What is the vertex of # y=2(x/2-2)^2-4 #?

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Answer 1 · 2018-07-10T06:01:51

Vertex #(4, -4)#

Given -

#y=2(x/2-2)^2-4#

#y=2(x^2/4-2x+4)-4#

#y=1/2 x^2-4x+8-4#

#y=1/2 x^2-4x+4#

Vertex -

#x=(-b)/(2a)=(-(-4))/(2 xx 1/2)=4/1=4#

At #x=4; y=2(4/2-2)^2-4=2(0)-4=-4#

Vertex #(4, -4)#

Answer 2 · 2018-07-10T06:20:14

#(4, -4)#

Given equation,

#y=2(x/2-2)^2-4#

#y=2(\frac{x-4}{2})^2-4#

#y=1/2(x-4)^2-4#

#y+4=1/2(x-4)^2#

#(x-4)^2=2(y+4)#

The above equation shows a vertical parabola with vertex

#(x-4=0, y+4=0)#

#(x=4, y=-4)#

#(4, -4)#