Question 760129
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=(y-1)^2/16+4}}}
This is an equation of a parabola that opens rightward with vertex at (4,1)