Name
5.4
Class
Date
Graphing Polynomial Functions
Essential Question: How do you sketch the graph of a polynomial function in intercept form?
Explore 1
Resource
Locker
Investigating the End Behavior of the
Graphs of Simple Polynomial Functions
Linear, quadratic, and cubic functions belong to a more general class of functions called polynomial
functions, which are categorized by their degree. Linear functions are polynomial functions of degree 1,
quadratic functions are polynomial functions of degree 2, and cubic functions are polynomial functions of degree 3.
In general, a polynomial function of degree n has the standard form p(x) = a nx n + a n-1x n-1 + ... + a 2x 2 + a 1x + a 0 ,
where a n, a n-1,..., a 2, a 1, and a 0 are real numbers called the coefficients of the expressions a nx n, a n-1x n - 1,..., a 2x 2, a 1x,
and a 0, which are the terms of the polynomial function. (Note that the constant term, a 0, appears to have no power of
x associated with it, but since x 0 = 1, you can write a 0 as a 0x 0 and treat a 0 as the coefficient of the term.)
A polynomial function of degree 4 is called a quartic function, while a polynomial function of degree 5 is
called a quintic function. After degree 5, polynomial functions are generally referred to by their degree, as in
“a sixth-degree polynomial function.”
A
Use a graphing calculator to graph the polynomial functions ƒ(x) = x, ƒ(x) = x 2, ƒ(x) = x 3, ƒ(x) = x 4,
ƒ(x) = x 5, and ƒ(x) = x 6. Then use the graph of each function to determine the function’s domain,
range, and end behavior. (Use interval notation for the domain and range.)
Function
Domain
Range
f(x) = x
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f(x) = x 2
f(x) = x 3
f(x) = x 4
f(x) = x 5
f(x) = x 6
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End Behavior
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
Lesson 4
B
Use a graphing calculator to graph the polynomial functions ƒ(x) = -x, ƒ(x) = -x 2, ƒ(x) = -x 3,
ƒ(x) = -x 4, ƒ(x) = -x 5, and ƒ(x) = -x 6. Then use the graph of each function to determine the
function’s domain, range, and end behavior. (Use interval notation for the domain and range.)
Function
Domain
Range
f(x) = -x
f(x) = - x 2
f(x) = - x 3
f(x) = - x 4
f(x) = - x 5
f(x) = - x 6
End Behavior
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
As x → +∞, f(x) →
.
As x → -∞, f(x) →
.
Reflect
1.
How can you generalize the results of this Explore for ƒ(x) = x n and ƒ(x) = -x n where n is positive whole
number?
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Module 5
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Lesson 4
Explore 2
Investigating the x-intercepts and Turning Points
of the Graphs of Polynomial Functions
The cubic function ƒ(x) = x 3 has three factors, all of which happen to be x. One or more of the x’s can be replaced
with other linear factors in x, such as x - 2, without changing the fact that the function is cubic.
In general, a polynomial function of the form p(x) = a(x - x 1)(x - x 2)...(x - x n) where a, x 1, x 2,..., and x n
are real numbers (that are not necessarily distinct) has degree n where n is the number of variable factors.
The graph of p(x) = a(x - x 1)(x - x 2)...(x - x n) has x 1, x 2,..., and x n as its x-intercepts, which is why the polynomial
is said to be in intercept form. Since the graph of p(x) intersects the x-axis only at its x-intercepts,
the graph must move away from and then move back toward the x-axis between each pair of successive x-intercepts,
which means that the graph has a turning point between those x-intercepts. Also, instead of crossing the x-axis at an
x-intercept, the graph can be tangent to the x-axis, and the point of tangency becomes a turning point because the
graph must move toward the x-axis and then away from it near the point of tangency.
