The question asks for the gradient of the tangent line to the curve $y = x^3 - 3x - 10$ at the point $(1, -12)$, and then asks to find the equation of the tangent line.

Steps to solve:

1. Find the gradient of the tangent line: The gradient of the tangent line can be found using the derivative of the function, since the derivative gives the slope of the tangent at any point.

   Given: $y = f(x) = x^3 - 3x - 10$ The derivative $f'(x)$ will give us the slope function.

   Differentiate $f(x)$: $f'(x) = 3x^2 - 3$

   Now, substitute $x = 1$ to find the gradient at the point $(1, -12)$: $f'(1) = 3(1)^2 - 3 = 3 - 3 = 0$ So, the gradient of the tangent line at the point $(1, -12)$ is 0.

2. Find the equation of the tangent line: The equation of a line can be written as: $y - y_1 = m(x - x_1)$ where $(x_1, y_1)$ is a point on the line and $m$ is the slope.

   Using the point $(1, -12)$ and the slope $m = 0$, we get: $y - (-12) = 0(x - 1)$ Simplifying: $y + 12 = 0$ $y = -12$

Thus, the equation of the tangent line is $y = -12$.

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Related Questions:

1. How do you find the tangent line to a curve at a different point?
2. What is the significance of a tangent line with zero slope?
3. How do you determine the derivative of more complex functions?
4. What are other methods to compute the gradient of a tangent line?
5. How would this process change for a different curve equation?

Tip:

For any cubic or higher-degree polynomial, the derivative will be a lower-degree polynomial, which simplifies finding the slope at any specific point.