* Recognize characteristics of graphs of polynomial functions. * Use factoring to find zeros of polynomial functions. * Identify zeros and their multiplicities. * Determine end behavior. * Understand the relationship between degree and turning points. * Graph polynomial functions. * Use the Intermediate Value Theorem.

1. 5.3 Graphs of Polynomial
Functions
Chapter 5 Polynomial and Rational Functions

2. Concepts and Objectives
⚫ Objectives for this section are
⚫ Recognize characteristics of graphs of polynomial
functions.
⚫ Use factoring to find zeros of polynomial functions.
⚫ Identify zeros and their multiplicities.
⚫ Determine end behavior.
⚫ Understand the relationship between degree and
turning points.
⚫ Graph polynomial functions.
⚫ Use the Intermediate Value Theorem.

3. Graph Characteristics
⚫ Polynomial functions of degree 2 or more have graphs
that do not have sharp corners; these types of graphs are
called smooth curves.
⚫ Polynomial graphs are also continuous because they
have no breaks.

4. Using Factoring to Find Zeros
⚫ If f is a polynomial function, the values of x for which
f(x) = 0 are called zeros of f. If the equation of the
polynomial function can be factored, we can set each
factor equal to zero and solve for the zeros.
⚫ Sometimes the factors will already be given to you;
otherwise, you will either need to factor it or use
Desmos to locate the x-intercepts (the zeros).
⚫ Example: Find the x-intercepts of ( ) ( )( )
= − +
3 1
f x x x x
0
x = 3 0
3
x
x
− =
=
1 0
1
x
x
+ =
= −

5. Using Factoring to Find Zeros
⚫ Example: Find the x-intercepts of
Set each factor equal to 0 and solve:
( ) 6 4 2
3 2
f x x x x
= − +
Re-write as equation
Factor out the GCF
Factor the trinomial
( )
2 4 2
3 1 0
x x x
− + =
( )( )
2 2 2
1 2 0
x x x
− − =
6 4 2
3 2 0
x x x
− + =
2
0
0
x
x
=
=
2
2
1 0
1
1
x
x
x
− =
=
= 
2
2
2 0
2
2
x
x
x
− =
=
= 

6. Using Factoring to Find Zeros
⚫ Example (cont):
You can also graph the function in Desmos, although
you lose some of the accuracy of the .
2


7. Finding the y-intercepts
⚫ Finding the y-intercepts of a polynomial graph is much
simpler: just plug in 0 for x and evaluate.
⚫ Example: Find the y-intercepts of
Substituting 0 for x makes everything 0 except for
the –6, so the y-intercept is (0, –6).
⚫ Example: Find the y-intercepts of
So the y-intercept is (0, 12)
( ) 3 2
4 6
g x x x x
= + + −
( ) ( ) ( )
2
2 2 3
h x x x
= − +
( ) ( ) ( )
( )
( )( )
2
0 0 2 2 0 3
4 3 12
h = − +
= =

8. Zeros and Multiplicities
⚫ The number of time a given factor appears in the
factored form of the equation of a polynomial is called
the multiplicity. For example, the polynomial
g has multiplicity 5.
⚫ The multiplicity of a zero and whether the multiplicity is
even or odd determines what the graph does at a zero.
⚫ A zero of multiplicity one crosses the x-axis.
⚫ A zero of even multiplicity turns or “bounces” at the x-axis
⚫ A zero of odd multiplicity greater than one crosses the x-
axis and “wiggles”.
⚫ The sum of the multiplicities is equal to the degree.
( ) ( )
5
4
x x
= −

9. Zeros and Multiplicities
⚫ Example: Use the graph of function of degree 6 to
identify the zeros of the function and their possible
multiplicities.

10. Zeros and Multiplicities
⚫ Example: Use the graph of function of degree 6 to
identify the zeros of the function and their possible
multiplicities.
From the left, the first zero is at
x = –3. The graph bounces, so it
is an even multiplicity.

11. Zeros and Multiplicities
⚫ Example: Use the graph of function of degree 6 to
identify the zeros of the function and their possible
multiplicities.
From the left, the first zero is at
x = –3. The graph bounces, so it
is an even multiplicity.
Next is a zero at x = –1. The
graph goes straight through, so
we assume a multiplicity of 1.

12. Zeros and Multiplicities
⚫ Example: Use the graph of function of degree 6 to
identify the zeros of the function and their possible
multiplicities.
From the left, the first zero is at
x = –3. The graph bounces, so it
is an even multiplicity.
Next is a zero at x = –1. The
graph goes straight through, so
we assume a multiplicity of 1.
Finally, we have a zero at x = 4
that “wiggles”, so we assume an
odd multiplicity.

13. Zeros and Multiplicities
⚫ Example: Use the graph of function of degree 6 to
identify the zeros of the function and their possible
multiplicities.
Because the degree of the
function is 6, the multiplicites
have to add up to 6.
So, even + 1 + odd = 6 means
that –3 has multiplicity 2, –1 has
multiplicity 1, and 4 has
multiplicity 3.

14. Turning Points
⚫ The point where a graph changes direction (“bounces”
or “wiggles”) is called a turning point of the function.
⚫ A function of degree n will have at most n – 1 turning
points, with at least one turning point between each
pair of adjacent zeros.

15. Intermediate Value Theorem
⚫ This means that if we plug in two numbers for x, and the
answers have different signs (one positive and one
negative), the function has to have crossed the x-axis
somewhere between the two values.
If f (x) defines a polynomial function with only real
coefficients, and if for real numbers a and b, the
values f (a) and f (b) are opposite in sign, then
there exists at least one real zero between a and b.

