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