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Chapter 2
Polynomials
and Rational
Functions
Copyright © 2014, 2010 Pearson Education, Inc.
SECTION 1.1
Section 2.1
1.
2.
3.
Quadratic Functions
Graph a quadratic function in standard form.
Graph any quadratic function.
Solve problems modeled by quadratic functions.
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QUADRATIC FUNCTION
A function of the form
f x   ax  bx  c,
2
where a, b, and c, are real numbers with a ≠ 0, is called a
quadratic function.
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THE STANDARD FORM OF
A QUADRATIC FUNCTION
The quadratic function
f x   a x  h   k ,
2
a0
is in standard form. The graph of f is a parabola with
vertex (h, k). The parabola is symmetric with respect to
the line x = h, called the axis of the parabola. If a > 0,
the parabola opens up, and if a < 0, the parabola
opens down.
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Example: Writing the Equation of a
Quadratic Function
Find the standard form of the quadratic function
whose graph has vertex (–3, 4) and passes through
the point (–4, 7).
Let y = f (x) be the quadratic function.
y  a  x  h  k
7  a  –4  3  4
y  a  x   3   4
a3
Hence,
2
2
y  a  x  3  4
2
2
y  3  x  3  4
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5
Example: Graphing a Quadratic Function
in Standard Form
Sketch the graph of f (x) = Sketch the graph of
2
a(x – h)2 + k
f  x   3 x  2   12.
Step 1 The graph is a
1. The graph of
parabola because it has the
f(x) = –3(x + 2)2 + 12
form f(x) = a(x – h)2 + k
Identify a, h, and k.
is the parabola: a = –3,
h= – 2, and k = 12.
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Example: Graphing a Quadratic Function
in Standard Form
Step 2 Determine how the 2. Since a = –3, a < 0, the
parabola opens. If a > 0,
parabola opens down.
the parabola opens up. If a
< 0, the parabola opens
down.
Step 3 Find the vertex.
The vertex is (h, k). If a > 0
(or a < 0), the function f
has a minimum (or a
maximum) value k at x = h.
3. The vertex (h,k) = (–2,
12). Since the parabola
opens down, the function f
has a maximum value of 12
at x = –2.
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Example: Graphing a Quadratic Function
in Standard Form
Step 4 Find the xintercepts (if any). Set f
(x) = 0 and solving the
equation a(x – h)2 + k = 0
for x. If the solutions are
real numbers, they are the
x-intercepts. If not, the
parabola lies above the xaxis (when a > 0) or below
the x-axis (when a < 0).
0  3  x  2   12
2
4.
12  3  x  2 
4   x  2
2
2
x  2  2
x  0 or x  4
x-intercepts: 0 and  4
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Example: Graphing a Quadratic Function
in Standard Form
Step 5
5.
Find the y-intercept.
Replace x with 0. Then
f (0) = ah2 + k is the
y-intercept.
f  0   3  0  2   12
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 3  4   12  0
y -intercept is 0 .
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Example: Graphing a Quadratic Function
in Standard Form
6. axis: x = –2
Step 6
Sketch the graph. Plot the
points found in Steps 3–5
and join them by a
parabola. Show the axis
x = h of the parabola by
drawing a dashed line.
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Example: Graphing a Quadratic Function
Graph f(x) = ax2 + bx + c,
a ≠ 0.
Sketch the graph of
f  x   2 x  8 x  10.
2
Step 1 Identify a, b, and c. 1. In the equation
y = f(x) = 2x2 + 8x – 10,
a = 2, b = 8, and c = – 10.
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Example: Graphing a Quadratic Function
Step 2 Determine how
2. Since a = 2 > 0, the
the parabola opens. If a > parabola opens up.
0, the parabola opens up.
If a < 0, the parabola
opens down.
3.
b
8
h

