10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
3.4 - 1
3.4
Polynomial Functions: Graphs,
Applications, and Models
Graphs of (x) = axn
Graphs of General Polynomial Functions
Turning Points and End Behavior
Graphing Techniques
Intermediate Value and Boundedness
Theorems
Approximating Real Zeros
Polynomial Models and Curve Fitting
3.4 - 2
Example 1
GRAPHING FUNCTIONS OF THE
FORM (x) = axn
Graph the function.
a. f ( x )  x 3
Solution Choose several values for x, and
find the corresponding values of (x), or y.
f (x)  x3
x
–2
–1
0
1
2
(x)
–8
–1
0
1
8
3.4 - 3
Example 1
GRAPHING FUNCTIONS OF THE
FORM (x) = axn
Graph the function.
b. g ( x )  x 5
Solution The graphs of (x) = x3 and g(x) = x5
are both symmetric with respect to the origin.
g( x )  x 5
x
–1.5
–1
0
1
1.5
g(x)
–7.6
–1
0
1
7.6
3.4 - 4
Example 1
GRAPHING FUNCTIONS OF THE
FORM (x) = axn
Graph the function.
c. f ( x )  x 4 , g ( x )  x 6
Solution The graphs of (x) = x3 and g(x) = x5
are both symmetric with respect to the origin.
f (x)  x
x
–2
–1
0
1
2
4
(x)
16
1
0
1
16
g( x )  x 6
x
–1.5
–1
0
1
1.5
g(x)
11.4
1
0
1
11.4
3.4 - 5
Graphs of General Polynomial
Functions
As with quadratic functions, the absolute
value of a in (x) = axn determines the width
of the graph. When a> 1, the graph is
stretched vertically, making it narrower, while
when 0 < a < 1, the graph is shrunk or
compressed vertically, so the graph is
broader. The graph of (x) = –axn is
reflected across the x-axis compared to the
graph of (x) = axn.
3.4 - 6
Graphs of General Polynomial
Functions
n
Compared with the graph of f ( x )  ax , the
n
f
(
x
)

ax
 k is translated (shifted) k
graph of
units up if k > 0 andkunits down if k < 0.
n
f
(
x
)

ax
,
Also, when compared with the graph of
the graph of (x) = a(x – h)n is translated h
units to the right if h > 0 and hunits to the left
if h < 0.
n
The graph of f ( x )  ax  k shows a
combination of these translations. The effects
here are the same as those we saw earlier with
quadratic functions.
3.4 - 7
Example 2
EXAMINING VERTICAL AND
HORIZONTAL TRANSLATIONS
Graph the function.
5
a. f ( x )  x  2
Solution The graph
will be the same as that
of (x) = x5, but
translated 2 units down.
3.4 - 8
Example 2
EXAMINING VERTICAL AND
HORIZONTAL TRANSLATIONS
Graph the function.
6
b. f ( x )  ( x  1)
Solution
In (x) = (x + 1)5,
function  has a graph
like that of
(x) = x6, but since
x + 1 = x – (–1), it is
translated 1 unit to the
left.
3.4 - 9
Example 2
EXAMINING VERTICAL AND
HORIZONTAL TRANSLATIONS
Graph the function.
3
c. f ( x )  2( x  1)  3
Solution
The negative sign in –2 causes
the graph of the function to be
reflected across the x-axis when
compared with the graph of (x)
= x3. Because  – 2> 1, the
graph is stretched vertically as
compared to the graph of
(x) = x3. It is also translated 1
unit to the right and 3 units up.
3.4 - 10
Unless otherwise restricted, the domain of a
polynomial function is the set of all real numbers.
Polynomial functions are smooth, continuous curves
on the interval (–, ). The range of a polynomial
function of odd degree is also the set of all real
numbers. Typical graphs of polynomial functions of
odd degree are shown in next slide. These graphs
suggest that for every polynomial function  of odd
degree there is at least one real value of x that
makes (x) = 0 . The real zeros are the x-intercepts
of the graph.
