To find the equation of the tangent line to the graph of $f(x) = x^3 + 8$ at the point $(-2, 0)$, we need to follow these steps:

1. Find the derivative of $f(x)$:

The derivative $f'(x)$ will give us the slope of the tangent line at any point on the curve.

$f(x) = x^3 + 8$

Differentiate with respect to $x$:

$f'(x) = 3x^2$

2. Find the slope at $x = -2$:

To get the slope of the tangent line at the point $(-2, 0)$, substitute $x = -2$ into $f'(x)$:

$f'(-2) = 3(-2)^2 = 3 \times 4 = 12$

So, the slope of the tangent line at $(-2, 0)$ is 12.

3. Use the point-slope form of the equation of a line:

The point-slope form of a line is:

$y - y_1 = m(x - x_1)$

where $m$ is the slope, and $(x_1, y_1)$ is the point on the line. We know:

• $m = 12$
• $(x_1, y_1) = (-2, 0)$

Substitute these values into the point-slope form:

$y - 0 = 12(x - (-2))$

Simplify:

$y = 12(x + 2)$

Final Answer:

The equation of the tangent line is:

$y = 12(x + 2)$

Would you like further details or have any questions?

Here are five related questions:

1. How do you find the derivative of a more complex function?
2. How does the point-slope form differ from other forms of linear equations?
3. What is the significance of the derivative at a point on a curve?
4. How do we confirm that $(-2, 0)$ lies on the graph of $f(x)$?
5. What would happen if we needed the tangent line at a different point?

Tip: When finding tangent lines, always verify that the point lies on the curve by substituting the $x$-value back into the original function!