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Section 3.2 Polynomial Functions

Before Starting this Section, Review

1 Graphs of equations (Section 2.2 , page 186)
2 Solving linear equations (Section 1.1 , page 83)
3 Solving quadratic equations (Section 1.3 , page 106)
4 Transformations (Section 2.7 , page 263)
5 Relative maximum and minimum values (Section 2.5 , page 237)

Objectives

1 Learn properties of the graphs of polynomial functions.
2 Determine the end behavior of polynomial functions.
3 Find the zeros of a polynomial function by factoring.
4 Identify the relationship between degrees, real zeros, and turning points.
5 Graph polynomial functions.

Kepler’s Wedding Reception

Kepler is best known as an astronomer who discovered the three laws of planetary motion. However, Kepler’s primary field was not astronomy, but mathematics. Legend has it that at his own wedding reception, Kepler observed that the Austrian vintners could quickly and mysteriously compute the volumes of a variety of wine barrels. Each barrel had a hole, called a bunghole, in the middle of its side. The vintner would insert a rod, called the bungrod, into the hole until it hit the far corner. Then he would announce the volume. During the reception, Kepler occupied his mind with the problem of finding the volume of a wine barrel with a bungrod.

In Example 11, we show how he solved the problem for barrels that are perfect cylinders. But barrels are not perfect cylinders; they are wider in the middle and narrower at the ends and in between, the sides make a graceful curve that appears to be an arc of a circle. What could be the formula for the volume of such a shape? Kepler tackled this problem in his important book Nova Stereometria Doliorum Vinariorum (1615) and developed a related mathematical theory.

Johannes Kepler (1571–1630)

Kepler was born in Weil in southern Germany and studied at the University of Tübingen. He studied with Michael Mastlin (professor of mathematics), who privately taught him the Copernican theory of the sun-centered universe while hardly daring to recognize it openly in his lectures. Kepler knew that to work out a detailed version of Copernicus’s theory, he had to have access to the observations of the Danish astronomer Tycho Brahe (1546–1601). Kepler’s correspondence with Brahe resulted in his appointment as Brahe’s assistant. Brahe died about 18 months after Kepler’s arrival. During those 18 months, Kepler learned enough about Brahe’s work to create his own major project. He produced the famous three laws of planetary motion published in 1609 in his book Astronomia Nova.

Polynomial Functions

1 Learn properties of the graphs of polynomial functions.

We review some basic terminology. A polynomial function of degree n is a function of the form

f (x) = a n x n + a n - 1 x n - 1 + \dots + a 2 x 2 + a 1 x + a 0,

$f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{2} x^{2} + a_{1} x + a_{0},$

where n is a nonnegative integer and the coefficients an,an−1, …,a2,a1,a0 $a_{n}, a_{n - 1}, \dots, a_{2}, a_{1}, a_{0}$ are real numbers with an≠0. $a_{n} \neq 0 .$ The term anxn $a_{n} x^{n}$ is called the leading term, the number an $a_{n}$ (the coefficient of xn $x^{n}$ ) is called the leading coefficient, and a0 $a_{0}$ is the constant term. A constant function f(x)=a(a≠0), $f (x) = a (a \neq 0),$ which may be written as f(x)=ax0, $f (x) = a x^{0},$ for x≠0 $x \neq 0$ is a polynomial of degree 0. Because the zero function f(x)=0 $f (x) = 0$ can be written in several ways, such as 0=0x6+0x=0x5+0=0x4+0x3+0, $0 = 0 x^{6} + 0 x = 0 x^{5} + 0 = 0 x^{4} + 0 x^{3} + 0,$ no degree is assigned to it.

In Section 2.6, we discussed linear functions, which are polynomial functions of degree 0 and 1. In Section 3.1, we studied quadratic functions, which are polynomial functions of degree 2. Polynomials of degree 3, 4, and 5 are also called cubic, quartic, and quintic polynomials, respectively. In this section, we concentrate on graphing polynomial functions of degree 3 or more.

Here are some common properties shared by all polynomial functions.

Example 1 Polynomial Functions

State which functions are polynomial functions. For each polynomial function, find its degree, the leading term, and the leading coefficient.

f(x)=5x4−2x+7 $f (x) = 5 x^{4} - 2 x + 7$
g(x)=7x2−x+1, 1≤x≤5 $g (x) = 7 x^{2} - x + 1, 1 \leq x \leq 5$

Solution

f(x)=5x4−2x+7 $f (x) = 5 x^{4} - 2 x + 7$ is a polynomial function. Its degree is 4, the leading term is 5x4, $5 x^{4},$ and the leading coefficient is 5.
g(x)=7x2−x+1, 1≤x≤5 $g (x) = 7 x^{2} - x + 1, 1 \leq x \leq 5$ is not a polynomial function because its domain is not (−∞, ∞). $(- \infty, \infty) .$

Practice Problem 1

Repeat Example 1 for each function.
1. f(x)=x2+1x−1 $f (x) = \frac{x^{2} + 1}{x - 1}$
2. g(x)=2x7+5x2−17 $g (x) = 2 x^{7} + 5 x^{2} - 17$

Power Functions

A function of the form f(x)=axn $f (x) = a x^{n}$ is the simplest nth-degree polynomial function and is called a power function.

The graph of a power function can be obtained by stretching or shrinking the graph of y=xn $y = x^{n}$ (and reflecting it about the x-axis if a<0 $a < 0$ ). So the basic shape of the graph of a power function f(x)=axn $f (x) = a x^{n}$ is similar to the shape of the graphs of the equations y=±xn. $y = \pm x^{n} .$

Power Function of Even Degree

The graph of y=xn $y = x^{n}$ (when n is even) has the same basic shape as the graph of y=x2. $y = x^{2} .$ See Figure 3.6(a). Notice that each of these graphs is symmetric about the y-axis and passes through the points (−1,1),(0,0), $(- 1, 1), (0, 0),$ and (1, 1). As the exponent n increases, the graph of f(x)=xn $f (x) = x^{n}$ becomes flatter around x=0 $x = 0$ and becomes almost vertical at x=−1 $x = - 1$ and x=1 $x = 1$ . The graphs of y=x2, y=x4, $y = x^{2}, y = x^{4},$ and y=x6 $y = x^{6}$ are compared in Figure 3.6(a). The graph of y=xn $y = x^{n}$ is always located in the shaded regions shown in Figure 3.6(b), and the larger the exponent n, the more closely the graph will resemble the dark red curve () $()$ shown in Figure 3.6(b).

Figure 3.6

Power functions of even degree

Finally, it is important to notice the following hierarchy of the power functions

x 2 > x 4 > x 6 > x 8 > \dots, for 0 < x < 1 x 2 < x 4 < x 6 < x 8 < \dots, for 1.

$\begin{array}{l} x^{2} > x^{4} > x^{6} > x^{8} > \dots, for 0 < x < 1 \\ x^{2} < x^{4} < x^{6} < x^{8} < \dots, for 1. \end{array}$

Observe that the hierarchy on [0, 1] is reversed on [1,∞) $[1, \infty)$ .

While it may appear that, for large n, the graph of y=xn $y = x^{n}$ coincides with the x-axis, that is not the case and merely reflects the shortcomings of the resolutions of plotting devices. In fact, the graph of y=xn $y = x^{n}$ only touches the x-axis at the origin.

