SOLUTION: How do you find the focus of y=1/20(x+5)^2-6

Question 715930: How do you find the focus of y=1/20(x+5)^2-6
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
How do you find the focus of y=1/20(x+5)^2-6
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Basic form of equation for a parabola that opens upward: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex
y=1/20(x+5)^2-6
(y+6)=(1/20)(x+5)^2
(x+5)^2=20(y+6)
vertex: (-5,-6)
axis of symmetry: y=-5
4p=20
p=5
focus: (-5,-1) (p-distance above vertex on the axis of symmetry)

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