Mathematics B30
Module 2
Lesson 7
Mathematics B30
Polynomial Functions
127
Lesson 7
Mathematics B30
128
Lesson 7
Polynomial Functions
Introduction
The terms, polynomial and function are familiar terms in mathematics. This lesson
defines these terms together and discusses the properties that are associated with
polynomial functions.
Once the characteristics of a given polynomial function are determined, it is possible to
use these characteristics to analyze the polynomial function and then to sketch the graph
of the function on the coordinate plane.
Technology gives us an opportunity to analyze these polynomial functions by a different
method. It is very important to realize that you must understand the process of analyzing
these polynomials before using a graphing calculator.
The last section shows the use of a graphing calculator in analyzing polynomial functions.
It is not necessary to use a graphing calculator, but it is strongly recommended. The
process without the calculator is taught in the first three sections and this is enough to
complete any evaluation.
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Lesson 7
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Lesson 7
Objectives
After completing this lesson you will be able to
Mathematics B30
•
define and illustrate polynomial functions.
•
sketch the graphs of polynomial functions with integral
coefficients, using calculators or computers.
•
analyze the characteristics of the graphs of polynomial functions.
•
identify the 'zeros' of these graphs.
131
Lesson 7
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Lesson 7
7.1 Review of Polynomials
A polynomial in a single variable x is an expression in the form
a n x n  a n 1 x n 1  a n 2 x n 2  ...  a1 x 1  a 0 ,
where:
•
a n , a n 1 , a n 2 ... a0 are real numbers called coefficients,
•
the exponents n, n  1 , ... are positive integers or zero,
•
the coefficient a n is called the leading coefficient,
•
a 0 is the constant term.
Some examples of polynomials are:
•
4 x3  8 x2  9 x  3
•
 x4  6 x2  7
4 5
1
x  2 x 4  12 x 3  5 x 2  x 
•
5
2
A familiar process is to determine the factors or the zeros of the polynomial. This can be
accomplished by using the factor theorem or the remainder theorem. Both of these
concepts were studied in Mathematics A30 and a review of this material is provided.
A factor of a polynomial is any term that divides evenly into the polynomial without a
remainder.
The zeros of a polynomial are any numbers that make a polynomial equal to zero.
If ab  0 , then either a = 0, b = 0 or both a and b are equal to zero.
To determine the factors of any polynomial, find a value x such that f  x   0 .
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Lesson 7
Factor Theorem
•
•
If f a   0 , then x  a  is a factor.
If x  a  is a factor, then f a   0 .
Example 1
Is x + 5 a factor of x 3  4 x 2  x  20 ?
Solution:
•
•
In the factor x + 5, a  5 .
Determine if f  5   0 .
f  x   x 3  4 x 2  x  20
Write the polynomial function.
f  5    5   4  5    5   20
f  5   125  100  5  20
f  5   0
Evaluate.
3
2
Yes, x + 5 is a factor of x 3  4 x 2  x  20 , because f  5   0 .
If the factor has a positive sign in it, the value that makes the polynomial
equal to zero will be negative.
•
If x + 5 is a factor, then f  5   0
The factor theorem is usually used when factoring polynomials with a degree of at least
three. Once the factor theorem has determined one of the factors of the polynomial, that
factor can then be divided into the original polynomial to determine the other factor(s).
Example 2
Factor the polynomial, x 3  7 x  6 and state the zeros of the polynomial.
Solution:
Read the problem.
Factor the polynomial, x 3  7 x  6 .
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Lesson 7
Determine a plan.
•
Find a factor of the expression by using the factor theorem and evaluating the
polynomial for f(1), f  1 , f(2), etc. until f(x) = 0.
•
Divide this factor into x 3  7 x  6 to determine the other factors.
Carry out the plan.
f x   x 3  7 x  6
Write the expression.
Evaluate f(1).
•
f 1   1 3  7 1   6
f 1  0
Therefore x  1 is a factor of x 3  7 x  6 because f(1) = 0.
Divide.
x2  x  6
x 1 x3  0 x2 7 x  6
Write 0x2 to take the place of
the missing x2 term.
x3  x2
x2 7 x
x2  x
6x 6
6x 6
0
•


