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To determine the point(s) at which the function \( f(x) = \frac{x^2}{x - 1} \) has a horizontal tangent, we need to follow these steps:

### Step 1: Find the derivative \( f'(x) \)
The horizontal tangent occurs when the derivative of the function is equal to zero, i.e., \( f'(x) = 0 \).

Given \( f(x) = \frac{x^2}{x - 1} \), we'll apply the **quotient rule** to find the derivative. The quotient rule is:

\[
\left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2}
\]

where \( u = x^2 \) and \( v = x - 1 \). Now, let's compute the derivatives:

- \( u' = \frac{d}{dx} (x^2) = 2x \)
- \( v' = \frac{d}{dx} (x - 1) = 1 \)

Now, applying the quotient rule:

\[
f'(x) = \frac{(x - 1) \cdot (2x) - (x^2) \cdot (1)}{(x - 1)^2}
\]
\[
f'(x) = \frac{2x(x - 1) - x^2}{(x - 1)^2}
\]
\[
f'(x) = \frac{2x^2 - 2x - x^2}{(x - 1)^2}
\]
\[
f'(x) = \frac{x^2 - 2x}{(x - 1)^2}
\]
\[
f'(x) = \frac{x(x - 2)}{(x - 1)^2}
\]

### Step 2: Set \( f'(x) = 0 \)
To find the points where the function has a horizontal tangent, we set the numerator of the derivative equal to zero:

\[
x(x - 2) = 0
\]

This gives us two solutions:

\[
x = 0 \quad \text{or} \quad x = 2
\]

### Step 3: Determine the corresponding \( y \)-values
To find the points of the horizontal tangents, substitute \( x = 0 \) and \( x = 2 \) back into the original function:

- For \( x = 0 \):
  \[
  f(0) = \frac{0^2}{0 - 1} = 0
  \]
  So, the point is \( (0, 0) \).

- For \( x = 2 \):
  \[
  f(2) = \frac{2^2}{2 - 1} = \frac{4}{1} = 4
  \]
  So, the point is \( (2, 4) \).

### Conclusion
The points at which the function has a horizontal tangent are:
\[
(0, 0) \quad \text{and} \quad (2, 4)
\]

Would you like more details or have any questions?

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Here are five related questions for further exploration:
1. How do we identify vertical asymptotes for the function \( f(x) = \frac{x^2}{x - 1} \)?
2. What is the significance of critical points in determining the shape of a graph?
3. Can you find where the function is increasing or decreasing using \( f'(x) \)?
4. How does the second derivative help in identifying concavity of the graph?
5. What happens to the function as \( x \to 1 \) and why?

**Tip:** Always simplify your derivative as much as possible before solving \( f'(x) = 0 \) to make your calculations easier!

Determine the point(s) at which the following function has a horizontal tangent. f(x) = x^2 / (x - 1)

Learn how to determine the points where the function f(x) = x^2 / (x - 1) has horizontal tangents. This step-by-step solution involves finding the derivative using the quotient rule and solving for where f'(x) = 0. Suitable for students in Grades 11-12 studying calculus.

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