Answer
vertex: $\quad (0,0) $
focus: $\quad (0,-1)$
directrix: $\quad y=1$
Work Step by Step
Comparing the form of the equation with table 2,
$\quad \quad \quad x^{2}=-4y\quad $has the form
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h,k-a)}&{y=k+a}&{(x-h)^{2}=-4a(y-k)}& down \\ \hline \end{array}$
$(x-0)^{2}=-4(1)(y-0)\quad \Rightarrow\quad h=0, k=0, a=1$
vertex: $\quad (0,0) $
focus: $\quad (h,k-a)=(0,0-1)=(0,-1)$
directrix: $\quad y=k+a\quad \Rightarrow\quad y=1$
To graph, find the endpoints of the latus rectum
(line segment parallel to the directrix, containing the focus)
by setting $y=-1$
$x^{2}=-4(-1)$
$x^{2}=4$
$x=\pm 2$
Endpoints of the latus rectum: $(-2,-1), (2,-1)$
