SOLUTION: (y-4)^2=16(x+1)which conic section does the equation describe?

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Question 654800: (y-4)^2=16(x+1)which conic section does the equation describe?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
(y-4)^2=16(x+1)which conic section does the equation describe?
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This is a parabola that opens rightwards:
Its standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
vertex: (4,-1)
axis of symmetry: y=4
4p=16
p=4 (distance from vertex to focus and directrix on the axis of symmetry)
see graph below:
y=4±4(x+1)^.5