 2
Step 3 Find the vertex.
2a
2  2
b
h
k  f  2 
2a
2

2

2
   8  2   10  18
 b 
k  f 

 h, k    2, 18 
 2a 
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Example: Graphing a Quadratic Function
Step 4 Find the xintercepts (if any). Set f
(x) = 0 and solving the
equation a(x – h)2 + k = 0
for x. If the solutions are
real numbers, they are the
x-intercepts. If not, the
parabola lies above the xaxis (when a > 0) or below
the x-axis (when a < 0).
4.
2 x 2  8 x  10  0
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2  x2  4 x  5  0
2  x  5  x  1  0
x  5  0 or x  1  0
x  5 x 1
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Example: Graphing a Quadratic Function
Step 5
5. Let x = 0.
Find the y-intercept.
Let x = 0. The result,
f (0) = c, is the y-intercept.
f  0   2  0   8  0   10
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y -intercept is  10 .
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Example: Graphing a Quadratic Function
Step 6
6.
The parabola is symmetric
with respect to its axis,
Axis of symmetry is x = –2.
The symmetric image of
(0, –10) with respect to the
axis x = –2 is (–4, –10).
b
x .
2a
Use this symmetry to find
additional points.
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Example: Graphing a Quadratic Function
Step 7
Draw a parabola through
the points found in Steps
b
3–6.
x .
7. Sketch the parabola
passing through the points
found in Steps 3–6.
2a
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Example: Identifying the Characteristics
of a Quadratic Function from Its Graph
The graph of f (x) = –2x2 +8x – 5 is shown.
a. Find the domain and range of f.
b. Solve the inequality –2x2 +8x – 5 > 0.
a. The domain is (–∞, ∞).
The range is (–∞, 3].
b. The graph is above the xaxis in the interval
4 6 4 6 
,

.
2 
 2
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Section 2.2
SECTION 1.1
1.
2.
3.
4.
5.
Polynomial Functions
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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Definitions
A polynomial function of degree n is a function of the
form
f x   an x n  an1 x n1  ...  a2 x 2  a1 x  a0 ,
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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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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Definitions
The expression
an x  an1 x
n
n1
 ...  a2 x  a1 x  a0
2
is a polynomial, the function f defined by
f x   an x n  an1 x n1  ...  a2 x 2  a1 x  a0
is a polynomial function, and the equation
an x  an1 x
n
n1
 ...  a2 x  a1 x  a0  0
2
is a polynomial equation.
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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: 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
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
n
Then
x 
n
x .
n
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 = x 2.
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POWER FUNCTIONS OF ODD DEGREE
Let f x   ax . If n is odd, then
n
Then
x 
n
 x .
n
f x   a x   ax   f x .
n
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
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
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
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
a<0
The graph rises to
the left and falls to
the right, similar to
y = –x3.
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Example: Understanding the End
Behavior of a Polynomial Function
Let P x   2x 3  5x 2  7x  11 be a polynomial
function of degree 3. Show that P x   2x 3
when |x| is very large.
5 7 11 

