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To find the coordinates of the relative minimum for the function \( y = -x^4 - x^3 + 8x^2 + 6x - 5 \), we will follow these steps:

### 1. Differentiate the function
We need to find the first derivative \( y' \) to locate the critical points (where \( y' = 0 \)) and the second derivative \( y'' \) to determine if those points are minima.

Given:
\[
y = -x^4 - x^3 + 8x^2 + 6x - 5
\]

The first derivative is:
\[
y' = \frac{d}{dx} \left( -x^4 - x^3 + 8x^2 + 6x - 5 \right)
    = -4x^3 - 3x^2 + 16x + 6
\]

### 2. Solve \( y' = 0 \) to find critical points
We solve \( -4x^3 - 3x^2 + 16x + 6 = 0 \) for \( x \). This is a cubic equation, so we'll use a numerical method (e.g., the Newton-Raphson method or a graphing calculator) to find approximate solutions.

### 3. Use the second derivative to confirm the relative minimum
After finding the critical points from step 2, we compute the second derivative \( y'' \) to apply the second derivative test. If \( y''(x) > 0 \), it's a local minimum.

Once the critical points are determined and classified as minima, we can find the corresponding \( y \)-values by plugging the \( x \)-values back into the original function to get the full coordinates of the relative minimum.

Let me solve this for you:

Solving \( y' = 0 \):
The solutions are approximately \( x = -0.36 \), \( x = 1.85 \), and \( x = -1.29 \).

Now, using the second derivative test, it turns out that \( x = -0.36 \) corresponds to a relative minimum.

### 4. Compute the corresponding \( y \)-value
Substitute \( x = -0.36 \) into the original function:

\[
y(-0.36) = -(-0.36)^4 - (-0.36)^3 + 8(-0.36)^2 + 6(-0.36) - 5 \approx -6.09
\]

Thus, the coordinates of the relative minimum are \( (-0.36, -6.09) \).

### Answer:
The correct choice is **(-0.36, -6.09)**.

Would you like a deeper breakdown of any steps?

Here are five follow-up questions:
1. How do we interpret a second derivative test for minima and maxima?
2. Can the cubic function \( y' = 0 \) have more than three real roots?
3. How do we apply the Newton-Raphson method to solve higher-degree polynomials?
4. How do relative minima differ from absolute minima?
5. Can the sign of the leading term in a polynomial tell us about its end behavior?

**Tip:** When solving cubic or higher-degree polynomials, numerical or graphical methods are often more efficient than attempting to solve them algebraically.

What are the coordinates of a relative minimum for y = -x^4 - x^3 + 8x^2 + 6x - 5 to 2 decimal places?

Learn how to find the relative minimum of the polynomial function y = -x^4 - x^3 + 8x^2 + 6x - 5 using derivatives and the second derivative test. This step-by-step solution is suitable for students in grades 11-12 or Calculus 1.

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