62
Sect 4.1 – Polynomial Functions and Models
In this section, we will explore polynomial functions in depth. We like to use
polynomial functions to model real world data since the graphs of
polynomial functions have no gaps or holes and contain no sharp corners
or points. Polynomial functions are also defined for all real numbers, so
they are extremely well behaved functions.
Objective #1 – Identify Polynomial Functions and Their Degree.
Polynomial Function:
A polynomial function is a function in the form of
f(x) = anxn + an – 1xn – 1 + … + a1x + a0
where an, an – 1, … , a1, a0 are real numbers and n is a whole number. As
stated before, the domain of a polynomial function is all real numbers. The
degree of a polynomial is the highest degree of the terms. If the polynomial
is a non-zero constant, then it has degree of 0. The zero polynomial is not
assigned a degree.
Determine if the following is a polynomial. If so, state its degree:
3
x3 −8
1a)
f(x) = 4 – 11x5
1d)
r(x) = 3
1e) p(x) = 9x2(x2 + 3)2
1f) H(x) = 0
Solution:
a)
f is a polynomial
€ of degree 5.
€
b)
g is not a polynomial function. It is a radical function.
c)
a is not a polynomial function. It is a rational function.
d)
r is a polynomial function of degree 0.
e)
Since 9x2(x2 + 3)2 = 9x2(x4 + 6x2 + 9) = 9x6 + 54x4 + 81x2, then
p is a polynomial function of degree 6.
f)
H is a polynomial function, but it is not assigned a degree.
1b)
g(x) =
x
1c)
a(x) =
x2 +16
Power Functions
A power function with degree n is a monomial of the form of
p(x) = axn
where a is a non-zero real number and n is a natural number.
Thus, a(x) = x, b(x) = x2, and c(x) = x3 are three power functions we have
already studied. Let's look at the graphs of some of the other power
functions. We will look at the cases when a = 1:
63
-3
b(x) = x2
d(x) = x4
f(x) = x6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
0
0
-2
-1 0
-1
1
2
3
-3
-2
1
2
3
-3
-2
-1 0
-1
c(x) = x3
e(x) = x5
g(x) = x7
3
3
3 2
2
2 1
1
1 0
-3
-1 0
-1
-2
-1 0
-1
2
3
-3
-2
-1 0
-1
2
3
1
2
3
0 0
1
1
1
2
3
‐3
‐2
‐1
0
‐1 -2
-2
‐2 -3
-3
‐3 Notice the pattern that is emerging; if n is even and greater than 2, the
graph looks like the x2, but "more squared out" while if n is odd and greater
than 3, the graph looks like the x3, but "more squared out."
Properties of Power Functions
If f(x) = xn and n is even natural number.
1)
f is an even function so it is symmetrical with respect to the y-axis.
2)
The domain is (– ∞, ∞) and the range is [0, ∞).
3)
The graph intersects the point (– 1, 1), (0, 0), and (1, 1).
4)
As n gets larger, the graph becomes more vertical for x < – 1 or x > 1,
more horizontal for – 1 < x < 1.
If f(x) = xn and n is odd natural number greater than 1.
1)
f is an odd function so it is symmetrical with respect to the origin.
2)
The domain is (– ∞, ∞) and the range is (– ∞, ∞).
3)
The graph intersects the point (– 1, – 1), (0, 0), and (1, 1).
4)
As n gets larger, the graph becomes more vertical for x < – 1 or x > 1,
more horizontal for – 1 < x < 1.
Objective #2: Graphing Power Functions Using Transformations
64
Graphing power functions work exactly the same way we used
transformations to graph quadratic and cube functions.
Graph the following using transformations:
Ex. 2
g(x) = 5 – x6
Ex. 3
h(x) = 2(x – 4)7 + 3
Solution:
Solution:
The graph of g is the
The graph of h is the graph
6
graph of x reflected
of x7 stretched by a factor of 2
across the x-axis and
and shifted right 4 and up 3.
shifted up 5 units.
8 6
7 5
6 4
5 3
4 2
3 1
2 1 0
-4
-3
-2
-1
-1
-2
0
1
2
3
4
0 ‐1
0
‐1 1
2
3
4
5
6
7
8
9
‐2 Objective #3:
Identifying the real zeros of a Polynomial Function and
Their Multiplicity.
If our polynomial function is already factored completely, we can use the
Zero-product property and find all the x-intercepts. Thus, if
f(x) = (x – 3)2(x + 2), then the zeros are x = 3 and x = – 2 and hence, the xintercepts are (3, 0) and (– 2, 0).
