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| Question 739485:  name the vertex, focus, and directrix of y=1/4(x-2)^2
 Answer by lwsshak3(11628)
      (Show Source): 
You can put this solution on YOUR website! 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)
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