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