Answer
$m=-\dfrac{2}{3}$
Work Step by Step
$f(x)=\dfrac{6}{x+1},$ at $(2,2)$
The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$
In this case, $a=2$
Find $f(a+h)$ by substituting $x$ by $2+h$ in $f(x)$ and simplifying:
$f(2+h)=\dfrac{6}{2+h+1}=\dfrac{6}{3+h}$
Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=2$
Substitute the known values into the formula that gives the slope of the tangent line and evaluate:
$m=\lim_{h\to0}\dfrac{\dfrac{6}{3+h}-2}{h}=\lim_{h\to0}\dfrac{\dfrac{6-2(3+h)}{3+h}}{h}=...$
$...=\lim_{h\to0}\dfrac{\dfrac{6-6-2h}{3+h}}{h}=\lim_{h\to0}\dfrac{\dfrac{-2h}{3+h}}{h}=...$
$...=\lim_{h\to0}\dfrac{-2h}{h(3+h)}=\lim_{h\to0}\dfrac{-2}{3+h}=\dfrac{-2}{3+0}=-\dfrac{2}{3}$
