$(x+3)^2=4(y+2)$, vertex $(-3,-2)$, focus $(-3,-1)$, directrix $y=-3$, see graph.
Work Step by Step
Step 1. Rewriting the equation as $x^2+6x+9=4y-1+9$ or $(x+3)^2=4(y+2)$, we have $4p=4$ and $p=1$ with the parabola opening upwards and vertex $(-3,-2)$. Step 2. We can find the focus at $(-3,-1)$ and directrix as $y=-3$ Step 3. We can graph the parabola as shown in the figure.