Two factors of x 3  7 x  6 are x  1 and x 2  x  6 .
Now factor x 2  x  6 .
x2  x  6
  x  3  x  2 
Write a concluding statement.
•
•
•
Using the factor theorem, x  1 is one factor.
From the quotient, x 2  x  6 , (x + 3), and  x  2  are also factors.
In factored form, x 3  7 x  6   x  1  x  3  x  2  .
•
The zeros are:
Mathematics B30
x  1 x  3
135
x2
Lesson 7
Synthetic Division
Synthetic Division is a simpler method of dividing a polynomial by another polynomial
that is in the form x  k where the coefficient of x is 1.
The problem is set up with only the coefficients of the dividend, and the divisor is the k
term from x  k .
All the terms that are missing in the dividend must be represented with a zero.
To divide ax 3  bx 2  cx  d by x  k , use the following pattern.
k a
b
c
d
ka
a
aa
•
•
•
r
Remainder
Vertical Pattern: Add terms.
Diagonal Pattern: Multiply by k.
The pattern for higher degree polynomials is similar.
Example 3
Use synthetic division to divide x 3  8 x 2  4 x  48 by x  4 .
Solution:
•
From x  4 , k = 4.
•
The coefficients of the dividend are:
x3
1

8x2
8
Mathematics B30

4x
4

48
48
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Lesson 7
Follow the pattern.
4
1
1
8
4
4
 16
4
 12
48
 48
0
Coefficients of Quotient
•
Zero Remainder
The quotient is 1 x 2  4 x  12
Note that the first term of the quotient is one degree less
than the first term of the dividend.
A written explanation of the steps for synthetic division for this question is:
•
•
•
•
•
•
•
•
•
•
Bring the 1 down and write it in the first column underneath the line.
Multiply 4 × 1 to get 4. Put this 4 underneath the  8 in the 2nd column.
Add  8 and 4. Put the answer,  4 , in the same column underneath the line.
Multiply 4 ×  12 to get  48 . Put this  48 underneath the 4 in the 3rd column.
Add 4 and  16 . Put the answer,  12 , in the same column underneath the line.
Multiply 4 ×  12 to get  48 . Put this  48 underneath the 48 in the 4th column.
Add 48 and  48 . Put the answer, 0, in the same column underneath the line.
The first three terms underneath the line are the three coefficients for the terms
in the answer.
There is no remainder because the last term is zero.
The quotient is 1 x 2  4 x  12 .
•
x 3  8 x 2  4 x  48 divided by x  4 is equal to 1 x 2  4 x  12 .
•
Therefore,
x 3  8 x 2  4 x  48
 x 2  4 x  12
x 4
The Remainder Theorem is used to find the remainder when a polynomial f(x) is divided
by another polynomial in the form x  a .
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Lesson 7
Remainder Theorem
If a polynomial, f(x), is divided by x  a ,
the remainder will be f(a).
Example 4
Find the remainder when f  x   x 2  6 x  5 is divided by x  2 .
Solution:
•
•
In the factor x  2 , a = 2.
It is then necessary to find f(2).
Write the expression.
f x   x 2  6 x  5
Evaluate f(2).
f 2   2   6 2   5
2
 4  12  5
 3
Evaluate.
This is the remainder.
Example 5
Find the remainder when f  x   x 3  2 x 2  6 is divided by x  3 .
Solution:
•
•
In the factor x + 3, a  3 .
It is then necessary to find f  3  .
f x   x 3  2 x 2  6
Write the expression.
Evaluate f  3  .
f  3    3   2  3   6
3
Evaluate.
2
 27  18  6
 39
Remainder
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Lesson 7
Exercise 7.1
1.
Determine if the binomial is a factor of the polynomial.
a.
b.
c.
d.
e.
f.
g.
2.
f  x   x 3  4 x 2  7 x  11; x  5
f  x   x 3  5 x 2  7 x  9; x  1
f  x   3 x 3  11 x 2  12 x  4; x  2
f  x   x 4  x 3  13 x 2  6 x  8; x  4
f  x   x 5  4 x 4  4 x 3  x 2  x  9; x  2
f  x   x 2  1; x  2
f  x   x 2  3 x  2; x  4
f  x   x 3  2 x 2  5 x  6; x  2
f  x   3 x 3  5 x 2  2 x  3; 3  x
f  x   2 x 3  3 x 2  4 x; x  2
f  x   x 4  5 x 3  6; x  1
Simplify using synthetic division.
a.
b.
c.
d.
e.
f.
g.
4.
f  x   3 x 2  8 x  6; x  3
Find the remainder when the polynomial is divided by the binomial.
a.
b.
c.
d.
e.
f.
3.
f  x   x 2  10 x  24 ; x  6
2 x  13 x  15   x  5 
x  5 x  7 x  3   x  1
x  4 x  6 x  12   x  2 
x  3 x  2   x  1
x  7 x  8   x  1
x  16   x  2 
12 x  17 x  8 x  40   x  2 
2
3
2
3
2
3
3
2
4
4
3
Factor each of the following polynomials.
a.
b.
c.
x3  2x2  x  2
x 3  3 x 2  10 x  24
