Answer
$(y+4)^2=-2(x-4)$, parabola, vertex $V(4, -4)$, focus $F(\frac{7}{2}, -4)$, directrix $x=\frac{9}{2}$
See graph.
Work Step by Step
Step 1. Rewrite the equation as $y^2+8y+16=-2x-8+16$ or $(y+4)^2=-2(x-4)$ which represents a parabola.
Step 2. The vertex can be found at $V(4, -4)$.
Step 3. To find the focus, compare the equation with a standard form to get $4p=-2$ and $p=-\frac{1}{2}$ and thus the focus is at $F(4-\frac{1}{2}, -4)$ or $F(\frac{7}{2}, -4)$
Step 4. The original directrix is $x=\frac{1}{2}$ and the shifted directrix is $x=4+\frac{1}{2}=\frac{9}{2}$
Step 5. See graph.
