Processing math: 100%
"

Chapter 5 Polynomial and Rational Functions

5.3 Graphs of Polynomial Functions

Learning Objectives

In this section, you will:

  • 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.

The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 1.

Year 2006 2007 2008 2009 2010 2011 2012 2013
Revenues 52.4 52.8 51.2 49.5 48.6 48.6 48.7 47.1

Table 1.

The revenue can be modeled by the polynomial function

R(t)=−0.037t4+1.414t3−19.777t2+118.696t−205.332

whereRrepresents the revenue in millions of dollars andtrepresents the year, witht=6corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general.

Recognizing Characteristics of Graphs of Polynomial Functions

Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.

Two graphs in which one has a polynomial function and the other has a function closely resembling a polynomial but is not.
Figure 1.

Recognizing Polynomial Functions

Which of the graphs in Figure 2 represents a polynomial function?

Four graphs where the first graph is of an even-degree polynomial, the second graph is of an absolute function, the third graph is an odd-degree polynomial, and the fourth graph is a disjoint function.
Figure 2.
Show Solution

The graphs offandhare graphs of polynomial functions. They are smooth and continuous.

The graphs ofgandkare graphs of functions that are not polynomials. The graph of functionghas a sharp corner. The graph of functionkis not continuous.

Do all polynomial functions have all real numbers as their domain?

Yes. Any real number is a valid input for a polynomial function.

Using Factoring to Find Zeros of Polynomial Functions

Recall that iffis a polynomial function, the values ofxfor whichf(x)=0are called zeros off.If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros.

We can use this method to findx-intercepts because at thex-intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases:

  1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring.
  2. The polynomial is given in factored form.
  3. Technology is used to determine the intercepts.

How To

Given a polynomial functionf,find the x-intercepts by factoring.

  1. Setf(x)=0.
  2. If the polynomial function is not given in factored form:
    1. Factor out any common monomial factors.
    2. Factor any factorable binomials or trinomials.
  3. Set each factor equal to zero and solve to find thex-intercepts.

Finding the x-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts off(x)=x6−3x4+2x2.

Show Solution

We can attempt to factor this polynomial to find solutions for f(x)=0.

x6−3x4+2x2=0Factor out the greatestcommon factor.x2(x4−3x2+2)=0Factor the trinomial.x2(x2−1)(x2−2)=0Set each factor equal to zero.
(x2−1)=0(x2−2)=0x2=0x2=1x2=2x=0x=±1x=±√2

This gives us five x-intercepts:(0,0),(1,0),(−1,0),(√2,0),and(−√2,0).See Figure 3. We can see that this is an even function because it is symmetric about the y-axis.

Graph of f(x)=x^6-3x^4+2x^2 with its five intercepts, (-sqrt(2), 0), (-1, 0), (0, 0), (1, 0), and (sqrt(2), 0).
Figure 3.

 

Finding the x-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts off(x)=x3−5x2−x+5.

Show Solution

Find solutions forf(x)=0
by factoring.

x3−5x2−x+5=0Factor by grouping.x2(x−5)−(x−5)=0Factor out the common factor.(x2−1)(x−5)=0Factor the difference of squares.(x+1)(x−1)(x−5)=0Set each factor equal to zero.
x+1=0x−1=0x−5=0x=−1x=1x=5
Graph of f(x)=x^3-5x^2-x+5 with its three intercepts (-1, 0), (1, 0), and (5, 0).
Figure 4.

There are three x-intercepts:(−1,0),(1,0),and(5,0).See Figure 4.

Finding the y– and x-Intercepts of a Polynomial in Factored Form

Find the y– and x-intercepts ofg(x)=(x−2)2(2x+3).

Show Solution

The y-intercept can be found by evaluatingg(0).

g(0)=(0−2)2(2(0)+3)=12

So the y-intercept is(0,12).

The x-intercepts can be found by solvingg(x)=0.

(x−2)2(2x+3)=0
(x−2)2=0(2x+3)=0x−2=0orx=−32x=2

So the x-intercepts are(2,0)and(−32,0).

