Download Chapter 3 Section 2 - Canton Local Schools

Chapter 3
Polynomial and
Rational
Functions
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All rights reserved
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1
SECTION 3.2
Polynomial Functions
OBJECTIVES
1
2
3
4
5
Learn properties of the graphs of polynomial
functions.
Determine the end behavior of polynomial
functions.
Find the zeros of a polynomial function by
factoring.
Identify the relationship between degrees, real
zeros, and turning points.
Graph polynomial functions.
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2
Definitions
A polynomial function of degree n is a
function of the form
f x   an x  an1 x
n
n1
 ...  a2 x  a1 x  a0 ,
2
where n is a nonnegative integer and the
coefficients an, an–1, …, a2, a1, a0 are real
numbers with an ≠ 0.
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3
Definitions
The term anxn is called the leading term.
The number an is called the leading
coefficient, and a0 is the constant term.
A constant function f (x) = a, (a ≠ 0) which
may be written as f (x) = ax0, is a polynomial
of degree 0.
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Definitions
The zero function f (x) = 0 has no degree
assigned to it.
Polynomials of degree 3, 4, and 5 are called
cubic, quartic, and quintic polynomials.
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5
COMMON PROPERTIES OF
POLYNOMIAL FUNCTIONS
1. The domain of a polynomial function is the
set of all real numbers.
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2. The graph of a polynomial function is a
continuous curve.
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3. The graph of a polynomial function is a
smooth curve.
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EXAMPLE 1
Polynomial Functions
State which functions are polynomial functions. For
each polynomial function, find its degree, the leading
term, and the leading coefficient.
a. f (x) = 5x4 – 2x + 7
b. g(x) = 7x2 – x + 1, 1  x  5
Solution
a. f (x) is a polynomial function. Its degree is 4, the
leading term is 5x4, and the leading coefficient is 5.
b. g(x) is not a polynomial function because its domain
is not (–, )
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POWER FUNCTION
A function of the form
f x   ax
n
is called a power function of degree n,
where a is a nonzero real number and n is a
positive integer.
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POWER FUNCTIONS OF EVEN DEGREE
Let f x   ax . If n is even, then x   x n .
n
n
Then
f x   a x   ax  f x .
n
n
The graph is symmetric with respect to the y-axis.
The graph of y = xn (n is even) is similar to the
graph of y = x2.
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POWER FUNCTIONS OF ODD DEGREE
Let f x   ax . If n is odd, then x   x n .
n
n
Then
f x   a x   ax n   f x .
n
The graph is symmetric with respect to the origin.
The graph of y = xn (n is odd) is similar to the
graph of y = x3.
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 1
n Even
a>0
The graph
rises to the left
and right,
similar to
y = x2.
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 2
n Even
a<0
The graph
falls to the left
and right,
similar to
y = –x2.
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 3
n Odd
a>0
The graph
rises to the
right and falls
to the left,
similar to
y = x3.
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 4
n Odd
a<0
The graph
rises to the left
and falls to the
right, similar
to y = –x3.
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EXAMPLE 2
Understanding the End Behavior of a
Polynomial Function
3
2
P
x

2x

5x
 7x  11 be a polynomial
Let  
3
P
x

2
x
function of degree 3. Show that  
when |x| is very large.
Solution
5 7 11 

P x   x  2   2  3 

x x
x 
5 7
11
When |x| is very large , 2 and 3 are
x x
x
close to 0.
3
Therefore, P x   x 2  0  0  0   2x .
3
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3
17
THE LEADING-TERM TEST
n
n1
f
x

