SOLUTION: Determine the vertex of the parabola whose equation is (y-1)^2=16(x-4)

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Question 760129: Determine the vertex of the parabola whose equation is (y-1)^2=16(x-4)
Found 2 solutions by MathLover1, lwsshak3:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
%28y-1%29%5E2=16%28x-4%29....it is already written in vertex form; h=4 and k=1
so, vertex is at (4,1)


Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Determine the vertex of the parabola whose equation is
(y-1)^2=16(x-4)
change equation to standard (vertex) form: x=A(y-k)^2+h, (h,k)=(x,y) coordinates of vertex, A is a coefficient that affects the slope or steepness of the curve.
(y-1)^2=16(x-4)
(y-1)^2=16x-64
(y-1)^2+64=16x
16x=(y-1)^2+64
x=%28y-1%29%5E2%2F16%2B4
This is an equation of a parabola that opens rightward with vertex at (4,1)