Answer
Graph
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2}}}{{{x^2} + 3}} \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& y = \frac{{{0^2}}}{{{0^2} + 3}} \cr
& y = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& 0 = \frac{{{x^2}}}{{{x^2} + 3}} \cr
& x = 0 \cr
& x{\text{ - intercept }}\left( {0,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{{x^2} + 3}}} \right] \cr
& y' = \frac{{\left( {{x^2} + 3} \right)\left( {2x} \right) - {x^2}\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& y' = \frac{{2{x^3} + 6x - 2{x^3}}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& y' = \frac{{6x}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{Let }}y' = 0 \cr
& \frac{{6x}}{{{{\left( {{x^2} + 3} \right)}^2}}} = 0 \cr
& x = 0 \cr
& \cr
& {\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ {\frac{{6x}}{{{{\left( {{x^2} + 3} \right)}^2}}}} \right] \cr
& y'' = \frac{{{{\left( {{x^2} + 3} \right)}^2}\left( 6 \right) - 6x\left( 2 \right)\left( {{x^2} + 3} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} \cr
& y'' = \frac{{\left( {{x^2} + 3} \right)\left( 6 \right) - 6x\left( 2 \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr
& y'' = \frac{{6{x^2} + 18 - 24{x^2}}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr
& y'' = \frac{{18 - 18{x^2}}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr
& {\text{Evaluate }}y''\left( 0 \right) \cr
& y''\left( 0 \right) = \frac{{18 - 18{{\left( 0 \right)}^2}}}{{{{\left( {{{\left( 0 \right)}^2} + 3} \right)}^3}}} = \frac{2}{3} > 0 \cr
& {\text{Relative minimum at }}\left( {0,f\left( 0 \right)} \right) \to \left( {0,0} \right) \cr
& \cr
& *{\text{Find the Inflection points}} \cr
& y'' = \frac{{18 - 18{x^2}}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr
& \frac{{18 - 18{x^2}}}{{{{\left( {{x^2} + 3} \right)}^3}}} = 0 \cr
& 18 - 18{x^2} = 0 \cr
& x = \pm 1 \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& y = \frac{{{x^2}}}{{{x^2} + 3}} \cr
& {x^2} + 3 = 0,{\text{ no real solutions, then}} \cr
& {\text{No vertical asymptotes }} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2}}}{{{x^2} + 3}}} \right) = 1 \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^2}}}{{{x^2} + 3}}} \right) = 1 \cr
& {\text{Horizontal asymptote }}y = 1 \cr
& \cr
& {\text{Graph}} \cr} $$
