Question

find and simplify the difference quotient. ( f(x)=x^{2}+8x + 7)
(2x + h+8)
(1)
(\frac{x^{2}+x + 2xh+h^{2}+h + 14}{h})
(2x+h + 7)

Explanation:

Step1: Recall difference - quotient formula

The difference - quotient formula for a function $y = f(x)$ is $\frac{f(x + h)-f(x)}{h}$, where $h
eq0$. Given $f(x)=x^{2}+8x + 7$, first find $f(x + h)$.

Step2: Calculate $f(x + h)$

$f(x + h)=(x + h)^{2}+8(x + h)+7$. Expand $(x + h)^{2}$ using the formula $(a + b)^{2}=a^{2}+2ab + b^{2}$, so $(x + h)^{2}=x^{2}+2xh+h^{2}$. Then $f(x + h)=x^{2}+2xh + h^{2}+8x+8h + 7$.

Step3: Calculate $f(x + h)-f(x)$

$f(x + h)-f(x)=(x^{2}+2xh + h^{2}+8x+8h + 7)-(x^{2}+8x + 7)$.
\[

$$\begin{align*} f(x + h)-f(x)&=x^{2}+2xh + h^{2}+8x+8h + 7 - x^{2}-8x - 7\\ &=2xh+h^{2}+8h \end{align*}$$

\]

Step4: Calculate the difference - quotient

$\frac{f(x + h)-f(x)}{h}=\frac{2xh+h^{2}+8h}{h}$. Factor out $h$ from the numerator: $\frac{h(2x + h+8)}{h}$. Since $h
eq0$, cancel out the $h$ terms, and we get $2x + h+8$.

Answer:

$2x + h+8$