3.3 Graphs of Polynomial Functions
181
Section 3.3 Graphs of Polynomial Functions
In the previous section, we explored the short run behavior of quadratics, a special case
of polynomials. In this section, we will explore the short run behavior of polynomials in
general.
Short run Behavior: Intercepts
As with any function, the vertical intercept can be found by evaluating the function at an
input of zero. Since this is evaluation, it is relatively easy to do it for a polynomial of any
degree.
To find horizontal intercepts, we need to solve for when the output will be zero. For
general polynomials, this can be a challenging prospect. While quadratics can be solved
using the relatively simple quadratic formula, the corresponding formulas for cubic and
4
th
degree polynomials are not simple enough to remember, and formulas do not exist for
general higher-degree polynomials. Consequently, we will limit ourselves to three cases:
1) The polynomial can be factored using known methods: greatest common
factor and trinomial factoring.
2) The polynomial is given in factored form.
3) Technology is used to determine the intercepts.
Other techniques for finding the intercepts of general polynomials will be explored in the
next section.
Example 1
Find the horizontal intercepts of
246
23)( xxxxf +−=
.
We can attempt to factor this polynomial to find solutions for f(x) = 0.
023
246
=+− xxx
Factoring out the greatest common factor
( )
023
242
=+− xxx
Factoring the inside as a quadratic in x
2
( )( )
021
222
=−− xxx
Then break apart to find solutions
0
0
2
=
=
x
x
or
( )
1
1
01
2
2
=
=
=−
x
x
x
or
( )
2
2
02
2
2
=
=
=−
x
x
x
This gives us 5 horizontal intercepts.
Chapter 3
182
Example 2
Find the vertical and horizontal intercepts of
)32()2()(
2
+−= tttg
The vertical intercept can be found by evaluating g(0).
12)3)0(2()20()0(
2
=+−=g
The horizontal intercepts can be found by solving g(t) = 0
0)32()2(
2
=+− tt
Since this is already factored, we can break it apart:
2
02
0)2(
2
=
=−
=−
t
t
t
or
2
3
0)32(
−
=
=+
t
t
We can always check our answers are reasonable by graphing the polynomial.
Example 3
Find the horizontal intercepts of
64)(
23
−++= tttth
Since this polynomial is not in factored form, has no
common factors, and does not appear to be factorable
using techniques we know, we can turn to technology to
find the intercepts.
Graphing this function, it appears there are horizontal
intercepts at t = -3, -2, and 1.
We could check these are correct by plugging in these
values for t and verifying that
( 3) ( 2) (1) 0h h h− = − = =
.
Try it Now
1. Find the vertical and horizontal intercepts of the function
24
4)( tttf −=
.
Graphical Behavior at Intercepts
If we graph the function
32
)1()2)(3()( +−+= xxxxf
,
notice that the behavior at each of the horizontal
intercepts is different.
At the horizontal intercept x = -3, coming from the
)3( +x
factor of the polynomial, the graph passes
directly through the horizontal intercept.
3.3 Graphs of Polynomial Functions
183
The factor
)3( +x
is linear (has a power of 1), so the behavior near the intercept is like
that of a line - it passes directly through the intercept. We call this a single zero, since the
zero corresponds to a single factor of the function.
At the horizontal intercept x = 2, coming from the
2
)2( −x
factor of the polynomial, the
graph touches the axis at the intercept and changes direction. The factor is quadratic
(degree 2), so the behavior near the intercept is like that of a quadratic – it bounces off
the horizontal axis at the intercept. Since
)2)(2()2(
2
−−=− xxx
, the factor is repeated
twice, so we call this a double zero. We could also say the zero has multiplicity 2.
At the horizontal intercept x = -1, coming from the
3
)1( +x
factor of the polynomial, the
graph passes through the axis at the intercept, but flattens out a bit first. This factor is
cubic (degree 3), so the behavior near the intercept is like that of a cubic, with the same
“S” type shape near the intercept that the toolkit
3
x
has. We call this a triple zero. We
could also say the zero has multiplicity 3.
By utilizing these behaviors, we can sketch a reasonable graph of a factored polynomial
function without needing technology.
