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Let's start by solving the first problem:

### Problem #90:
For the function \( f(x) = 4x - 2x^2 \), find \( f(a + h) - f(a) \).

1. **Find \( f(a + h) \)**:
   Substitute \( x = a + h \) into the function \( f(x) = 4x - 2x^2 \):

   \[
   f(a + h) = 4(a + h) - 2(a + h)^2
   \]

   Now, expand the expression:

   \[
   f(a + h) = 4a + 4h - 2[(a + h)(a + h)]
   \]
   \[
   f(a + h) = 4a + 4h - 2(a^2 + 2ah + h^2)
   \]
   \[
   f(a + h) = 4a + 4h - 2a^2 - 4ah - 2h^2
   \]
   \[
   f(a + h) = 4a - 2a^2 + 4h - 4ah - 2h^2
   \]

2. **Find \( f(a) \)**:
   Now, substitute \( x = a \) into the original function:

   \[
   f(a) = 4a - 2a^2
   \]

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

   \[
   f(a + h) - f(a) = (4a - 2a^2 + 4h - 4ah - 2h^2) - (4a - 2a^2)
   \]

   Simplifying:

   \[
   f(a + h) - f(a) = 4a - 2a^2 + 4h - 4ah - 2h^2 - 4a + 2a^2
   \]
   \[
   f(a + h) - f(a) = 4h - 4ah - 2h^2
   \]

Thus, the final answer is:

\[
f(a + h) - f(a) = 4h - 4ah - 2h^2
\]

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Would you like me to solve the other problems as well? Let me know if you want more details or clarifications.

### Related Questions:
1. What is the significance of the difference quotient in calculus?
2. How can the expression \( f(a + h) - f(a) \) be used to find the derivative of a function?
3. What happens when \( h \to 0 \) in this problem?
4. How do we simplify expressions involving binomials like \( (a + h)^2 \)?
5. How can this process be extended to more complex functions?

**Tip:** The difference \( f(a + h) - f(a) \) is a crucial part of the definition of the derivative, which is the rate of change of the function at \( x = a \).

For the function f(x) = 4x - 2x^2, find f(a + h) - f(a).

Solve the problem of finding f(a + h) - f(a) for the function f(x) = 4x - 2x^2. This step-by-step solution covers key algebraic concepts including binomial expansion and function operations. Suitable for students in Grades 10-12.

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