Section 2.3
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In 1980, U.S. doctors diagnosed 41 cases of a rare
form of cancer, Kaposi’s sarcoma, which involved
skin lesions, pneumonia, and severe
immunological deficiencies. All cases involved
gay men ranging in age from 26 to 51. By the end
of 2005, more than one million Americans,
straight and gay, male and female, old and young,
were infected with HIV.
Modeling AIDS-related data and making
predictions about the epidemic’s havoc is serious
business. This graph shows the number of AIDS
cases diagnosed in the U.S. from 1983 to 2005.

Changing circumstances and unforeseen events
can result in models for AIDS-related data that
are not particularly useful over long periods of
time. For example, the function
3
2
f x   49x  806x  3776x  2503
models the number of AIDS cases diagnosed in
the U.S. x years after 1983 to 2005.
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Polynomial functions are f(x) = anxn + an-1xn-1 + … +
a1x + a0 where an an-1 … a1 a0 are all real numbers
in the function.
A constant function f(x) = c, where c ≠ 0, is a
polynomial function of degree 0.
A linear function f(x) = mx + b where m ≠ 0, is a
polynomial function degree of 1.
A quadratic function f(x) = ax2 + bx + c, where a ≠ 0,
is a polynomial function degree of 2.
In this section, we focus on polynomial functions of
degree 3 or higher.
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Polynomial functions of degree 2 or higher have graphs that are smooth
and continuous
By smooth, we mean that the graphs
contain only rounded curves with no
sharp corners.
Notice there are no breaks and smooth, rounded curves.

Graph with sharp turns are non-polynomial
functions.
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By continuous, we mean that the graphs have no
breaks and can be drawn without lifting your
pencil.

By discontinuous, we mean that the graphs
have breaks and cannot be drawn without
lifting your pencil.
Notice the breaks and lack of smooth curves.
Page 312 Problems 1 – 14
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This graph shows the function
f x   49x 3  806x 2  3776x  2503
which models the number of U.S.
AIDS diagnosed from 1983
through 2005. What happens
by year 21 (2004)?

Do you think this is
appropriate to model AIDS
cases for the future? Why or
why not?
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
Use the equation to predict the amount of cases that were
reported in 2010.
3
2
49  27   806  27   3776  27   2503  272438
Reported cases in 2010 were 40,950. Source from CDC
So what do you think of that?
How does our prediction compare to the actual number?
The Actual equation is… f(x) = 22.14x3 – 1130x2 + 16310x – 1070
http://www.cdc.gov/nchhstp/newsroom/docs/2012/HIV-Infections-2007-2010.pdf
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The behavior of the graph of the function to the far right or
the far left is called its end behavior.
Many polynomials may have several intervals where they
increase or decrease, but the graph will eventually rise or fall
without stopping.
The Leading Coefficient Test determines how to find out the
end behavior of a function by examining different parts of the
equation or graphed function.
First we examine the degree; there are two scenarios. Even
and Odd.
If the leading coefficient is POSITIVE,
the graph will fall to the left and rise
to the right.
If the leading coefficient is NEGATIVE,
the graph will rise to the left and fall
to the right.
If the leading coefficient is POSITIVE,
the graph will rise to the left and rise
to the right.
If the leading coefficient is NEGATIVE,
the graph will fall to the left and fall to
the right.
Example 1
Determine end behavior f(x)= x3 + 3x2 – x – 3
Degree
Coefficient
Left Side
Right side
Example 2
Determine the end behavior f(x)= -4x3 (x-1)2 (x + 5)
Degree
Coefficient
Left Side
Right side
Example 3
Determine the end behavior f(x)= -49x3 + 806x2 + 3776x + 2503
Degree
Coefficient
Left Side
Right side
Example 4
Determine the end behavior f(x)= x4 + 8x3 + 4x2 + 2
Degree
Coefficient
Left Side
Right side
Page 312 Problems 15 – 24
Think About it…
Use the Leading Coefficient Test to determine the end behavior of the graph
of f(x)= - .08x4 – 9x3 + 7x2 + 4x + 7
This is the graph that you get with the standard viewing window. How do
you know that you need to change the window to see the end behavior of the
function? What viewing window will allow you to see the end behavior?
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If f is a polynomial function, then the values of x for which
f(x) = 0 are called the zeros of f.
These values of x are the roots, or solutions (factors), of
the polynomial equation f(x) = 0. Each real root of the
polynomial equation appears as an x-intercept of the graph
of the polynomial function. We usually find these by
factoring.
By definition, the zeros are the values of x for which f (x) is
equal to 0. Thus we set f (x) equal to 0 and solve for x as follows:


x 3  4 x 2   3x  12  0
x 2  x  4   3 x  4   0


2
x

4
x

 3  0
x  4  0 x2  3  0
x  4
x 3
2
x  3
Example 5
Find all zeros of f (x) = -x4 + 4x3 - 4x2
  x2  x2  4x  4
  x 2  x  2  x  2 
  x  x  2
zeros : 0 and 2
2
Example 6
Find all zeros of f (x) = x4 – 4x2
 x2  x2  4
 x 2  x  2  x  2 
zeros : 0 and  2 and 2
2

