Answer
$m=\frac{1}{6}$,
$y-3= \frac{1}{6}(x-8)$
Work Step by Step
$f(x)=\sqrt {x+1}$
$f'(x)=\frac{d}{dx}(\sqrt {x+1})=\frac{1}{2\sqrt {x+1}}$
$m=f'(8)=\frac{1}{2\sqrt {8+1}}=\frac{1}{6}$
Equation of a straight line passing through a given point $(x_{0},y_{0})$ having finite slope m is given by
$y-y_{0}=m(x-x_{0})$
Therefore, equation of the tangent at (8,3) to the curve $f(x)= \sqrt {x+1}$ is given by
$y-3= \frac{1}{6}(x-8)$