## Calculus with Applications (10th Edition)

$$\left( { - 2,20} \right){\text{ and }}\left( { - 4, - 6} \right)$$
\eqalign{ & f\left( x \right) = {x^3} + 9{x^2} + 19x - 10 \cr & {\text{find the derivative of the function}} \cr & f'\left( x \right) = {D_x}\left( {{x^3} + 9{x^2} + 19x - 10} \right) \cr & {\text{use the power rule for derivatives}} \cr & f'\left( x \right) = 3{x^2} + 9\left( {2x} \right) + 19\left( 1 \right) - 0 \cr & f'\left( x \right) = 3{x^2} + 18x + 19 \cr & \cr & {\text{Find the points of }}x{\text{ where the slope of the tangent line is }} - {\text{5}}{\text{, set }}f'\left( x \right) = - 5 \cr & {\text{Then}} \cr & - 5 = 3{x^2} + 18x + 19 \cr & 3{x^2} + 18x + 19 = - 5 \cr & 3{x^2} + 18x + 24 = 0 \cr & {\text{factoring}} \cr & 3\left( {{x^2} + 6x + 8} \right) = 0 \cr & 3\left( {x + 4} \right)\left( {x + 2} \right) = 0 \cr & {\text{solving we obtain}} \cr & x = - 4{\text{ and }}x = - 2 \cr & \cr & {\text{Find the point on the graph where the slope is }} - {\text{5}}{\text{, evaluate }}f\left( x \right){\text{ at }}x = - 2, - 4 \cr & f\left( { - 2} \right) = {\left( { - 2} \right)^3} + 9{\left( { - 2} \right)^2} + 19\left( { - 2} \right) - 10 = - 20 \cr & {\text{The first point is }}\left( { - 2,20} \right) \cr & and \cr & f\left( { - 4} \right) = {\left( { - 4} \right)^3} + 9{\left( { - 4} \right)^2} + 19\left( { - 4} \right) - 10 = - 6 \cr & {\text{The second point is }}\left( { - 4, - 6} \right) \cr & {\text{then}}{\text{, the points are }}\left( { - 2,20} \right){\text{ and }}\left( { - 4, - 6} \right) \cr}