Graphical Behavior of Polynomials at Horizontal Intercepts
If a polynomial contains a factor of the form
p
hx )( −
, the behavior near the horizontal
intercept h is determined by the power on the factor.
p = 1 p = 2 p = 3
Single zero Double zero Triple zero
Multiplicity 1 Multiplicity 2 Multiplicity 3
For higher even powers 4,6,8 etc.… the graph will still bounce off the horizontal axis
but the graph will appear flatter with each increasing even power as it approaches and
leaves the axis.
For higher odd powers, 5,7,9 etc… the graph will still pass through the horizontal axis
but the graph will appear flatter with each increasing odd power as it approaches and
leaves the axis.
Chapter 3
184
Example 4
Sketch a graph of
)5()3(2)(
2
−+−= xxxf
.
This graph has two horizontal intercepts. At x = -3, the factor is squared, indicating the
graph will bounce at this horizontal intercept. At x = 5, the factor is not squared,
indicating the graph will pass through the axis at this intercept.
Additionally, we can see the leading term, if this polynomial were multiplied out, would
be
3
2x−
, so the long-run behavior is that of a vertically reflected cubic, with the
outputs decreasing as the inputs get large positive, and the inputs increasing as the
inputs get large negative.
To sketch this we consider the following:
As
−→x
the function
→)(xf
so we know the graph starts in the 2
nd
quadrant
and is decreasing toward the horizontal axis.
At (-3, 0) the graph bounces off the horizontal axis and so the function must start
increasing.
At (0, 90) the graph crosses the vertical axis at the vertical intercept.
Somewhere after this point, the graph must turn back down or start decreasing toward
the horizontal axis since the graph passes through the next intercept at (5,0).
As
→x
the function
−→)(xf
so we know the
graph continues to decrease and we can stop drawing
the graph in the 4
th
quadrant.
Using technology we can verify the shape of the
graph.
Try it Now
2. Given the function
xxxxg 6)(
23
−−=
use the methods that we have learned so far to
find the vertical & horizontal intercepts, determine where the function is negative and
positive, describe the long run behavior and sketch the graph without technology.
3.3 Graphs of Polynomial Functions
185
Solving Polynomial Inequalities
One application of our ability to find intercepts and sketch a graph of polynomials is the
ability to solve polynomial inequalities. It is a very common question to ask when a
function will be positive and negative. We can solve polynomial inequalities by either
utilizing the graph, or by using test values.
Example 5
Solve
0)4()1)(3(
2
−++ xxx
As with all inequalities, we start by solving the equality
0)4()1)(3(
2
=−++ xxx
,
which has solutions at x = -3, -1, and 4. We know the function can only change from
positive to negative at these values, so these divide the inputs into 4 intervals.
We could choose a test value in each interval and evaluate the function
)4()1)(3()(
2
−++= xxxxf
at each test value to determine if the function is positive or
negative in that interval
On a number line this would look like:
From our test values, we can determine this function is positive when x < -3 or x > 4, or
in interval notation,
),4()3,( −−
We could have also determined on which intervals the function was positive by sketching
a graph of the function. We illustrate that technique in the next example
Interval
Test x in interval
f( test value)
>0 or <0?
x < -3
-4
72
> 0
-3 < x < -1
-2
-6
< 0
-1 < x < 4
0
-12
< 0
x > 4
5
288
> 0
0
0
0
positive
negative
negative
positive
Chapter 3
186
Example 6
Find the domain of the function
2
56)( tttv −−=
.
A square root is only defined when the quantity we are taking the square root of, the
quantity inside the square root, is zero or greater. Thus, the domain of this function will
be when
056
2
−− tt
.
We start by solving the equality
056
2
=−− tt
. While we could use the quadratic
formula, this equation factors nicely to
0)1)(6( =−+ tt
, giving horizontal intercepts t =
1 and t = -6.
Sketching a graph of this quadratic will allow us to
determine when it is positive.
From the graph we can see this function is positive
for inputs between the intercepts. So
056
2
−− tt
for
16 − t
, and this will be the domain of the v(t)
function.
Writing Equations using Intercepts
Since a polynomial function written in factored form will have a horizontal intercept
where each factor is equal to zero, we can form a function that will pass through a set of
horizontal intercepts by introducing a corresponding set of factors.
