Answer
$y=\displaystyle \frac{1}{4}x+\frac{1}{2}$
Work Step by Step
For x=$0$, the point on the graph is $(0, f(0))$.
The slope of the tangent at x=$0$ is $m=f^{\prime}(0).$
The point-slope equation of the tangent line is
$y-y_{1}=m(x-x_{1})$
$y-f(0)=f^{\prime}(0)(x-0)$
$f(0)=\displaystyle \frac{0+1}{0+2}=\frac{1}{2}$
For $f^{\prime}(x)$, use the quotient rule,
$\displaystyle \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$= \displaystyle \frac{(x+1)^{\prime}(x+2)-(x+1)(x+2)^{\prime}}{(x+2)^{2}}$
$f^{\prime}(x)= \displaystyle \frac{(1)(x+2)-(x+1)(1)}{(x+2)^{2}}=\frac{1}{(x+2)^{2}}$
... evaluate at x=$0$ ...
$f^{\prime}(0)=\displaystyle \frac{1}{4}$
An equation for the tangent line is
$y-\displaystyle \frac{1}{2}=\frac{1}{4}(x-0)$
$y=\displaystyle \frac{1}{4}x+\frac{1}{2}$
