Given the function \( f(x) = 44 - 15x \), we want to find the expression for the difference quotient:

\[
\frac{f(x+h) - f(x)}{h}
\]

Let's compute it step by step.

1. **Substitute** \( f(x) \) into the difference quotient:

   \[
   \frac{f(x+h) - f(x)}{h}
   \]

2. **Calculate \( f(x+h) \)**:

   \[
   f(x+h) = 44 - 15(x+h)
   \]
   Expanding this:
   \[
   f(x+h) = 44 - 15x - 15h
   \]

3. **Subtract \( f(x) \) from \( f(x+h) \)**:

   \[
   f(x+h) - f(x) = \left(44 - 15x - 15h\right) - \left(44 - 15x\right)
   \]
   Simplifying:
   \[
   f(x+h) - f(x) = -15h
   \]

4. **Divide by \( h \)**:

   \[
   \frac{f(x+h) - f(x)}{h} = \frac{-15h}{h}
   \]

5. **Simplify** the expression:

   \[
   \frac{f(x+h) - f(x)}{h} = -15
   \]

So, the difference quotient simplifies to \(-15\), which is independent of \(x\) and \(h\).

Would you like more details or have any questions?

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Here are 5 related questions you might find interesting:
1. What is the derivative of \( f(x) = 44 - 15x \) using first principles?
2. How would the difference quotient change if \( f(x) \) was a quadratic function?
3. Can you find the slope of the tangent line to the curve \( y = 44 - 15x \) at any point \( x \)?
4. How does the difference quotient relate to the concept of the derivative?
5. What happens to the difference quotient as \( h \) approaches 0?

**Tip:** The difference quotient is a foundational concept in calculus, used to define the derivative, which represents the rate of change of a function.

Explore the difference quotient for the linear function f(x) = 44 - 15x. This problem breaks down the steps to find the expression (f(x+h) - f(x)) / h and simplifies it to a constant value. Suitable for Grades 11-12, this explanation covers key algebraic concepts and the fundamentals of calculus.

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