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The problem is asking for the equation of the tangent line to the function \( f(x) = x^3 - 12 \) at \( x = -5 \).

### Step 1: Find the derivative of the function
The derivative of a function gives the slope of the tangent line at any point. Let's find the derivative of \( f(x) = x^3 - 12 \):

\[
f'(x) = \frac{d}{dx}(x^3 - 12) = 3x^2
\]

### Step 2: Calculate the slope at \( x = -5 \)
Now, substitute \( x = -5 \) into the derivative to find the slope of the tangent line at that point:

\[
f'(-5) = 3(-5)^2 = 3(25) = 75
\]

So, the slope of the tangent line at \( x = -5 \) is 75.

### Step 3: Find the point on the curve at \( x = -5 \)
Substitute \( x = -5 \) into the original function \( f(x) = x^3 - 12 \) to find the corresponding \( y \)-coordinate:

\[
f(-5) = (-5)^3 - 12 = -125 - 12 = -137
\]

So, the point of tangency is \( (-5, -137) \).

### Step 4: Use the point-slope form of a line
The equation of a line in point-slope form is:

\[
y - y_1 = m(x - x_1)
\]

Where \( m \) is the slope, and \( (x_1, y_1) \) is the point on the line. Substituting \( m = 75 \), \( x_1 = -5 \), and \( y_1 = -137 \):

\[
y - (-137) = 75(x - (-5))
\]

Simplifying:

\[
y + 137 = 75(x + 5)
\]
\[
y + 137 = 75x + 375
\]
\[
y = 75x + 375 - 137
\]
\[
y = 75x + 238
\]

### Final Answer:
The equation of the tangent line at \( x = -5 \) is:

\[
y = 75x + 238
\]

Would you like further details or have any questions?

Here are five related questions:
1. How do you find the equation of a tangent line for more complex functions?
2. What happens to the tangent line equation if the slope is zero?
3. Can you find the tangent line for trigonometric functions like \( \sin(x) \)?
4. How does the second derivative relate to the curvature of the tangent line?
5. How would the process change if you were working with a parametric equation?

**Tip:** The slope of a tangent line at any point on a curve represents the instantaneous rate of change of the function at that point.

For the function f(x) = x^3 - 12, find the equation of the tangent line at x = -5.

Learn how to find the equation of the tangent line for the function f(x) = x^3 - 12 at x = -5 using derivatives. This solution includes finding the derivative, calculating the slope, and applying the point-slope formula. Ideal for Grades 11-12 or early college calculus students.

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