To find \(\frac{f(x+h) - f(x)}{h}\) where \(f(x) = -x^2 + 1\), we can follow these steps:

### Step 1: Find \(f(x+h)\)
Given \(f(x) = -x^2 + 1\), we substitute \(x+h\) into the function:
\[
f(x+h) = -(x+h)^2 + 1
\]
Expanding \((x+h)^2\):
\[
(x+h)^2 = x^2 + 2xh + h^2
\]
Substitute this into the expression for \(f(x+h)\):
\[
f(x+h) = -(x^2 + 2xh + h^2) + 1 = -x^2 - 2xh - h^2 + 1
\]

### Step 2: Find \(f(x+h) - f(x)\)
Now, subtract \(f(x)\) from \(f(x+h)\):
\[
f(x+h) - f(x) = \left(-x^2 - 2xh - h^2 + 1\right) - \left(-x^2 + 1\right)
\]
Simplify the expression:
\[
f(x+h) - f(x) = -x^2 - 2xh - h^2 + 1 + x^2 - 1
\]
\[
f(x+h) - f(x) = -2xh - h^2
\]

### Step 3: Divide by \(h\)
Finally, divide the result by \(h\):
\[
\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2}{h}
\]
Simplify by dividing each term by \(h\):
\[
\frac{f(x+h) - f(x)}{h} = -2x - h
\]

### Final Answer:
\[
\frac{f(x+h) - f(x)}{h} = -2x - h
\]

Would you like further details or have any questions?

Here are some related questions:

1. What is the significance of this expression in calculus?
2. How does this relate to the derivative of \(f(x)\)?
3. What would \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\) represent?
4. How does the function change if we consider \(f(x) = x^2 + 1\) instead?
5. What is the role of the \(h\) term in the final expression?
6. How can this expression be used to find the slope of a tangent line?
7. What happens if \(h = 0\) in the final expression?
8. How does this relate to the concept of a difference quotient?

**Tip:** The expression \(\frac{f(x+h) - f(x)}{h}\) is known as the difference quotient, and it forms the basis for the definition of a derivative in calculus.

Explore how to calculate the difference quotient for the function f(x) = -x^2 + 1 in calculus. This detailed explanation covers step-by-step calculations and the significance of the difference quotient in finding derivatives. Suitable for advanced high school students and calculus enthusiasts.

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