Power Function of Odd Degree

The graph of y=xn $y = x^{n}$ (when n is odd) has the same basic shape as the graph of y=x3 $y = x^{3}$ . See Figure 3.7(a). Notice that each of these graphs is symmetric about the origin and passes through the points (−1,−1) $(- 1, - 1)$ , (0,0) $(0, 0)$ , and (1, 1). As the exponent n increases, the graph of f(x)=xn $f (x) = x^{n}$ becomes flatter around x=0 $x = 0$ and becomes almost vertical at x=−1 $x = - 1$ and x=1 $x = 1$ . The graphs of y=x3, y=x5, $y = x^{3}, y = x^{5},$ and y=x7 $y = x^{7}$ are compared in Figure 3.7(a). The graph of y=xn $y = x^{n}$ is always located in the shaded regions depicted in Figure 3.7(b). The larger the exponent n, the more closely the graph will resemble the dark red curve shown in Figure 3.7(b).

Figure 3.7

Power functions of odd degree

Finally, it is important to notice the following hierarchy of the power functions

x > x 3 > x 5 > x 7 > \dots, for 0 < x < 1 x < x 3 < x 5 < x 7 < \dots, for x > 1.

$\begin{array}{l} x > x^{3} > x^{5} > x^{7} > \dots, for 0 < x < 1 \\ x < x^{3} < x^{5} < x^{7} < \dots, for x > 1. \end{array}$

Observe that the hierarchy on [0, 1] is reversed on [1,∞) $[1, \infty)$ .

End Behavior of Polynomial Functions

2 Determine the end behavior of polynomial functions.

The polynomial function y=f(x) $y = f (x)$ may oscillate quite significantly. However, its behavior for large values of |x| $| x |$ is predictable and is dominated by its leading term. To describe this behavior more precisely, we use the −∞ $- \infty$ and ∞ $\infty$ notations. The notation x→∞ $x \to \infty$ (read “x approaches infinity”) means that x is increasing (i.e., getting larger and larger-think x=1,10,100,1000, …) $x = 1, 10, 100, 1000, \dots)$ without bound. Similarly, the notation x→−∞ $x \to - \infty$ (read “x approaches negative infinity”) means that x is decreasing (i.e., getting smaller and smaller-think x=−1,−10,−100,−1000, … $x = - 1, - 10, - 100, - 1000, \dots$ ) without bound.

The behavior of a polynomial function y=f(x) $y = f (x)$ for very large values of |x| $| x |$ , that is, as x→∞ $x \to \infty$ or x→−∞ $x \to - \infty$ , is called the end behavior of the function.

Symbol	Meaning
y=f(x)→∞, $y = f (x) \to \infty,$ as x→∞ $x \to \infty$	the values of `f`(`x`) increase without bound when `x` increases without bound
y=f(x)→−∞, $y = f (x) \to - \infty,$ as x→∞ $x \to \infty$	the values of `f`(`x`) decrease without bound when `x` increases without bound
y=f(x)→∞, $y = f (x) \to \infty,$ as x→−∞ $x \to - \infty$	the values of `f`(`x`) increase without bound when `x` decreases without bound
y=f(x)→−∞, $y = f (x) \to - \infty,$ as x→−∞ $x \to - \infty$	the values of `f`(`x`) decrease without bound when `x` decreases without bound

In the next box, we describe the behavior of the power function y=f(x)=axn $y = f (x) = a x^{n}$ as x→∞ $x \to \infty$ or x→−∞. $x \to - \infty .$ Every polynomial function exhibits end behavior similar to one of the four functions y=x2, y=−x2, y=x3 $y = x^{2}, y = - x^{2}, y = x^{3}$ , or y=−x3 $y = - x^{3}$ .

Side Note

Here is a scheme for remembering the axis directions associated with the symbols −∞ $- \infty$ and ∞. $\infty .$

x \to - \infty Left x \to \infty Right y \to - \infty Down y \to - \infty Up

$\begin{array}{l} x \to - \infty Left \\ x \to \infty Right \\ y \to - \infty Down \\ y \to - \infty Up \end{array}$

End Behavior for the Graph of y=f(x)=axn $y = f (x) = a x^{n}$

n Even

n Odd

Case 1

a>0 $a > 0$

The graph rises to the left and right, similar to y=x2. $y = x^{2} .$

Case 2

a<0 $a < 0$

The graph falls to the left and right, similar to y=−x2. $y = - x^{2} .$

Case 3

a>0 $a > 0$

The graph rises to the right and falls to the left, similar to y=x3. $y = x^{3} .$

Case 4

a<0 $a < 0$

The graph rises to the left and falls to the right, similar to y=−x3. $y = - x^{3} .$

In the next example, we show that the end behavior of the given polynomial function is determined by its leading term. If a polynomial P(x) has approximately the same values as f(x)=axn $f (x) = a x^{n}$ when |x| $| x |$ is very large, we write P(x)≈axn $P (x) \approx a x^{n}$ when |x| $| x |$ is very large.

Example 2 Understanding the End Behavior of a Polynomial Function

Let P(x)=2x3+5x2−7x+11 $P (x) = 2 x^{3} + 5 x^{2} - 7 x + 11$ be a polynomial function of degree 3. Show that P(x)≈2x3 $P (x) \approx 2 x^{3}$ when |x| $| x |$ is very large.

Solution

P (x) = = 2 x 3 + 5 x 2 - 7 x + 11 x 3 (2 + 5 x - 7 x 2 + 11 x 3) Given polynomial Distributive property

$\begin{array}{rcll} P (x) & = & 2 x^{3} + 5 x^{2} - 7 x + 11 & Given polynomial \\ = & x^{3} (2 + \frac{5}{x} - \frac{7}{x^{2}} + \frac{11}{x^{3}}) & Distributive property \end{array}$

When |x| $| x |$ is very large, the terms 5x, −7x2, $\frac{5}{x}, - \frac{7}{x^{2}},$ and 11x3 $\frac{11}{x^{3}}$ are close to 0. Table 3.1 shows some values of 5x $\frac{5}{x}$ as x increases. (You should compute the values of −7x2 $- \frac{7}{x^{2}}$ and 11x3 $\frac{11}{x^{3}}$ for x=10, 102, 103, $x = 10, 10^{2}, 10^{3},$ and 104.) $10^{4} .)$ When |x| $| x |$ is very large, we have

P (x) = x 3 (2 + 5 x - 7 x 2 + 11 x 3) \approx = x 3 (2 + 0 - 0 + 0) 2 x 3 .

$\begin{array}{rcl} P (x) = x^{3} (2 + \frac{5}{x} - \frac{7}{x^{2}} + \frac{11}{x^{3}}) & \approx & x^{3} (2 + 0 - 0 + 0) \\ = & 2 x^{3} . \end{array}$

So the polynomial P(x)≈2x3 $P (x) \approx 2 x^{3}$ when |x| $| x |$ is very large.

Table 3.1

`x`	5x $\frac{5}{x}$
10	0.5
102 $10^{2}$	0.05
103 $10^{3}$	0.005
104 $10^{4}$	0.0005

Practice Problem 2

Let P(x)=4x3+2x2+5x−17. $P (x) = 4 x^{3} + 2 x^{2} + 5 x - 17 .$ Show that P(x)≈4x3 $P (x) \approx 4 x^{3}$ when |x| $| x |$ is very large.

It is possible to use the technique of Example 2 to show that the end behavior of any polynomial function is determined by its leading term.

The end behaviors of polynomial functions are shown in the following graphs.

The Leading-Term Test

Let f(x)=anxn+an−1xn−1+⋯+a1x+a0(an≠0) $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} (a_{n} \neq 0)$ be a polynomial function. Its leading term is anxn. $a_{n} x^{n} .$ The behavior of the graph of y=f(x) $y = f (x)$ as x→∞ $x \to \infty$ or x→−∞ $x \to - \infty$ has end behavior similar to one of the following four graphs and is described as shown in each case.

n Even

n Odd

Case 1: similar to y=x2 $y = x^{2}$

an>0 $a_{n} > 0$

Case 2: similar to y=−x2 $y = - x^{2}$

an<0 $a_{n} < 0$

Case 3: similar to y=x3 $y = x^{3}$

an>0 $a_{n} > 0$

Case 4: similar to y=−x3 $y = - x^{3}$

an<0 $a_{n} < 0$

This test does not describe the middle portion of each graph, shown by the dashed lines.