3.4 - 11
Odd Degree
3.4 - 12
A polynomial function of even degree has a
range of the form (–, k] or [k, ) for some
real number k. Here are two typical graphs
of polynomial functions of even degree.
Even Degree
3.4 - 13
Recall that a zero c of a polynomial function
has as multiplicity the exponent of the factor
x – c. Determining the multiplicity of a zero
aids in sketching the graph near that zero. If
the zero has multiplicity one, the graph
crosses the x-axis at the corresponding xintercept as seen here.
3.4 - 14
If the zero has even multiplicity, the graph is
tangent to the x-axis at the corresponding
x-intercept (that is, it touches but does not
cross the x-axis there).
3.4 - 15
If the zero has odd multiplicity greater than
one, the graph crosses the x-axis and is
tangent to the x-axis at the corresponding
x-intercept. This causes a change in
concavity, or shape, at the x-intercept and
the graph wiggles there.
3.4 - 16
Turning Points and End Behavior
The previous graphs show that
polynomial functions often have turning
points where the function changes from
increasing to decreasing or from
decreasing to increasing.
3.4 - 17
Turning Points
A polynomial function of degree n has
at most n – 1 turning points, with at
least one turning point between each
pair of successive zeros.
3.4 - 18
End Behavior
The end behavior of a polynomial graph is
determined by the dominating term, that is,
the term of greatest degree. A polynomial of
the form
f ( x )  an x n  an 1x n 1   a0
n
f
(
x
)

a
x
has the same end behavior as
.
n
3.4 - 19
End Behavior
For instance,
f ( x )  2x  8 x  9
3
2
has the same end behavior as f ( x )  2 x .
It is large and positive for large positive
values of x and large and negative for
negative values of x with large absolute
value.
3
3.4 - 20
End Behavior
The arrows at the ends of the graph look like those
of the graph shown here; the right arrow points up
and the left arrow points down.
The graph shows that as x takes on larger and
larger positive values, y does also. This is
symbolized as x  , y  ,
read “as x approaches infinity, y
approaches infinity.”
3.4 - 21
End Behavior
For the same graph, as x takes on negative values
of larger and larger absolute value, y does also:
as x  ,
y  ,
3.4 - 22
End Behavior
For this graph, we have
as
x  ,
and as x  ,
y  ,
y  .
3.4 - 23
End Behavior of Polynomials
Suppose that axn is the dominating term of a
polynomial function  of odd degree.
1. If a > 0, then as x  , f ( x )  , and as
x  , f ( x )  .
Therefore, the end behavior of the graph is
of the type that looks like the figure shown
here.
We symbolize it as
.
3.4 - 24
End Behavior of Polynomials
Suppose that axn is the dominating term of a
polynomial function  of odd degree.
2. If a < 0, then as x  , f ( x )  , and as
x  , f ( x )  .
Therefore, the end behavior of the graph
looks like the graph shown here.
We symbolize it as
.
3.4 - 25
End Behavior of Polynomials
Suppose that axn is the dominating term of a
polynomial function  of even degree.
1. If a > 0, then as x  , f ( x )  .
Therefore, the end behavior of the
graph looks like the graph shown here.
We symbolize it as
.
3.4 - 26
End Behavior of Polynomials
Suppose that is the dominating term of a
polynomial function  of even degree.
2. If a < 0, then as x  , f ( x )  .
Therefore, the end behavior of the graph
looks like the graph shown here.
We symbolize it as
.
3.4 - 27
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
f ( x )  x  x  5x  4
4
A.
B.
2
C.
D.
Solution Because  is of even degree with
positive leading coefficient, its graph is C.
3.4 - 28
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
g( x )  x  x  3x  4
6
A.
B.
2
C.
D.
Solution Because g is of even degree with
negative leading coefficient, its graph is A.
3.4 - 29
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
h( x )  3 x  x  2 x  4
3
A.
B.
2
C.
D.