Writing an Equation for a Polynomial from Intercepts
If we know the horizontal intercepts and the behavior or multiplicity at those
intercepts, we can write a polynomial of minimal degree with those intercepts.
1. Determine the horizontal intercepts
12
, , ,
k
x x x x=
2. Examine the behavior at each intercept to determine the corresponding multiplicity
of each intercept,
12
, , ,
k
p p p
3. Write the polynomial in factored form
12
12
( ) ( ) ( ) ( )
k
p
pp
k
f x a x x x x x x= − − −
4. Use another point on the graph to solve for the stretch factor a
Notice the degree of the polynomial will be the sum of the multiplicities
i
p
.
3.3 Graphs of Polynomial Functions
187
Example 7
Write a formula for the polynomial function
graphed here.
This graph has three horizontal intercepts: x = -3,
2, and 5. At x = -3 and 5 the graph passes through
the axis, suggesting the corresponding factors of
the polynomial will be linear. At x = 2 the graph
bounces at the intercept, suggesting the
corresponding factor of the polynomial will be 2
nd
degree (quadratic).
Together, this gives us:
)5()2)(3()(
2
−−+= xxxaxf
To determine the stretch factor, we can utilize another point on the graph. Here, the
vertical intercept appears to be (0,-2), so we can plug in those values to solve for a:
30
1
602
)50()20)(30(2
2
=
−=−
−−+=−
a
a
a
The graphed polynomial appears to represent the function
)5()2)(3(
30
1
)(
2
−−+= xxxxf
.
Try it Now
3. Given the graph, write a formula for the function shown.
Chapter 3
188
Estimating Extrema
With quadratics, we were able to algebraically find the maximum or minimum value of
the function by finding the vertex. For general polynomials, finding these turning points
is not possible without more advanced techniques from calculus. Even then, finding
where extrema occur can still be algebraically challenging. For now, we will estimate the
locations of turning points using technology to generate a graph.
Example 8
An open-top box is to be constructed by cutting out squares from each corner of a 14cm
by 20cm sheet of plastic then folding up the sides. Find the size of squares that should
be cut out to maximize the volume enclosed by the box.
We will start this problem by drawing a picture, labeling the
width of the cut-out squares with a variable, w.
Notice that after a square is cut out from each end, it leaves a
)214( w−
cm by
)2120( w−
cm rectangle for the base of the
box, and the box will be w cm tall. This gives the volume:
32
468280)220)(214()( wwwwwwwV +−=−−=
Using technology to sketch a graph allows us to estimate the maximum value for the
volume, restricted to reasonable values for w: values from 0 to 7.
From this graph, we can estimate the maximum value is around 340, and occurs when
the squares are about 2.75cm square. To improve this estimate, we could use advanced
features of our technology, if available, or simply change our window to zoom in on our
graph.
w
w
3.3 Graphs of Polynomial Functions
189
From this zoomed-in view, we can refine our estimate for the max volume to about 339,
when the squares are 2.7cm square.
Try it Now
4. Use technology to find the maximum and minimum values on the interval [-1, 4] of the
function
)4()1()2(2.0)(
23
−+−−= xxxxf
.
Important Topics of this Section
Short Run Behavior
Intercepts (Horizontal & Vertical)
Methods to find Horizontal intercepts
Factoring Methods
Factored Forms
Technology
Graphical Behavior at intercepts
Single, Double and Triple zeros (or multiplicity 1, 2, and 3 behaviors)
Solving polynomial inequalities using test values & graphing techniques
Writing equations using intercepts
Estimating extrema
Chapter 3
190
Try it Now Answers
1. Vertical intercept (0, 0).
24
40 tt −=
factors as
( )
( )( )
2240
222
+−=−= ttttt
Horizontal intercepts (0, 0), (-2, 0), (2, 0)
2. Vertical intercept (0, 0),
Horizontal intercepts (-2, 0), (0, 0), (3, 0)
The function is negative on (
−
, -2) and (0, 3)
The function is positive on (-2, 0) and (3,
)
The leading term is
3
x
so as
−→x
,
−→)(xg
and as
→x
,
→)(xg
3. Double zero at x=-1, triple zero at x=2. Single zero at x=4.
)4()1()2()(
23
−+−= xxxaxf
. Substituting (0,-4) and solving for a,
32
1
( ) ( 2) ( 1) ( 4)
8
f x x x x= − − + −
4. The minimum occurs at approximately the point (0, -6.5), and the maximum occurs at
approximately the point (3.5, 7).