x 4  2 x 3  7 x 2  8 x  12
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Lesson 7
7.2 Polynomial Functions
A polynomial in a single variable x is an expression in the form
a n x n  a n 1 x n 1  a n 2 x n 2  ...  a1 x 1  a 0 ,
where:
•
a n , a n 1 , a n 2 ... a 0 are real numbers called coefficients,
•
the exponents n, n  1 ,... are positive integers or zero,
•
the coefficient a n is called the leading coefficient,
•
a 0 is the constant term.
The degree of a polynomial is the highest exponent of a variable in the polynomial. In the
above definition, the degree of the polynomial would be n, since n is the highest exponent.
If a function f is a polynomial in one variable, then f is a polynomial function.
Not all polynomials are real polynomial functions.
Example 1
Which of the following expressions are real polynomial functions?
3
a.
y  x3  x 4  9
b.
y  6 x 2  5 x
c.
y  7x  3x
d.
y
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1 5
4
x   2 x  25
3
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Lesson 7
Solution:
None of the expressions are real polynomial functions because:
a.
3
4
y  x  x  9
3
There is a rational exponent,
be positive or zero.
b.
y  6 x 2  5 x
c.
y  7x  3x
d.
y
3
, and the exponents must
4
There is a negative exponent,  1 , and the exponents
must be positive or zero.
3 x does not have an integral exponent.
1 5
x   2 x 4  25
3
The coefficients must be real, and the coefficient
 2 is
not real.
There are many steps in analyzing a polynomial function. Determining the degree of the
polynomial and the leading coefficient is the first step.
Example 2
Find the coefficients, constant term, degree and leading coefficient of the polynomial
function, f  x   9 x 4  3 x 3  x 2  7 x  8 .
Solution:
Coefficients
Constant term
Degree
Leading Coefficient
a 4  9 ; a 3  3 ; a 2  1 ; a1  7 ; a 0  8
a 0  8
The degree is 4 since the highest exponent of any variable is 4 in
the term 9 x 4 .
The leading coefficient is 9 since a 4  9 .
The next step is to identify the zeros and the multiplicity of each factor.
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Lesson 7
Consider the quadratic function f  x   x 2  9 x  20 . The roots of this function can be found
by setting the function equal to zero and factoring the quadratic equation.
x 2  9 x  20  0
 x  4  x  5   0
x4 0
x 5  0
x5
x4
The zeros of the function f  x   x 2  9 x  20 are 4 and 5. This means that f 4   0 and
f 5   0 . Since both these zeros occur only once, each has a multiplicity of one. If a zero
occurs twice, it has a multiplicity of two. If it occurs three times, it has a multiplicity of
three, etc.
Example 3
Find the zeros of the function f  x   x 2  6 x  9 and identify their multiplicity.
Solution:
x2  6x  9  0
x  3 2  0
 x  3 x  3   0
x=3
x=3
Zeros
The zero of the function f  x   x 2  6 x  9 is 3.
Multiplicity
Since this zero occurs twice it has a multiplicity of 2.
Example 4
Find the zeros of the function g  x   x 2  49 and identify their multiplicity.
Solution:
x 2  49  0
2
x  49
x    49
x  7 i
Zeros
The zeros of the function g  x   x 2  49 are 7 i,  7 i .
Multiplicity
Since each of the zeros occurs only once, each has a multiplicity of 1.
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Lesson 7
Example 5
Find the zeros of the function h  x   x 3  x  3  x  2 2 x 2  5  and identify their
multiplicity.
Solution:

h  x   x 3  x  3  x  2 2 x 2  5

Set each factor to zero.
x 3 0
x  3
x3  0
x 0
x2 0
x2
x2  5  0
x 2  5
x   5
x  5
The zeros of the function h  x   x 3  x  3  x  2  x 2  5  are x  0 ,  3, 2,  5 i
Zeros
2
Determine the multiplicity of each factor.
•
•
•
•
x  0 has a multiplicity of 3 because the factor is cubed.
x  3 has a multiplicity of 1.
x  2 has a multiplicity of 2 because the factor is squared.
x  5 i has a multiplicity of 1, and  5 i has a multiplicity of 1.
Exercise 7.2
1.
State the degree and leading coefficient of the following polynomials.
a.
b.
c.
d.
2.
3x2  2x
5  4 x4  x3
7x  5x2  3x4  4
x6  3x4  2x2
For each of the following functions identify its zeros and indicate their multiplicity.
a.
b.
c.
d.
ax   x  2 x  3 x  5 
b x    x  4   x  1 
2
d x   x 2 x  5  x  4 
3