Note that even-degree polynomial functions have the same end behavior on both ends of the x-axis. Odd-degree polynomial functions have the opposite end behavior on opposite ends of the x-axis.

Example 3 Using the Leading-Term Test

Use the leading-term test to determine the end behavior of the graph of

y = f (x) = - 2 x 3 + 3 x + 4 .

$y = f (x) = - 2 x^{3} + 3 x + 4 .$

Solution

Here the degree n=3 $n = 3$ (odd) and the leading coefficient an=−2<0. $a_{n} = - 2 < 0 .$ Case 4 applies. The graph of f rises to the left and falls to the right. See Figure 3.8. This end behavior is described as y→∞ $y \to \infty$ as x→−∞ $x \to - \infty$ and y→−∞ $y \to - \infty$ as x→∞. $x \to \infty .$

Figure 3.8

End behavior of f( x)=−2x3+3x+4 $f (x) = - 2 x^{3} + 3 x + 4$

Practice Problem 3

What is the end behavior of f(x)=−2x4+5x2+3? $f (x) = - 2 x^{4} + 5 x^{2} + 3 ?$

Zeros of a Function

3 Find the zeros of a polynomial function by factoring.

Let f be a function. An input c in the domain of f that produces output 0 is called a zero of the function f. For example, if f(x)=(x−3)2, $f (x) = {(x - 3)}^{2},$ then 3 is a zero of f because f(3)=(3−3)2=0. $f (3) = {(3 - 3)}^{2} = 0 .$ In other words, a number c is a zero of f if f(c)=0. $f (c) = 0 .$ The zeros of a function f can be obtained by solving the equation f(x)=0. $f (x) = 0 .$ One way to solve a polynomial equation f(x)=0 $f (x) = 0$ is to factor f(x) and then use the zero-product property.

If c is a real number and f(c)=0, $f (c) = 0,$ then c is called a real zero of f. Geometrically, this means that the graph of f has an x-intercept at x=c. $x = c .$

Example 4 Finding the Zeros of a Polynomial Function

Find all real zeros of each polynomial function.

f(x)=x4−2x3−3x2 $f (x) = x^{4} - 2 x^{3} - 3 x^{2}$
g(x)=x3−2x2+x−2 $g (x) = x^{3} - 2 x^{2} + x - 2$

Solution

We factor f(x) and then solve the equation f(x)=0. $f (x) = 0 .$

$f (x) x 2 (x + 1) (x - 3) x 2 = 0, x + 1 x = 0, x = = = = = = x 4 - 2 x 3 - 3 x 2 x 2 (x 2 - 2 x - 3) x 2 (x + 1) (x - 3) 0 0, or x - 3 = 0 - 1, or x = 3 Given function Factor out the common factor, x 2 . Factor (x 2 - 2 x - 3) . Set f (x) = 0. Zero-product property Solve each equation .$ $\begin{array}{rcll} f (x) & = & x^{4} - 2 x^{3} - 3 x^{2} & Given function \\ = & x^{2} (x^{2} - 2 x - 3) & Factor out the common factor, x^{2} . \\ = & x^{2} (x + 1) (x - 3) & Factor (x^{2} - 2 x - 3) . \\ x^{2} (x + 1) (x - 3) & = & 0 & Set f (x) = 0. \\ x^{2} = 0, x + 1 & = & 0, or x - 3 = 0 & Zero-product property \\ x = 0, x & = & - 1, or x = 3 & Solve each equation . \end{array}$

The zeros of f(x) are 0, −1, $0, - 1,$ and 3. (See the calculator graph of f in the margin.)
By grouping terms, we factor g(x) to obtain

$g (x) 0 x - 2 = = = = = x 3 - 2 x 2 + x - 2 x 2 (x - 2) + 1 \cdot (x - 2) (x - 2) (x 2 + 1) (x - 2) (x 2 + 1) 0, or x 2 + 1 = 0 Group terms . Factor out x - 2. Set g (x) = 0. Zero-product property$ $\begin{array}{rcll} g (x) & = & x^{3} - 2 x^{2} + x - 2 \\ = & x^{2} (x - 2) + 1 \cdot (x - 2) & Group terms . \\ = & (x - 2) (x^{2} + 1) & Factor out x - 2. \\ 0 & = & (x - 2) (x^{2} + 1) & Set g (x) = 0. \\ x - 2 & = & 0, or x^{2} + 1 = 0 & Zero-product property \end{array}$

Solving for x, we obtain 2 as the only real zero of g(x), since x2+1>0 $x^{2} + 1 > 0$ for all real numbers x. The calculator graph of g in the margin supports our conclusion.

Practice Problem 4

Find all real zeros of f(x)=2x3−3x2+4x−6. $f (x) = 2 x^{3} - 3 x^{2} + 4 x - 6 .$

Finding the zeros of a polynomial of degree 3 or more is one of the most important problems of mathematics. You will learn more about this topic in Sections 3.4 and 3.5. For now, we will approximate the zeros by using an Intermediate Value Theorem. Recall that the graph of a polynomial function f(x) is a continuous curve. Suppose you locate two numbers a and b such that the function values f(a) and f(b) have opposite signs (one positive and the other negative). Then the continuous graph of f connecting the points (a, f(a)) and (b, f(b)) must cross the x-axis (at least once) somewhere between a and b. In other words, the function f has a zero between a and b. See Figure 3.10.

Figure 3.10

Each function has at least one zero between `a` and `b`

Notice that Figure 3.10(c) shows more than one zero of f. That is why this Intermediate Value Theorem says that f has at least one zero between a and b.

Example 5 Using an Intermediate Value Theorem

Show that the function f(x)=−2x3+4x+5 $f (x) = - 2 x^{3} + 4 x + 5$ has a real zero between 1 and 2.

Solution

f (1) f (2) = = = = - 2 (1) 3 + 4 (1) + 5 - 2 + 4 + 5 = 7 - 2 (2) 3 + 4 (2) + 5 - 2 (8) + 8 + 5 = - 3 Replace x with 1 in f (x) . Simplify . Replace x with 2. Simplify .

$\begin{array}{rcll} f (1) & = & - 2 {(1)}^{3} + 4 (1) + 5 & Replace x with 1 in f (x) . \\ = & - 2 + 4 + 5 = 7 & Simplify . \\ f (2) & = & - 2 {(2)}^{3} + 4 (2) + 5 & Replace x with 2. \\ = & - 2 (8) + 8 + 5 = - 3 & Simplify . \end{array}$

Since f(1) is positive and f(2) is negative, they have opposite signs. So by this Intermediate Value Theorem, the polynomial function f(x)=−2x3+4x+5 $f (x) = - 2 x^{3} + 4 x + 5$ has a real zero between 1 and 2. See Figure 3.11.

Figure 3.11

A zero lies between 1 and 2

Practice Problem 5

Show that f(x)=2x3−3x−6 $f (x) = 2 x^{3} - 3 x - 6$ has a real zero between 1 and 2.

In Example 5, if you want to find the zero between 1 and 2 more accurately, you can divide the interval [1, 2] into tenths and find the value of the function at each of the points 1, 1.1, 1.2, …, 1.9, and 2. You will find that

f (1.8) = 0.536 and f (1.9) = - 1.118 .

$f (1.8) = 0.536 and f (1.9) = - 1.118 .$

So by this Intermediate Value Theorem, f has a zero between 1.8 and 1.9. If you divide the interval [1.8, 1.9] into tenths, you will find that f(1.83)>0 $f (1.83) > 0$ and f(1.84)<0, $f (1.84) < 0,$ so f has a zero between 1.83 and 1.84. By continuing this process, you can find the zero of f between 1.8 and 1.9 to any desired degree of accuracy.

