$(x+4)^2=4(y+2)$, vertex $(-4,-2)$, focus $(-4,-1)$, directrix $y=-3$, see graph.
Work Step by Step
Step 1. Rewriting the equation as $x^2+8x+16=4y-8+16$ or $(x+4)^2=4(y+2)$, we have $4p=4$ and $p=1$ with the parabola opening upwards and vertex $(-4,-2)$. Step 2. We can find the focus at $(-4,-1)$ and directrix as $y=-3$ Step 3. We can graph the parabola as shown in the figure.