Answer
$$y = - \frac{4}{5}x - \frac{{13}}{5}$$
Work Step by Step
$$\eqalign{
& y = - \sqrt {8x + 1} ,\,\,\,\,\,x = 3 \cr
& {\text{find the derivative of }}y \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ { - \sqrt {8x + 1} } \right] \cr
& \frac{{dy}}{{dx}} = - \frac{d}{{dx}}\left[ {{{\left( {8x + 1} \right)}^{1/2}}} \right] \cr
& {\text{by using the chain rule}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{2}{\left( {8x + 1} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ {8x + 1} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{1}{2}{\left( {8x + 1} \right)^{ - 1/2}}\left( 8 \right) \cr
& \frac{{dy}}{{dx}} = - \frac{4}{{{{\left( {8x + 1} \right)}^{1/2}}}} \cr
& \frac{{dy}}{{dx}} = - \frac{4}{{\sqrt {8x + 1} }} \cr
& {\text{Find the slope of the tangent line at }}x = 3 \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 3}} \cr
& m = - \frac{4}{{\sqrt {8\left( 3 \right) + 1} }} \cr
& m = - \frac{4}{5} \cr
& \cr
& {\text{Evaluate the function at }}x = 3 \cr
& y\left( 3 \right) = - \sqrt {8\left( 3 \right) + 1} \cr
& y\left( 3 \right) = - 5 \cr
& {\text{we know the point }}\left( {3, - 5} \right){\text{ and the slope }}m = - \frac{8}{5} \cr
& {\text{find the equation of the tangent line using the point - slope form of a line}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - \left( { - 5} \right) = - \frac{4}{5}\left( {x - 3} \right) \cr
& {\text{simplifying}} \cr
& y + 5 = - \frac{4}{5}x + \frac{{12}}{5} \cr
& y = - \frac{4}{5}x + \frac{{12}}{5} - 5 \cr
& y = - \frac{4}{5}x - \frac{{13}}{5} \cr} $$
