Answer
Graph
Work Step by Step
$$\eqalign{
& y = - \frac{1}{3}\left( {{x^3} - 3x + 2} \right) \cr
& {\text{Domain: }}\left( { - \infty ,\infty } \right) \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& y = - \frac{1}{3}\left( {{{\left( 0 \right)}^3} - 3\left( 0 \right) + 2} \right) \cr
& y = - \frac{2}{3} \cr
& y{\text{ - intercept }}\left( {0, - \frac{2}{3}} \right) \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& 0 = - \frac{1}{3}\left( {{x^3} - 3x + 2} \right) \cr
& {\text{Solving by a calculator}} \cr
& x = - 2,{\text{ }}x = 1 \cr
& x{\text{ - intercepts }}\left( { - 2,0} \right),\left( {1,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ { - \frac{1}{3}\left( {{x^3} - 3x + 2} \right)} \right] \cr
& y' = \frac{d}{{dx}}\left[ { - \frac{1}{3}{x^3} + x - \frac{2}{3}} \right] \cr
& y' = - {x^2} + 1 \cr
& {\text{Let }}y' = 0{\text{ to find critical points}} \cr
& - {x^2} + 1 = 0 \cr
& x = \pm 1 \cr
& *{\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ { - {x^2} + 1} \right] \cr
& y'' = - 2x \cr
& {\text{Evaluate }}y''\left( x \right){\text{ at }}x = \pm 1 \cr
& y''\left( { - 1} \right) = 2 > 0,{\text{ relative minimum at }}x = - 1 \cr
& f\left( { - 1} \right) = - \frac{4}{3} \to {\text{Relative minimum at }}\left( { - 1, - \frac{4}{3}} \right) \cr
& y''\left( 1 \right) = - 2 < 0,{\text{ relative maximum at }}x = 1 \cr
& f\left( 1 \right) = 0 \to {\text{Relative maximum at }}\left( {1,0} \right) \cr
& \cr
& {\text{Let }}y''\left( x \right) = 0 \cr
& - 2x = 0 \cr
& x = 0 \cr
& x = \pm \sqrt {\frac{{27}}{2}} \left( {{\text{These vales are not in the domain}}} \right),{\text{ }}x = 0 \cr
& x = 0 \cr
& {\text{Inflection point }}\left( {0,f\left( 0 \right)} \right) \cr
& y = - \frac{1}{3}\left( {{{\left( 0 \right)}^3} - 3\left( 0 \right) + 2} \right) \cr
& y = - \frac{2}{3} \cr
& {\text{Inflection point}} \to \left( {0, - \frac{2}{3}} \right) \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& {\text{No vertical asymptotes, the denominator is 1}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {2 - x - {x^3}} \right) = - \infty \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {2 - x - {x^3}} \right) = + \infty \cr
& {\text{No horizontal asymptotes}} \cr
& \cr
& {\text{Graph}} \cr} $$