3.3 Graphs of Polynomial Functions
191
Section 3.3 Exercises
Find the C and t intercepts of each function.
1.
( ) ( )( )
2 4 1 ( 6)C t t t t= − + −
2.
( ) ( )( )
3 2 3 ( 5)C t t t t= + − +
3.
( ) ( )
2
4 2 ( 1)C t t t t= − +
4.
( ) ( )( )
2
2 3 1C t t t t= − +
5.
( )
4 3 2
2 8 6C t t t t= − +
6.
( )
4 3 2
4 12 40C t t t t= + −
Use your calculator or other graphing technology to solve graphically for the zeros of the
function.
7.
( )
32
7 4 30f x x x x= − + +
8.
( )
32
6 28g x x x x= − + +
Find the long run behavior of each function as
t →
and
t → −
9.
( ) ( ) ( )
33
3 5 3 ( 2)h t t t t= − − −
10.
( ) ( ) ( )
23
2 3 1 ( 2)k t t t t= − + +
11.
( ) ( )( )
2
2 1 3p t t t t= − − −
12.
( ) ( )( )
3
4 2 1q t t t t= − − +
Sketch a graph of each equation.
13.
( ) ( )
2
3 ( 2)f x x x= + −
14.
( ) ( )( )
2
41g x x x= + −
15.
( ) ( ) ( )
32
13h x x x= − +
16.
( ) ( ) ( )
32
32k x x x= − −
17.
( ) ( )
2 1 ( 3)m x x x x= − − +
18.
( ) ( )
3 2 ( 4)n x x x x= − + −
Solve each inequality.
19.
( )( )
2
3 2 0xx− −
20.
( )( )
2
5 1 0xx− +
21.
( )( )( )
1 2 3 0x x x− + −
22.
( )( )( )
4 3 6 0x x x− + +
Find the domain of each function.
23.
( )
2
42 19 2f x x x= − + −
24.
( )
2
28 17 3g x x x= − −
25.
( )
2
45h x x x= − +
26.
( )
2
2 7 3k x x x= + +
27.
( ) ( )( )
2
32n x x x= − +
28.
( ) ( )
2
1 ( 3)m x x x= − +
29.
( )
2
1
28
pt
tt
=
+−
30.
( )
2
4
45
qt
xx
=
−−
Chapter 3
192
Write an equation for a polynomial the given features.
31. Degree 3. Zeros at x = -2, x = 1, and x = 3. Vertical intercept at (0, -4)
32. Degree 3. Zeros at x = -5, x = -2, and x = 1. Vertical intercept at (0, 6)
33. Degree 5. Roots of multiplicity 2 at x = 3 and x = 1, and a root of multiplicity 1 at
x = -3. Vertical intercept at (0, 9)
34. Degree 4. Root of multiplicity 2 at x = 4, and a roots of multiplicity 1 at x = 1 and
x = -2. Vertical intercept at (0, -3)
35. Degree 5. Double zero at x = 1, and triple zero at x = 3. Passes through the point
(2, 15)
36. Degree 5. Single zero at x = -2 and x = 3, and triple zero at x = 1. Passes through the
point (2, 4)
Write a formula for each polynomial function graphed.
37. 38. 39.
40. 41. 42.
43. 44.
3.3 Graphs of Polynomial Functions
193
Write a formula for each polynomial function graphed.
45. 46.
47. 48.
49. 50.
51. A rectangle is inscribed with its base on the x axis and its upper corners on the
parabola
2
5yx=−
. What are the dimensions of such a rectangle that has the greatest
possible area?
52. A rectangle is inscribed with its base on the x axis and its upper corners on the curve
4
16yx=−
. What are the dimensions of such a rectangle that has the greatest
possible area?
