Answer
vertex: $\quad (0,0) $
focus: $\quad (2,0)$
directrix: $\quad x=-2$
Work Step by Step
Comparing the form of the equation with table 2,
$x^{2}=4y \quad $has the form
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right\\ \hline \end{array}$
$(y-0)^{2}=4(2)(x-0)\quad \Rightarrow\quad h=0, k=0, a=2$
vertex: $\quad (0,0) $
focus: $\quad (h+a, k)=(0+2,0)=(2,0)$
directrix: $\quad x=h-a\quad \Rightarrow x=-2$
To graph, find the endpoints of the latus rectum
(line segment parallel to the directrix, containing the focus)
by setting $x=2$
$y^{2}=8\cdot 2$
$y^{6}=16$
$y=\pm 4$
Endpoints of the latus rectum: $(2,-4), (2,4)$