Complete problems 4, 6, 12, 20, 22, 26, 30 on page 312
Example 7
Find all zeros of f (x) = 3 (x + 5) (x + 2)4
zeros :  5 and  2
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Open a NEW graph document. Graph
the equation on your calculator.
Hit Menu and select option 6: Analyze
Graph. Select option 1: Zeros.
Use the vertical line to move to the
Lower bound of the x-intercept and
hit Enter (or click the mouse pad).
Now move it to the Upper bound of
the SAME x-intercept and click again.
The ZERO should appear on the graph.
Repeat for each zero of the function.
What are the zeros of the function?
f x   x 3  2x 2  4x  8
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Grab your calculator.
Now follow the technology tips to
find the zeros.
Try Example 8 on your own.
Then go back and check your
homework from page 312
problems 25-31 odds. You have 9
minutes to do all of this.
Example 8
Find all zeros of f (x) = x3 + x2 + 1
Page 312 Problems 25 – 32 only find the ZEROS
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We can use the results of factoring to express a polynomial as a
product of factors. For instance, in Example 5, we used our
factoring to express the function as follows: f (x)  -x 2 (x  2)2
Notice that each factor occurs twice. If the same factor occurs
multiple times, we call this a zero with multiplicity of 2.
For the polynomial function f (x), 0 and 2 each have a multiplicity
of 2.
Multiplicity means the amount of times a zero repeats itself in
the completely factored form.

If the multiplicity is even, the graph “touches” the x-axis.

If the multiplicity is odd, the graph “crosses” the x-axis.
Example 9
2
1
Find all zeros and multiplicity of f x  x  12x  3
2
Zeros
Multiplicity
3
1,
2
1,
2
crosses touches
Example 10
2
3

1
Find all zeros and multiplicity of f x  4  x   x  5
2

1
Zeros
 ,
2
Multiplicity 2,
5
3
touches crosses
Example 11
Find all zeros of f (x) = x3 + 4x2 + 4x
 x  x  2
zeros : 0 and  2
multiplicity : 1 and 2
cross and touch
2
Example 12
Find all zeros of f (x) = 3 (x + 5) (x + 2)4
zeros :  5 and  2
multiplicity : 1 and 4
cross and touch
Page 312 Problems 25 – 32 find the zeros and state the Multiplicity
Example 10c.
Find the zeros of f (x) = (x- 3)2(x-1)3 and give the multiplicity of
each zero. State whether the graph crosses the x-axis or touches
the x-axis and turns around at each zero.
zeros : 3 and 1
multiplicity : 2 and 3
touch and cross
Continued on the next slide.
Example 10c.
Find the zeros of x3  2 x 2  4 x  8  0
x
3
 2 x 2    4 x  8   0
x2  x  2  4( x  2)  0
 x  2  x2  4  0
 x  2 x  2 x  2  0
2 has a multiplicity of 2, and 2 has a multiplicity of 1. Notice how the graph
touches at -2 (even multiplicity), but crosses at 2 (odd multiplicity).
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
The Intermediate Value Theorem
tells us of the existence of real zeros.
The idea behind the theorem is
illustrated in Figure 2.22.
The figure shows that if (a, f (a)) lies
below the x-axis and (b, f (b)) lies
above the x-axis the smooth,
continuous graph of the function
must cross the x-axis at some value c
between a and b.
This value c is a real zero for the
function. This also occurs vice versa.
Example 11
Show that the polynomial function f (x) = x3 – 2x + 9 has a real
zero between - 3 and - 2.
f (-3)  (-3)3  2(3)  9  12
 3,  12 
f (-2)  (-2)3  2(2)  9  5
 2, 5
Example 12
Show that the function y = x3 – x + 5 has a zero between -2 and -1.
f (-2)  (-2)3  (2)  5  1
 2,  1
f (-1)  (-1)3  (1)  5  5
 1, 5
Page 312 Problems 33 – 40
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The graph of f (x) = x5 – 6x3 + 8x + 1 is
shown.
The graph has four smooth turning
points.
The polynomial is of degree 5. Notice
that the graph has four turning points.
In general, if the function is a
polynomial function of degree n, then
the graph has at most n - 1 turning
points.
Conversely, if the graph has m amount
of turns, the degree of the polynomial is
m+1
Example 15
f (x) = x3 + 4x2 + 4x
Example 16
f (x) = 3 (x + 5) (x + 2)4
Example 17
Graph f (x) = x4 – 2x2 + 1.
Step 1 Determine
end behavior
(max turning points)
y





Step 2 Find zeros
x

 1 x  1
 x  1 x  1 x  1 x  1
2
2
x

1
x

1
     1 and 1
2
2

x




















Step 3 Find y-intercept

y   0  2  0  1
y 1
4
2


Step 4 Use even/odd symmetry to draw
Page 312-313 Problems 41 – 64
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
Example 18
Now graph this function on your calculator. f (x) = (x- 3)2(x-1)3
Step 1 Determine end behavior
Step 2 Find zeros
zeros : 3 and 1
Step 3 Find y-intercept
y   0  3  0  1
y  9
2
3
Step 4 Use even/odd symmetry to draw
Example
Graph f(x) = x3 – 9x2
Step 1 Determine end behavior
Step 2 Find zeros
 x2  x  9
zeros : 0 and 9
Step 3 Find y-intercept
y   0  9  0
y0
3
Step 4 Use even/odd symmetry to draw
Add these problems to your notes paper to
help you review!
Additional Practice problems can be found on page 313 - 315 problems 65-99
Example 1
Use the Leading Coefficient Test to determine the end behavior
of the graph of the polynomial function f(x) = x3 – 9x2 + 27
(a)
falls left, rises right
(b)
rises left, falls right
(c)
rises left, rises right
(d)
falls left, falls right
Example 2
State whether the graph crosses the x-axis, or touches the
x-axis and turns around at the zeros of 1, and - 3.
f(x) = (x-1)2(x+3)3
(a) -3 touches, 1 touches
(b) -3 crosses, 1 crosses
(c) -3 touches, 1 crosses
(d) -3 crosses, 1 touches