of x or y is chosen for each of the n factors of (x + y). Thus, each term x(n−k) y k in the binomial expansion is obtained by choosing y from any k of the n factors. n Therefore, there are ways to select k items from the entire set of n items, which k explains the reason for the binomial coefficients in the binomial expansion.

Exercise 11.4.1: Binomial Expansion 1. What is the coefficient of the term x10 y 5 in the expansion of (x + y)15 ?

2. Find the fifth term of the expansion of (a + b)11 .

3. Expand (5x + y)6 using the Binomial Theorem.

4. Expand (3 + i)5 , where i is the imaginary number, using the Binomial Theorem.

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Conclusion It is my sincere hope that the Exercises contained within this text have enticed you to think deeply about some of the mathematics you’ve encountered previously, about new ideas presented, and about the connections between the two. It is also my hope that the inquiry-based “open forum” methodology used to deliver this course has broadened your perspective and enjoyment of the course topics and content contained within.

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