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