Analysis

Graph of g(x)=(x-2)^2(2x+3) with its two x-intercepts (2, 0) and (-3/2, 0) and its y-intercept (0, 12).
Figure 5.

We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial, as shown in Figure 5.

Finding the x-Intercepts of a Polynomial Function Using a Graph

Find the x-intercepts ofh(x)=x3+4x2+x−6.

Show Solution

This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown in Figure 6, it appears that there are x-intercepts atx=−3,−2,and1.

Graph of h(x)=x^3+4x^2+x-6.
Figure 6.

We can check whether these are correct by substituting these values forxand verifying that

h(−3)=h(−2)=h(1)=0

Sinceh(x)=x3+4x2+x−6,we have:

h(−3)=(−3)3+4(−3)2+(−3)−6=−27+36−3−6=0h(−2)=(−2)3+4(−2)2+(−2)−6=−8+16−2−6=0h(1)=(1)3+4(1)2+(1)−6=1+4+1−6=0

Each x-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.

h(x)=x3+4x2+x−6=(x+3)(x+2)(x−1)

Try It

Find the y– and x-intercepts of the functionf(x)=x4−19x2+30x.

Show Solution

y-intercept(0,0);x-intercepts(0,0),(–5,0),(2,0),and(3,0)

Identifying Zeros and Their Multiplicities

Graphs behave differently at various x-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and “bounce” off.

Suppose, for example, we graph the function shown.

f(x)=(x+3)(x−2)2(x+1)3

Notice in Figure 7 that the behavior of the function at each of the x-intercepts is different.

Graph of f(x)=(x+3)(x-2)^2(x+1)^3.
Figure 7.

The x-interceptx=−3 is the solution of equation(x+3)=0.The graph passes directly through the x-intercept atx=−3.The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.

The x-interceptx=2is the repeated solution of equation(x−2)2=0.The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.

(x−2)2=(x−2)(x−2)

The factor is repeated, that is, the factor(x−2)appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor,x=2,has multiplicity 2 because the factor(x−2)occurs twice.

The x-interceptx=−1is the repeated solution of factor(x+1)3=0.The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit functionf(x)=x3.We call this a triple zero, or a zero with multiplicity 3.

For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3.

3 graphs showing a single zero, zero with multiplicity 2, and multiplicity 3.
Figure 8.

For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis, but for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis.

For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis.

Graphical Behavior of Polynomials at x-Intercepts

If a polynomial contains a factor of the form(x−h)p,the behavior near thex-intercepthis determined by the powerp.We say thatx=his a zero of multiplicityp.

The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities.

The sum of the multiplicities is the degree of the polynomial function.

How To

Given a graph of a polynomial function of degreen,identify the zeros and their multiplicities.

  1. If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero.
  2. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity.
  3. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity.
  4. The sum of the multiplicities isn.

Identifying Zeros and Their Multiplicities

Graph of an even-degree polynomial with a positive leading coefficient.
Figure 9.

Use the graph of the function of degree 6 in Figure 9 to identify the zeros of the function and their possible multiplicities.

Show Solution

The polynomial function is of degree 6. The sum of the multiplicities must be 6.

Starting from the left, the first zero occurs atx=−3.The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of−3most likely has multiplicity2.

The next zero occurs atx=−1.The graph looks almost linear at this point. This is a single zero of multiplicity 1.

The last zero occurs atx=4.The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is 6.

Try It

Graph of an odd-degree polynomial with a positive leading coefficient. Note that as x goes to positive infinity, f(x) goes to negative infinity, and as x goes to negative infinity, f(x) goes to negative infinity.
Figure 10.

Use the graph of the function of degree 9 in Figure 10. to identify the zeros of the function and their multiplicities.

Show Solution

The graph has a zero of –5 with multiplicity 3, a zero of -1 with multiplicity 2, and a zero of 3 with multiplicity 4.

Determining End Behavior

As we have already learned, the behavior of a graph of a polynomial function of the form

f(x)=anxn+an−1xn−1+...+a1x+a0

will either ultimately rise or fall asx increases without bound and will either rise or fall asx decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000.

Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function,anxn,is an even power function, asxincreases or decreases without bound,f(x)increases without bound. When the leading term is an odd power function, asxdecreases without bound,f(x)also decreases without bound; asxincreases without bound,f(x)also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 11 summarizes all four cases.

 

Showing the distribution for the leading term.
Figure 11.

Understanding the Relationship between Degree and Turning Points

In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial functionf(x)=x4−x3−4x2+4xin Figure 12. The graph has three turning points.

Graph showing turning points and places the graph is increasing and decreasing.
Figure 12.

This functionf is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.

Interpreting Turning Points

A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).

A polynomial of degreenwill have at mostn−1turning points.

Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function

Find the maximum number of turning points of each polynomial function.

  1. f(x)=−x3+4x5−3x2+1
  2. f(x)=−(x−1)2(1+2x2)
Show Solution
  1. First, rewrite the polynomial function in descending order:f(x)=4x5−x3−3x2+1Identify the degree of the polynomial function. This polynomial function is of degree 5.

    The maximum number of turning points is5−1=4.

  2. First, identify the leading term of the polynomial function if the function were expanded. 

    Graph of f(x)=x^4-x^3-4x^2+4x which denotes where the function increases and decreases and its turning points.

    Then, identify the degree of the polynomial function. This polynomial function is of degree 4.

    The maximum number of turning points is4−1=3.

Graphing Polynomial Functions

We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.

How To

Given a polynomial function, sketch the graph.

  1. Find the intercepts.
  2. Check for symmetry. If the function is an even function, its graph is symmetrical about they-axis, that is,f(−x)=f(x).
    If a function is an odd function, its graph is symmetrical about the origin, that is,f(−x)=−f(x).
  3. Use the multiplicities of the zeros to determine the behavior of the polynomial at thex-intercepts.
  4. Determine the end behavior by examining the leading term.
  5. Use the end behavior and the behavior at the intercepts to sketch a graph.
  6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
  7. Optionally, use technology to check the graph.

Sketching the Graph of a Polynomial Function

Sketch a graph off(x)=−2(x+3)2(x−5).

Show Solution

This graph has two x-intercepts. Atx=−3,the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-intercept. Atx=5,the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.

The y-intercept is found by evaluatingf(0).

f(0)=−2(0+3)2(0−5)=−2⋅9⋅(−5)=90

The y-intercept is(0,90).

Additionally, we can see the leading term, if this polynomial were multiplied out, would be−2x3,
so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 13.

Graph of the end behavior for the function f(x)=-2(x+3)^2(x-5).
Figure 13.

To sketch this, we consider that:

  • Asx→−∞the functionf(x)→∞,so we know the graph starts in the second quadrant and is decreasing toward thex-axis.
  • Sincef(−x)=−2(−x+3)2(−x–5)
    is not equal tof(x),the graph does not display symmetry.
  • At(−3,0),the graph bounces off of the x-axis, so the function must start increasing.At(0,90),the graph crosses the y-axis at the y-intercept. See Figure 14.
Graph of the end behavior and intercepts, (-3, 0) and (0, 90), for the function f(x)=-2(x+3)^2(x-5).
Figure 14.

Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at(5,0).See Figure 15.

Graph of the end behavior and intercepts, (-3, 0), (0, 90) and (5, 0), for the function f(x)=-2(x+3)^2(x-5).
Figure 15.

Asx→∞the functionf(x)→−∞,so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.

The complete graph of the polynomial function f(x)=−2(x+3)^2(x−5)
Figure 16.

Using technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like our sketch in Figure 15.

Using the Intermediate Value Theorem

In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial functionfwhose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbersaandbin the domain off, if a<b and f(a)≠f(b), then the functionftakes on every value betweenf(a)andf(b).(While the theorem is intuitive, the proof is actually quite complicated and requires higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous functionfatx=alies above thex-axis and another point atx=blies below thex-axis, there must exist a third point betweenx=aandx=bwhere the graph crosses thex-axis. Call this point(c, f(c)).This means that we are assured there is a solutioncwhere f(c)=0.

In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross thex-axis. Figure 17 shows that there is a zero betweena andb.

Graph of an odd-degree polynomial function that shows a point f(a) that’s negative, f(b) that’s positive, and f(c) that’s 0.
Figure 17.

