To find the point(s) where the graph of the function $f(x) = \frac{x^2}{x - 6}$ has a horizontal tangent line, we need to follow these steps:

Step 1: Find the derivative of $f(x)$

The derivative $f'(x)$ gives us the slope of the tangent line. For horizontal tangent lines, the slope must be zero, meaning $f'(x) = 0$.

We will use the quotient rule to differentiate $f(x) = \frac{x^2}{x - 6}$. The quotient rule states:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) u'(x) - u(x) v'(x)}{[v(x)]^2}$

where $u(x) = x^2$ and $v(x) = x - 6$.

• $u'(x) = 2x$
• $v'(x) = 1$

Now, applying the quotient rule:

$f'(x) = \frac{(x - 6)(2x) - (x^2)(1)}{(x - 6)^2}$

Simplifying the numerator:

$f'(x) = \frac{2x(x - 6) - x^2}{(x - 6)^2}$ $= \frac{2x^2 - 12x - x^2}{(x - 6)^2}$ $= \frac{x^2 - 12x}{(x - 6)^2}$ $= \frac{x(x - 12)}{(x - 6)^2}$

Step 2: Set $f'(x) = 0$

To find the points where the tangent line is horizontal, we set $f'(x) = 0$:

$\frac{x(x - 12)}{(x - 6)^2} = 0$

This equation is equal to zero when the numerator is zero, so:

$x(x - 12) = 0$

Thus, $x = 0$ or $x = 12$.

Step 3: Determine the corresponding $y$-coordinates

Now we find the $y$-coordinates for these $x$-values by substituting them into the original function $f(x) = \frac{x^2}{x - 6}$:

• For $x = 0$: $f(0) = \frac{0^2}{0 - 6} = 0$ So the point is $(0, 0)$.

• For $x = 12$: $f(12) = \frac{12^2}{12 - 6} = \frac{144}{6} = 24$ So the point is $(12, 24)$.

Conclusion

The graph of $f(x) = \frac{x^2}{x - 6}$