Answer
$f'(x) =\frac{4}{(1-x)^{2}}$
$slope = f'(-2) = \frac{4}{9}$
Equation of tangent line:
$y = \frac{4}{9}(x+2)+\frac{1}{3}$
Work Step by Step
$f(x) = \frac{x+3}{1-x}$
$f(-2) = \frac{1}{3}$
Now we know the point on the graph is: $(-2,\frac{1}{3})$
$f'(x) = \lim\limits_{h \to 0}\frac{\frac{(x+h)+3}{1-(x+h)}-\frac{x+3}{1-x}}{h}$
$f'(x) = \lim\limits_{h \to 0}\frac{4h}{h(1-x)(1-x-h)}$
$f'(x) =\frac{4}{(1-x)^{2}}$
Slope at x = -2:
$slope = f'(-2) = \frac{4}{9}$
Equation of tangent line:
$\frac{y- \frac{1}{3}}{x +2} = \frac{4}{9}$
$ y = \frac{4}{9}(x+2)+\frac{1}{3}$