f x   xx  3  x 2  9
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3

143
Lesson 7
3.
Create a function f x  written in factored form which has the following zeros:
(Assume each zero has a multiplicity of one unless otherwise stated.)
a.
b.
c.
d.
4.
1,  3 and 4
8 with a multiplicity of one and 1 with a multiplicity of two.
5, 2 and  i .
0 with a multiplicity of two,  2 with a multiplicity of three and  4 i .
Find the zeros of each of the following functions. If the zero has a multiplicity other
than one, state its multiplicity.
a.
f  x   x 3  x 2  12 x
b.
c.
f  x   x 3  4 x 2  9 x  36
d.
e.
f x   x 3  3 x 2  6 x  8
f  x   2 x 3  5 x 2  9 x  18
f x   x 4  2 x 3  3 x 2  4 x  4
7.3 Analyzing Polynomial Functions
In the first section of this lesson it was shown that when you are given a polynomial
function, the following characteristics can be determined:
•
Leading coefficient
•
Zeros
•
Degree
•
Multiplicity
It is now possible to analyze these characteristics and use them to sketch the graph of the
polynomial function.
The following activity is an investigation to determine how the degree of a polynomial
function and the leading coefficient of the polynomial function affect where a graph begins
and ends.
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Lesson 7
Activity 7.3
y
This activity is to be handed in with
Assignment 4.
II
I
• Complete the tables provided in this activity
to organize the information that is collected.
Use a table of values or a graphing calculator.
• Answer the questions in each part and make
III
IV
a conclusion as to how the leading coefficient
and degree of a polynomial function affect the
graph of the function.
• The graph of a function is read from left to right.
• The quadrants in a coordinate plane have been identified in the diagram.
Function
Degree
Leading
Coefficient
Begins in
Quadrant
x
Ends in
Quadrant
y  2x 7
y  x  2
y  3x2  4
y  2 x 2  8
y  x3  4 x2
y  2 x 3  6 x 2  7
y  x4  3x2
y  2 x 4  5 x  3
y  4 x5  6 x3  7 x
y  x 5  3 x 4
y  2x6 7 x3  9
y  3 x 6  x 4  5 x 2
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Lesson 7
Use the following key strokes on the TI-83 Graphing Calculator for the
first function y  2 x  7 .
CLEAR Y= 2 X,T,  ,n + 7 ENTER GRAPH
Use the following key strokes on the TI-83 Graphing Calculator for the
last function y  3 x 6  x 4  5 x 2 .
CLEAR Y= (-) 3 X,T,  ,n ^ 6 + X,T,  ,n ^ 4 + 5 X,T,  ,n
^ 2 ENTER GRAPH
•
Additional help, if necessary, is given in the graphing calculator
appendix.
Analysis
1. List all the functions that begin in quadrant three and end in quadrant one.
Function
Degree
Leading
Coefficient
a)
How are the degrees of these functions related?
b)
How are the leading coefficients of these functions related?
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Lesson 7
2. List all the functions that begin in quadrant two and end in quadrant four.
Function
Degree
Leading
Coefficient
a)
How are the degrees of these functions related?
b)
How are the leading coefficients of these functions related?
3. List all the functions that begin in quadrant two and end in quadrant one.
Function
Degree
Leading
Coefficient
a)
How are the degrees of these functions related?
b)
How are the leading coefficients of these functions related?
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Lesson 7
4. List all the functions that begin in quadrant three and end in quadrant four.
Function
Degree
Leading
Coefficient
a)
How are the degrees of these functions related?
b)
How are the leading coefficients of these functions related?
Conclusion
5. From the information gathered, complete the table for any polynomial
function, showing where the graph begins and where it ends.
Degree
Leading
Coefficient
Begins in
Quadrant
Ends in
Quadrant
3
1
3
4
2
4
2
1
6. Given the function y  2 x 8  9 x 5  5 x 4  7 , state the quadrant where this
function will begin. State the quadrant where this function will end.

