Solving Polynomial Equations

We want to find the zeros and sketch the graph of the given polynomial function. $$\begin{array}{c}f(x)=\text{-}{x}^3-x^2+9x+9\end{array}$$ Let's do these things one at a time.

Zeros of the Function

To find the zeros, we need to find the values of $x$ for which $f(x)=0.$ $$\begin{array}{c}f(x)=0\iff \text{-}{x}^3-x^2+9x+9=0\end{array}$$ Since the function is not written in factored form, we will begin by factoring the equation. Note that in the process of factoring the equation, we used the difference of two squares. Now, we can apply the Zero Product Property. We found that the zeros of the function are $x=\text{-}3,$ $x=\text{-}1,$ and $x=3.$

Graph

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

$x$	$\text{-}{x}^3-x^2+9x+9$	$f(x)=\text{-}{x}^3-x^2+9x+9$
$\text{-}3.5$	$\text{-}(\text{-}3.5{)}^3-(\text{-}3.5{)}^2+9(\text{-}3.5)+9$	$8.125$
$\text{-}2$	$\text{-}(\text{-}2{)}^3-(\text{-}2{)}^2+9(\text{-}2)+9$	$\text{-}5$
$1$	$\text{-}{1}^3-1^2+9(1)+9$	$16$

The points $(\text{-}3.5,8.125),$ $(\text{-}2,\text{-}5),$ and $(1,16)$ are on the graph of the function. Now, we will determine the leading coefficient and degree of the polynomial function. $$\begin{array}{c}f(x)=\text{-}{x}^3-x^2+9x+9\\ \Updownarrow \\ f(x)=\text{-}1x^3-x^2+9x+9\end{array}$$ We can see now that the leading coefficient is $\text{-}1,$ which is a negative number. Also, the degree is $3,$ which is an odd number. Therefore, the end behavior is up and down. With this in mind, we will plot the zeros and the obtained points, and draw the graph of the function.