
## SectionShort-Run Behavior

Last section, we categorized polynomials by the long-run behavior. We now take a closer look at how these functions behave for smaller $x$-values. In particular, we see how the polynomial behaves around its roots.

### Subsection$x$-Intercepts (roots) and Short-Run Behavior

We have seen before that the $x$-intercepts of a quadratic polynomial, $f(x) = ax^2 + bx + c\text{,}$ occur at values of $x$ for which $f(x) = 0\text{,}$ that is, at the real-valued solutions of the equation $ax^2 + bx + c = 0\text{.}$ The same holds true for polynomials of higher degree.

Solutions of the equation $P(x) = 0$ are called zeros of the polynomial $P\text{.}$ In Example316, we graphed the cubic polynomial $P(x) = x^3 - 4x\text{.}$ Its $x$-intercepts are the solutions of the equation $x^3 - 4x = 0\text{,}$ which we can solve by factoring the polynomial $P(x)\text{.}$

\begin{align*} x^3 - 4x \amp= 0\\ x(x - 2)(x + 2) \amp= 0 \end{align*}

The zeros of $P$ are $0\text{,}$ $2\text{,}$ and $-2\text{.}$ Each zero of $P$ corresponds to a linear factor of $P(x)\text{.}$ This result suggests the following theorem, which holds for any polynomial $P\text{.}$

###### Factor Theorem

Let $P(x)$ be a polynomial with real number coefficients. Then $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0\text{.}$

Because a polynomial function of degree $n$ can have at most $n$ linear factors of the form $(x - a)\text{,}$ it follows that $P$ can have at most $n$ distinct zeros.

Another way of saying this is that if $P(x)$ is a polynomial of $n$th degree, the equation $P(x) = 0$ can have at most $n$ distinct solutions, some of which may be complex numbers. (We do not cover complex numbers in this course but there are plenty of reasonable resources available online if you wish to know more.)

Because only real-valued solutions appear on the graph as $x$-intercepts, we have the following corollary to the factor theorem.

###### $x$-Intercepts of Polynomials

A polynomial of degree $n$ can have at most $n$ $x$-intercepts.

If some of the zeros of $P$ are complex numbers, they will not appear on the graph, so a polynomial of degree $n$ may have fewer than $n$ $x$-intercepts.

###### Example322

Find the real-valued zeros of each of the following polynomials, and list the $x$-intercepts of its graph.

1. $f(x) = x^3 + 6x^2 + 9x$
2. $g(x) = x^4 - 3x^2 - 4$
Solution
1. Factor the polynomial to obtain

\begin{align*} f(x) \amp= x(x^2 + 6x + 9)\\ \amp= x(x + 3)(x + 3). \end{align*}

By the factor theorem, the zeros of $f$ are $0\text{,}$ $-3\text{,}$ and $-3\text{.}$ (We say that $f$ has a zero of multiplicity two at $-3\text{.}$) Because all of these are real numbers, all will appear as $x$-intercepts on the graph. Thus, the $x$-intercepts occur at $(0, 0)$ and at $(-3, 0)\text{.}$

2. Factor the polynomial to obtain

\begin{align*} g(x) \amp= (x^2 - 4)(x^2 + 1)\\ \amp= (x - 2)(x + 2)(x^2 + 1). \end{align*}

Because $x^2 + 1$ cannot be factored in real numbers, the graph has only two $x$-intercepts, at $(-2, 0)$ and $(2, 0)\text{.}$ Hence, $g(x)$ has only $x=-2$ and $x=2$ as its zeros. The graphs of both polynomials are shown in Figure323.

###### Example324
1. Find the real-valued zeros of $P(x) = -x^4 + x^3 + 2x^2$ by factoring.

2. Sketch a rough graph of $y = P(x)$ by hand.

Solution
1. Factor the polynomial to obtain

\begin{equation*} \begin{aligned} P(x) \amp= -x^4+x^3+2x^2 \\ \amp= -x^2(x-2)(x+1) \end{aligned} \end{equation*}

By the factor theorm, the zeros of $P$ are $0\text{,}$ $2\text{,}$ and $-1\text{.}$

### SubsectionZeros of Multiplicity Two or Three

The appearance of the graph near an $x$-intercept is determined by the multiplicity of the zero there. For example, both real zeros of the polynomial $g(x) = x^4 - 3x^2 - 4$ in Example322b are of multiplicity one, and the graph crosses the $x$-axis at each intercept.

However, the polynomial $f(x) = x^3 + 6x^2 + 9x$ in Example322a has a zero of multiplicity two at $x = -3\text{.}$ The graph of $f$ just touches the $x$-axis and then reverses direction without crossing the axis.

To understand what happens in general, compare the graphs of the three polynomials in Figure325.

1. In Figure325a, $L(x) = x - 2$ has a zero of multiplicity one at $x = 2\text{,}$ and its graph crosses the $x$-axis there.

2. In Figure325b, $Q(x) = (x - 2)^2$ has a zero of multiplicity two at $x = 2\text{,}$ and its graph touches the $x$-axis there but changes direction without crossing.

3. In Figure325c, $C(x) = (x - 2)^3$ has a zero of multiplicity three at $x = 2\text{.}$ In this case, the graph makes an S-shaped curve at the intercept, like the graph of $y = x^3\text{.}$

Near its $x$-intercepts, the graph of a polynomial takes one of the characteristic shapes illustrated in Figure325.

###### Multiplicities of Roots

Let $f(x)$ be a polynomial function.

1. At a root of even multiplicity the graph of $f(x)$ bounces off the $x$-axis.

2. At a root of odd multiplicity the graph of $f(x)$ crosses the $x$-axis.

###### Note326

Although we will not consider zeros of multiplicity greater than three, they correspond to similar behavior in the graph: At a zero of odd multiplicity, the graph has an inflection point at the intercept; its graph makes an S-shaped curve. At a zero of even multiplicity, the graph has a turning point; it changes direction without crossing the $x$-axis.

###### Example327

Graph the polynomial

\begin{equation*} f(x) = (x + 2)^3(x - 1)(x - 3)^2. \end{equation*}
Solution

The polynomial has degree six, an even number, so its graph starts at the upper left and extends to the upper right. The $y$-coordinate of the $y$-intercept is

\begin{equation*} f(0) = (2)^3(-1)(-3)^2 = -72. \end{equation*}

Note that $f$ has a zero of multiplicity three at $x = -2\text{,}$ a zero of multiplicity one at $x = 1\text{,}$ and a zero of multiplicity two at $x = 3\text{.}$ The graph has an S-shaped curve at $x = -2\text{,}$ crosses the $x$-axis at $x = 1\text{,}$ touches the $x$-axis at $x = 3\text{,}$ and then changes direction, as shown in Figure328.