a
x

a
x
 ...  a1 x  a0
Let   n
n1
be a polynomial function.
Its leading term is anxn.
an  0
The behavior of the graph of f as x → ∞ or as
x → –∞ is similar to one of the following four
graphs and is described as shown in each case.
The middle portion of each graph, indicated by
the dashed lines, is not determined by this test.
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THE LEADING-TERM TEST
Case 1
n Even
an > 0
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THE LEADING-TERM TEST
Case 2
n Even
an < 0
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THE LEADING-TERM TEST
Case 3
n Odd
an > 0
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THE LEADING-TERM TEST
Case 4
n Odd
an < 0
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EXAMPLE 3
Using the Leading-Term Test
Use the leading-term test to determine the end
behavior of the graph of
y  f x   2x  3x  4.
3
2
Solution
Here n = 3 (odd) and an = –2 < 0. Thus, Case 4
applies. The graph of f (x) rises to the left and
falls to the right. This behavior is described as
y
r∞
as x
r –∞
and y
r –∞
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as x
r ∞.
23
REAL ZEROS OF POLYNOMIAL FUNCTIONS
If f is a polynomial function and c is a real
number, then the following statements are
equivalent.
1. c is a zero of f .
2. c is a solution (or root) of the equation f
(x) = 0.
3. c is an x-intercept of the graph of f . The
point (c, 0) is on the graph of f .
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EXAMPLE 4
Finding the Zeros of a Polynomial Function
Find all zeros of each polynomial function.
3
2
a. f  x   x  2 x  x  2
b. g  x   x 3  2 x 2  x  2
Solution
Factor f (x) and then solve f (x) = 0.
a. f  x   x 3  2 x 2  x  2
 x 2  x  2   1 x  2 
  x  2   x  1
2
  x  2  x  1 x  1
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EXAMPLE 4
Finding the Zeros of a Polynomial Function
Solution continued
 x  2  x  1 x  1  0
x  2  0 or x  1  0 or x  1  0
x  2 or x  1 or x  1
The zeros of f(x) are –2, –1, and 1.
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EXAMPLE 4
Finding the Zeros of a Polynomial Function
Solution continued
b. f  x   x 3  2 x 2  x  2
 x 2  x  2   1 x  2 
  x  2   x  1
2
0   x  2   x  1
2
x  2  0 or x  1  0
x2
The only zero of g(x) is 2, since x2 + 1 > 0 for
all real numbers x.
2
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AN INTERMEDIATE VALUE THEOREM
Let a and b be two numbers such that a < b. If
f is a polynomial function such that f (a) and
f(b) have opposite signs, then there is at least
one number c, with a < c < b, for which
f(c) = 0.
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AN INTERMEDIATE VALUE THEOREM
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AN INTERMEDIATE VALUE THEOREM
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EXAMPLE 5
Using the Intermediate Value Theorem
Show that the function f (x) = –2x3 + 4x + 5 has
a real zero between 1 and 2.
Solution
f (x) = –2x3 + 4x + 5
3
f 1  2 1  4 1  5  7
f  2   2  2   4  2   5  3
3
f(1) and f(2) have opposite signs, so, by the
Intermediate Value Theorem, f(x) has a real
zero between 1 and 2.
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REAL ZEROS OF POLYNOMIAL FUNCTIONS
A polynomial function of degree n with real
coefficients has, at most, n real zeros.
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EXAMPLE 6
Finding the Number of Real Zeros
Find the number of distinct real zeros of the
following polynomial functions of degree 3.
a. f  x    x  1 x  2  x  3
b. g  x    x  1  x  1 c. h  x    x  3  x  1
2
2
Solution
a. Solve f(x) = 0.
f  x    x  1 x  2  x  3  0
x  1  0 or x  2  0 or x  3  0
x  1 or x  2 or x  3
f(x) has three real zeros: 1, –2, and 3.
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EXAMPLE 6
Finding the Number of Real Zeros
Solution continued
b. g  x    x  1  x 2  1  0
x  1  0 or x 2  1  0
x  1
x 2 +1 has no real solutions
g(x) has only one real zero: –1.
c. h  x    x  3  x  1
2
x  3  0 or x  1  0
x  3 or x  1
h(x) has two distinct real zeros: –1 and 3.
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MULTIPLICITY OF A ZERO
If c is a zero of a polynomial function f (x)
and the corresponding factor (x – c) occurs
exactly m times when f (x) is factored, then c
is called a zero of multiplicity m.
1. If m is odd, the graph of f crosses the
x-axis at x = c.
2. If m is even, the graph of f touches but
does not cross the x-axis at x = c.
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MULTIPLICITY OF A ZERO
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MULTIPLICITY OF A ZERO
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EXAMPLE 7
Finding the Zeros and Their Multiplicity
Find the zeros of the polynomial function
f (x) = x2(x + 1)(x – 2), and give the multiplicity
of each zero.
