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Let's analyze the problem:

### Given:
- Function: \( f(x) = 2x^3 \)
- Value: \( a = -1 \)

### Part (a) - Find \( f'(a) \)

1. To find \( f'(a) \), we first compute the derivative of \( f(x) = 2x^3 \).
   \[
   f'(x) = \frac{d}{dx} (2x^3) = 6x^2
   \]
2. Now, substitute \( a = -1 \) into the derivative:
   \[
   f'(-1) = 6(-1)^2 = 6
   \]

So, \( f'(-1) = 6 \).

---

### Part (b) - Find the equation of the tangent line at \( (a, f(a)) \)

1. The slope of the tangent line is given by \( f'(a) = 6 \), which we just found.
2. To find the point of tangency, calculate \( f(-1) \):
   \[
   f(-1) = 2(-1)^3 = -2
   \]
   Thus, the point of tangency is \( (-1, -2) \).
3. The equation of a line with slope \( m \) passing through the point \( (x_1, y_1) \) is given by:
   \[
   y - y_1 = m(x - x_1)
   \]
   Substituting \( m = 6 \), \( x_1 = -1 \), and \( y_1 = -2 \):
   \[
   y - (-2) = 6(x - (-1))
   \]
   Simplifying:
   \[
   y + 2 = 6(x + 1)
   \]
   \[
   y + 2 = 6x + 6
   \]
   \[
   y = 6x + 4
   \]

So, the equation of the tangent line is \( y = 6x + 4 \).

---

Would you like any further details or explanations?

Here are 5 related questions:
1. How do you find the derivative of higher-degree polynomials?
2. What is the significance of the slope in the tangent line equation?
3. How does the tangent line relate to the behavior of the function near \( x = a \)?
4. What happens if we choose a different value for \( a \) in this problem?
5. How does the power rule work for derivatives of polynomial functions?

**Tip**: Always double-check your point of tangency by evaluating both the function and its derivative at the specified point.

For the function f(x) = 2x^3 and a = -1, find f'(a) and determine the equation of the line tangent to the graph of f at (a, f(a)).

Find the derivative of f(x) = 2x^3 and determine the equation of the tangent line at x = -1. This problem involves calculating the derivative using the power rule and applying the point-slope form to find the tangent line.

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