Question 724026
Identify the vertex, focus, and directrix of the parabola. 
x=-1/4(y+1)^2+2
rewrite equation in basic form 
x-2=-1/4(y+1)^2
(y+1)^2=-4(x-2)
Basic form of equation for a parabola that opens leftward: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation:
vertex: (2-1)
axis of symmetry: y=-1
4p=4
p=1
focus: (1,-1) (p-distance to left of vertex on the axis of symmetry)
directrix: x=3 (p-distance to right of vertex on the axis of symmetry)