The y-coordinate of each turning point is a maximum or minimum value of the function at least near that turning
point. A maximum or minimum value is called global or absolute if the function never takes on a value that is greater
than the maximum or less than the minimum. A local maximum or local minimum, also called a relative maximum or
relative minimum, is a maximum or minimum within some interval around the turning point that need not be (but
may be) a global maximum or global minimum.
A
Use a graphing calculator to graph the cubic functions ƒ(x) = x 3, ƒ(x) = x 2(x - 2), and
ƒ(x) = x(x - 2)(x + 2). Then use the graph of each function to answer the questions in
the table.
Function
f (x) = x 2(x - 2)
f (x) = x 3
f (x) = x(x - 2)(x + 2)
How many distinct factors
does f(x) have?
What are the graph’s
x-intercepts?
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Is the graph tangent to the
x-axis or does it cross the
x-axis at each x-intercept?
How many turning points
does the graph have?
How many global maximum
values? How many local
maximum values that are not
global?
How many global minimum
values? How many local
minimum values that are not
global?
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Lesson 4
B
Use a graphing calculator to graph the quartic functions ƒ(x) = x 4, ƒ(x) = x 3(x - 2),
ƒ(x) = x 2(x - 2)(x + 2), and ƒ(x) = x(x - 2)(x + 2)(x + 3).Then use the graph of
each function to answer the questions in the table.
Function
f (x) = x 4
f (x) = x 3(x - 2)
How many distinct
factors?
f (x) = x 2(x - 2)
(x + 2)
f (x) = x(x - 2)
(x + 2)(x + 3)
What are the
x-intercepts?
Tangent to or cross
the x-axis
at x-intercepts?
How many turning
points?
How many global
maximum values?
How many local
maximum values
that are not
global?
How many global
minimum values?
How many local
minimum values
that are not
global?
Reflect
What determines how many x-intercepts the graph of a polynomial function in intercept form has?
3.
What determines whether the graph of a polynomial function in intercept form crosses the x-axis or is
tangent to it at an x-intercept?
4.
Suppose you introduced a factor of -1 into each of the quartic functions in Step B. (For instance, ƒ(x) = x 4
becomes ƒ(x) = -x 4.) How would your answers to the questions about the functions and their graphs
change?
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Lesson 4
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2.
Explain 1
Sketching the Graph of Polynomial Functions
in Intercept Form
Given a polynomial function in intercept form, you can sketch the function’s graph by using the end behavior,
the x-intercepts, and the sign of the function values on intervals determined by the x-intercepts. The sign of the
function values tells you whether the graph is above or below the x-axis on a particular interval. You can find the
sign of the function values by determining the sign of each factor and recognizing what the sign of the product of
those factors is.
Example 1
A
Sketch the graph of the polynomial function.
ƒ(x) = x(x + 2)(x − 3)
Identify the end behavior. For the function p(x) = a(x − x 1)(x − x 2) . . . (x − x n), the end behavior is
determined by whether the degree n is even or odd and whether the constant factor a is positive or
negative. For the given function f(x), the degree is 3 and the constant factor a, which is 1, is positive, so
ƒ(x) has the following end behavior:
As x → +∞, ƒ(x) → +∞.
As x → -∞, ƒ(x) → -∞.
Identify the graph’s x-intercepts, and then use the sign of ƒ(x) on intervals determined by the x-intercepts
to find where the graph is above the x-axis and where it’s below the x-axis.
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The x-intercepts are x = 0, x = -2, and x = 3. These three x-intercepts divide the x-axis into four
intervals: x < -2, -2 < x < 0, 0 < x < 3, and x > 3.
Interval
Sign of the
Constant
Factor
Sign of x
Sign of
x+2
Sign of
x-3
x < -2
+
-
-
-
-2 < x < 0
+
-
+
-
+
0< x <3
+
+
+
-
-
x>3
+
+
+
+
+
Sign of
f (x) = x(x +2)(x - 3)
-
So, the graph of ƒ(x) is above the x-axis on the intervals -2 < x < 0 and x > 3, and it’s below the x-axis
on the intervals x < -2 and 0 < x < 3.
