Answer
Vertex: $(-3,4)$
Foci: $(0,4)$
Directrix: $x=-3$
Axis of symmetry: $x$
Domain of the parabola:$[-3,\infty ]$
Range of the parabola: $[-\infty,\infty ]$
Work Step by Step
The first thing we see is that the equation has the form:
$$(y-k)^2=4p(x-h)$$ where the vertex is $$(h,k)$$
This means the parabola opens left or right, and has axis of symmetry of $x$ and the directrix $x=-p$, so let's find $p$.
If we match the equations we find:
$(y-4)^2=12(x+3)$
$4p=12$
$p=\frac{12}{4}=3$
$k=-(-4)=4$
$h=-(3)=-3$
So the vertex is:
$(-3,4)$,
and the directrix:
$y=-p$
$y=-3$
In order to find the foci, we know that the foci is in the same axis as the vertex and is $p$ units away of it, so:
Foci $(h+p,p)$
Foci $(-3+3,4)$
Foci $(0,4)$
To obtain the domain and the range of the parabola we see from the graph, $y$ can take any value so, it goes from $-\infty$ to $\infty$ this is the range.
For the domain: we know the vertex is the biggest or the smallest value of the parabola, so this means all values start from there, so from $-3$ to $\infty $.