Intermediate Value Theorem

Letf be a polynomial function. The Intermediate Value Theorem states that iff(a) andf(b) have opposite signs, then there exists at least one valuec betweena andb for whichf(c)=0.

 

Using the Intermediate Value Theorem

Show that the functionf(x)=x3−5x2+3x+6 has at least two real zeros betweenx=1 andx=4.

Show Solution

As a start, evaluatef(x)
at the integer valuesx=1,2,3,and 4.See Table 2.

x 1 2 3 4
f(x) 5 0 –3 2

Table 2.

We see that one zero occurs atx=2.Also, sincef(3)is negative andf(4)is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.

We have shown that there are at least two real zeros betweenx=1 and x=4.

Analysis

Graph of f(x)=x^3-5x^2+3x+6 and shows, by the Intermediate Value Theorem, that there exists two zeros since f(1)=5 and f(4)=2 are positive and f(3) = -3 is negative.
Figure 18.

We can also see on the graph of the function in Figure 18 that there are two real zeros betweenx=1 andx=4.

Try It

Show that the functionf(x)=7x5−9x4−x2has at least one real zero between x=1andx=2.

Show Solution

Becausefis a polynomial function and sincef(1)is negative andf(2)is positive, there is at least one real zero betweenx=1andx=2.

Writing Formulas for Polynomial Functions

Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors.

Factored Form of Polynomials

If a polynomial of lowest degreep has horizontal intercepts atx=x1,x2,…,xn, then the polynomial can be written in the factored form:f(x)=a(x−x1)p1(x−x2)p2⋯(x−xn)pn where the powerspi on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factora can be determined given a value of the function other than the x-intercept.

 

How To

Given a graph of a polynomial function, write a formula for the function.

  1. Identify the x-intercepts of the graph to find the factors of the polynomial.
  2. Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor.
  3. Find the polynomial of least degree containing all the factors found in the previous step.
  4. Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor.

Writing a Formula for a Polynomial Function from the Graph

Graph of a positive even-degree polynomial with zeros at x=-3, 2, 5 and y=-2.
Figure 19.

Write a formula for the polynomial function shown in Figure 19.

Show Solution

This graph has three x-intercepts:x=−3,2,and5.The y-intercept is located at(0,2).Atx=−3andx=5, the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. Atx=2,the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us

f(x)=a(x+3)(x−2)2(x−5)

To determine the stretch factor, we utilize another point on the graph. We will use they-intercept(0,–2)to solve fora.

f(0)=a(0+3)(0−2)2(0−5)−2=a(0+3)(0−2)2(0−5)−2=−60aa=130

The graphed polynomial appears to represent the functionf(x)=130(x+3)(x−2)2(x−5).

Try It

Graph of a negative even-degree polynomial with zeros at x=-1, 2, 4 and y=-4.
Figure 20.

Given the graph shown in Figure 20, write a formula for the function shown.

Show Solution

f(x)=−18(x−2)3(x+1)2(x−4)

Using Local and Global Extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function.

Local and Global Extrema

A local maximum or local minimum atx=a(sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval aroundx=a.If a function has a local maximum ata,thenf(a)≥f(x)for allxin an open interval aroundx=a.If a function has a local minimum ata,thenf(a)≤f(x)for allxin an open interval aroundx=a.

A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at a,thenf(a)≥f(x)for allx.If a function has a global minimum ata,thenf(a)≤f(x)for allx.

Graph of an even-degree polynomial that denotes the local maximum and minimum and the global maximum.
Figure 21.

Do all polynomial functions have a global minimum or maximum?

No. Only polynomial functions of even degree have a global minimum or maximum. For example,f(x)=xhas neither a global maximum nor a global minimum.

Using Local Extrema to Solve Applications

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 squares that should be cut out to maximize the volume enclosed by the box.

Show Solution

We will start this problem by drawing a picture like that in Figure 22, labeling the width of the cut-out squares with a variable,w.

Diagram of a rectangle with four squares at the corners.
Figure 22.