Zeros and Turning Points

4 Identify the relationship between degrees, real zeros, and turning points.

In Section 3.5, you will learn the Fundamental Theorem of Algebra. One of its consequences is the following fact:

Side Note

A polynomial of degree n can have no more than n real zeros. In Example 6, parts b and c, we see two polynomials of degree 3 each having fewer than 3 zeros.

Example 6 Finding the Number of Real Zeros

Find the number of distinct real zeros of the following polynomial functions of degree 3.

f(x)=(x−1)(x+2)(x−3) $f (x) = (x - 1) (x + 2) (x - 3)$
g(x)=(x+1)(x2+1) $g (x) = (x + 1) (x^{2} + 1)$
h(x)=(x−3)2(x+1) $h (x) = {(x - 3)}^{2} (x + 1)$

Solution

f(x)x−1x===(x−1)(x+2)(x−3)=00,orx+2=0,orx−3=01orx=−2orx=3Set f(x)=0.Zero-product propertySolve for x. $\begin{array}{rcll} f (x) & = & (x - 1) (x + 2) (x - 3) = 0 & Set f (x) = 0. \\ x - 1 & = & 0, or x + 2 = 0, or x - 3 = 0 & Zero-product property \\ x & = & 1 or x = - 2 or x = 3 & Solve for x . \end{array}$

So f(x) has three real zeros: 1, −2, $1, - 2,$ and 3.
g(x)x+1x===(x+1)(x2+1)=00 or x2+1=0−1Set g(x)=0.Zero-product propertySolve for x. $\begin{array}{rcll} g (x) & = & (x + 1) (x^{2} + 1) = 0 & Set g (x) = 0. \\ x + 1 & = & 0 or x^{2} + 1 = 0 & Zero-product property \\ x & = & - 1 & Solve for x . \end{array}$

Because x2+1>0 $x^{2} + 1 > 0$ for all real x, the equation x2+1=0 $x^{2} + 1 = 0$ has no real solution. So −1 $- 1$ is the only real zero of g(x).
h(x)(x−3)2x===(x−3)2(x+1)=00,orx+1=03,orx=−1Set h(x)=0.Zero-product propertySolve for x. $\begin{array}{rcll} h (x) & = & {(x - 3)}^{2} (x + 1) = 0 & Set h (x) = 0. \\ {(x - 3)}^{2} & = & 0, or x + 1 = 0 & Zero-product property \\ x & = & 3, or x = - 1 & Solve for x . \end{array}$

The real zeros of h(x) are 3 and −1. $- 1 .$ Thus, h(x) has two distinct real zeros.

Practice Problem 6

Find the distinct real zeros of

$f (x) = (x + 1) 2 (x - 3) (x + 5) .$ $f (x) = {(x + 1)}^{2} (x - 3) (x + 5) .$

In Example 6c, the factor (x−3) $(x - 3)$ appears twice in h(x). We say that 3 is a zero repeated twice, or that 3 is a zero of multiplicity 2. Notice that the graph of y=h(x) $y = h (x)$ in Figure 3.12 touches the x-axis at x=3, $x = 3,$ where h has a zero of multiplicity 2 (an even number). In contrast, the graph of h crosses the x-axis at x=−1, $x = - 1,$ where it has a zero of multiplicity 1 (an odd number). We next consider the geometric significance of the multiplicity of a zero.

Figure 3.12

Graph of h(x)=(x−3)2(x+1) $h (x) = {(x - 3)}^{2} (x + 1)$

The multiplicity of a zero determines the manner in which the graph of a polynomial function crosses or touches the x-axis. See the side note. This observation allows us to draw the graph of a polynomial function that is already fully factored with enhanced precision.

Side Note

As the multiplicity m of a zero c of a polynomial function increases, the graph of f becomes more curved around x=c $x = c$ , as illustrated below

If m=1 $m = 1$

If m=3 $m = 3$

Example 7 Finding Zeros and Their Multiplicities

Find the zeros of the polynomial function f(x)=x2(x+1)(x−2) $f (x) = x^{2} (x + 1) (x - 2)$ and give the multiplicity of each zero.

Solution

The polynomial f(x) is already in factored form.

f (x) x 2 x = = = x 2 (x + 1) (x - 2) = 0 0, or x + 1 = 0, or x - 2 = 0 0 or x = - 1 or x = 2 Set f (x) = 0. Zero-product property Solve for x .

$\begin{array}{rcll} f (x) & = & x^{2} (x + 1) (x - 2) = 0 & Set f (x) = 0. \\ x^{2} & = & 0, or x + 1 = 0, or x - 2 = 0 & Zero-product property \\ x & = & 0 or x = - 1 or x = 2 & Solve for x . \end{array}$

The function has three distinct zeros: 0, −1, $0, - 1,$ and 2. The zero x=0 $x = 0$ has multiplicity 2, and each of the zeros −1 $- 1$ and 2 has multiplicity 1. The graph of f is given in Figure 3.13. Notice that the graph of f in Figure 3.13 crosses the x-axis at −1 $- 1$ and 2 (the zeros of odd multiplicity), while it touches but does not cross the x-axis at x=0 $x = 0$ (the zero of even multiplicity).

Practice Problem 7

Write the zeros of f(x)=(x−1)2(x+3)(x+5) $f (x) = {(x - 1)}^{2} (x + 3) (x + 5)$ and give the multiplicity of each.

Turning Points

The graph of f(x)=2x3−3x2−12x+5 $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 5$ is shown in Figure 3.14. The value f(−1)=12 $f (- 1) = 12$ is a local (or relative) maximum value of f. Similarly, the value f(2)=−15 $f (2) = - 15$ is a local (or relative) minimum value of f. (See Section 2.5 for definitions.) The graph points corresponding to the local maximum and local minimum values of a polynomial function are the turning points. At each turning point, the graph changes direction from increasing to decreasing or from decreasing to increasing. One of the successes of calculus is finding the maximum and minimum values for many functions.

Example 8 Using a Graphing Device to Find the Number of Turning Points

Use a graphing calculator and the window −10≤x≤10; −30≤y≤30 $- 10 \leq x \leq 10; - 30 \leq y \leq 30$ to find the number of turning points of the graph of each polynomial function.

f(x)=x4−7x2−18 $f (x) = x^{4} - 7 x^{2} - 18$
g(x)=x3+x2−12x $g (x) = x^{3} + x^{2} - 12 x$
h(x)=x3−3x2+3x−1 $h (x) = x^{3} - 3 x^{2} + 3 x - 1$

Solution

The graphs of f, g, and h are shown in Figure 3.15.

Figure 3.15

Turning points on a calculator graph

Function f has three turning points: two local minima and one local maximum.
Function g has two turning points: one local maximum and one local minimum.
Function h has no turning points: it is increasing on the interval (−∞, ∞). $(- \infty, \infty) .$

Practice Problem 8

Find the number of turning points of the graph of f(x)=−x4+3x2−2. $f (x) = - x^{4} + 3 x^{2} - 2 .$

Side Note

Notice that in part (c) of Figure 3.15 the graph seems to “turn” but the graph is always increasing. There is no change from increasing to decreasing, so it has no turning points.

Graphing a Polynomial Function

5 Graph polynomial functions.

You can graph a polynomial function by constructing a table of values for a large number of points, plotting the points, and connecting the points with a smooth curve. (In fact, a graphing calculator uses this method.) Generally, however, this process is a poor way to graph a function by hand because it is tedious and error-prone, and so may give misleading results. Similarly, a graphing calculator may give misleading results if the viewing window is not chosen appropriately. Next, we discuss a better strategy for graphing a polynomial function.

