## Calculus with Applications (10th Edition)

$$\left( { - \frac{1}{2}, - \frac{{19}}{2}} \right)$$
\eqalign{ & f\left( x \right) = 6{x^2} + 4x - 9 \cr & {\text{find the derivative of the function}} \cr & f'\left( x \right) = {D_x}\left( {6{x^2} + 4x - 9} \right) \cr & {\text{use the power rule for derivatives}} \cr & f'\left( x \right) = 6\left( {2x} \right) + 4\left( 1 \right) + 0 \cr & f'\left( x \right) = 12x + 4 \cr & \cr & {\text{Find the points of }}x{\text{ where the slope of the tangent line is }} - {\text{2}}{\text{, set }}f'\left( x \right) = - 2 \cr & {\text{Then}} \cr & 12x + 4 = - 2 \cr & 12x = - 6 \cr & x = - \frac{1}{2} \cr & \cr & {\text{Find the point on the graph where the slope is }} - 2,{\text{ evaluate }}f\left( x \right){\text{ at }}x = - \frac{1}{2} \cr & f\left( { - \frac{1}{2}} \right) = 6{\left( { - \frac{1}{2}} \right)^2} + 4\left( { - \frac{1}{2}} \right) - 9 = - \frac{{19}}{2} \cr & {\text{then}}{\text{, the points is }}\left( { - \frac{1}{2}, - \frac{{19}}{2}} \right) \cr}