Notice that after a square is cut out from each end, it leaves a(14−2w)cm by(20−2w)cm rectangle for the base of the box, and the box will bewcm tall. This gives the volume

V(w)=(20−2w)(14−2w)w=280w−68w2+4w3

Notice, since the factors arew,20–2wand14–2w,the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so valueswmay take on are greater than zero or less than 7. This means we will restrict the domain of this function to[latex]\,07.\,[/latex]Using technology to sketch the graph ofV(w)on this reasonable domain, we get a graph like that in Figure 23. We can use this graph to estimate the maximum value for the volume, restricted to values forwthat are reasonable for this problem—values from 0 to 7.

Graph of V(w)=(20−2w)(14−2w)w
Figure 23.

From this graph, we turn our focus to only the portion on the reasonable domain,[0, 7].We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 24.

Graph of V(w)=(20-2w)(14-2w)w where the x-axis is labeled w and the y-axis is labeled V(w) on the domain [2.4, 3].
Figure 24.

From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side.

Try It

Use technology to find the maximum and minimum values on the interval[−1,4]of the functionf(x)=−0.2(x−2)3(x+1)2(x−4).

Show Solution

The minimum occurs at approximately the point(0,−6.5),and the maximum occurs at approximately the point(3.5,7).

 

Key Concepts

  • Polynomial functions of degree 2 or more are smooth, continuous functions.
  • To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero.
  • Another way to find thex-intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses thex-axis.
  • The multiplicity of a zero determines how the graph behaves at thex-intercepts.
  • The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity.
  • The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity.
  • The end behavior of a polynomial function depends on the leading term.
  • The graph of a polynomial function changes direction at its turning points.
  • A polynomial function of degreen has at mostn−1 turning points.
  • To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at mostn−1 turning points.
  • Graphing a polynomial function helps to estimate local and global extremas.
  • The Intermediate Value Theorem tells us that iff(a)andf(b) have opposite signs, then there exists at least one valuec betweena andb for whichf(c)=0.

Section Exercises

Verbal

  1. What is the difference between anx-intercept and a zero of a polynomial functionf?
Show Solution

Thex-intercept is where the graph of the function crosses thex-axis, and the zero of the function is the input value for whichf(x)=0.

2. If a polynomial function of degreen hasn distinct zeros, what do you know about the graph of the function?

3. Explain how the Intermediate Value Theorem can assist us in finding a zero of a function.

Show Solution

If we evaluate the function ata and atb and the sign of the function value changes, then we know a zero exists betweena andb.

4. Explain how the factored form of the polynomial helps us in graphing it.

5. If the graph of a polynomial just touches the x-axis and then changes direction, what can we conclude about the factored form of the polynomial?

Show Solution

There will be a factor raised to an even power.

Algebraic

For the following exercises, find thex- or t-intercepts of the polynomial functions.

6. C(t)=2(t−4)(t+1)(t−6)

7. C(t)=3(t+2)(t−3)(t+5)

Show Solution

(−2,0),(3,0),(−5,0)

8. C(t)=4t(t−2)2(t+1)

9. C(t)=2t(t−3)(t+1)2

Show Solution

(3,0),(−1,0),(0,0)

10. C(t)=2t4−8t3+6t2

11. C(t)=4t4+12t3−40t2

Show Solution

(0,0), (−5,0), (2,0)

12. f(x)=x4−x2

13. f(x)=x3+x2−20x

Show Solution

(0,0), (−5,0), (4,0)

14. f(x)=x3+6x2−7x

15. f(x)=x3+x2−4x−4

Show Solution

(2,0), (−2,0), (−1,0)

16. f(x)=x3+2x2−9x−18

17. f(x)=2x3−x2−8x+4

Show Solution

(−2,0),(2,0),(12,0)

18. f(x)=x6−7x3−8

19. f(x)=2x4+6x2−8

Show Solution

(1,0), (−1,0)

20. f(x)=x3−3x2−x+3

21. f(x)=x6−2x4−3x2

Show Solution

(0,0),(√3,0),(−√3,0)

22. f(x)=x6−3x4−4x2

23. f(x)=x5−5x3+4x

Show Solution

(0,0), (1,0), (−1,0), (2,0), (−2,0)

For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.

