Polynomial functions have the following important property:

“*Every polynomial function of degree $n$ has at most $n$ real zeros*.”

This property is called the *Fundamental Theorem of Algebra*.

As an application of this property, we see that a polynomial function of degree $n$ can have at most $n$ $x$-intercepts and at most $(n-1)$ turning points (local maximum and local minimum values).

*Example*. The function $f(x)=x^4-7x^3+12x^2+4x-16$ has three turning points that are two local minimum values and one local maximum value.

**Graphing a Polynomial Function** $p(x)$

- First find all zeros of $p(x)$
- Considering the even or odd multiplicity of each factor of $p(x)$, we can see the graph is crossing or touching the $x$-axis at each zero.
- Use the leading-term test to determine the end behavior.
- Use the $y$-intercept.

*Example*. Consider $f(x)=x^4-7x^3+12x^2+4x-16$. It can be factored as $f(x)=(x+1)(x-2)^2(x-4)$. (At this moment you don’t have to worry about how we get the facotring. We will discuss this in Sections 3.3 and 3.4.) So we find three zeros $-1$, $2$ (with multiplicity 2), and $4$. We can tell that the graph crosses at $-1$ and $4$ and touches the graph without crossing at $2$. Since the degree is $4$, an even number the graph goes up when $x\to -\infty$ and $x\to\infty$. These findings are all featured in the above graph.

**The Intermediate Value Theorem**

Let $p(x)$ be a polynomial (with real coefficients). Suppose that $p(a)$ and $p(b)$ have different signs for two distinct numbers $a$ and $b$. Then the graph of $p(x)$ must cross the $x$-axis between $a$ and $b$, i.e. $p(x)$ must have a zero between $a$ and $b$. This property is called the *Intermediate Value Theorem*.

*Example*. (a) Use the Intermediate Value Theorem to determine if

$$f(x)=x^3+3x^2-9x-13$$

has a zero between $a=1$ and $b=2$.

*Solution*. All you have to do is to evaluate $f(x)$ at $x=1$ and $x=2$, i.e. calculate $f(1)$ and $f(2)$ and see if they are different.

\begin{align*}

f(1)&=(1)^3+3(1)^2-9(1)-13=-18,\\

f(2)&=(2)^3+3(2)^2-9(2)-13=-11.

\end{align*}

$f(1)$ and $f(2)$ have the same sign, so the Intermediate Value Theorem won’t tell if $f(x)$ has a zero between $1$ and $2$. The following graph shows that it actually does not.

(b) Does $f(x)$ has a zero between $a=-5$ and $b=-4$?

*Solution*. $f(-5)=-18$ and $f(-4)=7$. Since their signs are different, by the Intermediate Value Theorem, there must be a zero between $-5$ and $-4$. The following graph confirms it.