Answer
$(x+3)^2=-12y$, vertex $V(-3,0)$, focus $F(-3, -3)$, directrix $y=3$
See graph.
Work Step by Step
Step 1. Make squares of the variables: $(x^2+6x+9)=-12y$, $(x+3)^2=-12y$
Step 2. We can identify that the equation represents a parabola with a vertex at $V(-3,0)$
Step 3. With $4p=-12$, we have $p=-3$, so the focus is at $F(-3, -3)$ and directrix $y=3$
Step 4. See graph.
