Answer
$$\left( { - 2, - 24} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} + 6{x^2} + 21x + 2 \cr
& {\text{find the derivative of the function}} \cr
& f'\left( x \right) = {D_x}\left( {{x^3} + 6{x^2} + 21x + 2} \right) \cr
& {\text{use the power rule for derivatives}} \cr
& f'\left( x \right) = 3{x^2} + 6\left( {2x} \right) + 21\left( 1 \right) + 0 \cr
& f'\left( x \right) = 3{x^2} + 12x + 21 \cr
& \cr
& {\text{Find the poins of }}x{\text{ where the slope of the tangent line is 9}}{\text{, set }}f'\left( x \right) = 9 \cr
& {\text{Then}} \cr
& 3{x^2} + 12x + 21 = 9 \cr
& 3{x^2} + 12x + 12 = 0 \cr
& {\text{factoring}} \cr
& 3\left( {{x^2} + 4x + 4} \right) = 0 \cr
& 6{\left( {x + 2} \right)^2} = 0 \cr
& {\text{solving we obtain}} \cr
& x = - 2 \cr
& \cr
& {\text{Find the point on the graph where the slope is }}12,{\text{ evaluate }}f\left( x \right){\text{ at }}x = - 2 \cr
& f\left( x \right) = {\left( { - 2} \right)^3} + 6{\left( { - 2} \right)^2} + 21\left( { - 2} \right) + 2 = - 24 \cr
& {\text{Then}}{\text{, the points is }}\left( { - 2, - 24} \right) \cr} $$
