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To determine the nature of the extrema for the given curve \( y = 4x^3 - 102x^2 + 360x - 270 \), we need to analyze the critical points by following these steps:

1. **Find the First Derivative**: The critical points occur where the derivative \( y' \) is zero or undefined.
   
   \[
   y' = \frac{d}{dx}(4x^3 - 102x^2 + 360x - 270) = 12x^2 - 204x + 360
   \]

2. **Set the Derivative to Zero**: Solve \( 12x^2 - 204x + 360 = 0 \) to find the critical points.

   Dividing through by 12:
   \[
   x^2 - 17x + 30 = 0
   \]
   Factoring gives:
   \[
   (x - 15)(x - 2) = 0
   \]
   So, the critical points are \( x = 15 \) and \( x = 2 \).

3. **Determine the Nature of Each Critical Point**: To classify each critical point, use the **second derivative test**.

   Compute the second derivative:
   \[
   y'' = \frac{d}{dx}(12x^2 - 204x + 360) = 24x - 204
   \]

   - For \( x = 2 \):
     \[
     y''(2) = 24(2) - 204 = 48 - 204 = -156
     \]
     Since \( y''(2) < 0 \), this indicates a **local maximum** at \( x = 2 \).

   - For \( x = 15 \):
     \[
     y''(15) = 24(15) - 204 = 360 - 204 = 156
     \]
     Since \( y''(15) > 0 \), this indicates a **local minimum** at \( x = 15 \).

4. **Conclusion**: Based on the analysis, the answer is:
   - **C: A local minimum at \( x = 15 \)**.

Would you like a more detailed explanation of any step?

Here are five related questions to consider:
1. How would the nature of extrema change if the coefficients in the function were altered?
2. What is the significance of the second derivative in determining concavity?
3. How could we use a graphing tool to visualize the local maxima and minima of this function?
4. What is the importance of local maxima and minima in real-world applications?
5. How does the second derivative test differ from the first derivative test?

**Tip:** When analyzing polynomials for maxima and minima, always start with the first derivative to locate critical points, then use the second derivative for classification.

The curve y = 4x^3 - 102x^2 + 360x - 270 has:
A. a local minimum at x = 2
B. a local maximum at x = 5
C. a local minimum at x = 15
D. a local maximum at x = 3
E. no local maxima or minima

Explore how to find local maxima and minima for the function y = 4x^3 - 102x^2 + 360x - 270. This problem demonstrates the use of the first and second derivatives to classify critical points, following the second derivative test. Ideal for calculus students in Grades 11-12.

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