the Difference Quotient for f(x) = -x^2-x+5

To find the **Difference Quotient** for the function \( f(x) = -x^2 - x + 5 \), we use the following formula:

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

### Step 1: Find \( f(x+h) \)
Substitute \( x+h \) into \( f(x) \):

\[
f(x+h) = -(x+h)^2 - (x+h) + 5
\]
Expanding this expression:
\[
f(x+h) = -(x^2 + 2xh + h^2) - (x + h) + 5
\]
\[
f(x+h) = -x^2 - 2xh - h^2 - x - h + 5
\]

### Step 2: Difference Quotient
Now, substitute \( f(x+h) \) and \( f(x) \) into the Difference Quotient formula:
\[
\frac{f(x+h) - f(x)}{h} = \frac{(-x^2 - 2xh - h^2 - x - h + 5) - (-x^2 - x + 5)}{h}
\]
Simplify the numerator:
\[
(-x^2 - 2xh - h^2 - x - h + 5) - (-x^2 - x + 5)
\]
This simplifies to:
\[
-2xh - h^2 - h
\]

### Step 3: Simplify the Difference Quotient
Now, divide the simplified expression by \( h \):
\[
\frac{-2xh - h^2 - h}{h} = -2x - h - 1
\]

Thus, the **Difference Quotient** for \( f(x) = -x^2 - x + 5 \) is:

\[
\boxed{-2x - h - 1}
\]

Would you like any further details on this process or any clarifications?

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Here are 5 related questions you might find helpful:

1. How is the Difference Quotient used to find the derivative of a function?
2. What is the Difference Quotient for a linear function?
3. Can the Difference Quotient be applied to trigonometric functions?
4. How does the Difference Quotient change for functions with higher powers of \( x \)?
5. How can we interpret the result of the Difference Quotient geometrically?

**Tip:** The Difference Quotient formula is the foundation of the derivative, which gives the slope of the tangent line at any point on a curve.

The query submitted by the user: the Difference Quotient for f(x) = -x^2 - x + 5

Explore the step-by-step solution for calculating the Difference Quotient of the quadratic function f(x) = -x^2 - x + 5. This solution explains key algebraic and calculus concepts, including expanding quadratic terms and simplifying the Difference Quotient formula. Suitable for Grades 11-12 pre-calculus and calculus students.

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