The problem you're working on asks to evaluate and simplify the expression

$\frac{f(a+h) - f(a)}{h}$

where $f(x) = -2x^2 + 3x - 4$.

Step-by-Step Solution:

1. Find $f(a+h)$: Substituting $a+h$ into $f(x) = -2x^2 + 3x - 4$, we get:

   $f(a+h) = -2(a+h)^2 + 3(a+h) - 4$ Now expand the terms: $(a+h)^2 = a^2 + 2ah + h^2$ So: $f(a+h) = -2(a^2 + 2ah + h^2) + 3(a + h) - 4$ Distribute the terms: $f(a+h) = -2a^2 - 4ah - 2h^2 + 3a + 3h - 4$

2. Find $f(a)$: Substitute $a$ into $f(x)$: $f(a) = -2a^2 + 3a - 4$

3. Calculate $f(a+h) - f(a)$: Now subtract $f(a)$ from $f(a+h)$: $f(a+h) - f(a) = (-2a^2 - 4ah - 2h^2 + 3a + 3h - 4) - (-2a^2 + 3a - 4)$ Distribute the negative sign: $f(a+h) - f(a) = -2a^2 - 4ah - 2h^2 + 3a + 3h - 4 + 2a^2 - 3a + 4$ Combine like terms: $f(a+h) - f(a) = -4ah - 2h^2 + 3h$

4. Divide by $h$: Now divide the expression by $h$: $\frac{f(a+h) - f(a)}{h} = \frac{-4ah - 2h^2 + 3h}{h}$ Simplify each term: $\frac{-4ah}{h} = -4a, \quad \frac{-2h^2}{h} = -2h, \quad \frac{3h}{h} = 3$ So the final simplified expression is: $\frac{f(a+h) - f(a)}{h} = -4a - 2h + 3$

This is the simplified form of the given difference quotient.

Would you like more details on any of these steps, or do you have additional questions?

Here are 5 related questions to explore further:

1. What is the significance of the difference quotient in calculus?
2. How would the result change if the function were cubic instead of quadratic?
3. Can you calculate the limit of this difference quotient as $h \to 0$?
4. How is the difference quotient related to the derivative of the function?
5. What happens if you substitute a specific value for $a$ into the final expression?

Tip: The difference quotient is fundamental in calculus as it leads directly to the definition of the derivative. Understanding how to manipulate and simplify these expressions will be key when you study more advanced topics.