Sketch the graph.
y
While you should be precise about where the graph crosses the x-axis,
you do not need to be precise about the y-coordinates of points on the
graph that aren’t on the x-axis. Your sketch should simply show where
the graph lies above the x-axis and where it lies below the x-axis.
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x
-4
0
2
4
Lesson 4
B
ƒ(x) = −(x − 4)(x − 1)(x + 1)(x + 2)
Identify the end behavior.
As x → +∞, ƒ(x) →
.
As x → -∞, ƒ(x) →
.
Identify the graph’s x-intercepts, and then use the sign of ƒ(x) on intervals determined by the x-intercepts
to find where the graph is above the x-axis and where it’s below the x-axis.
The x-intercepts are x =
Interval
, x=
Sign of the
Constant
Factor
, x=
Sign
of
x-4
, x=
Sign
of
x-1
Sign
of
x+1
.
Sign
of
x+2
-
-
-
<x<
-
-
+
<x<
-
+
+
<x<
-
+
+
-
+
+
x<
x>
Sign of
f(x) = -(x - 4)(x - 1)
(x + 1)(x + 2)
So, the graph of ƒ(x) is above the x-axis on the intervals
<x<
and
<x<
it’s below the x-axis on the intervals x <
and x >
y
, and
<x<
,
,
x
-4
.
-2
0
2
4
© Houghton Mifflin Harcourt Publishing Company
Sketch the graph.
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Lesson 4
Your Turn
Sketch the graph of the polynomial function.
y
ƒ(x) = -x 2(x - 4)
5.
x
-4
Interval
Explain 2
Sign of the
Constant
Factor
Sign of x 2
Sign of
x-4
-2
0
2
4
Sign of
f(x) = x 2(x - 4)
Modeling with a Polynomial Function
You can use cubic functions to model real-world situations. For example, you find the volume
of a box (a rectangular prism) by multiplying the length, width, and height. If each dimension
of the box is given in terms of x, then the volume is a cubic function of x.
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Example 2
To create an open-top box out of a sheet of cardboard that is 9 inches long and 5 inches wide,
you make a square flap of side length x inches in each corner by cutting along one of the flap’s
sides and folding along the other side. (In the first diagram, a solid line segment in the interior
of the rectangle indicates a cut, while a dashed line segment indicates a fold.) After you fold up
the four sides of the box (see the second diagram), you glue each flap to the side it overlaps.
To the nearest tenth, find the value of x that maximizes the volume of the box.
x
x
5 in.
9 in.
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Lesson 4
Analyze Information
Identify the important information.
A square flap of side length x inches is made in each corner of a rectangular
sheet of cardboard.
The sheet of cardboard measures 9 inches by 5 inches.
Formulate a Plan
Find the dimensions of the box once the flaps have been made and the sides
have been folded up. Create a volume function for the box, graph the function
on a graphing calculator, and use the graph to find the value of x that maximizes
the volume.
Solve
1. Write expressions for the dimensions of the box.
Length of box: 9 Width of box: 5 Height of box:
2. Write the volume function and determine its domain.
(
V(x) = 9 -
)(5 - )
Because the length, width, and height of the box must all be positive, the volume
function’s domain is determined by the following three constraints:
9 - 2x > 0, or x <
5 - 2x > 0, or x <
Taken together, these constraints give a domain of 0 < x <
© Houghton Mifflin Harcourt Publishing Company
x>0
.
3. Use a graphing calculator to graph the volume function on its domain.
Adjust the viewing window so you can see the maximum. From the
graphing calculator’s CALC menu, select 4: maximum to locate the
point where the maximum value occurs.
So, V(x) ≈ 21.0 when x ≈
, which means that the box has a
maximum volume of about 21 cubic inches when square flaps with a side
length of 1 inch are made in the corners of the sheet of cardboard.
