Answer
Graph
Work Step by Step
$$\eqalign{
& y = x + \frac{{32}}{{{x^2}}} \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& y = 0 + \frac{{32}}{{{0^2}}} \cr
& {\text{No }}y{\text{ - intercept}}{\text{.}} \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& 0 = x + \frac{{32}}{{{x^2}}} \cr
& - \frac{{32}}{{{x^2}}} = x \cr
& {x^3} = - 32 \cr
& x = - \root 3 \of {32} \cr
& x{\text{ - intercept }}\left( { - \root 3 \of {32} ,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {x + \frac{{32}}{{{x^2}}}} \right] \cr
& y' = 1 + 32\left( { - 2{x^{ - 3}}} \right) \cr
& y' = 1 - \frac{{64}}{{{x^3}}} \cr
& {\text{Let }}y' = 0 \cr
& 1 - \frac{{64}}{{{x^3}}} = 0 \cr
& {x^3} = 64 \cr
& x = 4 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ {1 - \frac{{64}}{{{x^3}}}} \right] \cr
& y'' = 0 - 64\left( { - 3{x^{ - 4}}} \right) \cr
& y'' = \frac{{192}}{{{x^4}}} \cr
& \frac{{192}}{{{x^4}}} = 0 \cr
& {\text{Evaluate }}y''\left( 4 \right) \cr
& y''\left( 4 \right) = \frac{{192}}{{{{\left( 4 \right)}^4}}} > 0 \cr
& {\text{Relative minimum at }}\left( {4,f\left( 4 \right)} \right) \to \left( {4,6} \right) \cr
& \frac{{192}}{{{x^4}}} = 0 \cr
& {\text{There are no values at which }}y'' = 0. \cr
& {\text{No inflection points}}{\text{.}} \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& x + \frac{{32}}{{{x^2}}} \cr
& x = 0 \cr
& {\text{Vertical asymptotes at }}x = 0 \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {x + \frac{{32}}{{{x^2}}}} \right) = \infty \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {x + \frac{{32}}{{{x^2}}}} \right) = - \infty \cr
& {\text{No horizontal asymptotes}} \cr
& \cr
& {\text{Graph}} \cr} $$