For a discussion of the reason for this conclusion go to p.167,
Lesson 7, Summary.
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Lesson 7
It can also be noted that the degree indicates the number of times the graph of the
function changes direction.
Degree of Polynomial
Function
1
Possible Graph
Number of Direction
Changes
y
1
x
2
y
2
x
3
y
3
x
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Lesson 7
y
4
4
x
5
5
y
x
One other factor that affects the shape of the graph of a polynomial function is the
multiplicity.
Multiplicity is the number of times each factor occurs in a polynomial function.
Follow what happens as the multiplicity of a simple polynomial function, y  x  2 
repeatedly increases by one.
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Lesson 7
Polynomial Function
Graph
y
y  x  2 
x
y
y   x  2 2
x
y
y   x  2 3
x
y
y   x  2 4
x
y   x  2 5
Mathematics B30
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Lesson 7
The pattern continues and the following conclusions can be made about the multiplicity of
a polynomial function:
•
•
•
A factor with a multiplicity of one passes through the x-axis at the point where the
factor is equal to zero.
A factor with an even multiplicity changes direction at the point on the x-axis
where the factor is equal to zero. It does not cross the x-axis.
A factor with an odd multiplicity greater than 1 crosses the x-axis at the point
where the factor is equal to zero, but as a distinct curve.
Example 1
Analyze and sketch the graph of f  x   x 3  4 x 2  3 x  18 .
Solution:
Factor the polynomial to determine the roots of the equation.
•
•
•
Is f(1) equal to zero?
Is f(2) equal to zero?
Is f(3) equal to zero?
NO
NO
YES
Therefore x  3 is a factor of f  x   x 3  4 x 2  3 x  18 .
Divide x  3 into f  x   x 3  4 x 2  3 x  18 using synthetic division or long division, if you
prefer.


•
x 3  4 x 2  3 x  18   x  3  x 2  x  6
•
x 2  x  6 = x  3 x  2 
In factored form, f  x   x 3  4 x 2  3 x  18
f x   x  3 x  3 x  2 
•
The zeros or x-intercepts are x  3 , x  2 .
Mathematics B30
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Lesson 7
Determine the leading coefficient and degree of the polynomial function.
Degree is 3.
f  x   x 3  4 x 2  3 x  18
Leading
Coefficient is 1
•
•
Positive:
Odd:
Leading Coefficien t 

Degree

The graph therefore begins in Quadrant
3 and ends in Quadrant 1.
Determine the multiplicity of each factor.
•
•
The factor x  2 has a multiplicity of one as it occurs only once.
The factor x  3 has a multiplicity of two as it occurs twice.
•
The graph of the function will cross the x-axis at x  2 and change direction at the
point x  3 .
Sketch the graph.
y
x
One more piece of information that would help in sketching the graph of a polynomial
function is the y-intercept.
In Example 1, to find the y-intercept, substitute x  0 into the original equation and then
solve for f(x).
f  x   x 3  4 x 2  3 x  18
f 0   18
The y-intercept is (0 , 18 ) .
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Lesson 7
Example 2
Analyze and sketch the graph of f  x    x 4  5 x 2  4 .
Solution:
Factor the polynomial to determine the roots of the equation.
•
Is f(1) equal to zero?
YES
Therefore x  1 is a factor of f  x    x 4  5 x 2  4 .
Divide x  1 into f  x    x 4  5 x 2  4 using synthetic division or long division, if you
prefer.
1
–1
0
–1
–1 –1
5 0
–1 4
4 4
–4
4
0
C
oefficients
ofQ
uotient
•

 x 4  5 x 2  4  x  1  x 3  x 2  4 x  4

Factor  x 3  x 2  4 x  4 in the same way by determining a factor and using synthetic
division or long division.
•
•
Is f(1) equal to zero?
Is f(2) equal to zero?
NO
YES
Therefore x  2 is a factor of  x 3  x 2  4 x  4 .
2
–1 –1 4 4
–2 –6 –4
–1 –3 –2 0
Coefficients
of Quotient
•

 x 3  x 2  4 x  4  x  2   x 2  3 x  2
Mathematics B30

154
Lesson 7
Mathematics B30
155
Lesson 7
Write the polynomial in factored form.

 x 4  5 x  4   x  1  x  2   x 2  3 x  2

 x  5 x  4   x  1  x  2  x  1  x  2 
4
 x 4  5 x  4  1 x  1  x  2  x  1  x  2 
The zeros or x-intercepts are x  2 , x  2 , x  1 , x  1 .
Determine the leading coefficient and degree of the polynomial function.
Degree is 4
f x    x 4  5 x 2  4
Leading
Coefficient is  1
•
•
Negative:
Even:
Leading Coefficien t 

Degree

The graph begins in Quadrant 3 and ends in
Quadrant 4.
Determine the multiplicity of each factor.
•
Each factor has a multiplicity of one as each occurs only once.
•
The graph of the function will cross the x-axis at x  2 , x  1 , x  1 , x  2 .
Determine the y-intercept.
y  x4  5 x  4
y  0 4  5 0   4
y  4
The y-intercept is 0,  4  .
Mathematics B30
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Lesson 7
Sketch the graph.
y
x
Example 3
Analyze and sketch the graph of f  x   x 3  3 x 2  3 x  4 .
Solution:
Factor the polynomial to determine the roots of the equation.
•
•
Is f(1) equal to zero?
Is f  4  equal to zero?
NO
YES
Therefore x  4 is a factor.
Divide x  4 into f  x   x 3  3 x 2  3 x  4 .


