https://r2.erweima.ai/mathbot/image/1725722973589c6wae5mw.jpg

To solve the problem in the image, let's break it down.

The question is asking for the expression that corresponds to the difference quotient of the given function, \( f(x) = (x - 3)^2 + 5 \).

### Step 1: Recall the difference quotient formula:
\[
\frac{\Delta y}{\Delta x} = \frac{f(x+h) - f(x)}{h}
\]

### Step 2: Apply it to the given function \( f(x) = (x - 3)^2 + 5 \).

- Find \( f(x+h) \):
\[
f(x+h) = ((x+h) - 3)^2 + 5 = (x+h-3)^2 + 5
\]
Expanding \( (x+h-3)^2 \):
\[
f(x+h) = (x - 3 + h)^2 + 5 = (x-3)^2 + 2(x-3)h + h^2 + 5
\]

- Find the difference \( f(x+h) - f(x) \):
\[
f(x+h) - f(x) = \left( (x-3)^2 + 2(x-3)h + h^2 + 5 \right) - \left( (x-3)^2 + 5 \right)
\]
Cancel out \( (x-3)^2 + 5 \):
\[
f(x+h) - f(x) = 2(x-3)h + h^2
\]

### Step 3: Divide by \( h \):
\[
\frac{f(x+h) - f(x)}{h} = \frac{2(x-3)h + h^2}{h} = 2(x-3) + h
\]

So, the difference quotient is:
\[
\frac{\Delta y}{\Delta x} = 2(x-3) + h
\]

### Step 4: Match the answer:
Now, looking at the multiple-choice options, the correct expression is \( 2x - 6 + h \).

This corresponds to option **C**.

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Do you have any questions or would you like further details?

Here are five follow-up questions you might find useful:
1. How would you compute the difference quotient for a cubic function?
2. What is the geometric interpretation of the difference quotient?
3. What happens to the difference quotient as \( h \) approaches zero?
4. How does the difference quotient relate to the derivative of a function?
5. Can you apply the difference quotient to a piecewise function?

**Tip**: The difference quotient is a stepping stone to finding the derivative of a function, which gives the slope of the tangent line at any point on the curve.

Given the function f(x) = (x - 3)^2 + 5, determine the correct expression for the difference quotient from the following options.

Explore the step-by-step solution for finding the difference quotient of the quadratic function f(x) = (x - 3)^2 + 5. This tutorial walks you through applying the formula and understanding the result. Suitable for high school calculus students.

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