Procedure in Action

Example 9 Graphing a Polynomial Function

OBJECTIVE	EXAMPLE
Sketch the graph of a polynomial function.	Sketch the graph of f(x)=−x3−4x2+4x+16. $f (x) = - x^{3} - 4 x^{2} + 4 x + 16 .$
Step 1 Determine the end behavior. Apply the leading-term test.	Since the degree, 3, of `f` is odd and the leading coefficient, −1, $- 1,$ is negative, the end behavior (Case 4, page 336) is similar to that of y=−x3 $y = - x^{3}$ as shown in this sketch.
Step 2 Find the real zeros of the polynomial function. Set f(x)=0. $f (x) = 0 .$ Solve the equation f(x)=0 $f (x) = 0$ by factoring. The real zeros with their multiplicities will give you the `x`-intercepts of the graph and determine whether or not the graph crosses the `x`-axis there.	−x3−4x2+4x+16=0−x2(x+4)+4(x+4)=0(x+4)(−x2+4)=0−(x+4)(x+2)(x−2)=0x=−4, x=−2, or x=2 $\begin{array}{r} - x^{3} - 4 x^{2} + 4 x + 16 = 0 \\ -^{} x^{2} (x + 4) + 4 (x + 4) = 0 \\ (x + 4) (- x^{2} + 4) = 0 \\ - (x + 4) (x + 2) (x - 2) = 0 \\ x = - 4, x = - 2, or x = 2 \end{array}$ There are three zeros, each of multiplicity 1, so the graph crosses the `x`-axis at each zero.
Step 3 Find the `y`-intercept by computing `f`(0).	The `y`-intercept is f(0)=16. $f (0) = 16 .$ The graph passes through the point (0, 16).
Step 4 Use symmetry (if any) Check whether the function is odd, even, or neither.	f(−x)==−(−x)3−4(−x)2+4(−x)+16x3−4x2−4x+16 $\begin{array}{rcl} f (- x) & = & - {(- x)}^{3} - 4 {(- x)}^{2} + 4 (- x) + 16 \\ = & x^{3} - 4 x^{2} - 4 x + 16 \end{array}$ Since f(−x)≠f(x), $f (- x) \neq f (x),$ there is no symmetry about the `y`-axis. Also, since f(−x)≠−f(x), $f (- x) \neq - f (x),$ there is no symmetry with respect to the origin.
Step 5 Determine the sign of `f`(`x`) by using arbitrarily chosen “test numbers” in the intervals defined by the `x`-intercepts. Find on which of these intervals the graph lies above or below the `x`-axis.	The three zeros −4, −2, $- 4, - 2,$ and 2 divide the `x`-axis into four intervals: (−∞, −4), (−4, −2), $(- \infty, - 4), (- 4, - 2),$ (−2, 2), $(- 2, 2),$ and (2, ∞). $(2, \infty) .$ We choose −5, −3, 0, $- 5, - 3, 0,$ and 3 as test numbers.
Step 6 Draw the graph using points from Steps 1–5. To check whether the graph is drawn correctly, use the fact that the number of turning points is less than the degree of the polynomial.	The graph connects the points we have found with a smooth curve. The number of turning points is 2, which is less than 3, the degree of `f`.

Practice Problem 9

Graph f(x)=−x4+5x2−4. $f (x) = - x^{4} + 5 x^{2} - 4 .$

Side Note

The leading term of a product of two polynomials f(x)⋅g(x) $f (x) \cdot g (x)$ is the product of the leading terms of f(x) and g(x).

Example 10 Graph a Polynomial Given in Factored Form

Graph the polynomial function f(x)=(1−x)(x+2)3(x−3)2. $f (x) = (1 - x) {(x + 2)}^{3} {(x - 3)}^{2} .$

Solution

Step 1 We can deduce the end behavior of the polynomial by writing its leading term as the product of the leading terms of each factor:

$f (x) = (1 - x) (x + 2) 3 (x - 3) 2 \approx (- x) (x) 3 (x) 2 = - x 6 .$ $f (x) = (1 - x) {(x + 2)}^{3} {(x - 3)}^{2} \approx (- x) {(x)}^{3} {(x)}^{2} = - x^{6} .$

The end behavior of f(x) (Case 2, page 336) is similar to that of y=−x2. $y = - x^{2} .$
Step 2 The polynomial f(x) is already in factored form

$f (x) = (1 - x) (x + 2) 3 (x - 3) 2 = 0 1 - x = 0 x = 1 or or x + 2 = 0 x = - 2 or or x - 3 = 0 x = 3 Set f (x) = 0. Zero-product property Solve for x .$ $\begin{array}{l} f (x) = (1 - x) {(x + 2)}^{3} {(x - 3)}^{2} = 0 & Set f (x) = 0. \\ \begin{array}{l} 1 - x = 0 & or & x + 2 = 0 & or & x - 3 = 0 \\ x = 1 & or & x = - 2 & or & x = 3 \end{array} & \begin{array}{l} Zero-product property \\ Solve for x . \end{array} \end{array}$

The function has three distinct zeros: −2, 1, $- 2, 1,$ and 3. The zero x=−2 $x = - 2$ has multiplicity 3, so the graph of f crosses the x-axis.

The zero x=1 $x = 1$ has multiplicity 1, so the graph of f crosses the x-axis.

The zero x=3 $x = 3$ has multiplicity 2, so the graph of f touches the x-axis.
Step 3 We compute the y-intercept:

$f (0) = (1 - 0) (0 + 2) 3 (0 - 3) 2 = 1 \cdot 23 \cdot (- 3) 2 = 1 \cdot 8 \cdot 9 = 72 .$ $f (0) = (1 - 0) {(0 + 2)}^{3} {(0 - 3)}^{2} = 1 \cdot 2^{3} \cdot {(- 3)}^{2} = 1 \cdot 8 \cdot 9 = 72 .$

The graph of f passes through the point (0, 72).
Step 4 f(−x)==(1−(−x))((−x)+2)3((−x)−3)2(1+x)(−x+2)3(−x−3)2 $\begin{array}{rcl} f (- x) & = & (1 - (- x)) {((- x) + 2)}^{3} {((- x) - 3)}^{2} \\ = & (1 + x) {(- x + 2)}^{3} {(- x - 3)}^{2} \end{array}$

So f(−x)≠f(x), $f (- x) \neq f (x),$ and f(−x)≠−f(x). $f (- x) \neq - f (x) .$ There are no symmetries.
Step 5 Although the information we obtained from the previous steps allows us to sketch the graph of the polynomial function f, we can further check our work using appropriately chosen “test numbers.” The three zeros −2, 1, $- 2, 1,$ and 3 divide the x-axis into four intervals (−∞,−2),(−2,1),(1,3), $(- \infty, - 2), (- 2, 1), (1, 3),$ and (3,∞) $(3, \infty)$ . We choose −3, 0, 2, $- 3, 0, 2,$ and 4 as test numbers. We evaluate f(−3)=−144<0,f(0)=72>0,f(2)=−64<0, $f (- 3) = - 144 < 0, f (0) = 72 > 0, f (2) = - 64 < 0,$ and f(4)=−648<0. $f (4) = - 648 < 0 .$ We gather all the information about the polynomial function below.
Step 6 We can sketch the graph of the polynomial function f by connecting the pieces of information we gathered above (End behavior and Zero behavior, see Figure 3.16(a) ) with a smooth curve (see Figure 3.16(b) ).

