https://file.aiquickdraw.com/mathbot/file/1727317067773d136dl6r.jpg

Let's analyze the problem from the image.

The function is given as:

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

### Part (A):
You are asked to evaluate:

\[
\frac{f(9) - f(5)}{9 - 5}
\]

First, compute \( f(9) \) and \( f(5) \).

\[
f(9) = 4 - 9^2 = 4 - 81 = -77
\]
\[
f(5) = 4 - 5^2 = 4 - 25 = -21
\]

Now, substitute these values into the expression:

\[
\frac{f(9) - f(5)}{9 - 5} = \frac{-77 - (-21)}{4} = \frac{-77 + 21}{4} = \frac{-56}{4} = -14
\]

So, the answer to Part (A) is:

\[
\boxed{-14}
\]

### Part (B):
You are asked to evaluate the following limit-like expression:

\[
\frac{f(5 + h) - f(5)}{h}
\]

We already know that \( f(5) = -21 \). Now, let's compute \( f(5 + h) \):

\[
f(5 + h) = 4 - (5 + h)^2 = 4 - (25 + 10h + h^2) = 4 - 25 - 10h - h^2 = -21 - 10h - h^2
\]

Now, substitute this into the expression:

\[
\frac{f(5 + h) - f(5)}{h} = \frac{-21 - 10h - h^2 - (-21)}{h} = \frac{-10h - h^2}{h} = -10 - h
\]

So, the answer to Part (B) is:

\[
\boxed{-10 - h}
\]

Would you like further clarifications or details?

Here are 5 related questions you can explore:

1. What is the derivative of the function \( f(x) = 4 - x^2 \)?
2. How do you interpret the result of Part (B) in terms of calculus (hint: it's a difference quotient)?
3. What happens to the expression in Part (B) as \( h \to 0 \)?
4. Can you generalize the difference quotient for \( f(x) \) at any point \( a \) instead of just 5?
5. What is the significance of negative values in the function \( f(x) = 4 - x^2 \)?

**Tip:** The expression in Part (B) is closely related to the concept of the derivative in calculus. Understanding how these difference quotients behave as \( h \to 0 \) is key to grasping differentiation.

Evaluate each of the following expressions, where h is a variable. Let f(x) = 4 - x^2.

Learn how to evaluate difference quotients for the function f(x) = 4 - x^2. This step-by-step solution involves key algebraic and precalculus concepts, such as quadratic functions and the difference quotient. Suitable for Grades 11-12 students preparing for calculus.

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