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.
 A: Consider $y=a(x+4)^2-2$. Let $Y=y+2$ and $X=x+4$. Then $Y=aX^2$.
This means that the curve $y=a(x+4)^2-2$ is a translation of the curve $Y=aX^2$: the origin $(0,0)$ in $XY$-space goes to the point $(-4,-2)$ in $xy$-space.
Thus, the vertex of $y=a(x+4)^2-2$ is at $(-4,-2)$, the symmetry axis is $x=-4$, etc.
A: Everything is right except for the x intercept(s). Try moving the -2 to the Left Side, and then simplyfying to get x in terms of square roots.
A: 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 $y=\frac{1}{4}\left( x+4\right) ^2-2$
$$\therefore \quad 0=\frac{1}{4}\left( x+4\right) ^2-2$$ 
$$ 2= \frac{1}{4}\left( x+4\right) ^2$$
$$ 8= \left( x+4\right) ^2$$
$$ x+4=\pm\sqrt8 $$
$$ x=-4\pm\sqrt8 $$
