Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& x{\text{ - intercepts: }}\left( {0,0} \right),\left( {1.4415,0} \right),\left( { - 2.7748,0} \right) \cr
& {\text{local maximum at }}\left( {0,0} \right) \cr
& {\text{local minimum at }}\left( {1, - 5} \right){\text{ and }}\left( { - 2, - 32} \right) \cr
& {\text{inflection points:}} \cr
& \left( {0.5485, - 2.6792} \right),\left( { - 1.2152, - 18.3565} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 3{x^4} + 4{x^3} - 12{x^2} \cr
& {\text{Domain: }}\left( { - \infty ,\infty } \right) \cr
& \cr
& {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr
& f\left( 0 \right) = 3{\left( 0 \right)^4} + 4{\left( 0 \right)^3} - 12{\left( 0 \right)^2} \cr
& f\left( 0 \right) = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr
& 3{x^4} + 4{x^3} - 12{x^2} = 0 \cr
& {x^2}\left( {3{x^2} + 4x - 12} \right) = 0 \cr
& {x_1} = 0,{\text{ and by the quadratic formula we obtain}} \cr
& {x_2} = \frac{{ - 2 + 2\sqrt {10} }}{3} \approx 1.4415,{\text{ }}{x_3} = \frac{{ - 2 - 2\sqrt {10} }}{3} \approx - 2.7748 \cr
& x{\text{ - intercepts: }}\left( {0,0} \right),\left( {1.4415,0} \right),\left( { - 2.7748,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^4} + 4{x^3} - 12{x^2}} \right] \cr
& f'\left( x \right) = 12{x^3} + 12{x^2} - 24x \cr
& {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr
& 12{x^3} + 12{x^2} - 24x = 0 \cr
& 12x\left( {{x^2} + x - 2} \right) = 0 \cr
& 12x\left( {x + 2} \right)\left( {x - 1} \right) = 0 \cr
& x = - 2,{\text{ }}x = 0,{\text{ }}x = 1 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {12{x^3} + 12{x^2} - 24x} \right] \cr
& f''\left( x \right) = 36{x^2} + 24x - 24 \cr
& \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = - 2,{\text{ }}x = 0,{\text{ }}x = 1 \cr
& *f''\left( { - 2} \right) = 36{\left( { - 2} \right)^2} + 24\left( { - 2} \right) - 24 = 72 > 0,{\text{ then}} \cr
& {\text{There is a local minimum at }}\left( { - 2,f\left( { - 2} \right)} \right) \cr
& f\left( { - 2} \right) = 3{\left( { - 2} \right)^4} + 4{\left( { - 2} \right)^3} - 12{\left( { - 2} \right)^2} = - 32 \cr
& \to {\text{local minimum at }}\left( { - 2, - 32} \right) \cr
& *f''\left( 0 \right) = 36{\left( 0 \right)^2} + 24\left( 0 \right) - 24 = - 24 < 0,{\text{ then}} \cr
& {\text{There is a local maximum at }}\left( {0,f\left( 0 \right)} \right) \cr
& f\left( 0 \right) = 3{\left( 0 \right)^4} + 4{\left( 0 \right)^3} - 12{\left( 0 \right)^2} = 0 \cr
& \to {\text{local maximum at }}\left( {0,0} \right) \cr
& *f''\left( 1 \right) = 36{\left( 1 \right)^2} + 24\left( 1 \right) - 24 = 36 > 0,{\text{ then}} \cr
& {\text{There is a local minimum at }}\left( {1,f\left( 1 \right)} \right) \cr
& f\left( 1 \right) = 3{\left( 1 \right)^4} + 4{\left( 1 \right)^3} - 12{\left( 1 \right)^2} = - 5 \cr
& \to {\text{local minimum at }}\left( {1, - 5} \right) \cr
& \cr
& {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr
& 36{x^2} + 24x - 24 = 0 \cr
& 12\left( {3{x^2} + 2x - 2} \right) = 0 \cr
& 12\left( {3{x^2} + 2x - 2} \right) = 0 \cr
& {\text{By the quadratic formula}} \cr
& {x_1} = \frac{{ - 1 + \sqrt 7 }}{3} \approx 0.5485{\text{ and }}{x_2} = \frac{{ - 1 - \sqrt 7 }}{3} \approx - 1.2152 \cr
& f\left( {0.5485} \right) = 3{\left( {0.5485} \right)^4} + 4{\left( {0.5485} \right)^3} - 12{\left( {0.5485} \right)^2} \cr
& f\left( {0.5485} \right) \approx - 2.6792 \cr
& f\left( { - 1.2152} \right) = 3{\left( { - 1.2152} \right)^4} + 4{\left( { - 1.2152} \right)^3} - 12{\left( { - 1.2152} \right)^2} \cr
& f\left( { - 1.2152} \right) \approx - 18.3565 \cr
& {\text{The inflection points are: }} \cr
& \left( {0.5485, - 2.6792} \right),\left( { - 1.2152, - 18.3565} \right) \cr
& \cr
& {\text{Graph}} \cr} $$
