The behavior of the graph of a polynomial function as x becomes very large $(x\to \infty )$ or very small $(x\to -\infty )$ is referred to as the end behavior of the graph. The leading term of a polynomial function determines its end behavior.

Using the graphs shown on the following page, let’s see if we can discover some general patterns by comparing the end behavior of even- and odd-degree functions. We also note the effect of positive and negative leading coefficients.

Even Degree

Odd Degree

We can summarize our observations as follows.

Let’s review the meaning of the real zeros of a function and their connection to the x-intercepts of the function’s graph.

The connection between the real-number zeros of a function and the x-intercepts of the graph of the function is easily seen in the preceding examples. If c is a real zero of a function (that is, $f(c)=0)$, then $(c,0)$ is an x-intercept of the graph of the function.

Example 2

Consider $P(x)=x^3+x^2-17x+15$. Determine whether each of the numbers 2 and −5 is a zero of $P(x)$.

Solution

We first evaluate $P(2):$

$$\begin{array}{ll}P(2)={(2)}^3+{(2)}^2-17(2)+15=-7.& \mathbf{\text{Substituting }}\mathbf{2}\mathbf{\text{ into}}\\ & \mathbf{\text{the polynomial}}\end{array}$$

Since $P(2)\ne 0$, we know that 2 is not a zero of the polynomial function.

We then evaluate $P(-5):$

$$\begin{array}{ll}P(-5)={(-5)}^3+{(-5)}^2-17(-5)+15=0.& \mathbf{\text{Substituting }}\mathbf{-}\mathbf{5}\mathbf{\text{ into}}\\ & \mathbf{\text{the polynomial}}\end{array}$$

Since $P(-5)=0$, we know that −5 is a zero of $P(x)$.

Now Try Exercise 23.

Let’s take a closer look at the polynomial function

(see Connecting the Concepts above). The factors of $h(x)$ are

and the zeros are

We note that when the polynomial is expressed as a product of linear factors, each factor determines a zero of the function. Thus if we know the linear factors of a polynomial function $f(x)$, we can easily find the zeros of $f(x)$ by solving the equation $f(x)=0$ using the principle of zero products.

Example 3

Find the zeros of

$$\begin{array}{rcl}f(x)& =& 5(x-2)(x-2)(x-2)(x+1)\\ & =& 5{(x-2)}^3(x+1).\end{array}$$

Solution

To solve the equation $f(x)=0$, we use the principle of zero products, solving $x-2=0$ and $x+1=0$. The zeros of $f(x)$ are 2 and −1.

Example 4

Find the zeros of

$$\begin{array}{rcl}g(x)& =& -(x-1)(x-1)(x+2)(x+2)\\ & =& -{(x-1)}^2{(x+2)}^2.\end{array}$$

Solution

To solve the equation $g(x)=0$, we use the principle of zero products, solving $x-1=0$ and $x+2=0$. The zeros of $g(x)$ are 1 and −2.

Let’s consider the occurrences of the zeros in the functions in Examples 3 and 4 and their relationship to the graphs of those functions. In Example 3, the factor $x-2$ occurs three times. In a case like this, we say that the zero we obtain from this factor, 2, has a multiplicity of 3. The factor $x+1$ occurs one time. The zero we obtain from this factor, −1, has a multiplicity of 1.

In Example 4, the factors $x-1$ and $x+2$ each occur two times. Thus both zeros, 1 and −2, have a multiplicity of 2.

Note, in Example 3, that the zeros have odd multiplicities and the graph crosses the x-axis at both −1 and 2. But in Example 4, the zeros have even multiplicities and the graph is tangent to (touches but does not cross) the x-axis at −2 and 1. This leads us to the following generalization.

Some polynomials can be factored by grouping. Then we use the principle of zero products to find their zeros.

Other factoring techniques can also be used.

Only the real-number zeros of a function correspond to the x-intercepts of its graph. For instance, the real-number zeros of the function in Example 6, $-\sqrt{5}$ and $\sqrt{5}$, can be seen on the graph of the function below, but the nonreal zeros, −3i and 3i, cannot.

This is often stated as follows: “Every polynomial function of degree n, with $n\ge 1$, has exactly n zeros.” This statement is compatible with the preceding statement, if one takes multiplicities into account.

Polynomial functions have many uses as models in science, engineering, and business. The simplest use of polynomial functions in applied problems occurs when we merely evaluate a polynomial function. In such cases, a model has already been developed.

Example 7

Ibuprofen in the Bloodstream. The polynomial function

$$M(t)=0.5t^4+3.45t^3-96.65t^2+347.7t$$

can be used to estimate the number of milligrams of the pain relief medication ibuprofen in the bloodstream t hours after 400 mg of the medication has been taken. Find the number of milligrams in the bloodstream at $t=0$, 0.5, 1, 1.5, and so on, up to 6 hr. Round the function values to the nearest tenth.

Solution

Using a calculator, we compute function values:

$M(0)=0$,

$M(0.5)=150.2$,

$M(1)=255$,

$M(1.5)=318.3$,

$M(2)=344.4$,

$M(2.5)=338.6$,

$M(3)=306.9$,

$M(3.5)=255.9$,

$M(4)=193.2$,

$M(4.5)=126.9$,

$M(5)=66$,

$M(5.5)=20.2$,

$M(6)=0$.

Now Try Exercise 49.

Recall that the domain of a polynomial function, unless restricted by a statement of the function, is $(-\infty ,\infty )$. The implications of the application in Example 7 restrict the domain of the function. If we assume that a patient had not taken any of the medication before, it seems reasonable that $M(0)=0;$ that is, at time 0, there is 0 mg of the medication in the bloodstream. After the medication has been taken, $M(t)$ will be positive for a period of time and eventually decrease back to 0 when $t=6$ and not increase again (unless another dose is taken). Thus the restricted domain is [0, 6].