Definition
If f is a function and r is a real number such that f(r) = 0, then r is a real
zero of f. The following are equivalent:
1)
r is a real zero of a polynomial function f.
2)
(r, 0) is an x-intercept of the graph of f.
3)
(x – r) is a factor of f
The power m of the factor of (x – r) is called multiplicity. Thus, we say that r
is a zero of multiplicity m of f.
65
For the function f(x) = (x – 3)2(x + 2), – 2 is a zero of multiplicity 1 of f and 3
is a zero of multiplicity 2 of f.
Find the zeros and Their Multiplicity:
Ex. 4
g(x) = – 3(x – 4)(x + 2)3(x – 1/3)4
Solution:
4 is a zero of multiplicity of 1 since the power of (x – 4) = 1.
– 2 is a zero of multiplicity of 3 since the power of (x + 2) = 3.
1/3 is a zero of multiplicity of 4 since the power of (x – 1/3) = 4.
Find a 3rd degree polynomial function with the following zeros:
Ex. 5
– 5, 3, and 4
Solution:
Since – 5, 3, and 4 are zeros, then (x + 5), (x – 3) and (x – 4) are
factors. Thus, a 3rd degree polynomial has to be a constant times the
product of the three factors. We can pick the constant to be one.
Thus, p(x) = 1(x + 5)(x – 3)(x – 4) = x3 – 2x2 – 23x + 60
With all polynomial functions at the x-intercepts, the graph of the function
must either touch or cross the x-axis. Thus, between any two consecutive
x-intercepts, the graph is either above the x-axis (f(x) is positive) or below
the x-axis (f(x) is negative). This means that once we locate all the zeros of
the polynomial function, we will have all the x-intercepts of the polynomial.
We will then pick a value of x between to consecutive x-intercepts and use
it to determine if the function is positive (above the x-axis) in the interval or
negative (below of the x-axis) in that interval.
5 4 3 f is positive
2 Crosses
Crosses
1 0 ‐6
‐5
‐4
‐3
‐2
‐1
0
‐1 ‐2 ‐3 ‐4 ‐5 1
2
3
4
Touches
f is negative
5
6
7
8
66
Graph the following Polynomial using its x-intercepts:
Ex. 6
g(x) = 0.5(x – 1)(x + 2)2
Solution:
First, we will find the intercepts.
x-intercepts (y = g(x) = 0)
0.5(x – 1)(x + 2)2 = 0
Thus x = 1 and x = – 2.
So, x = 1 is a zero of multiplicity 1 (graph crosses the x-axis) and
x = – 2 is a zero of multiplicity 2 (graph touches the x-axis).
Thus, the x-intercepts are (1, 0) and (– 2, 0)
y-intercepts (x = 0)
g(0) = 0.5(0 – 1)(0 + 2)2 = – 2
So, (0, – 2) is the y-intercept.
If we plot the zeros on a number line, it will divide the real number line
into three intervals. We will pick a value of x that is clearly in each
interval and find the function value. This will tell us if the function is
positive or negative in each interval.
-4
-3
-2
Interval (– ∞, – 2)
Pick an x – 3
Value of f f(– 3) = – 2
Point
(– 3, – 2)
-1
0
(– 2, 1)
–1
f(– 1) = – 1
(– 1, – 1)
1
2
3
(1, ∞)
2
f(2) = 8
(2, 8)
Now plot the points and construct a graph:
Notice that this suggests the
9 following:
8 Theorem
7 Let r be a zero of a
6 polynomial function f(x) with
5 multiplicity m.
4 1) If m is even, the sign of
3 f(x) does not change when
2 crossing at r and the graph
1 touches the x–axis at x = r.
0 2) If m is odd, the sign of
‐6
‐5
‐4
‐3
‐2
‐1
‐1 0
1
2
3
4
5
6
f(x) does change when
‐2 crossing at r and the graph
‐3 crosses the x–axis at x = r.
67
Objective #4:
Behavior a Near Zero, Turning Points, and End Behavior.
