Answer
Vertex: $(3,2)$
$x$-intercepts: $2$; $4$;
$y$-intercept: $-16$
Work Step by Step
The function is in the vertex form $f(x)=a(x-h)^2+k$:
$$\begin{align*}
f(x)&=-2(x-3)^2+2.
\end{align*}$$
Identify the constants $a$, $h$, $k$:
$$\begin{align*}
a&=-2\\
h&=3\\
k&=2.
\end{align*}$$
Determine the vertex of the function:
$$(h,k)=(3,2).$$
The axis of symmetry is $x=3$. It opens downward because $a<0$.
Determine the $x$-intercepts by solving the equation $f(x)=0$:
$$\begin{align*}
-2(x-3)^2+2&=0\\
(x-3)^2&=1\\
x-3&=\pm 1\\
x-3=-1&\text{ or }x-3=1\\
x=2&\text{ or }x=4.
\end{align*}$$
The $x$-intercepts are $2$ and $4$.
Determine the $y$-intercept:
$$f(0)=-2(0-3)^2+2=-16.$$
The $y$-intercept is $-16$.
We plot the vertex $(3,2)$ and the intercept points $(2,0)$,
$(4,0)$ and $(0,-16)$, then join them to sketch the graph of the function:
