given the function f(x)=x^2 + 2x, determine the slope of the tangent line of f at x = 8 using the limit…

Explanation:

Step1: Recall the limit - definition of the derivative

The slope of the tangent line of a function $y = f(x)$ at $x=a$ is given by $m=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a = 8$ and $f(x)=x^{2}+2x$.

Step2: Find $f(8 + h)$ and $f(8)$

First, find $f(8 + h)$: [ \begin{align*} f(8 + h)&=(8 + h)^{2}+2(8 + h)\ &=64+16h+h^{2}+16 + 2h\ &=h^{2}+18h + 80 \end{align*} ] Next, find $f(8)$: [ f(8)=8^{2}+2\times8=64 + 16=80 ]

Step3: Substitute into the limit - formula

[ \begin{align*} \lim_{h\rightarrow0}\frac{f(8 + h)-f(8)}{h}&=\lim_{h\rightarrow0}\frac{(h^{2}+18h + 80)-80}{h}\ &=\lim_{h\rightarrow0}\frac{h^{2}+18h}{h} \end{align*} ]

Answer:

$\lim_{h\rightarrow0}\frac{(8 + h)^{2}+2(8 + h)-(8^{2}+2\times8)}{h}=\lim_{h\rightarrow0}\frac{h^{2}+18h}{h}$