16. Intermediate Value Theorem
⚫ Example: Show that has a real
zero between 2 and 3.
( )= − − +
3 2
2 1
f x x x x

17. Intermediate Value Theorem
⚫ Example: Show that has a real
zero between 2 and 3.
Since the function went from negative at 2 to
positive at 3, it must have crossed 0 somewhere
between.
( )= − − +
3 2
2 1
f x x x x
( ) ( )
3 2
2 2 2 2 2 1
8 8 2 1 1
f = − − +
= − − + = −
( ) ( )
3 2
3 3 2 3 3 1
27 18 3 1 7
f = − − +
= − − + =

18. Intermediate Value Theorem
⚫ If f (a) and f (b) are not opposite in sign, it does not
necessarily mean that there is no zero between a and b.
Consider the function, , at –1 and 3:
( )= − −
2
2 1
f x x x
f (–1) = 2 > 0 and f (3)= 2 >0
This would imply that there is no
zero between –1 and 3, but we can
see that f has two zeros between
those points.

19. Putting It All Together
⚫ Given a graph of a polynomial function, we can use
everything we’ve learned to write a formula for the
function.
⚫ To do this:
1. Identify the x-intercepts to find the factors.
2. Determine the multiplicity of the zeros.
3. Write a polynomial of least degree that fits.
4. Use any other point on the graph (the y-intercept
may be easiest) to write the exact function.

20. Putting It All Together
⚫ Example: Write a formula for the graph.
What we know:
• The graph has zeros at –3, 2,
and 5.
• –3 and 5 have multiplicity 1,
• 2 has an even multiplicity.
Since we are assuming a
polynomial of least degree, we
will set the multiplicity at 2.
• The y-intercept is at (0, –2)

21. Putting It All Together
⚫ Example: Write a formula for the graph.
This gives us a starting point of
Now we plug in the y-intercept:
( ) ( )( ) ( )
2
3 2 5
f x a x x x
= + − −
( )( ) ( )
( )( )( )
2
0 3 0 2 0 5 2
3 4 5 2
60 2
1
30
a
a
a
a
+ − − = −
− = −
− = −
=

22. Putting It All Together
⚫ Example: Write a formula for the graph.
So the graph appears to represent
the function
( ) ( )( ) ( )
2
1
3 2 5
30
f x x x x
= + − −

23. Local and Global Extrema
⚫ With quadratics, we were able to find the maximum or
minimum value of the function by finding the vertex,
either algebraically or by graphing.
⚫ To do this algebraically for functions of higher degree
requires more advanced techniques from calculus, so for
our purposes, we will estimate the locations of turning
points using graphs.
⚫ Each turning point represents a local maximum or
minimum. Sometimes a turning point is the highest or
lowest point on the entire graph. In these cases, the
turning point is the global maximum/minimum.

24. Using Local Extrema
⚫ An open-top box is to be constructed by cutting out
squares from each corner of a 14 cm by 20 cm sheet
of plastic and then folding up the sides. Find the size of
the squares that should be cut out to maximize the
volume enclosed by the box.

25. Using Local Extrema
⚫ An open-top box is to be constructed by cutting out
squares from each corner of a 14 cm by 20 cm sheet of
plastic and then folding up the sides. Find the size of the
squares that should be cut out to maximize the volume
enclosed by the box.
1. Draw a picture! It’s a
lot easier to see
what’s going on if
you sketch it out.
x
x
20 cm
14
cm

26. Using Local Extrema (cont.)
2. Notice that after the squares are cut out, it leaves a
(20-2x) cm by (14-2x) cm rectangle for the base of the
box, and the box will be x cm tall.
x
x
20 cm
14
cm
20-2x
14-2x
3. This gives the volume
4. The three zeros are 0,
10, and 7. A height of 0
doesn’t make sense, so
we will consider only
10 and 7.
( ) ( )( )
20 2 14 2
V x x x x
= − −

27. Using Local Extrema (cont.)
5. If we use x = 7, the side that is 14-2x becomes 0, so we
have to restrict the domain of our function to 0 < x < 7.
6. Now, let’s look at the graph of this function in Desmos:

28. Using Local Extrema (cont.)
5. If we use x = 7, the side that is 14-2x becomes 0, so we
have to restrict the domain of our function to 0 < x < 7.
6. Now, let’s look at the graph of this function in Desmos:
7.
7. Looking at the graph
between 0 and 7, we
can see that the
maximum value
around x = 2.75.

29. Using Local Extrema (cont.)
5. If we use x = 7, the side that is 14-2x becomes 0, so we
have to restrict the domain of our function to 0 < x < 7.
6. Now, let’s look at the graph of this function in Desmos:
7.
7. Looking at the graph
between 0 and 7, we
can see that the
maximum value
around x = 2.75.
8. Clicking on the local
maximum, we get a
more precise figure.

30. Using Local Extrema (cont.)
9. The local maximum gives us the value of (2.7, 339).
10. This means that if we cut a square of 2.7 cm from each
end, we will get a maximum volume of 339 cm2.

31. Classwork
⚫ College Algebra 2e
⚫ 5.3: 8-28 (×4); 5.2: 26-38 (even); 5.1: 66-72 (even)
⚫ 5.3 Classwork Check
⚫ Quiz 5.2
⚫ I have reposted the notes from 5.2 in case you need to
refresh your memory.