Answer
vertex $(-4,-2)$, focus $(-4,2)$, directrix $y=-6$
See graph.
Work Step by Step
1. We can identify that $(x+4)^2=16(y+2)$ is a parabola opens upwards,
2. Compare with the standard form to get:
vertex $(-4,-2)$, $p=4$, focus $(-4,2)$, directrix $y=-6$
3. Use symmetry and a few test points such as at $x=-4,-3,-2,-1,0$ to get the graph.
