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To find the points on the graph of \( f(x) = x^3 - 12x + 1 \) where the function has a horizontal tangent line, we need to calculate the derivative of the function and set it equal to zero. This will give us the values of \( x \) where the tangent line is horizontal.

1. Compute the derivative \( f'(x) \):
   \[
   f'(x) = 3x^2 - 12
   \]

2. Set the derivative equal to zero and solve for \( x \):
   \[
   3x^2 - 12 = 0
   \]
   \[
   3x^2 = 12
   \]
   \[
   x^2 = 4
   \]
   \[
   x = \pm 2
   \]

3. Substitute \( x = 2 \) and \( x = -2 \) into \( f(x) \) to find the corresponding \( y \)-values:
   \[
   f(2) = 2^3 - 12 \times 2 + 1 = 8 - 24 + 1 = -15
   \]
   \[
   f(-2) = (-2)^3 - 12 \times (-2) + 1 = -8 + 24 + 1 = 17
   \]

Thus, the points on the graph where the function has a horizontal tangent line are \( (2, -15) \) and \( (-2, 17) \).

Would you like any further details on this solution or have any other questions?

### Relative Questions:
1. How would you determine the nature of these points (i.e., maximum, minimum, or saddle point)?
2. What are the implications of finding horizontal tangents on the behavior of the function?
3. How does changing the coefficients in the function affect the position and number of horizontal tangents?
4. Could you explain the significance of the second derivative in this context?
5. How can you graph the function \( f(x) = x^3 - 12x + 1 \) to visually confirm these points?

### Tip:
Remember, the derivative of a function gives you the slope of the tangent line at any point along the curve, and setting this derivative to zero helps in finding critical points where the slope is zero (horizontal tangents).

Find the points 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 horizontal tangent lines. This problem involves finding the derivative, solving quadratic equations, and determining the coordinates of the critical points.

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