Power Functions and Polynomial Functions

Learning Objectives

In this section, you will:

• Identify power functions.
• Identify end behavior of power functions.
• Identify polynomial functions.
• Identify the degree and leading coefficient of polynomial functions.

Figure 1 (credit: Jason Bay, Flickr)

Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in Table 1.

Year	20092009	20102010	20112011	20122012	20132013
Bird Population	800800	897897	992992	1,0831,083	1,1691,169

The population can be estimated using the function P(t)=−0.3t3+97t+800,𝑃(𝑡)=−0.3𝑡3+97𝑡+800, where P(t)𝑃(𝑡) represents the bird population on the island t𝑡 years after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we will examine functions that we can use to estimate and predict these types of changes.

Identifying Power Functions

Before we can understand the bird problem, it will be helpful to understand a different type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number.

As an example, consider functions for area or volume. The function for the area of a circle with radius r𝑟 is

A(r)=πr2𝐴(𝑟)=𝜋𝑟2

and the function for the volume of a sphere with radius r𝑟 is

V(r)=43πr3𝑉(𝑟)=43𝜋𝑟3

Both of these are examples of power functions because they consist of a coefficient, π𝜋 or 43π,43𝜋, multiplied by a variable r𝑟 raised to a power.

POWER FUNCTION

A power function is a function that can be represented in the form

f(x)=kxp𝑓(𝑥)=𝑘𝑥𝑝

where k𝑘 and p𝑝 are real numbers, and k𝑘 is known as the coefficient.

Q&A

Is f(x)=2x𝑓(𝑥)=2𝑥 a power function?

No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function.

EXAMPLE 1

Identifying Power Functions

Which of the following functions are power functions?

f(x)f(x)f(x)f(x)f(x)f(x)f(x)f(x)========1xx2x31x1x2x−−√x−−√3Constant functionIdentity functionQuadratic functionCubic functionReciprocal functionReciprocal squared functionSquare root functionCube root function𝑓(𝑥)=1Constant function𝑓(𝑥)=𝑥Identity function𝑓(𝑥)=𝑥2Quadratic function𝑓(𝑥)=𝑥3Cubic function𝑓(𝑥)=1𝑥Reciprocal function𝑓(𝑥)=1𝑥2Reciprocal squared function𝑓(𝑥)=𝑥Square root function𝑓(𝑥)=𝑥3Cube root function

Solution

All of the listed functions are power functions.

The constant and identity functions are power functions because they can be written as f(x)=x0𝑓(𝑥)=𝑥0 and f(x)=x1𝑓(𝑥)=𝑥1 respectively.

The quadratic and cubic functions are power functions with whole number powers f(x)=x2𝑓(𝑥)=𝑥2 and f(x)=x3.𝑓(𝑥)=𝑥3.

The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as f(x)=x−1𝑓(𝑥)=𝑥−1 and f(x)=x−2.𝑓(𝑥)=𝑥−2.

The square and cube root functions are power functions with fractional powers because they can be written as f(x)=x12𝑓(𝑥)=𝑥12 or f(x)=x13.𝑓(𝑥)=𝑥13.

TRY IT #1

Which functions are power functions?

f(x)g(x)h(x)===2x⋅4×3−x5+5x32x5−13×2+4𝑓(𝑥)=2𝑥⋅4𝑥3𝑔(𝑥)=−𝑥5+5𝑥3ℎ(𝑥)=2𝑥5−13𝑥2+4

Identifying End Behavior of Power Functions

Figure 2 shows the graphs of f(x)=x2,g(x)=x4𝑓(𝑥)=𝑥2,𝑔(𝑥)=𝑥4 and h(x)=x6,ℎ(𝑥)=𝑥6, which are all power functions with even, positive integer powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin.

Figure 2 Even-power functions

To describe the behavior as numbers become larger and larger, we use the idea of infinity. We use the symbol ∞∞ for positive infinity and −∞−∞ for negative infinity. When we say that “ x𝑥 approaches infinity,” which can be symbolically written as x→∞,𝑥→∞, we are describing a behavior; we are saying that x𝑥 is increasing without bound.