Solution Because function h has odd
degree and the dominating term is positive,
its graph is in B.
3.4 - 30
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
k( x )  x  x  4
7
A.
B.
C.
D.
Solution Because function k has odd
degree and a negative dominating term, its
graph is in D.
3.4 - 31
Graphing Techniques
We have discussed several characteristics of the graphs of
polynomial functions that are useful for graphing the
function by hand. A comprehensive graph of a polynomial
function will show the following characteristics:
1. all x-intercepts (zeros)
2. the y-intercept
3. the sign of (x) within the intervals formed by the xintercepts, and all turning points
4. enough of the domain to show the end behavior.
In Example 4, we sketch the graph of a polynomial function
by hand. While there are several ways to approach this,
here are some guidelines.
3.4 - 32
Graphing a Polynomial
Function
n
n 1
f
(
x
)

a
x

a
x
  a1x  a0 , an  0,
Let
n
n 1
be a polynomial function of degree n. To
sketch its graph, follow these steps.
Step 1 Find the real zeros of . Plot them as
x-intercepts.
Step 2 Find (0). Plot this as the y-intercept.
3.4 - 33
Graphing a Polynomial
Function
Step 3 Use test points within the intervals
formed by the x-intercepts to
determine the sign of (x) in the
interval. This will determine whether
the graph is above or below the xaxis in that interval.
3.4 - 34
Graphing a Polynomial
Function
Use end behavior, whether the graph
crosses, bounces on, or wiggles through the
x-axis at the x-intercepts, and selected points
as necessary to complete the graph.
3.4 - 35
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 1 The possible rational zeros are 1,
2, 3, 6, ½, and 3/2. Use synthetic
division to show that 1 is a zero.
12
2
5 1  6
2
7
6
7
6
0
(1) = 0
3.4 - 36
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 1 Thus,
f ( x )  ( x  1)(2 x 2  7 x  6)
 ( x  1)(2 x  3)( x  2).
Factor 2x2 + 7x + 6.
Set each linear factor equal to 0, then solve for x to
find real zeros. The three real zeros of  are 1, – 3/2,
and – 2.
3.4 - 37
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 2 f (0)  6, so plot (0, 6).
3.4 - 38
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
The x-intercepts divide the x-axis into four
intervals: (–, –2), (–2, –3/2), (–3/2, 1), and
(1, ). Because the graph of a polynomial
function has no breaks, gaps, or sudden
jumps, the values of (x) are either always
positive or always negative in any given
interval.
3.4 - 39
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
To find the sign of (x) in each interval, select
an x-value in each interval and substitute it
into the equation for (x) to determine
whether the values of the function are
positive or negative in that interval.
3.4 - 40
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
When the values of the function are negative,
the graph is below the x-axis, and when (x)
has positive values, the graph is above the xaxis.
3.4 - 41
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
A typical selection of test points and the
results of the tests are shown in the table on
the next slide. (As a bonus, this procedure
also locates points that lie on the graph.)
3.4 - 42
GRAPHING A POLYNOMIAL
FUNCTION
Example 4
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
Graph Above
Value
or Below xof (x) Sign of (x)
Axis
Interval
Test
Point
(–, –2)
–3
–12
Negative
Below
(–2, –3/2)
–7/4
Positive
Above
(–3/2, 1)
0
11/32
–6
Negative
Below
(1, )
2
28
Positive
Above
3.4 - 43
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
Plot the test points and
join the x-intercepts, yintercept, and test
points with a smooth
curve to get the graph.
3.4 - 44
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
Because each zero has
odd multiplicity (1), the
graph crosses the x-axis
each time. The graph
has two turning points,
the maximum number for
a third-degree polynomial
function.
3.4 - 45
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
The sketch could be improved
by plotting the points found in
each interval in the table. Notice
that the left arrow points down
and the right arrow points up.
This end behavior is correct
since the dominating term of the
polynomial is 2x3.
