Answer
One Horizontal tangent at $(0, 0).$
One Horizontal tangent at $(2, 4).$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x^2 u'(x)=2x$
$v(x)=x-1; v'(x)=1$
$f'(x)=\frac{(2x)(x-1)-1(x^2)}{(x-1)^2}=\frac{x(x-2)}{(x-1)^2}$
$f'(x)=0 \rightarrow (x-2)x=0 \rightarrow x=0$ or $x=2$
$f(0)=\frac{0^2}{0-1}=0 \rightarrow $ The point is $(0, 0)$.
$f(2)=\frac{2^2}{2-1}=4 \rightarrow $ The point is $(2, 4).$
