Answer
$$\left( { - 1,6} \right){\text{ and }}\left( {4, - 59} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2{x^3} - 9{x^2} - 12x + 5 \cr
& {\text{find the derivative of the function}} \cr
& f'\left( x \right) = {D_x}\left( {2{x^3} - 9{x^2} - 12x + 5} \right) \cr
& {\text{use the power rule for derivatives}} \cr
& f'\left( x \right) = 2\left( {3{x^2}} \right) - 9\left( {2x} \right) - 12\left( 1 \right) + 0 \cr
& f'\left( x \right) = 6{x^2} - 18x - 12 \cr
& \cr
& {\text{Find the points of }}x{\text{ where the slope of the tangent line is }}12,{\text{ set }}f'\left( x \right) = 12 \cr
& {\text{Then}} \cr
& 6{x^2} - 18x - 12 = 12 \cr
& 6{x^2} - 18x - 24 = 0 \cr
& {\text{factoring}} \cr
& 6\left( {{x^2} - 3x - 4} \right) = 0 \cr
& 6\left( {x - 4} \right)\left( {x + 1} \right) = 0 \cr
& {\text{solving we obtain}} \cr
& x = - 1{\text{ and }}x = 4 \cr
& \cr
& {\text{Find the point on the graph where the slope is }}12,{\text{ evaluate }}f\left( x \right){\text{ at }}x = - 1,4 \cr
& f\left( { - 1} \right) = 2{\left( { - 1} \right)^3} - 9{\left( { - 1} \right)^2} - 12\left( { - 1} \right) + 5 = 6 \cr
& {\text{The first point is }}\left( { - 2,6} \right) \cr
& and \cr
& f\left( 4 \right) = 2{\left( 4 \right)^3} - 9{\left( 4 \right)^2} - 12\left( 4 \right) + 5 = - 59 \cr
& {\text{The second point is }}\left( { - 4, - 6} \right) \cr
& {\text{then}}{\text{, the points are }}\left( { - 1,6} \right){\text{ and }}\left( {4, - 59} \right) \cr} $$