24. f(x)=x3−9x, betweenx=−4 andx=−2.

25. f(x)=x3−9x, betweenx=2 andx=4.

Show Solution

f(2)=–10 andf(4)=28.
Sign change confirms.

26. f(x)=x5−2x, betweenx=1 andx=2.

27. f(x)=−x4+4, betweenx=1 andx=3

Show Solution

f(1)=3 andf(3)=–77. Sign change confirms.

28. f(x)=−2x3−x, betweenx=–1 andx=1.

29. f(x)=x3−100x+2, betweenx=0.01 andx=0.1

Show Solution

f(0.01)=1.000001 andf(0.1)=–7.999. Sign change confirms.

For the following exercises, find the zeros and give the multiplicity of each.

30. f(x)=(x+2)3(x−3)2

31. f(x)=x2(2x+3)5(x−4)2

Show Solution

0 with multiplicity 2,−32 with multiplicity 5, 4 with multiplicity 2

32. f(x)=x3(x−1)3(x+2)

33. f(x)=x2(x2+4x+4)

Show Solution

0 with multiplicity 2, –2 with multiplicity 2

34. f(x)=(2x+1)3(9x2−6x+1)

35. f(x)=(3x+2)5(x2−10x+25)

Show Solution

−23withmultiplicity5,5withmultiplicity2

36. f(x)=x(4x2−12x+9)(x2+8x+16)

37. f(x)=x6−x5−2x4

Show Solution

0withmultiplicity4,2withmultiplicity1,–1withmultiplicity1

38. f(x)=3x4+6x3+3x2

39. f(x)=4x5−12x4+9x3

Show Solution

32 with multiplicity 2, 0 with multiplicity 3

40. f(x)=2x4(x3−4x2+4x)

41. f(x)=4x4(9x4−12x3+4x2)

Show Solution

0withmultiplicity6,23withmultiplicity2

Graphical

For the following exercises, graph the polynomial functions. Notex- andy-intercepts, multiplicity, and end behavior.

42. f(x)=(x+3)2(x−2)

43. g(x)=(x+4)(x−1)2

Show Solution

x-intercepts, (1,0) with multiplicity 2, (–4,0) with multiplicity 1, y-intercept (0,4). As x→−∞,f(x)→−∞,asx→∞,f(x)→∞.Graph of g(x)=(x+4)(x-1)^2.

44. h(x)=(x−1)3(x+3)2

45. k(x)=(x−3)3(x−2)2

Show Solution

x-intercepts(3,0)with multiplicity 3,(2,0)with multiplicity 2,y-intercept(0,–108). As x→−∞,f(x)→−∞,asx→∞,f(x)→∞.Graph of k(x)=(x-3)^3(x-2)^2.

46. m(x)=−2x(x−1)(x+3)

47. n(x)=−3x(x+2)(x−4)

Show Solution

x-intercepts(0,0), (–2,0), (4,0) with multiplicity 1,y-intercept(0,0). As x→−∞,f(x)→∞,asx→∞,f(x)→−∞.

Graph of n(x)=-3x(x+2)(x-4).

For the following exercises, use the graphs to write the formula for a polynomial function of least degree.

48. Write the equation:
Graph of a positive odd-degree polynomial with zeros at x=-2, 1, and 3.
49. Write the equation:
Graph of a negative odd-degree polynomial with zeros at x=-3, 1, and 3.
Show Solution

f(x)=−29(x−3)(x+1)(x+3)

50. Write the equation:
Graph of a negative odd-degree polynomial with zeros at x=-1, and 2.
51. Write the equation:
Graph of a positive odd-degree polynomial with zeros at x=-2, and 3.
Show Solution

f(x)=14(x+2)2(x−3)

52. Write the equation:
Graph of a negative even-degree polynomial with zeros at x=-3, -2, 3, and 4.

For the following exercises, use the graph to identify zeros and multiplicity.