P x   x  2   2  3 

x x
x 
5 7
11
When |x| is very large , 2 and 3 are close to 0.
x x
x
3
Therefore,
P x   x 3 2  0  0  0   2x 3.
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THE LEADING–TERM TEST
Let f  x   an x  an1 x
n
n1
 ...  a1 x  a0
an  0
be a polynomial function.
Its leading term is anxn.
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
an > 0
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THE LEADING–TERM TEST
Case 2
an < 0
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THE LEADING–TERM TEST
Case 3
an > 0
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THE LEADING–TERM TEST
Case 4
an < 0
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Example: Using the Leading-Term
Test
Use the leading-term test to determine the end
behavior of the graph of
y  f x   2x 3  3x 2  4.
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 → ∞ as
x → –∞ and y → –∞ as x → ∞.
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REAL ZEROS OF POLYNOMIAL FUNCTIONS
If f is a polynomial function and c is a real
number, then the following statements are
equivalent.
1.
2.
3.
c is a zero of f .
c is a solution (or root) of the equation f (x) = 0.
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: Finding the Zeros of a
Polynomial Function
Find all zeros of each polynomial function.
a. f  x   x 4  2 x 3  3 x 2
b. g  x   x  2 x  x  2
3
2
Factor f (x) and then solve f (x) = 0.
a. f  x   x 4  2 x 3  3 x 2
 x 2  x 2  2 x  3
 x 2  x  1 x  3
x 2  0, x  1  0, x  3  0
x  0, x  1, x  3
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Example: Finding the Zeros of a
Polynomial Function
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 2  1
x  2  0 or x 2  1  0
x2
The only zero of g(x) is 2, since x2 + 1 > 0 for all real
numbers x.
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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: 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
2
2
 x  1
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: –2, 1, and 3.
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Example: Finding the Number of Real
Zeros
b. g  x    x  1  x 2  1  0
x  1  0 or x 2  1  0
x  1
No Real Solutions
g(x) has only one real zero: –1.
c. h  x    x  3
2
 x  1
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: Finding the Zeros and Their
Multiplicities
Find the zeros of the polynomial function
f (x) = x2(x + 1)(x – 2), and give the multiplicity of
each zero.
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: 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  7 x  18
4
2
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: 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: Finding the Number of
Turning Points
b. g x   x 3  x 2  12x
g has two total turning points; one local maximum
and one local minimum.
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Example: Finding the Number of
Turning Points
c. h x   x 3  3x 2  3x  1
h has no turning points, it is increasing on the interval
(–∞, ∞).
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Example: Graphing a Polynomial
Function
Sketch the graph of a
polynomial function
Sketch the graph of
f  x    x3  4 x 2  4 x  16.
Step 1 Determine the end 1. Degree = 3
behavior. Apply the
Leading coefficient = –1
leading-term test.
End behavior shown.
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Example: Graphing a Polynomial
Function
Step 2
Find the zeros
of the polynomial
function.
2.
 x3  4 x 2  4 x  16  0
  x  4  x  2  x  2   0
x  4, x  2, or x  2
Set f (x) = 0 and solve. The
zeros give the x-intercepts. Each zero has multiplicity 1,
the graph crosses the x-axis
at each zero.
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Example: Graphing a Polynomial
Function
Step 3 Find the
y-intercept by computing
f (0).
Step 4 Use symmetry to
check whether the function
is odd, even, or neither.
3. The y-intercept is f(0) = 16.
the graph passes through the
point (0,16).
4.
f   x      x   4   x   16
3
 x3  4 x 2  4 x  16
There is no symmetry in the
y-axis nor with respect to the
origin.
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Example: Graphing a Polynomial
Function
Step 5 Determine the sign 5.
of f(x) by using arbitrarily
chosen “test numbers” in
the intervals defined by the
x-intercepts. Find the
intervals on which the graph
lies above or below the xaxis.
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Example: Graphing a Polynomial
Function
Step 6 Draw the graph.
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.
6. Draw the graph.
The number of turning points
is 2, which is less than 3, the
degree of f.
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Section 2.3
SECTION 1.1
1.
2.
3.
Dividing Polynomials and the
Rational Zeros Test
Learn the Division Algorithm.
Use the Remainder and Factor Theorems.
Use the Rational Zeros Test.
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POLYNOMIAL FACTOR
A polynomial D(x) is a factor of a polynomial F(x) if
there is a polynomial Q(x) such that F(x) = D(x) ∙
Q(x).
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THE DIVISION ALGORITHM
If a polynomial F(x) is divided by a polynomial D(x), with
D(x) ≠ 0, there are unique polynomials Q(x) and R(x)
such that
Either R(x) is the zero polynomial, or the degree of
R(x) is less than the degree of D(x).
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Example: Using Long Division and
Synthetic Division
Use long division and synthetic division to find the
quotient and remainder when
2x4 + x3 − 16x2 + 18 is divided by x + 2.
We will start by performing long division.
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Example: Using Long Division and
Synthetic Division
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Example: Using Long Division and
Synthetic Division
quotient: 2x3 – 3x2 – 10x + 20
remainder: –22
The result is
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68
THE REMAINDER THEOREM
If a polynomial F(x) is divided by x – a, then the
remainder R is given by
R  F  a .
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69
Example: Using the Remainder
Theorem
Find the remainder when the polynomial
F x   2x 5  4x 3  5x 2  7x  2
is divided by x  1.
By the Remainder Theorem, F(1) is the remainder.
F 1  2 1  4 1  5 1  7 1  2
5
3
2
 2  4  5  7  2  2
The remainder is –2.
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70
Example: Using the Remainder
Theorem
Let f  x   x  3x  5 x  8 x  75. Find f 3.
4
3
2
One way is to evaluate f (x) when x = –3.
f  3   3  3 3  5  3  8  3  75  6
4
3
2
Another way is to use synthetic division.
3 1
1
3
5
8
75
3 0 15 69
0 5 23
6
Either method yields a value of 6.
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71
THE FACTOR THEOREM
A polynomial F(x) has (x – a) as a factor if and
only if F(a) = 0.
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72
Example: Using the Factor Theorem
Given that 2 is a zero of the function
f x   3x 2  2x 2  19x  6,
solve the polynomial equation
3x  2x  19x  6  0.
2
2
Since 2 is a zero of f (x), f (2) = 0 and (x – 2) is a
factor of f (x). Perform synthetic division by 2.
2
3 2
6
3 8
19
16
3
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6
6
0
73
Example: Using the Factor Theorem
Since the remainder is 0,
f  x   3x 2  2 x 2  19 x  6   x  2   3x 2  8 x  3
To find other zeros, solve the depressed equation.
3x 2  8 x  3  0
 3x  1 x  3  0
3 x  1  0 or x  3  0
1
x
or
x  3
3
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74
Example: Using the Factor Theorem
Including the original zero of 2, the solution set is
1 