Solution
f (x) is already in factored form.
f (x) = x2(x + 1)(x – 2) = 0
x2 = 0, or x + 1 = 0, or x – 2 = 0
x = 0 or
x = –1 or
x=2
The zero x = 0 has multiplicity 2, while each of
the zeros –1 and 2 have multiplicity 1.
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TURNING POINTS
A local (or relative) maximum value of f is
higher than any nearby point on the graph.
A local (or relative) minimum value of f is lower
than any nearby point on the graph.
The graph points corresponding to the local
maximum and local minimum values are called
turning points. At each turning point the graph
changes from increasing to decreasing or vice
versa.
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TURNING POINTS
The graph of f
has turning points
at (–1, 12) and at
(2, –15).
f x   2x 3  3x 2  12x  5
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NUMBER OF TURNING POINTS
If f (x) is a polynomial of degree n, then the
graph of f has, at most, (n – 1) turning points.
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EXAMPLE 8
Finding the Number of Turning Points
Use a graphing calculator and the window
–10  x  10; –30  y  30 to find the number of
turning points of the graph of each polynomial.
a. f  x   x 4  7 x 2  18
b. g  x   x 3  x 2  12 x
c. h  x   x 3  3 x 2  3 x  1
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EXAMPLE 8
Solution
Finding the Number of Turning Points
a. f x   x  7x  18
4
2
f has three total turning points; two local
minimum and one local maximum.
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EXAMPLE 8
Finding the Number of Turning Points
Solution continued
b. g x   x  x  12x
3
2
g has two total turning points; one local
maximum and one local minimum.
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EXAMPLE 8
Finding the Number of Turning Points
Solution continued
c. h x   x  3x  3x  1
3
2
h has no turning points, it is increasing on the
interval (–∞, ∞).
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GRAPHING A POLYNOMIAL FUNCTION
Step 1 Determine the end behavior. Apply
the leading-term test.
Step 2 Find the zeros of the polynomial
function. Set f (x) = 0 and solve. The
zeros give the x-intercepts.
Step 3 Find the y-intercept by computing
f (0).
Step 4 Use symmetry to check whether the
function is odd, even, or neither.
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Step 5 Determine the value of f(x) at any
number less than the smallest zero or
greater than the largest zero as a
starting point for sketching the graph.
Step 6 Draw the graph. Use the multiplicities
of each zero to decide whether the
graph crosses the x-axis.
Use the fact that the number of turning
points is less than the degree of the
polynomial to check whether the graph
is drawn correctly.
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EXAMPLE 9
Graphing a Polynomial Function
Sketch the graph of f  x    x3  4 x 2  4 x  16.
Solution
Step 1 Determine end
behavior.
Degree = 3
Leading coefficient = –1
End behavior shown in
sketch.
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EXAMPLE 9
Graphing a Polynomial Function
Solution continued
Step 2 Find the zeros by setting f (x) = 0.
Each zero has
multiplicity 1, the
graph crosses the
x-axis at each
zero.
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EXAMPLE 9
Graphing a Polynomial Function
Solution continued
Step 3 Find the y-intercept by computing f (0).
The y-intercept is f (0) = 16. The graph passes
through (0, 16).
Step 4 Use symmetry.
There is no symmetry in the y-axis nor with
respect to the origin.
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EXAMPLE 9
Graphing a Polynomial Function
Solution continued
Step 5 Determine the value of f (x). The
three zeros of f are –4, –2, and 2. We
choose –5 as a convenient number
less than the smallest zero of f. Then
we find
f(–5) = –(–5)3 – 4(–5)2 + 4(–5) + 16
= 21.
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EXAMPLE 9
Graphing a Polynomial Function
Solution continued
Step 6 Draw the
graph.
The number of
turning points is 2,
which is less than
3, the degree of f.
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EXAMPLE 10 Volume of a Wine Barrel
Express the volume V of the wine barrel as a
function of x.
height
2z z

  2.
Assume that
diameter 2r r
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53
EXAMPLE 10 Volume of a Wine Barrel
Solution
We have the following relationships:
2
V  2 r z
z
 2
r
2
2
x
z
2
2
2
x  4r  z

 4 2  426
2
r
r
2
2
x
x
x
2
6
 r 
 r
2
r
6
6
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54
EXAMPLE 10 Volume of a Wine Barrel
Solution continued
V  2 r 2 z
x
z
r
and
 2  z  2r
r
6
2
x
x
x
x
2
z 2

, so z 
, r 
6
6
3
3
2
V  2 r z
 x2   x 
 3
3
V  2   

x  0.6046 x

 6  3  3 3
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