Behavior Near a Zero
To approximate the behavior of function near a zero r, replace x by r in all
the other factors of the the polynomial except for the factor that r came from
and simplify. In the last example, if we want to approximate the behavior
near – 2, we will replace x by – 2 in the factor (x – 1) and simplify:
g(x) = 0.5(x – 1)(x + 2)2
≈ 0.5(– 2 – 1)(x + 2)2 = – 1.5(x + 2)2
So, near x = – 2, the function is behaving like a parabola that has been
reflected across the x-axis, stretched by a factor of 1.5 and shifted to the
left by two units. To approximate the behavior near the zero of 1, we will
replace x by 1 in the factor (x + 2) and simplify"
g(x) = 0.5(x – 1)(x + 2)2
≈ 0.5(x – 1)(1 + 2)2 = 4.5(x – 1) = 4.5x – 4.5
Thus, near x = 1, the function is behaving like a line with slope of 4.5 and yintercept of (0, – 4.5). Thus, we can sketch in these two pieces near their
respective zeros:
9 8 7 6 5 4 3 2 1 0 ‐6
‐5
‐4
‐3
‐2
‐1
‐1 0
1
2
3
4
5
6
‐2 ‐3 This will help us get a more accurate picture of the graph.
Turning Points
The point that a polynomial changes direction (from increasing to
decreasing or vice-versi) is called a "turning point". Think of a turning point
as being the top of a hill or the bottom of a valley. For a quadratic function,
there was only one turning point which was the vertex. In Calculus, these
turning points are called relative maxima or relative minima. The maximum
68
number of turning points of a polynomial function is always one less than
the degree of the polynomial.
Turning Point Theorem
1)
A polynomial with degree n can have at most n – 1 turning points.
2)
If the graph of a polynomial has n – 1 turning points, then the degree
of the polynomial has to be at least n.
Recall the follow to functions we graphed earlier:
Ex. 3
h(x) = 2(x – 4)7 + 3
Ex. 6
g(x) = 0.5(x – 1)(x + 2)2
8 9 7 8 7 6 6 5 5 4 4 3 3 2 2 1 1 0 0 ‐1
0
‐1 ‐2 1
2
3
4
5
6
7
8
9
‐6
‐5
‐4
‐3
‐2
‐1
‐1 0
1
2
3
4
5
6
‐2 ‐3 Even though the degree of the polynomial in example 3 was 7, it had no
turning points. The Theorem only guarentees that the maximum number of
turning points is one less than 7 or 6. In example 6, the degree of the
polynomial was 3. Notice that it has two turning points (at (– 2, 0) and at
(0, – 2)) which is the maximum number that it can have.
End Behavior
End behavior describes the behavior of the polynomial function when |x|
gets very large. When this happens the leading term (the term with the
highest degree) will dominate the value of the other terms and the
polynomial function will approximate the behavior of a power function.
End Behavior Theorem
For large positive and negative values of x, the graph of the polynomial
function p(x) = anxn + an – 1xn – 1 + … + a1x + a0 will approximate the power
y = a nx n.
69
When the degree of the polynomial is greater than or equal to 2, get four
cases for the end behavior of a polynomial function:
Case 1
Case 2
Case 3
Case 4
n is even
n is even
n is odd
n is odd
a<0
a>0
a<0
a>0
Determine which graph could be the graph of the following
polynomial:
Ex. 7
f(x) = x4 – 4x3 + 3x2 + 4x – 4
a)
b)
7
4
6
3
5
2
4
1
3
0
-5 -4 -3 -2 -1
-1 0
2
1
-2
0
-3
-5 -4 -3 -2 -1
-1 0
1
2
3
4
5
4
0 ‐4
0
-5
3
2 1
-4
5
4 2
-3
4
6 d)
3
-2
3
-5
4
-5 -4 -3 -2 -1
-1 0
2
-4
-2
c)
1
1
2
3
4
5
‐3
‐2
‐1
‐2 0
‐4 ‐6 ‐8 ‐10 ‐12 ‐14 1
2
70
Solution:
We will do this by a process of elimination. Since the function is even
and a > 0, then the end behavior will look like a 4th degree power
equation. The graph for c has the end behavior of an odd function so
we can cross that choice out. Since we have a fourth degree
polynomial, the maximum number of turning points the graph can
have is 3. The graph for d has 5 turning points, so we can cross out
d. Since f(0) = – 4, then the y-intercept is (0, – 4). The graph for a has
a y-intercept of (0, 4), so we can eliminate a. The only choice left is b
so that is our graph.
We would go through the same process if the only information given
in the problem was that we had a fourth degree polynomial with yintercept (0, – 4).
Objective #5:
Analyze the graph of a Polynomial Function
Let's outline a procedure for analyzing the graph of polynomial functioons:
1)
a) Evaluate f(0) to find the y-intercept.
b) Set f(x) = 0 and solve to find the x-intercepts.
2)
Determine whether f crosses (odd multiplicity) or touches (even
multiplicity) the x-axis at each x-intercept.
3)
Find the end behavior (for large |x|, f(x) ≈ anxn)
4)
Determine the maximum number of turning points (n – 1) on the
graph of f.
