Question 739485
name the vertex, focus, and directrix of 
y=1/4(x-2)^2
(x-2)^2=4y
This is an equation of a parabola that opens upward.
Its basic form: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex
For given equation:
vertex: (2,0)
axis of symmetry: x=2
4p=4
p=1
focus: (2,1) (p-distance above vertex on the axis of symmetry)
directrix: y=-1 (p-distance below vertex on the axis of symmetry)