Question

find the difference quotient $\frac{f(x + h)-f(x)}{h}$, where $h
eq0$, for the function below.

$f(x)=-x^{2}-2x + 5$

simplify your answer as much as possible.

$\frac{f(x + h)-f(x)}{h}=$

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$:
\[

$$\begin{align*} f(x + h)&=-(x + h)^2-2(x + h)+5\\ &=-(x^{2}+2xh+h^{2})-2x-2h + 5\\ &=-x^{2}-2xh - h^{2}-2x-2h + 5 \end{align*}$$

\]

Step2: Calculate $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=(-x^{2}-2xh - h^{2}-2x-2h + 5)-(-x^{2}-2x + 5)\\ &=-x^{2}-2xh - h^{2}-2x-2h + 5 + x^{2}+2x - 5\\ &=-2xh - h^{2}-2h \end{align*}$$

\]

Step3: Find the difference quotient $\frac{f(x + h)-f(x)}{h}$

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{-2xh - h^{2}-2h}{h}\\ &=\frac{h(-2x - h - 2)}{h}\\ &=-2x - h - 2 \end{align*}$$

\]

Answer:

$-2x - h - 2$