Question 28969: use completing the square to find the vertex of the quadratic function
y=-2x^2+16x+4
This is what i tried
y=-2(x^2-8x)+ 4
Y= -2(x-4)^2 +16-16+4
Y=-2(x-4)^2-12
vertex= (-4,-12) the answer is supposed to be vertex= (4,36)
Thanks for the help in advance
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! use completing the square to find the vertex of the quadratic function
y=-2x^2+16x+4
This is what i tried
y=-2x^2+16x+4
y =-2[x^2-8x-2]
y = -2{[(x-4)^2-16] -2}
(perfecting the square for which we needed 16 and hence added and subtracted 16)
y = -2{(x-4)^2-18)
y = -2(x-4)^2 +36
(y-36) = -2(x-4)^2
(-1/2)(y-36)= (x-4)^2 (multiplying by (-1/2) )
That is (x-4)^2= -4(1/8)(y-36)
which is of the form (x-h)^2 = -4a(y-k)
Square in x and linear in y represents a parabola with axis vertical
and since the coefficient of y is negative,the parabola looks downwards (that is the union symbol inverted)
Vertex = A(h,k) = (4,36)
a = distance of the focus from the vertex
=(1/8)
That is the focus is the point S(4,36-1/8) = S(4,287/8) which is a point
on the line of symmetry, x=4
and (1/8) unit below the vetex
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