Answer
The slope of the tangent line is $3$.
Work Step by Step
Let $m$ be the slope of the equation.
The point-slope form can be written as:
$y-y_1=m(x-x_1)$
This gives: $y-8=3(x-1)y \\ y=3x+5$
Consider the tangent line that contains the point $(x_1,y_1)$. The slope the tangent line to the graph of $f(x)$ at $(x_1,y_1)$ can be written as $m_{tan} =\lim\limits_{x \to x_1} \dfrac{f(x)-f(x_1)}{x-x_1}$
For a given point $(1,8)$, we have:
$m =\lim\limits_{x \to 1} \dfrac{(3x+5)-(3+5)}{x-1} \\=\lim\limits_{x \to 1} \dfrac{3(x-1)}{x-1} \\ =3 $