Module 5
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Lesson 4
Justify and Evaluate
Making square flaps with a side length of 1 inch means that the box will be 7 inches long, 3 inches
wide, and 1 inch high, so the volume is 21 cubic inches. As a check on this result, consider making
square flaps with a side length of 0.9 inch and 1.1 inches:
V(0.9) = (9 - 1.8)(5 - 1.8)(0.9) =
V(1.1) = (9 - 2.2)(5 - 2.2)(1.1) =
Both volumes are slightly less than 21 cubic inches, which suggests that 21 cubic
inches is the maximum volume.
Reflect
6.
Discussion Although the volume function has three constraints on its domain, the domain involves
only two of them. Why?
Your Turn
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7.
x
To create an open-top box out of a sheet of cardboard that is 25 inches long and x
13 inches wide, you make a square flap of side length x inches in each corner by
13 in.
cutting along one of the flap’s sides and folding along the other. (In the diagram,
a solid line segment in the interior of the rectangle indicates a cut, while a
25 in.
dashed line segment indicates a fold.) Once you fold up the four sides of the
box, you glue each flap to the side it overlaps. To the nearest tenth, find the value of x that maximizes the
volume of the box.
Elaborate
8.
Compare and contrast the domain, range, and end behavior of ƒ(x) = x n when n is even and when n is odd.
9.
Essential Question Check-In For a polynomial function in intercept form, why is the constant factor
important when graphing the function?
Module 5
301
Lesson 4
Evaluate: Homework and Practice
t0OMJOF)PNFXPSL
t)JOUTBOE)FMQ
t&YUSB1SBDUJDF
Use a graphing calculator to graph the polynomial function. Then use
the graph to determine the function’s domain, range, and end behavior.
(Use interval notation for the domain and range.)
1.
ƒ(x) = x 7
2.
ƒ(x) = -x 9
3.
ƒ(x) = x 10
4.
ƒ(x) = -x 8
Use a graphing calculator to graph the function. Then use the graph to determine the
number of turning points and the number and type (global, or local but not global) of
any maximum or minimum values.
ƒ(x) = x(x + 1)(x + 3)
6.
2
ƒ(x) = (x + 1) (x - 1)(x - 2)
7.
ƒ(x) = -x(x - 2)
8.
ƒ(x) = -(x - 1)(x + 2)
Module 5
2
302
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5.
3
Lesson 4
Sketch the graph the polynomial function.
9.
2
ƒ(x) = x (x - 2)
y
Interval
Sign of f(x) = x 2(x - 2)
x
-4
-2
0
2
4
10. ƒ(x) = -(x + 1)(x - 2)(x - 3)
y
Interval
Sign of
f(x) = -(x + 1)(x - 2)(x - 3)
x
© Houghton Mifflin Harcourt Publishing Company
-4
-2
0
2
4
2
11. ƒ(x) = x(x + 2) (x - 1)
y
x
-4
Module 5
-2
0
2
Interval
Sign of f(x) = x(x + 2) 2(x - 1)
4
303
Lesson 4
12. To create an open-top box out of a sheet of cardboard that is 6 inches long and
3 inches wide, you make a square flap of side length x inches in each corner
by cutting along one of the flap’s sides and folding along the other. Once you
fold up the four sides of the box, you glue each flap to the side it overlaps. To
the nearest tenth, find the value of x that maximizes the volume of the box.
13. The template shows how to create a box from a square sheet of cardboard
that has a side length of 36 inches. In the template, solid line segments
indicate cuts, dashed line segments indicate folds, and grayed rectangles
indicate pieces removed. The vertical strip that is 2 inches wide on the left
side of the template is a flap that will be glued to the side of the box that it
overlaps when the box is folded up. The horizontal strips that are __2x inches
wide at the top and bottom of the template are also flaps that will overlap
to form the top and bottom of the box when the box is folded up. Write a
volume function for the box in terms of x only. (You will need to determine
a relationship between x and y first.) Then, to the nearest tenth, find the
dimensions of the box with maximum volume.
x
x
3 in.
