Answer
the critical points are: (0,0), (4,8)
Work Step by Step
Given $y=\frac{(x)^2}{x-2}$
The critical point occurs where the first derivative is zero:
so $\frac{dy}{dx}=\frac{d{\frac{(x)^2}{x-2})}}{dx}$
${\frac{dy}{dx}}=\frac{{2(x)}({x-2})-x^2}{(x-2)^2}$
${\frac{dy}{dx}}=\frac{((x^2)-4x)}{(x-2)^2}$
so the critical point is: ${\frac{dy}{dx}}=0$
$f'(x)=\frac{((x^2)-4x)}{(x-2)^2}=0$
$x^2-4x=0$
since the denominator is undefined at $x=2$, this is not a critical point.
$x=0,x=4$
the critical value of y can be obtained by putting in the value of x:
$y=\frac{((x^2))}{(x-2)}=0$
$y=\frac{((x^2))}{(x-2)}=8$
thus the critical points are: (0,0), (4,8)