3, , 2  .
3 

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75
THE RATIONAL ZEROS TEST
If F  x   an x  an1 x
n
n1
 ...  a2 x  a1x  a0
2
is a polynomial function with integer coefficients
p
(an ≠ 0, a0 ≠ 0) and
is a rational number in
q
lowest terms that is a zero of F(x), then
1. p is a factor of the constant term a0;
2. q is a factor of the leading coefficient an.
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Example: Using the Rational Zeros
Test
Find all the rational zeros of
F  x   2 x3  5 x 2  4 x  3.
List all possible zeros
p
Factors of the constant term,  3

q Factors of the leading coefficient, 2
Factors of  3 :  1,  3
Factors of 2 :
 1,  2
1
3
Possible rational zeros are:  1,  ,  ,  3.
2
2
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77
Example: Using the Rational Zeros
Test
Begin testing with 1. if it is not a rational zero, then try
another possible zero.
1 2
2
5
2
7
4
7
3
3
3
0
The remainder of 0 tells us that (x – 1) is a factor of
F(x). The other factor is 2x2 + 7x + 3. To find the other
zeros, solve 2x2 + 7x + 3 = 0.
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Example: Using the Rational Zeros
Test
2x  7x  3  0
2
 2 x  1 x  3  0
2 x  1  0 or x  3  0
1
x
or x  3
2
1


The solution set is 1,  ,  3 .
2


1
The rational zeros of F are 3,  , and 1.
2
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79
Section 2.4
SECTION 1.1
1.
2.
Zeros of a Polynomial Function
Learn basic facts about the complex zeros of polynomials.
Use the Conjugate Pairs Theorem to find zeros of
polynomials.
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Definitions
If we extend our number system to allow the coefficients
of polynomials and variables to represent complex
numbers, we call the polynomial a complex
polynomial. If P(z) = 0 for a complex number z we say
that z is a zero or a complex zero of P(x).
In the complex number system, every nth-degree
polynomial equation has exactly n roots and every nthdegree polynomial can be factored into exactly n linear
factors.
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FUNDAMENTAL THEOREM OF ALGEBRA
Every polynomial
P  x   an x n  an 1 x n 1  ...  a1 x  a0
 n  1, an  0 
with complex coefficients an, an – 1, …, a1, a0 has at
least one complex zero.
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FACTORIZATION THEOREM FOR POLYNOMIALS
If P(x) is a complex polynomial of degree n ≥ 1, it can
be factored into n (not necessarily distinct) linear
factors of the form
P  x   a  x  r1  x  r2  ...  x  rn  ,
where a, r1, r2, … , rn are complex numbers.
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83
Example: Constructing a Polynomial
Whose Zeros Are Given
Find a polynomial P(x) of degree 4 with a leading
coefficient of 2 and zeros –1, 3, i, and –i. Write P(x)
a. in completely factored form;
b. by expanding the product found in part a.
a. Since P(x) has degree 4, we write
P  x   a  x  r1  x  r2   x  r3   x  r4 
 2  x   1   x  3 x  i   x   i  
 2  x  1 x  3 x  i  x  i 
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84
Example: Constructing a Polynomial
Whose Zeros Are Given
b. Expand the product found in part a.
P x   2 x  1x  3x  i x  i 