Figure 3.16

Practice Problem 10

Graph the polynomial function f(x)=x(x+2)3(x−1)2. $f (x) = x {(x + 2)}^{3} {(x - 1)}^{2} .$

Volume of a Wine Barrel

Kepler first analyzed the problem of finding the volume of a cylindrical wine barrel. (Review the introduction to this section and see Figure 3.17.) The volume V of a cylinder with radius r and height h is given by the formula V=πr2h. $V = π r^{2} h .$ In Figure 3.17, h=2z, $h = 2 z,$ so V=2πr2z. $V = 2 π r^{2} z .$ Let x represent the value measured by the rod. Then, by the Pythagorean Theorem,

x 2 = 4 r 2 + z 2 (2 r) 2 = 4 r 2

$x^{2} = 4 r^{2} + z^{2} {(2 r)}^{2} = 4 r^{2}$

Kepler’s challenge was to calculate V, given only x—in other words, to express V as a function of x. To solve this problem, Kepler observed the key fact that Austrian wine barrels were all made with the same height-to-diameter ratio. Kepler assumed that the winemakers would have made a smart choice for this ratio, one that would maximize the volume for a fixed value of r. He calculated the height-to-diameter ratio to be 2–√. $\sqrt{2} .$

Example 11 Volume of a Wine Barrel

Express the volume V of the wine barrel in Figure 3.17 as a function of x. Assume that heightdiameter=2z2r=zr=2–√. $\frac{height}{diameter} = \frac{2 z}{2 r} = \frac{z}{r} = \sqrt{2} .$

Solution

We have the following relationships:

V z r x 2 x 2 r 2 x 2 r 2 = = = = = 2 π r 2 z 2 - \sqrt 4 r 2 + z 2 4 + z 2 r 2 4 + 2 = 6 Volume with radius r and height 2 z Given ratio Pythagorean Theorem Divide both sides by r 2 . Since z r = 2 - \sqrt, z 2 r 2 = (z r) 2 = (2 - \sqrt) 2 = 2.

$\begin{array}{rcll} V & = & 2 π r^{2} z & Volume with radius r and height 2 z \\ \frac{z}{r} & = & \sqrt{2} & Given ratio \\ x^{2} & = & 4 r^{2} + z^{2} & Pythagorean Theorem \\ \frac{x^{2}}{r^{2}} & = & 4 + \frac{z^{2}}{r^{2}} & Divide both sides by r^{2} . \\ \frac{x^{2}}{r^{2}} & = & 4 + 2 = 6 & Since \frac{z}{r} = \sqrt{2}, \frac{z^{2}}{r^{2}} = {(\frac{z}{r})}^{2} = {(\sqrt{2})}^{2} = 2. \end{array}$

From x2r2=6, r2=x26, $\frac{x^{2}}{r^{2}} = 6, r^{2} = \frac{x^{2}}{6},$ so r=x6–√(r>0). $r = \frac{x}{\sqrt{6}} (r > 0) .$ Using zr=2–√, $\frac{z}{r} = \sqrt{2},$ we obtain

z = 2 - \sqrt r = = 2 - \sqrt x 6 - \sqrt x 3 - \sqrt Replace r with x 6 - \sqrt . 2 - \sqrt 6 - \sqrt = 2 - \sqrt 2 - \sqrt \cdot 3 - \sqrt = 1 3 - \sqrt

$\begin{array}{rcll} z = \sqrt{2} r & = & \sqrt{2} \frac{x}{\sqrt{6}} & Replace r with \frac{x}{\sqrt{6}} . \\ = & \frac{x}{\sqrt{3}} & \frac{\sqrt{2}}{\sqrt{6}} = \frac{\sqrt{2}}{\sqrt{2} \cdot \sqrt{3}} = \frac{1}{\sqrt{3}} \end{array}$

Now

V = = = \approx 2 π r 2 z 2 π (x 2 6) (x 3 - \sqrt) π 3 3 - \sqrt x 3 0.6046 x 3 Replace z x 3 - \sqrt and r 2 with x 2 6 . Simplify . Use a calculator .

$\begin{array}{rcll} V & = & 2 π r^{2} z \\ = & 2 π (\frac{x^{2}}{6}) (\frac{x}{\sqrt{3}}) & Replace z \frac{x}{\sqrt{3}} and r^{2} with \frac{x^{2}}{6} . \\ = & \frac{π}{3 \sqrt{3}} x^{3} & Simplify . \\ \approx & 0.6046 x^{3} & Use a calculator . \end{array}$

Consequently, Austrian vintners could easily calculate the volume of a wine barrel by using the formula V=0.6046x3 $V = 0.6046 x^{3}$ where x was the length measured by the rod. For a quick estimate, use V≈0.6x3. $V \approx 0.6 x^{3} .$

Practice Problem 11

Use Kepler’s formula to compute the volume of wine (in liters) in a barrel with x=70 $x = 70$ cm. Note: 1 decimeter (dm)=10 cm $(dm) = 10 cm$ and 1 liter=(1 dm)3. $1 liter = {(1 dm)}^{3} .$

Section 3.2 Exercises

Concepts and Vocabulary

Consider the polynomial function f(x)=2x5−3x4+x−6. $f (x) = 2 x^{5} - 3 x^{4} + x - 6 .$ The degree of this polynomial is , its leading term is its leading coefficient is , and its constant term is .
A number c for which f(c)=0 $f (c) = 0$ is called a(n) of the polynomial function f.
If c is a zero of even multiplicity for a polynomial function f, then the graph of f .
If c is a zero of odd multiplicity for a polynomial function f, then the graph of f .
True or False. A curve that has a sharp corner can be a graph of a polynomial function.
True or False. The end behavior of any polynomial function is determined by its leading term.
True or False. The polynomial function of degree n has, at most, n−1 $n - 1$ zeroes.
True or False. The graph of a polynomial function of degree n has, at most, n−1 $n - 1$ turning points.

Building Skills

In Exercises 9–14, for each polynomial function, find the degree, the leading term, and the leading coefficient.

f(x)=2x5−5x2 $f (x) = 2 x^{5} - 5 x^{2}$
f(x)=3−5x−7x4 $f (x) = 3 - 5 x - 7 x^{4}$
f(x)=2x3+7x3 $f (x) = \frac{2 x^{3} + 7 x}{3}$
f(x)=−7x+11+2–√x3 $f (x) = - 7 x + 11 + \sqrt{2} x^{3}$
f(x)=πx4+1−x2 $f (x) = π x^{4} + 1 - x^{2}$
f(x)=5 $f (x) = 5$

In Exercises 15–22, explain why the given function is not a polynomial function.

f(x)=x2+3|x|−7 $f (x) = x^{2} + 3 | x | - 7$
f(x)=2x3−5x,0≤x≤2 $f (x) = 2 x^{3} - 5 x, 0 \leq x \leq 2$
f(x)=x2−1x+5 $f (x) = \frac{x^{2} - 1}{x + 5}$
f(x)={2x+3,1,x≠0x=0 $f (x) = {\begin{array}{l} 2 x + 3, & x \neq 0 \\ 1, & x = 0 \end{array}$
f(x)=5x2−7x−−√ $f (x) = 5 x^{2} - 7 \sqrt{x}$
f(x)=x4.5−3x2+7 $f (x) = x^{4.5} - 3 x^{2} + 7$
f(x)=x2−1x−1,x≠1 $f (x) = \frac{x^{2} - 1}{x - 1}, x \neq 1$
f(x)=3x2−4x+7x−2 $f (x) = 3 x^{2} - 4 x + 7 x^{- 2}$

In Exercises 23–28, explain why the given graph cannot be the graph of a polynomial function.

In Exercises 29–34, match the polynomial function with its graph. Use the leading-term test and the y-intercept.

