Answer
Vertex: $(-1,4)$
No $x$-intercept
$y$-intercept: $5$
Work Step by Step
The function is in the vertex form $f(x)=a(x-h)^2+k$:
$$\begin{align*}
f(x)&=(x+1)^2+4.
\end{align*}$$
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 upward because $a>0$.
Determine the $x$-intercepts by solving the equation $f(x)=0$:
$$\begin{align*}
(x+1)^2+4&=0\\
(x-1)^2&=-4.
\end{align*}$$
There is no $x$-intercept.
Determine the $y$-intercept:
$$f(0)=(0+1)^2+4=5.$$
The $y$-intercept is $5$.
We plot the vertex $(-1,4)$, the $y$-intercept point $(0,5)$ and two other points $(-3,8)$ and $(1,8)$, then join them to sketch the graph of the function:
