Answer
$(y+2)^{2}=4(x+1)$
The latus rectum has endpoints $(0,-4)$ and $(0,0).$
Work Step by Step
1.$\quad $The vertex and focus lie on the line $y=-2\quad \Rightarrow\quad $parabola opens left or right.
The focus is right of the vertex$\quad \Rightarrow\quad $opens right.
2.$\quad $From Table 2,
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h+a, k)}&{x=h-a}&{(y-k)^{2}=4a(x-h)}& right\\ \hline \end{array}$
$(h,k)=(-1,-2)$
The equation is $\quad (y+2)^{2}=4a(x+1)\quad $
3.$\quad $Since the focus is at $\quad (h+a, k)$
and we are given the focus is at$\quad (0,-2)$, we find $a$
$h+a=0$
$-1+a=0$
$a=1$
So, the equation is$\quad (y+2)^{2}=4(x+1)$
4. $\quad $For the latus rectum, set $x=0, \quad $
$(y+2)^{2}=4(0+1)$
$(y+2)^{2}=4$
$y+2=\pm 2$
$y=-2\pm 2$
The latus rectum (line segment parallel to the directrix, containing the focus)
has endpoints $(0,-4)$ and $(0,0).$
5. $\quad $We have enough details for the graph.
Graph the directrix, $\quad x=-1$