3.4 - 46
Graphing Polynomial Functions
Note If a polynomial function is given in
factored form, such as
2
f ( x )  (  x  1)( x  3)( x  2) ,
Step 1 of the guidelines is easier to perform,
since real zeros can be determined by
inspection. For this function, we see that 1
and 3 are zeros of multiplicity1, and –2 is a
zero of multiplicity 2.
3.4 - 47
Graphing Polynomial Functions
Note Since the
dominating term is
 x ( x )( x 2 )   x 4 , the
end behavior of the
graph is
.
The y-intercept is
f (0)  1( 3)(2)  12.
2
3.4 - 48
Graphing Polynomial Functions
Note
The graph intersects the
x-axis at 1 and 3 but
bounces at –2.
This information is
sufficient to quickly
sketch the graph of (x).
3.4 - 49
Important Relationships
We emphasize the important relationships
among the following concepts.
1. the x-intercepts of the graph of y = (x)
2. the zeros of the function 
3. the solutions of the equation (x) = 0
4. the factors of (x)
3.4 - 50
x-Intercepts, Zeros, Solutions,
and Factors
If a is an x-intercept of the graph of
y  f ( x ), then a is a zero of , a is a
solution of (x) = 0, and x – a is a
factor of (x).
3.4 - 51
Intermediate Value Theorem
for Polynomials
If (x) defines a polynomial function
with only real coefficients, and if for
real numbers a and b, the values (a)
and (b) are opposite in sign, then
there exists at least one real zero
between a and b.
3.4 - 52
Example 5
LOCATING A ZERO
Use synthetic division and a graph to show
that (x) = x3 – 2x2 – x + 1 has a real zero
between 2 and 3.
Solution Since
(2) is negative and
(3) is positive, by
the intermediate
value theorem there
must be a real zero
between 2 and 3.
2 1 2
2
1
0
1 1
02
 1  1  f ( 2)
3 1  2 1 1
3
3 6
1
1
2 7  f (3 )
3.4 - 53
Caution Be careful how you interpret the
intermediate value theorem.
If (a) and (b) are not opposite in sign, it does not
necessarily mean that there is no zero between a
and b.
In the graph shown here, for
example, (a) and (b) are both
negative, but –3 and –1, which
are between a and b, are zeros
of (x).
3.4 - 54
Boundedness Theorem
Let (x) be a polynomial function of degree
n ≥ 1 with real coefficients and with a
positive leading coefficient. If (x) is divided
synthetically by x – c, and
(a) if c > 0 and all numbers in the bottom
row of the synthetic division are
nonnegative, then (x) has no zero
greater than c;
3.4 - 55
Boundedness Theorem
(b) if c < 0 and the numbers in the bottom
row of the synthetic division alternate in
sign (with 0 considered positive or
negative, as needed), then (x) has no
zero less than c.
3.4 - 56
Proof
We outline the proof of part (a). The proof for
part (b) is similar. By the division algorithm, if
(x) is divided by x – c , then for some q(x)
and r,
f ( x )  ( x  c )q( x )  r
where all coefficients of q(x) are nonnegative,
r ≥ 0, and c > 0. If x > c, then x – c > 0. Since
q(x) > 0 and r ≥ 0,
f ( x )  ( x  c )q( x )  r  0
This means that (x) will never be 0 for x > c.
3.4 - 57
Example 6
USING THE BOUNDEDNESS
THEOREM
Show that the real zeros of
(x) = 2x4 – 5x3 + 3x +1 satisfy the given
conditions.
a. No real zero is greater than 3.
Solution Since (x) has real coefficients
and the leading coefficient, 2, is positive,
use the boundedness theorem. Divide (x)
synthetically by x – 3.
32
2
5 0 3 1
6 3 9 36
1 3 12 37
All are nonnegative.
3.4 - 58
Example 6
USING THE BOUNDEDNESS
THEOREM
a. No real zero is greater than 3.
Solution
32
2
5 0 3 1
6 3 9 36
1 3 12 37
All are nonnegative.