With the positive even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as x𝑥 approaches positive or negative infinity, the f(x)𝑓(𝑥) values increase without bound. In symbolic form, we could write

as x→±∞, f(x)→∞as 𝑥→±∞, 𝑓(𝑥)→∞

Figure 3 shows the graphs of f(x)=x3,g(x)=x5,𝑓(𝑥)=𝑥3,𝑔(𝑥)=𝑥5, and h(x)=x7,ℎ(𝑥)=𝑥7, which are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin.

Figure 3 Odd-power functions

These examples illustrate that functions of the form f(x)=xn𝑓(𝑥)=𝑥𝑛 reveal symmetry of one kind or another. First, in Figure 2 we see that even functions of the form f(x)=xn, n𝑓(𝑥)=𝑥𝑛, 𝑛 even, are symmetric about the y-𝑦- axis. In Figure 3 we see that odd functions of the form f(x)=xn, n𝑓(𝑥)=𝑥𝑛, 𝑛  odd, are symmetric about the origin.

For these odd power functions, as x𝑥 approaches negative infinity, f(x)𝑓(𝑥) decreases without bound. As x𝑥 approaches positive infinity, f(x)𝑓(𝑥) increases without bound. In symbolic form we write

as x→−∞, f(x)→−∞ as x→∞, f(x)→∞as 𝑥→−∞, 𝑓(𝑥)→−∞ as 𝑥→∞, 𝑓(𝑥)→∞

The behavior of the graph of a function as the input values get very small ( x→−∞𝑥→−∞ ) and get very large ( x→∞𝑥→∞ ) is referred to as the end behavior of the function. We can use words or symbols to describe end behavior.

Figure 4 shows the end behavior of power functions in the form f(x)=kxn𝑓(𝑥)=𝑘𝑥𝑛 where n𝑛 is a non-negative integer depending on the power and the constant.

Figure 4

HOW TO

Given a power function f(x)=kxn𝑓(𝑥)=𝑘𝑥𝑛 where n𝑛 is a positive integer, identify the end behavior.

1. Determine whether the power is even or odd.
2. Determine whether the constant is positive or negative.
3. Use Figure 4 to identify the end behavior.

EXAMPLE 2

Identifying the End Behavior of a Power Function

Describe the end behavior of the graph of f(x)=x8.𝑓(𝑥)=𝑥8.

Solution

The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As x𝑥 approaches infinity, the output (value of f(x)𝑓(𝑥) ) increases without bound. We write as x→∞,f(x)→∞.𝑥→∞,𝑓(𝑥)→∞. As x𝑥 approaches negative infinity, the output increases without bound. In symbolic form, as x→−∞, f(x)→∞.𝑥→−∞, 𝑓(𝑥)→∞. We can graphically represent the function as shown in Figure 5.

Figure 5

EXAMPLE 3

Identifying the End Behavior of a Power Function.

Describe the end behavior of the graph of f(x)=−x9.𝑓(𝑥)=−𝑥9.

Solution

The exponent of the power function is 9 (an odd number). Because the coefficient is –1–1 (negative), the graph is the reflection about the x-𝑥- axis of the graph of f(x)=x9.𝑓(𝑥)=𝑥9. Figure 6 shows that as x𝑥 approaches infinity, the output decreases without bound. As x𝑥 approaches negative infinity, the output increases without bound. In symbolic form, we would write

as x→−∞, f(x)→∞ as x→∞, f(x)→−∞as 𝑥→−∞, 𝑓(𝑥)→∞ as 𝑥→∞, 𝑓(𝑥)→−∞

Figure 6

Analysis

We can check our work by using the table feature on a graphing utility.

x𝑥	f(x)𝑓(𝑥)
–10	1,000,000,000
–5	1,953,125
0	0
5	–1,953,125
10	–1,000,000,000

We can see from Table 2 that, when we substitute very small values for x,𝑥, the output is very large, and when we substitute very large values for x,𝑥, the output is very small (meaning that it is a very large negative value).

TRY IT #2

Describe in words and symbols the end behavior of f(x)=−5×4.𝑓(𝑥)=−5𝑥4.

Identifying Polynomial Functions

An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil slick by combining two functions. The radius r𝑟 of the spill depends on the number of weeks w𝑤 that have passed. This relationship is linear.

r(w)=24+8w𝑟(𝑤)=24+8𝑤

We can combine this with the formula for the area A𝐴 of a circle.

A(r)=πr2𝐴(𝑟)=𝜋𝑟2

Composing these functions gives a formula for the area in terms of weeks.

