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Let's solve the given problem step-by-step.

### Problem:  
We need to **find the difference quotient**  
\[
\frac{f(x+h) - f(x)}{h}, \quad h \neq 0
\]
for the function \( f(x) = -5x + 8 \).

---

### Step 1: Calculate \( f(x + h) \)
Since \( f(x) = -5x + 8 \), substitute \( x + h \) in place of \( x \):
\[
f(x + h) = -5(x + h) + 8
= -5x - 5h + 8
\]

---

### Step 2: Find the Difference \( f(x + h) - f(x) \)
Now, subtract \( f(x) \) from \( f(x + h) \):
\[
f(x + h) - f(x) = (-5x - 5h + 8) - (-5x + 8)
\]
Simplify:
\[
f(x + h) - f(x) = -5x - 5h + 8 + 5x - 8
= -5h
\]

---

### Step 3: Form the Difference Quotient
Now, plug the result of \( f(x + h) - f(x) = -5h \) into the difference quotient:
\[
\frac{f(x+h) - f(x)}{h} = \frac{-5h}{h}
\]
Since \( h \neq 0 \), we can cancel \( h \):
\[
\frac{-5h}{h} = -5
\]

---

### Final Answer:
\[
\frac{f(x+h) - f(x)}{h} = -5
\]

This means the slope of the linear function \( f(x) = -5x + 8 \) is \( -5 \), which is consistent with the derivative of this linear function.

---

Do you need more clarification or have any further questions?

### Five Related Questions:
1. How does the difference quotient relate to the derivative of a function?
2. What would change if the function were \( f(x) = -5x + 8 + x^2 \)?
3. How does the slope interpretation change for non-linear functions?
4. Why do we cancel \( h \) in the difference quotient only when \( h \neq 0 \)?
5. How is the difference quotient useful in calculus beyond derivatives?

### Tip:
For linear functions \( f(x) = mx + b \), the difference quotient always gives the slope \( m \), which is also the derivative.

Find the difference quotient \(\frac{f(x+h) - f(x)}{h}\), where h ≠ 0, for the function \( f(x) = -5x + 8 \).

Learn how to find the difference quotient \(\frac{f(x+h) - f(x)}{h}\) for the linear function \(f(x) = -5x + 8\). This detailed step-by-step solution explains key algebraic concepts and formulas. Suitable for students in Grades 9-12.

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