•
x 3  3 x 2  3 x  4  x  4  x 2  x  1
•
When factoring x 2  x  1 , it is found that there are two complex roots, x 
1i 3
.
2
Write the function in factored form.

x 3  3 x 2  3 x  4  x  4  x 2  x  1

The zero or x-intercept is x  4 .
Mathematics B30
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Lesson 7
Determine the leading coefficient and degree of the polynomial function.
Degree is 3

x 3  3 x 2  3 x  4  x  4  x 2  x  1

Leading
Coefficient is 1
•
•
Positive:
Odd:
Leading Coefficient 

Degree

The graph therefore beings in
Quadrant 3 and ends in Quadrant 1.
Determine the multiplicity of each factor.
•
Each factor has a multiplicity of one as each occurs only once.
•
The graph of the function will cross the x-axis only at x   4 because the other two
roots are complex.
Determine the y-intercept.
y  x3  3x2  3x  4
y  0 3  3 0 2  3 0   4
y4
The y-intercept is 0, 4 .
Sketch the graph.
y
x
The direction of the graph will change three times because the degree of the
function is equal to three.
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Lesson 7
Exercise 7.3
1.
List the major characteristics of each of the graphs representing polynomial
functions.
a.
y
x
b.
y
x
c.
y
x
Mathematics B30
159
Lesson 7
d.
y
x
2.
Given the following functions, analyze the major characteristics and then sketch the
graph of the function.
The major characteristics are:
•
degree
•
leading coefficient
•
factors, zeros
•
multiplicity
•
y-intercept
a.
b.
c.
d.
e.
f x   x 3  2 x 2  9
f x    x 3  4 x 2  x  6
f x   x 3  x 2  x  1
f  x    x  2   x  1  x  5 
2
f  x   2 x 4  x 3  17 x 2  16 x  12
7.4 Graphing Polynomial Functions Using a
Graphing Calculator
A graphing calculator can also be used to analyze a polynomial function. It must be
stressed that it is important to first understand the concepts that were explained in the
first three sections of this lesson before using a graphing calculator.
An explanation of how to graph a polynomial was given in Section 7.1. A review of this
material can also be found in the appendix at the end of the course.
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Lesson 7
From the graph of a function two properties can be determined:
•
the zeros of the function and
•
the y-intercept.
A sketch of the graph can then be made by using these two properties and by transferring
the shape of the graph.
Example 1
Analyze the polynomial function f  x   x 3  5 x 2  6 x using a graphing calculator.
Solution:
Use the following key stroke pattern on the TI-83 Graphing Calculator:
CLEAR Y= X,T,  ,n ^ 3 - 5 X,T,  ,n ^ 2 + 6
X,T,  ,n ENTER GRAPH
Determine the zeros of the function.
•
•
Access the CALC function key by pressing 2nd TRACE.
Arrow down on the menu to select 2:zero. Press ENTER.
•
Use the left-right () arrow keys to set the boundaries around
an x-intercept.
Select any value to the left of an x-intercept and press ENTER.
Select any value to the right of the same x-intercept and press
ENTER. Press ENTER a third time to display the exact zero
value or x-intercept.
Two arrows will appear on the screen. These arrows must point
towards each other. ()
•
•
•
•
Repeat this procedure for each x-intercept.
The x-intercepts or zeros of the function are x  0 , x  2 , x  3 .
Mathematics B30
161
Lesson 7
Determine the y-intercept.
•
•
•
Access the CALC function key by pressing 2nd TRACE.
Arrow down on the menu to select 1:value. Press ENTER.
Press 0 ENTER. (This shows the value of y when x  0 .
The y-intercept is y  0 or (0, 0).
Sketch the graph.
y
x
Example 2
Analyze the polynomial function f  x    x 3  x 2  6 using a graphing calculator.
Solution:
Use the following key stroke pattern on the TI-83 Graphing Calculator:
CLEAR Y= – X,T,  ,n ^ 3 + X,T,  ,n ^ 2 - 6 ENTER GRAPH
Mathematics B30
162
Lesson 7
Determine the zeros of the function.
•
•
Access the CALC function key by pressing 2nd TRACE.
Arrow down on the menu to select 2:zero. Press ENTER.
•
Use the left-right () arrow keys to set the boundaries
around
