Find the point(s), if any, at which the graph of *f* has a horizontal tangent line. *f*(*x*) = *(x^2)/(x* − 6)

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

Find the point(s), if any, at which the graph of f has a horizontal tangent line. f(x) = (x^2)/(x − 6)

Learn how to find the points where the function f(x) = (x^2)/(x - 6) has horizontal tangent lines by using the derivative and the quotient rule. This problem includes calculus concepts like derivatives and solving for zero slopes. Suitable for students in Grades 11-12.

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