**Solution:**

Answer: Vertex $(-4,2)$, focus $(-\frac{127}{32},2)$, axis $y=2$, directrix $x=-\frac{1}{32}$

We will complete the square.

$y^{2}-4y-8x-28=(y-2)^{2}-4=8x+28\Longrightarrow (y-2)^{2}=8x+32$

So the equation of the parabola is $\frac{1}{8}(y-2)^{2}=(x+4)$

The vertex is at $(-4,2)$, $p=\frac{1}{32}$ The focus is $(-4+p,2)=(-\frac{127}{32},2)$,

the axis is $y=2$ and directrix is $x=-p=-\frac{1}{32}$