## Calculus with Applications (10th Edition)

$$y = - \frac{3}{4}x - \frac{9}{4}$$
\eqalign{ & y = \frac{3}{{x - 1}},\,\,\,\,\,x = - 1 \cr & \cr & {\text{find the derivative of }}y \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{3}{{x - 1}}} \right] \cr & \frac{{dy}}{{dx}} = - \frac{3}{{{{\left( {x - 1} \right)}^2}}} \cr & {\text{Find the slope of the tangent line at }}x = - 1 \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = - 1}} \cr & m = - \frac{3}{{{{\left( { - 1 - 1} \right)}^2}}} \cr & m = - \frac{3}{4} \cr & \cr & {\text{Evaluate the function at }}x = - 1 \cr & y\left( { - 1} \right) = \frac{3}{{ - 1 - 1}} \cr & y\left( { - 1} \right) = - \frac{3}{2}{\text{ }} \cr & {\text{we know the point }}\left( { - 1, - \frac{3}{2}} \right){\text{ and the slope }}m = - \frac{3}{4} \cr & {\text{find the equation of the tangent line using the point - slope form of a line}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \left( { - \frac{3}{2}} \right) = - \frac{3}{4}\left( {x + 1} \right) \cr & {\text{simplifying}} \cr & y + \frac{3}{2} = - \frac{3}{4}x - \frac{3}{4} \cr & y + \frac{3}{2} = - \frac{3}{4}x - \frac{3}{4} - \frac{3}{2} \cr & y = - \frac{3}{4}x - \frac{9}{4} \cr}