Chapter 7

Polynomial and Rational Functions with Applications

Section 7.1

Section 7.1Polynomial Functions and Their

Graphs• Definition of a Polynomial Function

• Identifying Graphs of Polynomial Functions

• Finding Zeros of Polynomial Functions

• Multiplicity of Zeros of Polynomial Functions

• Factor Theorem

• Sketching Graphs of Polynomial Functions

Definition of a Polynomial Function

A polynomial function of degree n is one of the form

where n is a nonnegative integer, an, an–1, …, a0 are the

coefficients of f(x), and an 0.

anxn leading term

an leading coefficient

a0 constant term

012

22

21

1 ...)( axaxaxaxaxaxf nn

nn

nn

Determine which of the following are polynomial functions. If the given function is polynomial, identify the degree, leading coefficient, and constant term.

a.

Yes. All powers of the variables are nonnegative integers.Degree: 4 Leading coefficient: –8 Constant term: 0

b.

Yes. All powers of the variables are nonnegative integers.Degree: 3 Leading coefficient: 1 Constant term: 1/9

c.

xxxxf3

24835)(

3637)( xxxxf

233)( xxf

No. Not all variables have powers that are nonnegative integers: The square root of x is equivalent to x1/2, and the exponent on the variable of the last term is –3.

Graphs of Polynomial Functions

Graphs of polynomial functions of different degrees will vary in:

Turning points

Where the graph turns and changes direction; relative maxima and minima.

Zeros

The x-intercepts

End behavior

How the graph appears as it extends farther in either direction; it is indicated by the positions of the arms (ends) of the graph, which may be pointed up or down.

Polynomial Function-End Behavior

The graphs of polynomial functions will display the following behavior:

Even-degree polynomials

Both arms (ends) of the graph will point up or down. Positive leading coefficient arms will point up Negative leading coefficient arms will point down

Odd-degree polynomials

One arm of the graph will point up, and the other will point down. Positive leading coefficient right-hand end will point up Negative leading coefficient right-hand end points down 

For each polynomial function, determine the number of turning points, the zeros, and the end behavior.

Two turning points Zeros: (–3, 0), (–1, 0), (1, 0) Left end points down and right-hand arm points up; this is an odd-degree polynomial with positive leading coefficient.

Three turning points Zeros: (–3, 0), (–1, 0), (2, 0) Both ends points down; this is an even-degree polynomial with negative leading coefficient.

-3 -2 -1 1 2

-6-4-2

246

x

y

y = f(x)

Zeros of Polynomial Functions

The graph of a polynomial function of degree n can have at most n x-intercepts. A zero occurring r number of times is called a zero with multiplicity r.

Example: Find the real zeros of f(x) = 12(x + 8)3(x – 1)2(x – 5) and statethe multiplicity for each zero. Without multiplying through, determine the degree.

We know the zeros occur when f(x) = 0. Letting the polynomial12(x + 8)3(x – 1)2(x – 5) = 0 and solving for x, the zeros are:

–8 with multiplicity 3, because the exponent on the factor is 3. 1 with multiplicity 2, because the exponent on the factor is 2. 5 with multiplicity 1, because the exponent on the factor is 1.

This is a polynomial function of degree 6 (sum of the degrees of each linear factor).

Multiple Zeros and Graphs of Polynomial Functions

What does the multiplicity tell us with respect to the graph?

If k is a zero of even multiplicity, the graph will touch the x-axis and turn around at k.

Example:

f(x) = (x – 3)2

3 is a zero of multiplicity 2

If k is a zero of odd multiplicity, the graph will cross the x-axis at k.

Example:

f(x) = (x – 3)3

3 is a zero of multiplicity 3

x

y

3

x

y

3

Factor Theorem

If f is a polynomial function, then (x – k) is a factor of f if and only if f(k) = 0. This implies that if f(k) = 0, then (x – k) is a factor of f.

Example: Determine if (x – 5) is a factor of the polynomial function f(x) = x3 + x2 – 32x + 10.

If (x – 5) is a factor of the polynomial, then f(5) = 0.

f(5) = (5)3 + (5)2 – 32(5) + 10 = 0.

Therefore, (x – 5) is a factor of the given f(x).

Optionally, we can verify graphically:  

Find all the zeros of f(x) = x4 – 7x2 – 18.

We know the zeros occur when f(x) = 0. Let the given polynomial x4 – 7x2 – 18 = 0, and solve for x using substitution.

