SOLUTION: what are the vertex, focus and directrix of parabola 12y=x^2-6x+45? have so far: (x-3)^2=12(y-3). vertex is (3,3) or (-3,-3)???? how do I do focus and directrix?

Question 710263: what are the vertex, focus and directrix of parabola 12y=x^2-6x+45? have so far: (x-3)^2=12(y-3). vertex is (3,3) or (-3,-3)???? how do I do focus and directrix?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
what are the vertex, focus and directrix of parabola
12y=x^2-6x+45?
x^2-6x=12y-45
complete the square:
(x^2-6x+9)=12y-45+9
(x-3)^2=12y-36
(x-3)^2=12(y-3)
This is an equation of a parabola that opens upwards.
Its standard form: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex
For given equation:(x-3)^2=12(y-3)
vertex: (3,3)
axis of symmetry: x=3
4p=12
p=3
focus: (3,6) (p-units above vertex on the axis of symmetry)
directrix: y=0 (p-units below vertex on the axis of symmetry)
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