Given $f(x)= \frac{1}{4}(x+4)^2-2$ Find vertex, $ y$ intercept etc. Given $f(x)= \frac{1}{4}(x+4)^2-2$
Find: vertex, $y$-intercept, $x$-intercepts (if any), axis of symmetry
What I have so far:
Vertex: $(-4,-2)$
$y$-intercept: $(0,2)$
$x$-intercept: $2$
Axis of symmetry $x=-4$ 
If you could please tell me if these are right or not that would be great. If they are not right please correct them and tell me why they are wrong.

Because the function is in parabola vertex-form the vertex is at $\left( -4,-2\right)$
The axis of symmetry is about the $x-$value of the vertex, because the function is a vertical parabola.
$$ \therefore x_{axis}=-4 $$
for the $y-$intercept you need $f\left( 0\right)$
$$ f\left( 0\right) = \frac{1}{4}\left( 0+4\right) ^2-2 $$
$$ f\left( 0\right) =\frac{4^2}{4}-2 $$
$$ f\left( 0\right) = 4-2 $$
$$ f\left( 0\right) = 2 $$
For $x-$intercepts the line $y=0$ intersects the function $y=\frac{1}{4}\left( x+4\right) ^2-2$
$$\therefore \quad 0=\frac{1}{4}\left( x+4\right) ^2-2$$ 
$$ 2= \frac{1}{4}\left( x+4\right) ^2$$
$$ 8= \left( x+4\right) ^2$$
$$ x+4=\pm\sqrt8 $$
$$ x=-4\pm\sqrt8 $$
