Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& x{\text{ - intercepts: }}\left( { - 3,0} \right){\text{ and }}\left( {0,0} \right) \cr
& {\text{Relative maximum at }}\left( { - 3,0} \right) \cr
& {\text{Relative minimum at }}\left( { - 1, - \root 3 \of 4 } \right) \cr
& {\text{Inflection points: }}\left( {0,0} \right) \cr
& {\text{No vertical asymptotes}} \cr
& {\text{No horizontal asymptotes}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{1/3}}{\left( {x + 3} \right)^{2/3}} \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& f\left( 0 \right) = {\left( 0 \right)^{1/3}}{\left( {0 + 3} \right)^{2/3}} \cr
& f\left( 0 \right) = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& {x^{1/3}}{\left( {x + 3} \right)^{2/3}} = 0 \cr
& x = 0,{\text{ }}x = - 3 \cr
& x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( { - 3,0} \right) \cr
& \cr
& {\text{*Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{x^{1/3}}{{\left( {x + 3} \right)}^{2/3}}} \right] \cr
& y' = \frac{2}{3}{x^{1/3}}{\left( {x + 3} \right)^{ - 1/3}} + \frac{1}{3}{x^{ - 2/3}}{\left( {x + 3} \right)^{2/3}} \cr
& {\text{Let }}y' = 0{\text{ to find critical points}} \cr
& \frac{2}{3}{x^{1/3}}{\left( {x + 3} \right)^{ - 1/3}} + \frac{1}{3}{x^{ - 2/3}}{\left( {x + 3} \right)^{2/3}} = 0 \cr
& \frac{1}{3}{x^{ - 2/3}}{\left( {x + 3} \right)^{ - 1/3}}\left[ {2x + \left( {x + 3} \right)} \right] = 0 \cr
& \frac{1}{3}{x^{ - 2/3}}{\left( {x + 3} \right)^{ - 1/3}}\left( {3x + 3} \right) = 0 \cr
& {x^{ - 2/3}}{\left( {x + 3} \right)^{ - 1/3}}\left( {x + 1} \right) = 0 \cr
& {\text{The derivative is not defined at }}x = - 3{\text{ and }}x = 0 \cr
& {\text{The derivative is }}0{\text{ at }}x = - 1 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ {{x^{ - 2/3}}{{\left( {x + 3} \right)}^{ - 1/3}}\left( {x + 1} \right)} \right] \cr
& y'' = \frac{d}{{dx}}\left[ {{{\left( {x + 3} \right)}^{ - 1/3}}\left( {{x^{1/3}} + {x^{ - 2/3}}} \right)} \right] \cr
& {\text{Using a CAS}} \cr
& y'' = - \frac{2}{{{x^{5/3}}{{\left( {x + 3} \right)}^{4/3}}}} \cr
& {\text{*Evaluate }}y''{\text{ at the critical point }}x = - 1 \cr
& {\text{ }}y''\left( { - 1} \right) > 0,{\text{ there is a relative minimum at }}f\left( { - 1} \right) \cr
& {\text{ }}f\left( { - 1} \right) = {\left( { - 1} \right)^{1/3}}{\left( { - 1 + 3} \right)^{2/3}} = - \root 3 \of 4 \cr
& {\text{Relative minimum at }}\left( { - 1, - \root 3 \of 4 } \right) \cr
& {\text{Find the extrema at }}x = - 3{\text{ and }}x = 0{\text{ using the first}} \cr
& {\text{derivative test, then}} \cr
& y'\left( { - 4} \right) > 0{\text{ and }}y'\left( { - 2} \right) < 0 \cr
& {\text{The derivative changes + to }} - {\text{ at }}x = - 3,{\text{ then there is}} \cr
& {\text{a relative maximum at }}x = - 3 \cr
& {\text{ }}f\left( { - 3} \right) = 0 \to {\text{relative maximum at }}\left( { - 3,0} \right) \cr
& y'\left( { - 0.5} \right) > 0{\text{ and }}y'\left( 1 \right) > 0 \cr
& {\text{The derivative does not change at }}x = 0,{\text{ no extrema at }}x = 0 \cr
& \cr
& {\text{Let }}y''\left( x \right) = 0 \cr
& - \frac{2}{{{x^{5/3}}{{\left( {x + 3} \right)}^{4/3}}}} = 0 \cr
& {\text{Undefined at }}x = 0{\text{ and }}x = - 3 \cr
& {\text{By the first derivative test, there is a point of inflection at}} \cr
& x = 0 \cr
& f\left( 0 \right) = 0 \cr
& {\text{The inflection:}} \cr
& \left( {0,0} \right) \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& {\text{No vertical asymptotes, the denominator is 1}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \infty } {x^{1/3}}{\left( {x + 3} \right)^{2/3}} = \infty \cr
& \mathop {\lim }\limits_{x \to - \infty } {x^{1/3}}{\left( {x + 3} \right)^{2/3}} = - \infty \cr
& {\text{No horizontal asymptotes}} \cr
& \cr
} $$
