Find the point(s), if any, at which the graph of f has a horizontal tangent line. $f ( x ) = \frac { x ^ { 2 } } { x - 1 }$
A) $( 0,0 ) , ( 2,4 )$
B) $( 0,2 ) , ( 0,4 )$
C) $( 4,0 ) , ( 2,0 )$
D) $( 0,4 ) , ( 2,0 )$
E) $( 0,0 ) , ( 4,2 )$

Question

Quizplus · Accepted Answer

To find the points where the graph of $f(x) = \frac{x^2}{x-1}$ has a horizontal tangent line, we need to find where the derivative of $f(x)$, $f'(x)$, equals zero. The derivative of $f(x)$ is $f'(x) = \frac{2x(x-1) - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2}$. Setting $f'(x) = 0$ gives $x^2 - 2x = 0$, which factors to $x(x - 2) = 0$. Thus, $x = 0$ and $x = 2$ are the points where the derivative is zero. Substituting these values back into $f(x)$ gives the points $(0, 0)$ and $(2, 4)$ where the graph has a horizontal tangent line.

Find the Point(s), If Any, at Which the Graph of F