an x-intercept.
Select any value to the left of an x-intercept and press ENTER.
Select any value to the right of the x-intercept and press
ENTER. Press ENTER a third time to display the exact zero
value or x-intercept.
Display: x = 1.537656 y = 0
•
•
•
The x-intercept or zero of the function is x  1 .54 .
•
The other roots must be complex as the graph changes direction but the graph
doesn't cross the x-axis again.
Determine the y-intercept.
•Access the CALC function key by pressing 2nd TRACE.
•Arrow down on the menu to select 1:value. Press ENTER.
•Press 0 ENTER. (This shows the value of y when x  0 .)
The y-intercept is y  6 or 0 ,  6  .
Sketch the graph.
y
x
Mathematics B30
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Lesson 7
Activity 7.4
The following activity is to be handed in with Assignment 4.
Using your graphing calculator, find a possible equation for each of the following
graphs.
Helpful Hints:
• Use the information gained from the entire lesson.
• The WINDOW on your graphing calculator may need to be adjusted to
accommodate some equations.
• You may need to use trial and error!
y
y
x
x
Please indicate on this sheet if you do not have access to a graphing calculator.
Your assignment will have another question for you to do in place of this
activity.
Mathematics B30
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Lesson 7
Mathematics B30
165
Lesson 7
Exercise 7.4
1.
Analyze the following polynomial functions using a graphing calculator or
computer. Then sketch the function. There may be cases where the x-intercepts
are not exact.
a.
f x   x 3  2 x 2  2 x  1
b.
f  x    x 3  4 x 2  3 x  18
c.
f x   x 3  4 x 2  7 x  6
d.
f x    x 3  4
e.
f  x   2 x 2  5 x  2
f.
f x   x 4  3 x 2  2 x  1
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Lesson 7
Self Evaluation
Analyze and sketch the graph of each of the following polynomial functions.
Use all the steps in your analysis.
a.
f x    x 2  6 x  5
b.
f x   x 3  7 x  6
c.
f  x   x 3  5 x 2  2 x  24
d.
f  x   x 4  x 3  26 x 2  24 x
e.
f  x    x 4  3 x 3  7 x 2  15 x  18
Mathematics B30
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Lesson 7
Summary – Lesson 7
•
Create a summary of this lesson to assist you come examination time.
•
Each summary is to be sent in with the assignment to be evaluated.
•
Items to include in a summary:
•
definitions
•
formulas
•
calculator “shortcuts”
Reason for Conclusion (p. 148)
The leading term of a Polynomial Function determines where the graph begins and
ends.
For example, y  x 3 7 x  6 has a leading term x 3 . For any large positive value of x the
x 3 term is positive and contributes the most to the value of y ; it dominates the other
terms. Therefore, the y value will be positive and the plotted points will be in
Quadrant 1. For large negative values of x , x 3 is negative and consequently y will be
negative. The plotted points will be in the third Quadrant.
This curve starts in Quadrant 3 and ends in Quadrant 1.
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Lesson 7
Mathematics B30
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Lesson 7
Answers to Exercises
Exercise 7.1 1.
a.
b.
c.
d.
e.
f.
g.
yes
no
no
no
yes
yes
no
2.
a.
b.
c.
d.
e.
f.
3
26
4
 45
 12
 10
3.
a.
5

f 6   0
f  3   3
f 5   1
f  1  6
f 2   0
f 4   0
f 2   7
2
 13
10
15
 15
2
3
0

 2 x 2  13 x  15   x  5   2 x  3 
4.
Mathematics B30
b.
c.
d.
e.
f.
g.
x2  4 x  3
x2  6x  6
x2  x  2
x2  8x  8
x3  2x2  4 x  8
12 x 3  7 x 2  14 x  20
a.
b.
c.
x  2 x  1x  1
x  2 x  4 x  3 
x  1x  2 x  3 x  2 
170
Lesson 7
Exercise 7.2 1.
a.
b.
c.
d.
Degree:
Degree:
Degree:
Degree:
2.
a.
b.
c.
 5 ,  2 and 3 each with a multiplicity of one.
 4 with a multiplicity of two and  1 with a multiplicity of one.
0 with a multiplicity of two, 5 with a multiplicity of three and
 4 with a multiplicity of one.
0, 3i,  3 i each with a multiplicity of one and  3 with a
multiplicity of three.
d.
3.
a.
b.
c.
4.
3
4
3
1
f x   x  1x  3 x  4 
f  x    x  8  x  1 
2

