We need to find the expression for \(\frac{f(3+h) - f(3)}{h}\) given the function \( f(x) = x^2 + 5 \).

### Step 1: Calculate \( f(3+h) \)
Substitute \( x = 3+h \) into the function \( f(x) = x^2 + 5 \):
\[
f(3+h) = (3+h)^2 + 5
\]
Expand \( (3+h)^2 \):
\[
(3+h)^2 = 9 + 6h + h^2
\]
So,
\[
f(3+h) = 9 + 6h + h^2 + 5 = h^2 + 6h + 14
\]

### Step 2: Calculate \( f(3) \)
Substitute \( x = 3 \) into \( f(x) = x^2 + 5 \):
\[
f(3) = 3^2 + 5 = 9 + 5 = 14
\]

### Step 3: Compute \( \frac{f(3+h) - f(3)}{h} \)
Now subtract \( f(3) \) from \( f(3+h) \):
\[
f(3+h) - f(3) = (h^2 + 6h + 14) - 14 = h^2 + 6h
\]
Now, divide by \( h \):
\[
\frac{f(3+h) - f(3)}{h} = \frac{h^2 + 6h}{h}
\]
Simplify the expression:
\[
\frac{f(3+h) - f(3)}{h} = h + 6
\]

### Final Answer:
\[
\boxed{h + 6}
\]

Would you like to explore this further or have any questions?

Here are some related questions you might find interesting:

1. How would you find the derivative of \( f(x) = x^2 + 5 \) using the limit definition?
2. What happens to the expression \( \frac{f(3+h) - f(3)}{h} \) as \( h \) approaches 0?
3. How do you calculate \( \frac{f(x+h) - f(x)}{h} \) for a general function \( f(x) \)?
4. What is the value of \( \frac{f(3+h) - f(3)}{h} \) when \( h = 1 \)?
5. How would the solution change if \( f(x) \) were \( x^2 - 5 \) instead of \( x^2 + 5 \)?

**Tip:** Expressions like \( \frac{f(x+h) - f(x)}{h} \) are fundamental in calculus, as they form the basis for understanding derivatives.

Learn how to calculate the difference quotient \( \frac{f(3+h) - f(3)}{h} \) for the function f(x) = x^2 + 5 with step-by-step solutions. Understand the concepts of quadratic functions and function evaluation. Suitable for Grades 10-12.

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