Let u = x2 and u2 = x4.

u2 – 7u – 18 = 0

(u – 9)(u + 2) = 0

u – 9 = 0 or u + 2 = 0

u = 9 or u = –2

We now use reverse substitution to solve the equation in terms of the original variable, x. Since u = x2 , we have

 

2x9x

2x9x 22

2,3 iZeros

Properties of Polynomial Graphs: Summary

Let f(x) be a polynomial function of degree n:

1. Domain is the set of all real numbers.

2. Graph has at most n – 1 turning points.

3. Graph has at most n zeros (x-intercepts).

4. End behavior for even-degree polynomials:

Both arms (ends) of the graph will point up or down.Positive leading coefficient arms point upNegative leading coefficient arms point down

5. End behavior for odd-degree polynomials:

One arm of the graph will point up; the other will point down.Positive leading coefficient right-hand end points upNegative leading coefficient right-hand end points down

Sketching Graphs of Polynomial Functions

Let f(x) be a polynomial function with real zeros:

1. Find and plot the y-intercept.

2. Determine the end behavior – check the degree of the polynomial and the sign of the leading coefficient.

3. Find the zeros and their multiplicities.

4. Find a test point between each real zero and connect the points with a smooth curve.

Note: Without applying calculus techniques, we can only approximate the turning points. Alternately, we can use the graphing calculator to find any maxima and minima of the polynomial function.

Sketch the graph of f(x) = (x + 1)3(x – 2)(x – 3)2. Label the axes, tick marks, and the y-intercept.

1.Find and plot the y-intercept.

f(0) = (0 + 1)3(0 – 2) (0 – 3)2 = –18. The y-intercept is (0, –18).

2. Determine the end behavior – check the degree of the polynomial and the sign of the leading coefficient.

This is polynomial of degree 6 and a positive leading coefficient. This means that both ends of the graph will point up.

3. Find the zeros and their multiplicities.

–1 with multiplicity 3; the graph will cross the x-axis at –1.

2 with multiplicity 1; the graph will cross the x-axis at 2.

3 with multiplicity 2; the graph will touch the x-axis and turn around at 3.

(continued on the next slide)

(Contd.)Sketch the graph of f(x) = (x + 1)3(x – 2)(x – 3)2. Label the axes, tick marks, and the y-intercept.

4. Find a test point between each real zero and connect the points with a smooth curve.

Using (1, –32) and (2.5, 5.36) as test points, a sketch of the graph is shown next.

-5 -4 -3 -2 -1 1 2 3 4 5

-45-40-35-30-25-20-15-10-5

5101520253035

x

y

y = f(x)

(1, -32)

(-0, 18)

(2.5, 5.36)

Find a polynomial function that could be represented by the given graph.

Both ends of the graph are pointing down, which means that this is an even-degree polynomial with a negative leading coefficient.

The real zeros are given at x = –4, x = –2, and x = 1. The graph touches the x-axis and turns around at –2, which means that this is a zero of even multiplicity. The zeros with odd multiplicity are –4 and 1 because the graph crosses the x-axis at these values.

(continued on the next slide)

-5 -4 -3 -2 -1 1 2 3

-50

-40

-30

-20

-10

10

20

30

x

y

y = f(x)(0, 16)

(Contd.)Find a polynomial function that could be represented by the given graph.

Let f(x) = a(x + 2)2(x + 4)(x – 1)

We know that the y-intercept is (0, 16), therefore we have

f(0) = a(0 + 2)2(0 +4)(0 – 1) = 16. Next, we solve for a:

a(2)2(4)( –1) = 16 –16a = 16 a = –1

Our polynomial function is f(x) = –(x + 2)2(x + 4)(x – 1)

 

-5 -4 -3 -2 -1 1 2 3

-50

-40

-30

-20

-10

10

20

30

x

y

y = f(x)(0, 16)

The number of earthquakes in the United States from 2000 to 2010 can be modeled by the polynomial function f(x) = 37.10587x3 – 480.62413x2 + 1696.18959x + 1824.05594, where x is the number of years after 2000.Sources: www.earthquake.usgs.gov; www.mapsofworld.com

a. Use the given function to estimate the number of earthquakes in the United States during the year 2008. Round to the nearest whole number. Verify your answer numerically.

Evaluating the given function at x = 8, we get

37.10587(8)3 – 480.62413(8)2 + 1696.18959(8) + 1824.05594 = 3631.8.

During the year 2008 there were approximately 3,632 earthquakes in the United States.

(continued on the next slide)

(Contd.)The number of earthquakes in the United States from 2000 to 2010 can be modeled by the polynomial function f(x) = 37.10587x3 – 480.62413x2 + 1696.18959x + 1824.05594, where x is the number of years after 2000.Sources: www.earthquake.usgs.gov; www.mapsofworld.com

b. Graph the given function for 0 ≤ x ≤ 10. Label the axes and show your window.

y: Number of earthquakes in U.S.

x: Year [0, 10, 1] by [0, 9000, 1000]

(continued on the next slide)

(Contd.)The number of earthquakes in the United States from 2000 to 2010 can be modeled by the polynomial function f(x) = 37.10587x3 – 480.62413x2 + 1696.18959x + 1824.05594, where x is the number of years after 2000.Sources: www.earthquake.usgs.gov; www.mapsofworld.com

c. Use your graph to determine the year with the lowest number of earthquakes in the U.S. between 2003 and 2010. State the year and number of earthquakes. Round each answer to the nearest whole number.

For 2003, x = 3; for 2010, x = 10. The graph that follows displays the minimum value for the corresponding period.

The lowest number of earthquakes in the U.S. between 2003 and 2010 was in 2006, with approximately 2,708 earthquakes.

Using your textbook, practice the problems assigned by your instructor to review the concepts from Section 7.1.