1. f(x)=−x4+3x2+4 $f (x) = - x^{4} + 3 x^{2} + 4$
2. f(x)=x6−7x4+7x2+15 $f (x) = x^{6} - 7 x^{4} + 7 x^{2} + 15$
3. f(x)=x4−10x2+9 $f (x) = x^{4} - 10 x^{2} + 9$
4. f(x)=x3+x2−17x+15 $f (x) = x^{3} + x^{2} - 17 x + 15$
5. f(x)=x3+6x2+12x+8 $f (x) = x^{3} + 6 x^{2} + 12 x + 8$
6. f(x)=−x3−2x2+11x+12 $f (x) = - x^{3} - 2 x^{2} + 11 x + 12$

In Exercises 35–42, describe the end behavior of the polynomial function f.

f(x)=x−x3 $f (x) = x - x^{3}$
f(x)=2x3−2x2+1 $f (x) = 2 x^{3} - 2 x^{2} + 1$
f(x)=4x4+2x3+1 $f (x) = 4 x^{4} + 2 x^{3} + 1$
f(x)=−x4+3x3+x $f (x) = - x^{4} + 3 x^{3} + x$
f(x)=(x+2)2(2x−1) $f (x) = {(x + 2)}^{2} (2 x - 1)$
f(x)=(x−2)3(2x+1) $f (x) = {(x - 2)}^{3} (2 x + 1)$
f(x)=(x+2)2(4−x) $f (x) = {(x + 2)}^{2} (4 - x)$
f(x)=(x+3)3(2−x) $f (x) = {(x + 3)}^{3} (2 - x)$

In Exercises 43–50, Find the real zeros of f and state the multiplicity for each zero. State whether the graph of f crosses or touches the x-axis at each zero.

f(x)=3(x−1)(x+2)(x−3) $f (x) = 3 (x - 1) (x + 2) (x - 3)$
$f (x) = - 5 (x + 1) (x + 2) (x - 3)$
$f (x) = {(x + 2)}^{2} (2 x - 1)$
$f (x) = {(x - 2)}^{3} (2 x + 1)$
$f (x) = x^{2} (x^{2} - 9) {(3 x + 2)}^{3}$
$f (x) = - x^{3} (x^{2} - 4) {(3 x - 2)}^{2}$
$f (x) = (x^{2} + 1) {(3 x - 2)}^{2}$
$f (x) = (x^{2} + 1) (x + 1) (x - 2)$

In Exercises 51–54, use the Intermediate Value Theorem to show that each polynomial P(x) has a real zero in the specified interval. Approximate this zero to two decimal places.

$P (x) = x^{4} - x^{3} - 10; [2, 3]$
$P (x) = x^{4} - x^{2} - 2 x - 5; [1, 2]$
$P (x) = x^{5} - 9 x^{2} - 15; [2, 3]$
$P (x) = x^{5} + 5 x^{4} + 8 x^{3} + 4 x^{2} - x - 5; [0, 1]$

In Exercises 55–62, use zeros, turning points, and end behavior to find the polynomial with a leading coefficient of either 1 or $- 1$ and with smallest possible degree that matches the given graph.

In Exercises 63–74, sketch the graph of the polynomial function f using the techniques described in this section.

$f (x) = 2 (x + 1) (x - 2) (x + 4)$
$f (x) = - (x - 1) (x + 3) (x - 4)$
$f (x) = {(x - 1)}^{2} (x + 3) (x - 4)$
$f (x) = - x^{2} (x + 1) (x - 2)$
$f (x) = - x^{2} {(x - 3)}^{2}$
$f (x) = {(x - 2)}^{2} {(x + 3)}^{2}$
$f (x) = {(x - 1)}^{2} {(x + 2)}^{3} (x - 3)$
$f (x) = {(x - 1)}^{3} {(x + 1)}^{2} (x - 3)$
$f (x) = - x^{2} (x^{2} - 1) (x + 1)$
$f (x) = x^{2} (x^{2} - 4) (x + 2)$
$f (x) = x^{2} (x^{2} + 1) (x - 2)$
$f (x) = x (x^{2} + 9) {(x - 2)}^{2}$

Applying the Concepts

Horse power. One very important feature of any car is the power of the engine. The power P (in horsepower HP) required to overcome its aerodynamic drag by a typical passenger car is a power function of its velocity v (in mph)

$P = \frac{0.04825}{746} v^{3} .$
1. Compute the horsepower required for a car to cruise on a highway at 40 mph.
2. Compute the horsepower required for a car to cruise on a highway at 65 mph.
3. Compute the horsepower required for a car to cruise on a highway at 80 mph.
4. Show that doubling the speed requires increasing the available power by a factor of 8.
J. B. S. Haldane wrote in his essay On Being the Right Size: “You can drop a small mouse down a thousand-yard mine shaft; and, on arriving at the bottom, it gets a slight shock and walks away, provided that the ground is fairly soft. A rat is killed, a man is broken, a horse splashes.”
Kinetic energy. If an object falls down through air and reaches its terminal velocity, we can estimate its kinetic energy E (in joules, J) at the time of contact with the surface of the earth by the formula

$E = 12800 \cdot L^{4}$

where L represents the length of the object (expressed in meters).
1. Compute the kinetic energy at the time of contact for a human, assuming $L_{H} = 1.75 m .$
2. Compute the kinetic energy at the time of contact for a cat, assuming $L_{C} = \frac{L_{H}}{3} = 0.5833 m$ (a third the length of a human). By what factor did the energy change?
3. Compute the kinetic energy at the time of contact for a mouse, assuming $L_{M} = \frac{L_{H}}{25} = 0.07 m$ (1/25 the length of a human). By what factor did the energy change?
4. Show that reducing the size of the animal by a factor of 2 reduces the impact energy by a factor of 16.
Drug reaction. Let $f (x) = 3 x^{2} (4 - x) .$
1. Find the zeros of f and their multiplicities.
2. Sketch the graph of $y = f (x) .$
3. How many turning points are in the graph of f?
4. A patient’s reaction R(x) $(R (x) \geq 0)$ to a certain drug, the size of whose dose is x, is observed to be $R (x) = 3 x^{2} (4 - x) .$ What is the domain of the function R(x)? What portion of the graph of f(x) constitutes the graph of R(x)?
In Exercise 77, use a graphing calculator to estimate the value of x for which R(x) is a maximum.
Cough and air velocity. The normal radius of the windpipe of a human adult at rest is approximately 1 centimeter. When you cough, your windpipe contracts to, say, a radius r. The velocity v at which the air is expelled depends on the radius r and is given by the formula $v = a (1 - r) r^{2},$ where a is a positive constant.
1. What is the domain of v?
2. Sketch the graph of v(r).
In Exercise 79, use a graphing calculator to estimate the value of r for which v is a maximum.
Maximizing revenue. A manufacturer of slacks has dis-covered that the number x of khakis sold to retailers per week at a price p dollars per pair is given by the formula

