Answer
vertex: $\quad (-1,2) $
focus: $\quad (1,2)$
directrix: $\quad x=-3$
Work Step by Step
Comparing the form of the equation with table 2,
$\quad \quad \quad (y-2)^{2}=8(x+1)\quad $has the form
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right\\ \hline \end{array}$
$(y-2)^{2}=4(2)(x+1)\quad \Rightarrow\quad h=-1, k=2, a=2$
vertex: $\quad (-1,2) $
focus: $\quad (h+a, k)=(-1+2,2)=(1,2)$
directrix: $\quad x=h-a\quad \Rightarrow\quad x=-3$
To graph, find the endpoints of the latus rectum
(line segment parallel to the directrix, containing the focus)
by setting $x=1$
$(y-2)^{2}=8(1+1)$
$(y-2)^{2}=16$
$y-2=\pm 4$
$y=2\pm 4$
Endpoints of the latus rectum: $(1,-2), (1,6)$
