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The problem asks you to evaluate two expressions for the given function \( f(x) = -4x^2 - x - 1 \).

### First Expression: \( f(x + h) \)

To find \( f(x + h) \), substitute \( x + h \) into the function for \( x \):

\[
f(x + h) = -4(x + h)^2 - (x + h) - 1
\]

Now, expand and simplify:

1. Expand \( (x + h)^2 \):
\[
(x + h)^2 = x^2 + 2xh + h^2
\]

2. Substitute this back into the function:
\[
f(x + h) = -4(x^2 + 2xh + h^2) - (x + h) - 1
\]

3. Distribute the constants:
\[
f(x + h) = -4x^2 - 8xh - 4h^2 - x - h - 1
\]

Thus, the simplified form of \( f(x + h) \) is:
\[
f(x + h) = -4x^2 - 8xh - 4h^2 - x - h - 1
\]

---

### Second Expression: \( \frac{f(x + h) - f(x)}{h} \)

Now, use the expressions for \( f(x + h) \) and \( f(x) \) to find this difference quotient:

\[
\frac{f(x + h) - f(x)}{h} = \frac{(-4x^2 - 8xh - 4h^2 - x - h - 1) - (-4x^2 - x - 1)}{h}
\]

Simplify the numerator:
\[
= (-4x^2 - 8xh - 4h^2 - x - h - 1) + 4x^2 + x + 1
\]

Cancel the like terms:
\[
= (-8xh - 4h^2 - h)
\]

So, the numerator simplifies to:
\[
-8xh - 4h^2 - h
\]

Now divide by \( h \):
\[
\frac{-8xh - 4h^2 - h}{h} = -8x - 4h - 1
\]

Thus, the simplified form of the difference quotient is:
\[
\frac{f(x + h) - f(x)}{h} = -8x - 4h - 1
\]

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Would you like more details on any part of the solution or clarification on a step? Here are some related questions you might find useful:

1. What is the geometric meaning of the difference quotient?
2. How does this function represent a parabola?
3. How would the result change if the function had a different degree?
4. Can you apply this method to a cubic function?
5. What happens to the difference quotient as \( h \) approaches 0?

**Tip:** The difference quotient is the foundation for finding the derivative of a function in calculus!

For the function f(x) = -4x^2 - x - 1, evaluate and fully simplify each of the following: f(x + h), (f(x + h) - f(x)) / h

Learn how to evaluate and simplify the function f(x + h) and the difference quotient (f(x + h) - f(x)) / h for the quadratic function f(x) = -4x^2 - x - 1. This problem involves polynomial expansion and difference quotient concepts, suitable for students in Grades 10-12.

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