A(w)===A(r(w))A(24+8w)π(24+8w)2𝐴(𝑤)=𝐴(𝑟(𝑤))=𝐴(24+8𝑤)=𝜋(24+8𝑤)2

Multiplying gives the formula.

A(w)=576π+384πw+64πw2𝐴(𝑤)=576𝜋+384𝜋𝑤+64𝜋𝑤2

This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power.

POLYNOMIAL FUNCTIONS

Let n𝑛 be a non-negative integer. A polynomial function is a function that can be written in the form

f(x)=anxn+…+a2x2+a1x+a0𝑓(𝑥)=𝑎𝑛𝑥𝑛+…+𝑎2𝑥2+𝑎1𝑥+𝑎0

This is called the general form of a polynomial function. Each ai𝑎𝑖 is a coefficient and can be any real number, but an≠0𝑎𝑛≠0. Each expression aixi𝑎𝑖𝑥𝑖 is a term of a polynomial function.

EXAMPLE 4

Identifying Polynomial Functions

Which of the following are polynomial functions?

f(x)g(x)h(x)===2×3⋅3x+4−x(x2−4)5x+2−−−−√𝑓(𝑥)=2𝑥3⋅3𝑥+4𝑔(𝑥)=−𝑥(𝑥2−4)ℎ(𝑥)=5𝑥+2

Solution

The first two functions are examples of polynomial functions because they can be written in the form f(x)=anxn+…+a2x2+a1x+a0,𝑓(𝑥)=𝑎𝑛𝑥𝑛+…+𝑎2𝑥2+𝑎1𝑥+𝑎0, where the powers are non-negative integers and the coefficients are real numbers.

• f(x)𝑓(𝑥) can be written as f(x)=6×4+4.𝑓(𝑥)=6𝑥4+4.
• g(x)𝑔(𝑥) can be written as g(x)=−x3+4x.𝑔(𝑥)=−𝑥3+4𝑥.
• h(x)ℎ(𝑥) cannot be written in this form and is therefore not a polynomial function.

Identifying the Degree and Leading Coefficient of a Polynomial Function

Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term.

TERMINOLOGY OF POLYNOMIAL FUNCTIONS

We often rearrange polynomials so that the powers are descending.

When a polynomial is written in this way, we say that it is in general form.

HOW TO

Given a polynomial function, identify the degree and leading coefficient.

1. Find the highest power of x𝑥 to determine the degree of the function.
2. Identify the term containing the highest power of x𝑥 to find the leading term.
3. Identify the coefficient of the leading term.

EXAMPLE 5

Identifying the Degree and Leading Coefficient of a Polynomial Function

Identify the degree, leading term, and leading coefficient of the following polynomial functions.

f(x)g(t)h(p)===3+2×2−4x35t5−2t3+7t6p−p3−2𝑓(𝑥)=3+2𝑥2−4𝑥3𝑔(𝑡)=5𝑡5−2𝑡3+7𝑡ℎ(𝑝)=6𝑝−𝑝3−2

Solution

For the function f(x),𝑓(𝑥), the highest power of x𝑥 is 3, so the degree is 3. The leading term is the term containing that degree, −4×3.−4𝑥3. The leading coefficient is the coefficient of that term, −4.−4.

For the function g(t),𝑔(𝑡), the highest power of t𝑡 is 5,5, so the degree is 5.5. The leading term is the term containing that degree, 5t5.5𝑡5. The leading coefficient is the coefficient of that term, 5.5.

For the function h(p),ℎ(𝑝), the highest power of p𝑝 is 3,3, so the degree is 3.3. The leading term is the term containing that degree, −p3.−𝑝3. The leading coefficient is the coefficient of that term, −1.−1.

TRY IT #3

Identify the degree, leading term, and leading coefficient of the polynomial f(x)=4×2−x6+2x−6.𝑓(𝑥)=4𝑥2−𝑥6+2𝑥−6.

Identifying End Behavior of Polynomial Functions

Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x𝑥 gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the power function consisting of the leading term. See Table 3.

