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To find the point(s) on the graph of \( f(x) = x^3 - 12x + 1 \) where the function has a horizontal tangent line, we need to find where the derivative of the function equals zero. This is because a horizontal tangent line occurs when the slope of the function is zero, and the derivative gives us the slope of the tangent line at any point on the graph.

### Step 1: Find the derivative of \( f(x) \).
The derivative of \( f(x) \) is:
\[
f'(x) = \frac{d}{dx}(x^3 - 12x + 1) = 3x^2 - 12
\]

### Step 2: Set the derivative equal to zero.
To find the points where the tangent line is horizontal, we solve for \( x \) when \( f'(x) = 0 \):
\[
3x^2 - 12 = 0
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]

### Step 3: Find the corresponding \( y \)-values.
Now, substitute \( x = 2 \) and \( x = -2 \) into the original function \( f(x) \) to find the corresponding points.

- For \( x = 2 \):
\[
f(2) = 2^3 - 12(2) + 1 = 8 - 24 + 1 = -15
\]
So, one point is \( (2, -15) \).

- For \( x = -2 \):
\[
f(-2) = (-2)^3 - 12(-2) + 1 = -8 + 24 + 1 = 17
\]
So, another point is \( (-2, 17) \).

### Final Answer:
The points on the graph where the function has a horizontal tangent line are:
\[
(2, -15) \quad \text{and} \quad (-2, 17)
\]

Would you like more details on any step or have further questions? 

Here are 5 questions to expand your understanding:
1. How does the derivative help us find points with horizontal tangents?
2. What is the geometric interpretation of a horizontal tangent line?
3. Why do we solve for when the derivative equals zero?
4. How would the solution change if the function had a higher degree term?
5. How do critical points relate to horizontal tangent lines?

**Tip:** Always check for both positive and negative solutions when solving quadratic equations like \( x^2 = 4 \).

Find the point(s) on the graph of f(x) = x^3 - 12x + 1 for which the function has a horizontal tangent line.

Learn how to find the points on the graph of f(x) = x^3 - 12x + 1 where the function has a horizontal tangent line. This step-by-step solution involves taking the derivative and solving for critical points where the slope is zero. Suitable for Grades 11-12 students learning calculus.

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