SOLUTION: find the vertex of the parabola: x^2-4x-4y+16=0

Question 462980: find the vertex of the parabola: x^2-4x-4y+16=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
find the vertex of the parabola: x^2-4x-4y+16=0
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Standard form of parabola: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex) A is a multiplier which affects the steepness of the curve.
x^2-4x-4y+16=0
4y=x^2-4x+16
divide by 4
y=(1/4)(x^2-4x+16)
completing the square
y=(1/4)(x^2-4x+4)+4-1
y=(1/4)(x-2)^2+3
coordinates of the vertex: (2,3)
see the graph below as a visual check on the answer
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