53. Write the equation:
Graph of a negative even-degree polynomial with zeros at x=-4, -2, 1, and 3.
Show Solution

–4, –2, 1, 3 with multiplicity 1

54. Write the equation:
Graph of a positive even-degree polynomial with zeros at x=-4, -2, and 3.
55. Write the equation:
Graph of a positive even-degree polynomial with zeros at x=-2,, and 3.
Show Solution

–2, 3 each with multiplicity 2

56. Write the equation:

Graph of a negative odd-degree polynomial with zeros at x=-3, -2, and 1.

For the following exercises, use the given information about the polynomial graph to write the equation.

57. Degree 3. Zeros atx=–2, x=1,andx=3.y-intercept at(0,–4).

Show Solution

f(x)=−23(x+2)(x−1)(x−3)

58. Degree 3. Zeros atx=–5, x=–2,andx=1.y-intercept at(0,6)

59. Degree 5. Roots of multiplicity 2 atx=3 andx=1 , and a root of multiplicity 1 atx=–3. y-intercept at(0,9)

Show Solution

f(x)=13(x−3)2(x−1)2(x+3)

60. Degree 4. Root of multiplicity 2 atx=4,and a roots of multiplicity 1 atx=1andx=–2.y-intercept at(0,–3).

61. Degree 5. Double zero atx=1,and triple zero atx=3. Passes through the point(2,15).

Show Solution

f(x)=−15(x−1)2(x−3)3

62. Degree 3. Zeros atx=4,x=3,andx=2.y-intercept at(0,−24).

63. Degree 3. Zeros atx=−3,
x=−2
andx=1.
y-intercept at(0,12).

Show Solution

f(x)=−2(x+3)(x+2)(x−1)

64. Degree 5. Roots of multiplicity 2 atx=−3 andx=2 and a root of multiplicity 1 atx=−2. y-intercept at(0,4).

65. Degree 4. Roots of multiplicity 2 atx=12and roots of multiplicity 1 atx=6andx=−2. y-intercept at(0,18).

Show Solution

f(x)=−32(2x−1)2(x−6)(x+2)

66. Double zero atx=−3 and triple zero atx=0. Passes through the point(1,32).

Technology

For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum.

67. f(x)=x3−x−1

Show Solution

local max(–.58, –.62),
local min(.58, –1.38)

68. f(x)=2x3−3x−1

69. f(x)=x4+x

Show Solution

global min(–.63, –.47)

70. f(x)=−x4+3x−2

71. f(x)=x4−x3+1

Show Solution

global min(.75, .89)

Extensions

For the following exercises, use the graphs to write a polynomial function of least degree.

72. Write the equation.
Graph of a positive odd-degree polynomial with zeros at x=--2/3, and 4/3 and y=8.
73. Write the equation.
Graph of a positive odd-degree polynomial with zeros at x=--200, and 500 and y=50000000.
Show Solution

f(x)=(x−500)2(x+200)

74.
Graph of a positive odd-degree polynomial with zeros at x=--300, and 100 and y=-90000.

Real-World Applications

For the following exercises, write the polynomial function that models the given situation.

75. A rectangle has a length of 10 units and a width of 8 units. Squares ofx byx units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a polynomial function in terms ofx.

Show Solution

f(x)=4x3−36x2+80x

76. Consider the same rectangle of the preceding problem. Squares of2x by2x units are cut out of each corner. Express the volume of the box as a polynomial in terms ofx.

77. A square has sides of 12 units. Squaresx +1
byx +1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms ofx.

Show Solution

f(x)=4x3−36x2+60x+100

78. A cylinder has a radius ofx+2 units and a height of 3 units greater. Express the volume of the cylinder as a polynomial function.

79. A right circular cone has a radius of3x+6 and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone isV=13πr2h for radiusr and heighth.

Show Solution

f(x)=9π(x3+5x2+8x+4)

Glossary

global maximum
highest turning point on a graph;f(a) wheref(a)≥f(x) for allx.
global minimum
lowest turning point on a graph;f(a) wheref(a)≤f(x) for allx.
Intermediate Value Theorem
for two numbersa andb in the domain off, if[latex]\,a
multiplicity
the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form(x−h)p,x=his a zero of multiplicityp.

Media Attributions

License

Icon for the Creative Commons Attribution 4.0 International License

College Algebra Copyright © 2024 by LOUIS: The Louisiana Library Network is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book