Answer
vertex: $\quad (0,0) $
focus: $\quad (-4,0)$
directrix: $\quad x=4$
Work Step by Step
Comparing the form of the equation with table 2,
$y^{2}=-16x\quad $has the form
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h-a,k)}&{x=h+a}&{(y-k)^{2}=-4a(x-h)} &left\\ \hline \end{array}$
$(y-0)^{2}=-4(4)(x-0)\quad \Rightarrow\quad h=0, k=0, a=4$
vertex: $\quad (0,0) $
focus: $\quad (h-a,k)=(0-4,0)=(-4,0)$
directrix: $\quad x=h+a\quad \Rightarrow\quad x=4$
To graph, find the endpoints of the latus rectum
(line segment parallel to the directrix, containing the focus)
by setting $x=-4$
$y^{2}=-16(-4)$
$y^{2}=64$
$y=\pm 8$
Endpoints of the latus rectum: $(-4,-8), (-4,8)$