 2 x  1x  3 x 2  1


 2x  3
 2 x  1 x 3  3x 2  x  3

 2 x 4  2x 3  2x 2
 2x  4 x  4 x  4 x  6
4
3
2
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85
CONJUGATE PAIRS THEOREM
If P(x) is a polynomial function whose coefficients
are real numbers and if z = a + bi is a zero of P,
then its conjugate, z  a  bi, is also a zero of P.
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86
ODD–DEGREE POLYNOMIALS
WITH REAL ZEROS
Any polynomial P(x) of odd degree with real
coefficients must have at least one real zero.
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87
Example: Using the Conjugate Pairs
Theorem
A polynomial P(x) of degree 9 with real
coefficients has the following zeros: 2, of
multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i.
Write all nine zeros of P(x).
Since complex zeros occur in conjugate pairs, the
conjugate 4 – 5i of 4 + 5i is a zero of multiplicity 2, and
the conjugate 3 + 7i of 3 – 7i is a zero of P(x). The
nine zeros of P(x) are:
2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i
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FACTORIZATION THEOREM FOR A POLYNOMIAL
WITH REAL COEFFICIENTS
Every polynomial with real coefficients can be uniquely
factored over the real numbers as a product of linear
factors and/or irreducible quadratic factors.
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89
Example: Finding the Complex Zeros
of a Polynomial
Given that 2 – i is a zero of
P x   x 4  6x 3  14x 2  14x  5,
find the remaining zeros.
The conjugate of 2 – i, 2 + i is also a zero.
So P(x) has linear factors:  x  2  i   x  2  i 
 x  2  i x  2  i 
 x  2   i  x  2   i 
 x  2   i  x  4 x  5
2
2
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2
90
Example: Finding the Complex Zeros
of a Polynomial
Divide P(x) by x2 – 4x + 5.
x 2  2x  1
2
4
3
2
x  4 x  5 x  6x  14 x  14 x  5
x  4 x  5x
4
3
2
2x 3  9x 2  14 x
2x 3  8x 2  10x
x
2
 4x  5
x
2
 4x  5
0
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Example: Finding the Complex Zeros
of a Polynomial
Therefore
P  x    x  2 x  1 x  4 x  5 
2
2
  x  1 x  1  x 2  4 x  5 
  x  1 x  1  x   2  i    x   2  i  
The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.
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92
Example: Finding the Zeros of a
Polynomial
Find all zeros of the polynomial
P(x) = x4 – x3 + 7x2 – 9x – 18.
Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18
Use synthetic division to find that 2 is a zero.
2 1 1 7
2 2
1 1 9
9 18
18 18
9
0
3
2
(x – 2) is a factor of P(x). Solve x  x  9x  9  0
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Example: Finding the Zeros of a
Polynomial
x  x  9x  9  0
3
2
 x  1  9  x  1  0
2
 x  9   x  1  0
x
2
x  1  0 or x  9  0
2
x  1 or x  9
x  1 or x  3i
2
The four zeros of P(x) are –1, 2, –3i, and 3i.
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Section 2.5
SECTION 1.1
1.
2.
3.
4.
5.
6.
7.
Rational Functions
Define a rational function.
Define vertical and horizontal asymptotes.
1
Graph translations of f ( x)  x .
Find vertical and horizontal asymptotes (if any).
Graph rational functions.
Graph rational functions with oblique asymptotes.
Graph a revenue curve.
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RATIONAL FUNCTION
A function f that can be expressed in the form
N x 
f x  
,
D x 
where the numerator N(x) and the denominator D(x)
are polynomials and D(x) is not the zero polynomial,
is called a rational function. The domain of f
consists of all real numbers for which D(x) ≠ 0.
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Example: Finding the Domain of a
Rational Function
Find the domain of each rational function.
3x 2  12
a. f x  
x 1
x2  4
c. h x  
x2
x
b. g x   2
x  6x  8
a. The domain of f (x) is the set of all real numbers
for which x – 1 ≠ 0; that is, x ≠ 1 .
In interval notation: ,1 U 1,  
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Example: Finding the Domain of a
Rational Function
b. Find the values of x for which the denominator
x2 – 6x + 8 = 0, then exclude those values from the
domain.
x  2 x  4   0
x  2  0 or x  4  0
x  2 or x  4
The domain of g (x) is the set of all real numbers
such that x ≠ 2 and x ≠ 4 .
In interval notation: , 2  U 2, 4  U 4,  
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Example: Finding the Domain of a
Rational Function
x2  4
c. The domain of h  x  
x 2
is the set of all real numbers for which x – 2 ≠ 0;
that is, x ≠ 2 .
The domain of g (x) is the set of all real numbers
such that x ≠ 2.
In interval notation: , 2  U 2,  
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VERTICAL ASYMPTOTES
The line with equation x = a is called a vertical
asymptote of the graph of a function f if
f  x    as x  a

or as x  a

or if f  x    as x  a or as x  a .

