Answer
parabola, vertex $(2,-1)$, focus $(2,-2)$, and directrix $y=0$.
Work Step by Step
1. Given $4x^2-16x+16y+32=0$ or $(x-2)^2=-4(y+1)$, we can identify it as a parabola opens downward with $p=1$.
2. We can find its vertex $(2,-1)$, focus $(2,-2)$, and directrix $y=0$.
