vertex $(-2,-2)$, focus $(-2,-4)$, directrix $y=0$, see graph.
Work Step by Step
Step 1. Given $(x+2)^2=-8(y+2)$, we have $4p=-8$ and $p=-2$ with the parabola opening downwards and vertex $(-2,-2)$. Step 2. We can find the focus at $(-2,-4)$ and directrix as $y=0$ Step 3. We can graph the parabola as shown in the figure.