What is the vertex of # y= 1/5x^2- (x/2-3)^2 #?

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Answer 1 · 2018-07-12T12:37:38

#(30,36)#.

We have, #y=1/5x^2-(x/2-3)^2#.

#:. y=x^2/5-(x^2/4-3x+9)#,

#=x^2/5-x^2/4+3x-9#,

#:. y=-x^2/20+3x-9

graph{-x^2/20+3x-9 [-150.1, 150.3, -75, 75]} #,

# or, y+9=-x^2/20+3x#.

#:. 20(y+9)=-x^2+60x#.

Completing square on the R.H.S., we get,

#20y+180=(-x^2+2xx30x-30^2)+30^2#.

#:. 20y+180-900=-x^2+60x-900,#

# i.e., 20y-720=-(x^2-60x+900)#,

# or, 20(y-36)=-(x-30)^2#.

# rArr (y-36)=-1/20(x-30)^2#.

Consequently, the vertex is #(30,36)#.