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### Given Problem:
We are asked to find the point(s) on the graph of \( f(x) = x^3 - 12x + 1 \) where the function has a **horizontal tangent line**.

### Step 1: Understand what a horizontal tangent line means
A horizontal tangent line occurs when the slope of the tangent line is zero. The slope of the tangent line is given by the derivative of the function. Therefore, we need to **find the derivative of \( f(x) \) and set it equal to zero**.

### Step 2: Differentiate the function
The function given is:
\[
f(x) = x^3 - 12x + 1
\]

We now take the derivative of \( f(x) \):
\[
f'(x) = \frac{d}{dx}(x^3 - 12x + 1)
\]

Using basic differentiation rules:

- The derivative of \( x^3 \) is \( 3x^2 \)
- The derivative of \( -12x \) is \( -12 \)
- The derivative of the constant \( 1 \) is 0

Thus, we get:
\[
f'(x) = 3x^2 - 12
\]

### Step 3: Set the derivative equal to zero
To find where the tangent line is horizontal, we set the derivative equal to zero:
\[
f'(x) = 0 \quad \Rightarrow \quad 3x^2 - 12 = 0
\]

### Step 4: Solve for \( x \)
We solve the equation \( 3x^2 - 12 = 0 \):

First, divide through by 3:
\[
x^2 - 4 = 0
\]

Now solve for \( x^2 \):
\[
x^2 = 4
\]

Take the square root of both sides:
\[
x = \pm 2
\]

So, we have two solutions for \( x \): \( x = 2 \) and \( x = -2 \).

### Step 5: Find the corresponding \( y \)-values
Now that we have the \( x \)-values where the tangent line is horizontal, we need to find the corresponding \( y \)-values by plugging these \( x \)-values into the original function \( f(x) \).

#### 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, the other 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 further clarifications or any additional details?

Here are 5 related questions for deeper understanding:
1. Why do we set the derivative equal to zero to find horizontal tangents?


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

Explore how to find points on the cubic function f(x) = x^3 - 12x + 1 where there are horizontal tangents by setting its first derivative to zero and solving for x. This guide includes detailed steps to derive the equation, solve for critical points, and determine the y-values for these points.

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