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
100
VERTICAL ASYMPTOTES
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101
VERTICAL ASYMPTOTES
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102
LOCATING VERTICAL ASYMPTOTES OF
RATIONAL FUNCTIONS
N x 
If f x  
D x 
is a rational function,
where the N(x) and D(x) do not have a common factor
and a is a real zero of D(x), then the line with equation
x = a is a vertical asymptote of the graph of f.
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Example: Finding Vertical
Asymptotes
Find all vertical asymptotes of the graph of each
rational function.
1
1
a. f x  
x 1
1
c. h x   2
x 1
b. g x  
x2  9
a. No common factors, zero of the denominator is
x = 1. The line with equation x = 1 is a vertical
asymptote of f (x).
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Example: Finding Vertical
Asymptotes
b. No common factors. Factoring
x2 – 9 = (x + 3)(x – 3), we see the zeros of the
denominator are –3 and 3. The lines with equations
x = – 3 and x = 3 are the two vertical asymptotes of
g (x).
c. The denominator x2 + 1 has no real zeros. Hence,
the graph of the rational function h (x) has no
vertical asymptotes.
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105
Example: Rational Function Whose
Graph Has a Hole
Find all vertical asymptotes of the graph of each
rational function.
x2  9
a. h  x  
x3
x2
b. g x   2
x 4
x  9  x  3 x  3
a. h  x  

x3
x3
 x  3, x  3
2
The graph is the line with equation y = x + 3, with a
gap (hole) corresponding to x = 3.
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Example: Rational Function Whose
Graph Has a Hole
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107
Example: Rational Function Whose
Graph Has a Hole
x2
x2
b. g x   2

x  4 x  2 x  2 
1

,
x2
x  2
The graph has a hole at x = –2. However, the graph of
g(x) also has a vertical asymptote at x = 2.
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Example: Rational Function Whose
Graph Has a Hole
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HORIZONTAL ASYMPTOTES
The line with equation y = k is called a horizontal
asymptote of the graph of a function f if
f x   k as x   or x  .
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RULES FOR LOCATING
HORIZONTAL ASYMPTOTES
N x 
Let f be a rational function given by f x  
D x 
an x  an1 x  ...  a2 x  a1 x  a0

, an  0,bn  0
m
m1
2
bm x  bm1 x  ...  b2 x  b1 x  b0
n
n1
2
To find whether the graph of f has one horizontal
asymptote or no horizontal asymptote, we compare
the degree of the numerator, n, with that of the
denominator, m:
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RULES FOR LOCATING
HORIZONTAL ASYMPTOTES
1. If n < m, then the x-axis (y = 0) is the horizontal
asymptote.
an
2. If n = m, then the line with equation y 
bm
is the horizontal asymptote.
3. If n > m, then the graph of f has no horizontal
asymptote.
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Example: Finding the Horizontal
Asymptote
Find the horizontal asymptote (if any) of the graph of
each rational function.
5x  2
a. f x  
1  3x
2x
b. g x   2
x 1
3x  1
c. h x  
x2
2
a. Numerator and denominator have degree 1.
5
5
y
  is the horizontal asymptote.
3
3
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Example: Finding the Horizontal
Asymptote
2x
b. g x   2
x 1
degree of denominator > degree of numerator y = 0
(the x-axis) is the horizontal asymptote.
3x 2  1
c. h x  
x2
degree of numerator > degree of denominator The
graph has no horizontal asymptote.
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114
Example: Graphing a Rational Function
f x 
N( x )
, where
D( x )
Graph
is in lowest terms.
f(x) Sketch the graph of
2 x2  2
f  x  2
.
x 9
2x 2  2  0
Step 1 Find the intercepts. 1.
Since f is in lowest terms,
2  x  1 x  1  0
the x-intercepts are found
x  1  0 or x  1  0
by solving the equation
x  1 or x  1
N(x) = 0. The y-intercept is,
The x-intercepts are 1 and 1,
if there is one, is f (0).
so the graph passes through
(1, 0) and (1, 0).
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Example: Graphing a Rational Function
Step 1 Find the intercepts. 1.
2
2 0  2 2
Since f is in lowest terms,
f 0 