6 in.
2 in. x
x
2
y
x
y
36 in.
x
2
36 in.
© Houghton Mifflin Harcourt Publishing Company
Module 5
304
Lesson 4
Write a cubic function in intercept form for the given graph, whose x-intercepts
are integers. Assume that the constant factor a is either 1 or -1.
14.
15.
y
y
x
-4
-2
0
2
x
-4
4
-2
0
2
4
Write a quartic function in intercept form for the given graph, whose x-intercepts
are integers. Assume that the constant factor a is either 1 or -1.
16.
17.
y
y
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x
-4
Module 5
-2
0
2
x
-4
4
305
0
2
4
Lesson 4
18. Multiple Response Select all statements that apply to the graph of ƒ(x) = (x - 1) (x + 2).
2
A. The x-intercepts are x = 1 and x = -2.
B. The x-intercepts are x = -1 and x = 2.
C. The graph crosses the x-axis at x = 1 and is tangent to the x-axis at x = -2.
D. The graph crosses the x-axis at x = -1 and is tangent to the x-axis at x = 2.
E. The graph is tangent to the x-axis at x = 1 and crosses the x-axis at x = -2.
F. The graph is tangent to the x-axis at x = -1 and crosses the x-axis at x = 2.
G. A local, but not global, minimum occurs on the interval -2 < x < 1, and a local, but
not global, maximum occurs at x = 1.
H. A local, but not global, maximum occurs on the interval -2 < x < 1, and a local, but
not global, minimum occurs at x = 1.
I. A local, but not global, minimum occurs on the interval -1 < x < 2, and a local, but
not global, maximum occurs at x = 2.
J. A local, but not global, maximum occurs on the interval -1 < x < 2, and a local, but
not global, minimum occurs at x = 2.
H.O.T. Focus on Higher Order Thinking
19. Explain the Error A student was asked to sketch the graph of the function
ƒ(x) = x 2(x - 3). Describe what the student did wrong. Then sketch the correct graph.
y
y
-4
Module 5
-2
0
2
4
© Houghton Mifflin Harcourt Publishing Company
x
x
-4
-2
306
0
2
4
Lesson 4
20. Make a Prediction Knowing the characteristics of the graphs of cubic and
quartic functions in intercept form, sketch the graph of the quintic function
2
ƒ(x) = x 2(x + 2)(x - 2) .
y
x
-4
-2
0
2
4
21. Represent Real-World Situations A rectangular piece of sheet metal is rolled
and riveted to form a circular tube that is open at both ends, as shown. The sheet
metal has a perimeter of 36 inches. Each of the two sides of the rectangle that form
the two ends of the tube has a length of x inches, and the tube has a circumference
of x - 1 inches because an overlap of 1 inch is needed for the rivets. Write a volume
function for the tube in terms of x. Then, to the nearest tenth, find the value of x that
maximizes the volume of the tube.
x- 1
© Houghton Mifflin Harcourt Publishing Company
x
Module 5
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Lesson 4
Lesson Performance Task
The template shows how to create a box with a lid from a sheet of card stock that is 10 inches wide
and 24 inches long. In the template, solid line segments indicate cuts, and dashed line segments
indicate folds. The square flaps, each with a side length of x inches, are glued to the sides they overlap
when the box is folded up. The box has a bottom and four upright sides. The lid, which is attached to
one of the upright sides, has three upright sides of its own. Assume that the three sides of the lid can
be tucked inside the box when the lid is closed.
x
x
10 in.
24 in.
a. Write a polynomial function that represents the volume of the box, and state its domain.
b. Use a graphing calculator to find the value of x that will produce the box with
maximum volume. What are the dimensions of that box?
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Module 5
308
Lesson 4