Since 3 > 0 and all numbers in the last row of
the synthetic division are nonnegative, (x)
has no real zero greater than 3.
3.4 - 59
Example 6
USING THE BOUNDEDNESS
THEOREM
b. No real zero is greater than –1.
Solution
1 2
2
5 0 3 1
2 7  7 4
7 7 4 5
Divide (x) by x + 1.
These numbers
alternate in sign.
Here –1 < 0 and the numbers in the last row
alternate in sign, so (x) has no real zero less
than –1.
3.4 - 60
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Approximate the real zeros of
(x) = x4 – 6x3 + 8x2 + 2x – 1.
Solution The greatest degree term is x4, so
the graph will have end behavior similar to
the graph of (x) = x4, which is positive for all
values of x with large absolute values. That
is, the end behavior is up at the left and the
right,
.
There are at most four real zeros, since the
polynomial is fourth-degree.
3.4 - 61
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Approximate the real zeros of
(x) = x4 – 6x3 + 8x2 + 2x – 1.
Solution Since (0) = –1, the y-intercept is –1.
Because the end behavior is positive
on the left and the right, by the intermediate
value theorem  has at least one zero on either
side of x = 0. To approximate the zeros, we use
a graphing calculator.
3.4 - 62
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Approximate the real zeros of
(x) = x4 – 6x3 + 8x2 + 2x – 1.
Solution The graph below shows that there are four
real zeros, and the table indicates that they are
between –1 and 0, 0 and 1, 2 and 3, and 3 and 4
because there is a sign change in (x) in each case.
3.4 - 63
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Solution Using the capability of the calculator,
we can find the zeros to a great degree of
accuracy. The graph shown here shows that the
negative zero is approximately – .4142136.
Similarly, we find that the other three zeros are
approximately
.26794919, 2.4142136,
and 3.7320508.
3.4 - 64
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Year
The table shows the
number of
transactions, in
millions, by users of
bank debit cards for
selected years.
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
3.4 - 65
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
a. Using x = 0 to
Year
represent 1995, x = 3
to represent 1998, and 1995
so on, use the
1998
regression feature of a 2000
calculator to determine 2004
the quadratic function
2009
that best fits the data.
Plot the data and
graph.
Transactions
(in millions)
829
3765
6797
14,106
22,120
3.4 - 66
Example 8
Solution
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Year
a. The best-fitting
quadratic function
for the data is
defined by
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
y  25.53 x 2  120 x  453.1
3.4 - 67
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
a. The regression
coordinates screen
is shown below.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
3.4 - 68
Example 8
Solution
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
a. The graph is shown
below.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
3.4 - 69
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
b. Repeat part (a) for a
cubic function.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
3.4 - 70
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Example 8
Solution
Year
b. The best-fitting cubic
function is shown
below and is defined
by
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
y  6.735 x  164.1x  543.1x  831.0
3
2
3.4 - 71
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
b. The graph of the bestfitting cubic function is
shown below.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
3.4 - 72
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
c. Repeat part (a) for a
quartic function.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
3.4 - 73
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
Year
c. The best-fitting quartic
function is defined by
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
y  .0576 x 4  5.198 x 3  151.9 x 2  571.4 x  829
3.4 - 74
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
c. The graph of the best- Year
fitting quartic function is
1995
shown below.
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
3.4 - 75
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
d. The correlation coefficient,
R, is a measure of the
Year
Transactions
strength of the relationship
(in millions)
between two variables. The
1995
829
2
values of R and R are used
1998
3765
to determine how well a
2000
6797
regression model fits a set
of data. The closer the
2004
14,106
value of R2 is to 1, the better
2009
22,120
2
the fit. Compare R for the
three functions to decide
which function fits the data.
Example 8
3.4 - 76
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
d. Find the correlation
values R2. See the graph
for the quadratic function.
The others are
.999999265 for the cubic
function and 1 for the
quartic function.
Therefore, the quartic
function provides the best
fit.
3.4 - 77