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

Question

1598 views around the world

You can reuse this answer
Creative Commons License

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 $(30,36)$ .

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