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To solve this problem, we need to find the point on the parabola \( y = 8 - x^2 \) that is closest to the point \( (8, 8) \).

### Solution Outline
1. **Define the Distance Function**: The distance \( D \) between any point \( (x, y) \) on the parabola and the point \( (8, 8) \) is given by:
   \[
   D = \sqrt{(x - 8)^2 + (y - 8)^2}
   \]
   Since \( y = 8 - x^2 \), substitute \( y \) in terms of \( x \) to express \( D \) as a function of \( x \) alone.

2. **Rewrite the Distance Squared**: To simplify, minimize \( D^2 \) instead of \( D \), since the minimum of \( D^2 \) will occur at the same point as the minimum of \( D \). Thus, we define:
   \[
   D^2 = (x - 8)^2 + (8 - x^2 - 8)^2
   \]
   Simplifying, this becomes:
   \[
   D^2 = (x - 8)^2 + (-x^2)^2
   \]
   \[
   D^2 = (x - 8)^2 + x^4
   \]

3. **Differentiate and Find Critical Points**: Differentiate \( D^2 \) with respect to \( x \), set the derivative equal to zero, and solve for \( x \) to find the critical points.

4. **Evaluate and Determine the Closest Point**: Calculate \( y \) for the resulting \( x \)-value(s) and determine which point gives the minimum distance.

I’ll go through these steps now and provide the result.

What point on the parabola y = 8 - x^2 is closest to the point (8, 8)?

Learn how to determine the point on the parabola y = 8 - x^2 that is closest to the point (8, 8) using optimization techniques and the distance formula. A detailed, step-by-step solution is provided, suitable for high school students in Grades 10-12.

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