Polynomial Function	Leading Term	Graph of Polynomial Function
f(x)=5×4+2×3−x−4𝑓(𝑥)=5𝑥4+2𝑥3−𝑥−4	5×45𝑥4
f(x)=−2×6−x5+3×4+x3𝑓(𝑥)=−2𝑥6−𝑥5+3𝑥4+𝑥3	−2×6−2𝑥6
f(x)=3×5−4×4+2×2+1𝑓(𝑥)=3𝑥5−4𝑥4+2𝑥2+1	3×53𝑥5
f(x)=−6×3+7×2+3x+1𝑓(𝑥)=−6𝑥3+7𝑥2+3𝑥+1	−6×3−6𝑥3

EXAMPLE 6

Identifying End Behavior and Degree of a Polynomial Function

Describe the end behavior and determine a possible degree of the polynomial function in Figure 7.

Figure 7

Solution

As the input values x𝑥 get very large, the output values f(x)𝑓(𝑥) increase without bound. As the input values x𝑥 get very small, the output values f(x)𝑓(𝑥) decrease without bound. We can describe the end behavior symbolically by writing

as x→−∞, f(x)→−∞ as x→∞, f(x)→∞as 𝑥→−∞, 𝑓(𝑥)→−∞ as 𝑥→∞, 𝑓(𝑥)→∞

In words, we could say that as x𝑥 values approach infinity, the function values approach infinity, and as x𝑥 values approach negative infinity, the function values approach negative infinity.

We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive.

TRY IT #4

Describe the end behavior, and determine a possible degree of the polynomial function in Figure 8.

Figure 8

EXAMPLE 7

Identifying End Behavior and Degree of a Polynomial Function

Given the function f(x)=−3×2(x−1)(x+4),𝑓(𝑥)=−3𝑥2(𝑥−1)(𝑥+4), express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function.

Solution

Obtain the general form by expanding the given expression for f(x).𝑓(𝑥).

f(x)===−3×2(x−1)(x+4)−3×2(x2+3x−4)−3×4−9×3+12×2𝑓(𝑥)=−3𝑥2(𝑥−1)(𝑥+4)=−3𝑥2(𝑥2+3𝑥−4)=−3𝑥4−9𝑥3+12𝑥2

The general form is f(x)=−3×4−9×3+12×2.𝑓(𝑥)=−3𝑥4−9𝑥3+12𝑥2. The leading term is −3×4;−3𝑥4; therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (–3), so the end behavior is

as x→−∞, f(x)→−∞ as x→∞, f(x)→−∞as 𝑥→−∞, 𝑓(𝑥)→−∞ as 𝑥→∞, 𝑓(𝑥)→−∞

TRY IT #5

Given the function f(x)=0.2(x−2)(x+1)(x−5),𝑓(𝑥)=0.2(𝑥−2)(𝑥+1)(𝑥−5), express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function.

Identifying Local Behavior of Polynomial Functions

In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing.

We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-intercept (0,a0).(0,𝑎0). The x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept. See Figure 9.

Figure 9

INTERCEPTS AND TURNING POINTS OF POLYNOMIAL FUNCTIONS

A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero.

HOW TO

Given a polynomial function, determine the intercepts.

1. Determine the y-intercept by setting x=0𝑥=0 and finding the corresponding output value.
2. Determine the x-intercepts by solving for the input values that yield an output value of zero.

EXAMPLE 8

Determining the Intercepts of a Polynomial Function

Given the polynomial function f(x)=(x−2)(x+1)(x−4),𝑓(𝑥)=(𝑥−2)(𝑥+1)(𝑥−4), written in factored form for your convenience, determine the y– and x-intercepts.

Solution

The y-intercept occurs when the input is zero so substitute 0 for x.𝑥.

f(0)===f(0)=(0−2)(0+1)(0−4)(−2)(1)(−4)8𝑓(0)=𝑓(0)=(0-2)(0+1)(0-4)=(-2)(1)(-4)=8

The y-intercept is (0, 8).

The x-intercepts occur when the output is zero.

0=(x−2)(x+1)(x−4)0=(𝑥−2)(𝑥+1)(𝑥−4)

x−2x==02ororx+1x==0−1ororx−4x==04𝑥−2=0or𝑥+1=0or𝑥−4=0𝑥=2or𝑥=−1or𝑥=4

The x-intercepts are (2,0),(–1,0),(2,0),(–1,0), and (4,0).(4,0).

We can see these intercepts on the graph of the function shown in Figure 10.

