Answer
Vertex: $(-4,-2)$
$x$-intercepts: $-4-\sqrt 2,-4+\sqrt 2$
$y$-intercept: $14$
Work Step by Step
The function is in vertex form $f(x)=a(x-h)^2+k$:
$$f(x)=(x+4)^2-2.$$
Identify the constants $a$, $h$, $k$:
$$\begin{align*}
a&=1\\
h&=-4\\
k&=-2.
\end{align*}$$
Determine the vertex of the function:
$$(h,k)=(-4,-2).$$
The axis of symmetry is $x=-4$. It opens upward because $a>0$.
Determine the $x$-intercepts by solving the equation $f(x)=0$:
$$\begin{align*}
(x+4)^2-2&=0\\
(x+4)^2&=2\\
x+4&=\pm \sqrt 2\\
x+4=-\sqrt 2&\text{ or }x+4=\sqrt 2\\
x=-4-\sqrt 2&\text{ or }x=-4+\sqrt 2.
\end{align*}$$
The $x$-intercepts are $-4-\sqrt 2$ and $-4+\sqrt 2$.
Determine the $y$-intercept:
$$f(0)=(0+4)^2-2=14.$$
The $y$-intercept is $14$.
We plot the vertex $(-4,-2)$ and the intercept points $(-4-\sqrt 2,0)$,
$(-4+\sqrt 2,0)$ and $(0,14)$, then join them to sketch the graph of the function:
