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The problem asks us to evaluate certain expressions for the function \( f(x) = 7 - 2x + 13x^2 \). Let's solve each part step-by-step:

### 1. Calculate \( f(a) \)
Substitute \( x = a \) into the function:

\[
f(a) = 7 - 2a + 13a^2
\]

### 2. Calculate \( f(a + h) \)
Substitute \( x = a + h \) into the function:

\[
f(a + h) = 7 - 2(a + h) + 13(a + h)^2
\]

Simplify the expression:

\[
f(a + h) = 7 - 2a - 2h + 13(a^2 + 2ah + h^2) 
\]

\[
= 7 - 2a - 2h + 13a^2 + 26ah + 13h^2
\]

### 3. Calculate \( \frac{f(a + h) - f(a)}{h} \) for \( h \neq 0 \)
Substitute the values of \( f(a + h) \) and \( f(a) \):

\[
\frac{f(a + h) - f(a)}{h} = \frac{(7 - 2a - 2h + 13a^2 + 26ah + 13h^2) - (7 - 2a + 13a^2)}{h}
\]

Simplify the numerator:

\[
= \frac{-2h + 26ah + 13h^2}{h}
\]

Factor out \( h \):

\[
= \frac{h(-2 + 26a + 13h)}{h}
\]

Since \( h \neq 0 \), cancel \( h \):

\[
= -2 + 26a + 13h
\]

### Final Answers:
1. \( f(a) = 7 - 2a + 13a^2 \)
2. \( f(a + h) = 7 - 2a - 2h + 13a^2 + 26ah + 13h^2 \)
3. \( \frac{f(a + h) - f(a)}{h} = -2 + 26a + 13h \)

Would you like any further details, or do you have any questions? Here are some related questions:

1. What is the derivative of \( f(x) \)?
2. How does the result of \( \frac{f(a + h) - f(a)}{h} \) relate to the derivative of \( f(x) \)?
3. How do you find the instantaneous rate of change of \( f(x) \) at \( x = a \)?
4. How would you evaluate these expressions if the function were different?
5. How does this process help in finding limits?

**Tip:** The expression \( \frac{f(a + h) - f(a)}{h} \) is crucial in calculus as it represents the difference quotient, which is used to find the derivative of a function.

Let f(x) = 7 - 2x + 13x^2. Calculate the following values: f(a), f(a + h), and (f(a + h) - f(a)) / h for h ≠ 0.

Learn how to calculate f(a), f(a + h), and the difference quotient (f(a + h) - f(a)) / h for the function f(x) = 7 - 2x + 13x^2. This problem covers key calculus concepts such as limits and the difference quotient. Suitable for students in Grades 11-12.

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