Figure 10

EXAMPLE 9

Determining the Intercepts of a Polynomial Function with Factoring

Given the polynomial function f(x)=x4−4×2−45,𝑓(𝑥)=𝑥4−4𝑥2−45, determine the y– and x-intercepts.

Solution

The y-intercept occurs when the input is zero.

f(0)==(0)4−4(0)2−45−45𝑓(0)=(0)4−4(0)2−45=−45

The y-intercept is (0,−45).(0,−45).

The x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial.

f(x)===x4−4×2−45(x2−9)(x2+5)(x−3)(x+3)(x2+5)𝑓(𝑥)=𝑥4−4𝑥2−45=(𝑥2−9)(𝑥2+5)=(𝑥−3)(𝑥+3)(𝑥2+5)

0=(x−3)(x+3)(x2+5)0=(𝑥−3)(𝑥+3)(𝑥2+5)

x−3x==03ororx+3x==0−3ororx2+5=0(no real solution)𝑥−3=0or𝑥+3=0or𝑥2+5=0𝑥=3or𝑥=−3or(no real solution)

The x-intercepts are (3,0)(3,0) and (–3,0).(–3,0).

We can see these intercepts on the graph of the function shown in Figure 11. We can see that the function is even because f(x)=f(−x).𝑓(𝑥)=𝑓(−𝑥).

Figure 11

TRY IT #6

Given the polynomial function f(x)=2×3−6×2−20x,𝑓(𝑥)=2𝑥3−6𝑥2−20𝑥, determine the y– and x-intercepts.

Comparing Smooth and Continuous Graphs

The degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function of nth𝑛th degree is the product of n𝑛 factors, so it will have at most n𝑛 roots or zeros, or x-intercepts. The graph of the polynomial function of degree n𝑛 must have at most n–1𝑛–1 turning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors.

A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth.

INTERCEPTS AND TURNING POINTS OF POLYNOMIALS

A polynomial of degree n𝑛 will have, at most, n𝑛 x-intercepts and n−1𝑛−1 turning points.

EXAMPLE 10

Determining the Number of Intercepts and Turning Points of a Polynomial

Without graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for f(x)=−3×10+4×7−x4+2×3.𝑓(𝑥)=−3𝑥10+4𝑥7−𝑥4+2𝑥3.

Solution

The polynomial has a degree of 10,10, so there are at most 10 x-intercepts and at most 9 turning points.

TRY IT #7

Without graphing the function, determine the maximum number of x-intercepts and turning points for f(x)=108−13×9−8×4+14×12+2×3.𝑓(𝑥)=108−13𝑥9−8𝑥4+14𝑥12+2𝑥3.

EXAMPLE 11

Drawing Conclusions about a Polynomial Function from the Graph

What can we conclude about the polynomial represented by the graph shown in Figure 12 based on its intercepts and turning points?

Figure 12

Solution

The end behavior of the graph tells us this is the graph of an even-degree polynomial. See Figure 13.

Figure 13

The graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4.

TRY IT #8

What can we conclude about the polynomial represented by the graph shown in Figure 14 based on its intercepts and turning points?

Figure 14

EXAMPLE 12

Drawing Conclusions about a Polynomial Function from the Factors

Given the function f(x)=−4x(x+3)(x−4),𝑓(𝑥)=−4𝑥(𝑥+3)(𝑥−4), determine the local behavior.

Solution

The y-intercept is found by evaluating f(0).𝑓(0).

f(0)==−4(0)(0+3)(0−40𝑓(0)=−4(0)(0+3)(0−4=0

The y-intercept is (0,0).(0,0).

The x-intercepts are found by determining the zeros of the function.

0=−4x(x+3)(x−4)0=−4𝑥(𝑥+3)(𝑥−4)

xx==00ororx+3x==0−3ororx−4x==04𝑥=0or𝑥+3=0or𝑥−4=0𝑥=0or𝑥=−3or𝑥=4

The x-intercepts are (0,0),(–3,0),(0,0),(–3,0), and (4,0).(4,0).

The degree is 3 so the graph has at most 2 turning points.

TRY IT #9

Given the function f(x)=0.2(x−2)(x+1)(x−5),𝑓(𝑥)=0.2(𝑥−2)(𝑥+1)(𝑥−5), determine the local behavior.