2
the x-intercepts are found
0  9 9
by solving the equation
N(x) = 0. The y-intercept is,
if there is one, is f (0).
2
The y-intercept is 9 , so the
graph of f passes through
the point (0, 2 ).
9
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116
Example: Graphing a Rational Function
Step 2 Find the vertical
asymptotes (if any).
Solve D(x) = 0 to find the
vertical asymptotes of the
graph. Sketch the vertical
asymptotes.
2.
Solve x  9  0,
2
x  3
The vertical asymptotes are
x = 3 and x = –3.
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Example: Graphing a Rational Function
Step 3 Find the horizontal 3. Since n = m = 2 the
asymptotes (if any).
horizontal asymptote is
Use the rules from the
previous slide.
Leading coefficient of N  x 
y
Leasing coefficient of D( x )
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
2
 2.
1
118
Example: Graphing a Rational Function
Step 4 Test for symmetry. 4.
2
2x  2
If f (–x) = f (x), then f is
f x 
2
symmetric with respect to
x  9
the y-axis. If f (–x) = –f (x),
2
2x  2
then f is symmetric with
 2
 f  x .
x 9
respect to the origin.
The graph of f is symmetric
in the y-axis. This is only
one symmetry.
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Example: Graphing a Rational Function
Step 5 Locate the graph
relative to the horizontal
asymptote (if any). Use the sign
graphs and test numbers
associated with the zeros of N(x)
and D(x), to determine where the
graph of f is above the x-axis
and where it is below the x-axis.
5.
2 x2  2
16
f  x  2
 2 2
x 9
x 9
R(x) = 16 has no zeros and
D(x) has zeros –3 and 3.
These zeros divide the xaxis into three intervals. We
choose test points –4, 0,
and 4 to test the sign of
16
.
2
x 9
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Example: Graphing a Rational Function
Step 6 Sketch the graph.
6.
Plot some points and graph the
asymptotes found in Steps 1-5;
use symmetry to sketch the
graph of f.
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Example: Graphing a Rational
Function
x2  2
.
Sketch the graph of f x  
x  2 x  1
Step 1
Since x2 + 2 > 0, no x-intercepts.
02  2
f  0 
 1 ; y-intercept is –1.
 0  2  0  1
Step 2
Solve (x + 2)(x – 1) = 0; x = –2, x = 1
Vertical asymptotes are x = –2 and x = 1.
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Example: Graphing a Rational
Function
Step 3
degree of den = degree of num
y = 1 is the horizontal asymptote
Step 4 Symmetry. None
Step 5 The zeros of the denominator –2 and 1 yield
the following figure:
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Example: Graphing a Rational
Function
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Example: Graphing a Rational
Function
Step 6 Sketch the
graph.
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Example: Graphing a Rational
Function
x2
.
Sketch a graph of f x   2
x 1
Step 1 Since f (0) = 0 and setting f (x) = 0 yields 0,
x-intercept and y-intercept are 0.
Step 2 Because x2 +1 > 0 for all x, the domain is the
set of all real numbers. Since there are no
zeros for the denominator, there are no vertical
asymptotes.
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Example: Graphing a Rational
Function
Step 3 degree of den = degree of num
y = 1 is the horizontal asymptote.
Step 4 Symmetry.
2
x

x
f x 
 2
 f  x
2
x  1 x  1
2
Symmetric with respect to the y-axis.
Step 5 The graph is always above the x-axis, except
at x = 0.
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Example: Graphing a Rational
Function
Step 6 Sketch the
graph.
x2
f x   2
x 1
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OBLIQUE ASUMPTOTES
Suppose
N x 
f x  
, and the degree of N(x)
D x 
is greater than the degree of D(x). Then
N x 
R x 
f x  
 Q x  
.
D x 
D x 
Thus, as x  , f x   Q x   0  Q x .
That is, the graph of f approaches the graph of
the oblique asymptote defined by Q(x).
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Example: Graphing a Rational
Function with an Oblique Asymptote
x 4
Sketch the graph of f x  
.
x 1
2
Step 1 Solve x2 – 4 = 0;
x-intercepts: –2, 2
04
f 0  
 4 ; y-intercept: –4.
0 1
Step 2 Solve x + 1 = 0; x = –1; domain is set of all
real numbers except –1.
Vertical asymptote is x = –1.
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Example: Graphing a Rational
Function with an Oblique Asymptote
Step 3
degree of numerator > degree of denominator
x2  4
3
f x  
 x 1
x 1
x 1
y = x – 1 is an oblique asymptote.
Step 4 Symmetry. None
Step 5 Use the intervals determined by the zeros of
the denominator.
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Example: Graphing a Rational
Function with an Oblique Asymptote
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Example: Graphing a Rational
Function with an Oblique Asymptote
Step 6 Sketch the
graph.
x2  4
f  x 
x 1
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