f  x    x  5  x  2  x 2  1



f  x   x 2  x  2  x 2  16
a.
b.
c.
 3 , 0, 4
 4 ,  1, 2
4, 3i,  3 i
3
 2, , 3
2
 1 and 2 each with a multiplicity of two.
e.
Mathematics B30
Leading Coefficient:
Leading Coefficient:
Leading Coefficient:
Leading Coefficient:
d.
d.
Exercise 7.3 1.
2
4
4
6
3
a.
Degree: 3
Leading Coefficient: Positive
x-intercepts: x  4 ,  1, 3
y-intercepts: y  3
Multiplicity: each factor has a multiplicity of 1.
b.
Degree: 4
Leading coefficient: Positive
x-intercepts: x  5 ,  2 , 2, 5
y-intercepts: y  4
Multiplicity: each factor has a multiplicity of 1.
171
Lesson 7
2.
c.
Degree: 3
Leading coefficient: Positive
x-intercepts: x  4 , 0
y-intercept: y  0
Multiplicity: the factor x  4  has a multiplicity of 1
the factor x  0 has a multiplicity of 2
d.
Degree: 5
Leading coefficient: Positive
x-intercepts: x  4 ,  1, 1, 4 , 7
y-intercepts: y  2
Multiplicity: all the factors have a multiplicity of 1.
a.
Degree: 3
Leading Coefficient: Positive
Factors: f  x    x  3  x 2  x  3  two complex roots
Multiplicity: All factors have a multiplicity of 1.
y-intercept: y  9


y
x
b.
Mathematics B30
Degree: 3
Leading Coefficient: Negative
Factors: f x   x  3 x  2 x  1
Multiplicity: All factors have a multiplicity of 1.
y-intercept: y  6
172
Lesson 7
y
x
Mathematics B30
173
Lesson 7
c.
Degree: 3
Leading Coefficient: Positive
Factors: f x   x  1x  1x  1
Multiplicity:
 x  1 2  multiplicity of 2
x  1  multiplicity of 1
y-intercept: y  1
y
x
d.
Degree: 4
Leading Coefficient: Negative
2
Factors: f  x    x  2   x  1  x  5 
Multiplicity:
 x  2 2  Multiplicity of 2
x  1 and x  5  have a multiplicity of 1.
y-intercept: y  20
y
x
Mathematics B30
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Lesson 7
e.
Degree: 4
Leading Coefficient: Positive
2
Factors: f  x    x  2  2 x  1  x  3 
Multiplicity:
 x  2 2  multiplicity of 2
2 x  1 & x  3  have a multiplicity of 1
y-intercept: y  12
y
x
Exercise 7.4 1.
a.
x-intercepts: x  1, 0 .38 , 2 .6
y-intercept: y  1
y
x
b.
x-intercepts: x  3, 2
y-intercept: y  18
y
x
Mathematics B30
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Lesson 7
c.
x-intercepts: x  2 (2 complex roots)
y-intercept: y  6
y
x
d.
x-intercepts: x  1 .6 (2 complex roots)
y-intercept: y  4
y
x
e.
1
,2
2
y-intercept: y  2
x-intercepts: x 
y
x
Mathematics B30
176
Lesson 7
f.
x-intercepts: x  2 .1, 1 .4 (2 complex roots)
y-intercept: y  1
y
x
Mathematics B30
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Lesson 7
Answers to Self Evaluation
a.
f x    x 2  6 x  5
y
Zeros: x = 1, 5
Leading Coefficient: Negative
Degree: Even
y-interceipt: y  5
b.
x
f x   x 3  7 x  6
y
Zeros: x =  3 , 1, 2
Leading coefficient: Positive
Degree: Odd
y-intercept: y  6
c.
x
f  x   x 3  5 x 2  2 x  24
y
Zeros: x =  2 , 3, 4
Leading Coefficient: Positive
Degree: Odd
y-intercept: y = 24
Mathematics B30
x
178
Lesson 7
d.
y
f  x   x 4  x 3  26 x 2  24 x
Zeros: x =  6 , 0, 1, 4
Leading Coefficient: Positive
Degree: Even
y-intercept: y = 0
e.
x
f  x    x 4  3 x 3  7 x 2  15 x  18
y
Zeros: x =  3 , 1, 2
Multiplicity: x  3 has a multiplicity
of 2
Leading Coefficient: Negative
Degree: Even
y-intercept: y  18
Mathematics B30
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x
Lesson 7
Mathematics B30
180
Lesson 7