Question

find the zeros of $h(x) = -x^3 - 2x^2 + 9x + 18$.
$x = -3$ $x = -4$ $x = -1$
$x = 2$ $x = 3$ $x = 1$
$x = -2$ $x = 4$ $x = 5$
identify the graph of the function.
four graphs of cubic functions are shown, with options to select the correct one based on zeros and shape

Explanation:

Step1: Factor by grouping

Group terms and factor out common factors:

$$\begin{align*} h(x)&=-x^3 - 2x^2 + 9x + 18\\ &=-x^2(x+2)+9(x+2)\\ &=(x+2)(-x^2+9) \end{align*}$$

Step2: Factor the quadratic term

Rewrite $-x^2+9$ as difference of squares:
$$-x^2+9=-(x^2-9)=-(x-3)(x+3)$$
So $h(x)=(x+2)(-1)(x-3)(x+3)=-(x+3)(x+2)(x-3)$

Step3: Find zeros by setting $h(x)=0$

Set each factor equal to 0:
$x+3=0 \implies x=-3$; $x+2=0 \implies x=-2$; $x-3=0 \implies x=3$

Step4: Match zeros to the graph

The correct graph has x-intercepts at $(-3,0)$, $(-2,0)$, and $(3,0)$, which is the bottom-left graph.

Answer:

Zeros: $x=-3$, $x=-2$, $x=3$
Correct graph: The bottom-left graph with x-intercepts $(-3,0)$, $(-2,0)$, $(3,0)$