## University Calculus: Early Transcendentals (3rd Edition)

$f'(x) =\frac{4}{(1-x)^{2}}$ $slope = f'(-2) = \frac{4}{9}$ Equation of tangent line: $y = \frac{4}{9}(x+2)+\frac{1}{3}$
$f(x) = \frac{x+3}{1-x}$ $f(-2) = \frac{1}{3}$ 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: $\frac{y- \frac{1}{3}}{x +2} = \frac{4}{9}$ $y = \frac{4}{9}(x+2)+\frac{1}{3}$