https://file.aiquickdraw.com/mathbot/file/17255049790682pjwucbe.jpg

You have provided an image containing the function \( f(x) = 2 + 2x^2 \) and you're asking to calculate the difference quotient:

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

Let's solve this step-by-step:

1. **Substitute \( f(x) \) into the difference quotient.** First, we need to find \( f(a+h) \) and \( f(a) \).

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

2. **Expand \( f(a+h) \).**

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

3. **Now, subtract \( f(a) \) from \( f(a+h) \):**

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

4. **Divide by \( h \):**

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

5. **Simplify the expression:**

\[
\frac{f(a+h) - f(a)}{h} = 4a + 2h
\]

Thus, the simplified form of the difference quotient is:

\[
\frac{f(a+h) - f(a)}{h} = 4a + 2h
\]

If you have any questions or need further clarification, feel free to ask!

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Here are five related questions to deepen understanding:

1. What is the difference quotient used for in calculus?
2. How would the derivative of \( f(x) = 2 + 2x^2 \) relate to this expression?
3. What happens to the difference quotient as \( h \to 0 \)?
4. Can this technique be applied to functions with more complex expressions, like polynomials of higher degrees?
5. What is the geometric interpretation of the difference quotient?

**Tip:** The difference quotient is the foundation of the derivative, which gives the slope of the tangent line to the function at any point!

Solve the difference quotient for the function f(x) = 2 + 2x^2. This problem breaks down key concepts in calculus such as limits, derivatives, and quadratic functions. Suitable for high school students studying calculus.

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