5)
Determine the behavior of the graph near each intercept. (replace x
by the zero in all the other factors of the the polynomial except for the
factor that the r came from and simplify
6)
Find some additional points and sketch the graph.
Sketch the graph of the following polynomials:
Ex. 8
f(x) = (x – 2)3(x + 5)2
Solution:
1)
a) f(0) = (0 – 2)3(0 + 5)2 = – 200
b) f(x) = (x – 2)3(x + 5)2 = 0
x = 2 or x = – 5
The y-intercept is (0, – 200) and the x-intercepts are (2, 0) and
(– 5, 0). We will have to rescale the y-axis.
2)
2 is a zero of multiplicity of 3 so it crosses the x-axis.
5 is a zero of multiplicity of 2 so it touches the x-axis.
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3)
4)
5)
For large |x|, f(x) ≈ x5.
The maximum turning points is 5 – 1 = 4.
Near x = 2, f(x) ≈ (x – 2)3(2 + 5)2 = 49(x – 2)3. It will like a cube
function shifted to the right by 2 units and stretched by a factor
of 49.
Near x = – 5, f(x) ≈ (– 5 – 2)3(x + 5)2 = – 343(x + 5)2. It will look
like a parabola shifted to the left 5 units, reflected across the xaxis and stretched by a factor of 343. Let's sketch out what we
currently have:
300 200 100 0 ‐6
‐5
‐4
‐3
‐2
‐1
0
1
2
‐100 ‐200 ‐300 3
4
We will need to find some
other points. Let's find
f(– 6), f(– 2), & f(3).
f(– 6) = (– 6 – 2)3(– 6 + 5)2
= – 512
f(– 2) = (– 2 – 2)3(– 2 + 5)2
= – 576
f(3) = (3 – 2)3(3 + 5)2 = 64
We will need to adjust our
scale on the y-axis and plot
the points we found. Then, we
will connect the dots to get
our graph.
200 200 100 100 0 0 ‐6
‐5
‐4
‐3
‐2
‐1
0
‐100 1
2
3
4
‐6
‐5
‐4
‐3
‐2
‐1
0
‐100 ‐200 ‐200 ‐300 ‐300 ‐400 ‐400 ‐500 ‐500 ‐600 ‐600 ‐700 ‐700 ‐800 ‐800 1
2
3
4
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Ex. 9
f(x) = – 0.5x4 + 2x2
Solution:
1)
a) f(0) = – 0.5(0)4 + 2(0)2= 0
b) f(x) = – 0.5x4 + 2x2 = 0
– 0.5x2(x2 – 4) = 0
– 0.5x2(x – 2)(x + 2) = 0
x = 0, x = 2 or x = – 2
The y-intercept is (0, 0) and the x-intercepts are (– 2, 0),
(0, 0), and (2, 0).
2)
– 2 is a zero of multiplicity of 1 so it crosses the x-axis.
0 is a zero of multiplicity of 2 so it touches the x-axis.
2 is a zero of multiplicity of 1 so it crosses the x-axis.
3)
For large |x|, f(x) ≈ – 0.5x4.
4)
The maximum turning points is 4 – 1 = 3.
5)
Near x = – 2, f(x) ≈ – 0.5(– 2)2((– 2) – 2)(x + 2) = 8(x + 2). It will
like a linear function with slope of 8 and y-intercept of (0, 16).
Near x = 0, f(x) ≈ – 0.5x2((0) – 2)(0 + 2) = 2x2. It will look
like a parabola stretched by a factor of 2.
Near x = 2, f(x) ≈ – 0.5(2)2(x – 2)((2) + 2) = – 8(x – 2). It will
like a linear function with slope of – 8 and y-intercept of (0, 16).
Let's sketch out what we currently have:
5 We will need to find some
other points. Let's find
f(– 1) and f(1).
f(– 1) = – 0.5(– 1)4 + 2(– 1)2
= 1.5
f(1) = – 0.5(1)4 + 2(1)2
= 1.5
4 3 2 1 0 ‐4
‐3
‐2
‐1
‐1 ‐2 ‐3 ‐4 ‐5 0
1
2
3
4
We will plot the points we
found. Then, we will connect
the dots to get our graph.
73
5 5 4 4 3 3 2 2 1 1 0 ‐4
‐3
‐2
‐1
‐1 0 0
1
2
3
4
‐4
‐3
‐2
‐1
‐1 ‐2 ‐2 ‐3 ‐3 ‐4 ‐4 ‐5 ‐5 0
1
2
3
4