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Let's break this down and solve each part step by step. The function given is:

\[
f(x) = 6x^2 + 5x
\]

### 1. \( f(a) \)
We need to substitute \( a \) into the function \( f(x) \):

\[
f(a) = 6a^2 + 5a
\]

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

\[
f(a + h) = 6(a + h)^2 + 5(a + h)
\]

Now expand each term:

\[
f(a + h) = 6(a^2 + 2ah + h^2) + 5(a + h)
\]
\[
f(a + h) = 6a^2 + 12ah + 6h^2 + 5a + 5h
\]

### 3. \( \frac{f(a + h) - f(a)}{h} \)
We now calculate \( \frac{f(a + h) - f(a)}{h} \):

\[
\frac{f(a + h) - f(a)}{h} = \frac{(6a^2 + 12ah + 6h^2 + 5a + 5h) - (6a^2 + 5a)}{h}
\]

Simplify the expression inside the numerator:

\[
= \frac{6a^2 + 12ah + 6h^2 + 5a + 5h - 6a^2 - 5a}{h}
\]
\[
= \frac{12ah + 6h^2 + 5h}{h}
\]

Now factor out \( h \) in the numerator:

\[
= \frac{h(12a + 6h + 5)}{h}
\]

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

\[
= 12a + 6h + 5
\]

### Final answers:
1. \( f(a) = 6a^2 + 5a \)
2. \( f(a + h) = 6a^2 + 12ah + 6h^2 + 5a + 5h \)
3. \( \frac{f(a + h) - f(a)}{h} = 12a + 6h + 5 \)

Let me know if you'd like more details or have any other questions!

Here are some related questions for practice:
1. What is \( f'(x) \) (the derivative of \( f(x) = 6x^2 + 5x \))?
2. What happens to \( \frac{f(a+h)-f(a)}{h} \) as \( h \to 0 \)?
3. How would you interpret the expression \( f(a+h)-f(a) \) geometrically?
4. How would \( f(x) = 6x^2 + 5x \) change if the coefficient of \( x^2 \) were different?
5. Can you compute \( f'(x) \) for the general function \( f(x) = ax^2 + bx \)?

**Tip**: Always expand and simplify carefully to avoid mistakes when calculating limits or differences in functions!

Given the function f(x) = 6x^2 + 5x, calculate the following values: f(a), f(a + h), and (f(a + h) - f(a)) / h.

Learn how to evaluate the function f(x) = 6x^2 + 5x for values f(a), f(a + h), and the difference quotient (f(a + h) - f(a)) / h. This involves expanding binomials and simplifying the expression. Suitable for students in Grades 10-12.

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