Answer
vertex: $\quad (-4,-2)$
focus: $\quad (-4,2)$
directrix: $\quad y=-6$
Work Step by Step
Comparing the form of the equation with table 2,
$\quad \quad \quad (x+4)^{2}=16(y+2)\quad $has the form
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h,k+a)}&{y=k-a}&{(x-h)^{2}=4a(y-k)}&up\\ \hline \end{array}$
$(x+4)^{2}=4(4)(y+2)\quad \Rightarrow\quad h=-4, k=-2, a=4$
vertex: $\quad (h,k)=(-4,-2)$
focus: $\quad (h,k+a)=(-4,-2+4)=(-4,2)$
directrix: $\quad y=k-a\quad \Rightarrow\quad y=-6$
To graph, find the endpoints of the latus rectum
(line segment parallel to the directrix, containing the focus)
by setting $y=2$
$(x+4)^{2}=16(2+2)$
$(x+4)^{2}=64$
$x+4=\pm 8$
$x=-4\pm 8$
Endpoints of the latus rectum: $(-12,2), (4,2)$