$p = 27 - {(\frac{x}{300})}^{2} .$
1. Write the weekly revenue R(x) as a function of x.
2. What is the domain of R(x)?
3. Graph $y = R (x) .$
1. In Exercise 81, use a graphing calculator to estimate the number of slacks sold to maximize the revenue.
2. What is the maximum revenue from part (a)?
Maximizing production. An orange grower in California hires migrant workers to pick oranges during the season. He has 12 employees, and each can pick 400 oranges per hour. He has discovered that if he adds more workers, the production per worker decreases due to lack of supervision. When x new workers (above the 12) are hired, each worker picks $400 - 2 x^{2}$ oranges per hour.
1. Write the number N(x) of oranges picked per hour as a function of x.
2. Find the domain of N(x).
3. Graph the function $y = N (x) .$
1. In Exercise 83, estimate the maximum number of oranges that can be picked per hour.
2. Find the number of employees involved in part (a).
Making a box. From a rectangular $8 \times 15$ piece of cardboard, four congruent squares with sides of length x are cut out, one at each corner. (See the accompanying figure.) The resulting crosslike piece is then folded to form a box.
1. Find the volume V of the box as a function of x.
2. Sketch the graph of $y = V (x) .$
In Exercise 85, use a graphing calculator to estimate the largest possible value of the volume of the box.
Mailing a box. According to Postal Service regulations, the girth plus the length of a parcel sent by Priority Mail may not exceed 108 inches.
1. Express, as a function of x, the volume V of a rectangular parcel with two square faces (with sides of length x inches) that can be sent by mail.
2. Sketch a graph of V(x).
In Exercise 87, use a graphing calculator to estimate the largest possible volume of the box that can be sent by Priority Mail.
Luggage design. AAA Airlines requires that the total outside dimensions $(length + width + height)$ of a piece of checked-in luggage not exceed 62 inches. You have designed a suitcase in the shape of a box whose height (x inches) equals its width. Your suitcase meets AAA’s requirements.
1. Write the volume V of your suitcase as a function of x.
2. Graph the function $y = V (x) .$
In Exercise 89, use a graphing calculator to find the approximate dimensions of the piece of luggage with the largest volume that you can check in on this airline.
Luggage dimensions. American Airlines requires that the total dimensions $(length + width + height)$ of a carry-on bag not exceed 45 inches. You have designed a rectangular boxlike bag whose length is twice its width x. Your bag meets American’s requirements.
1. Write the volume V of your bag as a function of x, where x is measured in inches.
2. Graph the function $y = V (x) .$
In Exercise 91, use a graphing calculator to estimate the dimensions of the bag with the largest volume that you can carry on an American flight. (Source: www.aa.com)
Worker productivity. An efficiency study of the morning shift at an assembly plant indicates that the number y of units produced by an average worker t hours after 6 a.m. may be modeled by the formula $y = - t^{3} + 6 t^{2} + 16 t .$ Sketch the graph of $y = f (t) .$
In Exercise 93, estimate the time in the morning when a worker is most efficient.
Texting. The graph shows the annual number of text messages (in trillions) sent in the United States, 2008–2014.

The cubic function $f (x) = 0.963 + 0.88 x - 0.192 x^{2} + 0.0117 x^{3}$ models these data, where x represents the number of years after 2008.
1. Use this function to estimate the number of text messages sent in 2015.
2. Use a graphing calculator to estimate the number of turning points and the number of zeros.
3. Use a graphing calculator to estimate in which year the number of text messages was at the maximum. Round to the nearest year and discuss whether it fits the original data.
Global warming. The graph shows global temperature anomalies (in Celsius) expressed as a deviation from the 1951–1980 global temperature mean.

Source: http://data.giss.nasa.gov/gistemp

The cubic function $f (x) = - 0.23 + 0.0081 x - 0.0002 x^{2} + 0.000002 x^{3}$ models these data, where x represents the number of years after 1905.
1. Use this function to estimate the temperature anomaly in 2025.
2. Use a graphing calculator to estimate the number of turning points and the number of zeros.
3. Use a graphing calculator to estimate in which year the anomaly was 0. Round to the nearest year and discuss whether it fits the original data.

Beyond the Basics

In Exercises 97–106, use transformations of the graph of $y = x^{4}$ or $y = x^{5}$ to graph each function f(x). Find the zeros of f(x) and their multiplicity.

$f (x) = {(x - 1)}^{4}$
$f (x) = - {(x + 1)}^{4}$
$f (x) = x^{4} + 2$
$f (x) = x^{4} + 81$
$f (x) = - {(x - 1)}^{4}$
$f (x) = \frac{1}{2} {(x - 1)}^{4} - 8$
$f (x) = x^{5} + 1$
$f (x) = {(x - 1)}^{5}$
$f (x) = 8 - \frac{{(x + 1)}^{5}}{4}$
$f (x) = 81 + \frac{{(x + 1)}^{5}}{3}$
Prove that if $0 < x < 1,$ then $0 < x^{4} < x^{2} .$ This shows that the graph of $y = x^{4}$ is flatter than the graph of $y = x^{2}$ on the interval (0, 1).
Prove that if $x > 1,$ then $x^{4} > x^{2} .$ This shows that the graph of $y = x^{4}$ is higher than the graph of $y = x^{2}$ on the interval $(1, \infty) .$
Sketch the graph of $y = x^{2} - x^{4} .$ Use a graphing calculator to estimate the maximum vertical distance between the graphs $y = x^{2}$ and $y = x^{4}$ on the interval (0, 1). Also estimate the value of x at which this maximum distance occurs.
Repeat Exercise 109 by sketching the graph of $y = x^{3} - x^{5} .$

In Exercises 111–114, write a polynomial with the given characteristics.

A polynomial of degree 3 that has exactly two x-intercepts
A polynomial of degree 3 that has three distinct x-intercepts and whose graph rises to the left and falls to the right
A polynomial of degree 4 that has exactly two distinct x-intercepts and whose graph falls to the left and right
A polynomial of degree 4 that has exactly three distinct x-intercepts and whose graph rises to the left and right

In Exercises 115–118, find the smallest possible degree of a polynomial with the given graph. Explain your reasoning.

For large values of x the behavior of a polynomial function is determined by its leading term, while for small values of x the behavior of a polynomial function is determined by the constant term, together with the term that has the smallest exponent. Use this argument to determine the behavior of the given polynomial function for large values of x as well as for small values of x and, using this information, sketch its graph.

$f (x) = 10 x^{4} + x + 1$
$f (x) = x^{3} - x + 1$
$f (x) = x^{4} - x^{2} + 1$
$f (x) = x^{3} - x^{2} + 1$

Critical Thinking / Discussion / Writing

Square-cube law. The principle was first described in 1638 by Galileo Galilei and states that if a solid is proportionally scaled up, then its surface area (or the area of any of its cross sections) increases by the square of the scaling factor, while its volume increases by the cube of the scaling factor.

Giants. The weight of a body is proportional to its volume, while the strength of any bone is proportional to its cross- sectional area. Based on the square-cube law, discuss what will happen if a human is scaled up by a factor of 20.
Lilliputians. Body heat production is proportional to the body’s volume, while heat loss is proportional to the body’s surface area. Based on the square-cube law, discuss what will happen if a human is scaled down by a factor of 70.
Is it possible for the graph of a polynomial function to have no y-intercepts? Explain.
Is it possible for the graph of a polynomial function to have no x-intercepts? Explain.
Is it possible for the graph of a polynomial function of degree 3 to have exactly one local maximum and no local minimum? Explain.
Is it possible for the graph of a polynomial function of degree 4 to have exactly one local maximum and exactly one local minimum? Explain.
Let P and Q be polynomial functions of degree m and n, respectively. Is the composition $P \circ Q$ a polynomial function? If so, what is the degree of $P \circ Q ?$ If not, provide an example where the composition fails to be a polynomial.
Let P and Q be polynomial functions of degree m and n with leading coefficients $a_{m}$ and $b_{n},$ respectively. Find the leading coefficient of the compositions $P \circ Q$ and $Q \circ P .$

Getting Ready for the Next Section

In Exercises 131–134, evaluate each function at the indicated values of x.

$f (x) = x^{3} - 3 x^{2} + 2 x - 9; x = - 2$
$g (x) = x^{4} + 2 x^{3} - 20 x - 5; x = 3$
$g (x) = (- 7 x^{5} + 2 x) (x - 13); x = 13$
$f (x) = (x^{8} - 15 x^{7} + 5 x^{3}) (x + 12); x = - 12$

In Exercises 135–138, factor each expression.

$x^{2} - 3 x - 10$
$- 2 x^{2} + x + 1$
$(2 x + 3) (x^{2} + 6 x - 7)$
$(x - 9) (6 x^{2} - 7 x - 3)$
$3 x^{3} - 6 x^{2} + 5 x - 10$

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Table of Contents for Section 3.1 Quadratic Functions

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Table of Contents for
Section 3.1 Quadratic Functions