Answer
Graph
Work Step by Step
$$\eqalign{
& y = \frac{{2{x^2}}}{{{x^2} - 4}} \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& y = \frac{{2{{\left( 0 \right)}^2}}}{{{{\left( 0 \right)}^2} - 4}} \to y = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& 0 = \frac{{2{x^2}}}{{{x^2} - 4}} \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{{2{x^2}}}{{{x^2} - 4}}} \right] \cr
& y' = \frac{{\left( {{x^2} - 4} \right)\left( {4x} \right) - 2{x^2}\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr
& y' = \frac{{4{x^3} - 16x - 4{x^4}}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr
& y' = \frac{{ - 16x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr
& {\text{Let }}y' = 0 \cr
& - 16x = 0 \cr
& x = 0 \cr
& {\text{Evaluate }}y'\left( { - 1} \right){\text{ and }}y'\left( 1 \right) \cr
& y\left( { - 1} \right)' = \frac{{ - 16\left( { - 1} \right)}}{{{{\left( {{{\left( { - 1} \right)}^2} - 4} \right)}^2}}} > 0 \cr
& y\left( 1 \right)' = \frac{{ - 16\left( 1 \right)}}{{{{\left( {{{\left( 1 \right)}^2} - 4} \right)}^2}}} < 0 \cr
& y\left( { - 1} \right)' > 0,{\text{ }}y\left( 0 \right)' = 0,{\text{ }}y\left( 1 \right)' < 0 \cr
& {\text{Then, there is a relative maximum at }}x = 0 \cr
& y\left( 0 \right) = \frac{{2{{\left( 0 \right)}^2}}}{{{{\left( 0 \right)}^2} - 4}} = 0 \cr
& {\text{Relative maximum at }}\left( {0,0} \right) \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& \frac{{x + 1}}{{{x^2} - 4}} \cr
& {x^2} - 4 = 0 \to x = \pm 2 \cr
& {\text{Vertical asymptotes at }}x = \pm 2 \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2}}}{{{x^2} - 4}} = 2 \cr
& \mathop {\lim }\limits_{x \to - \infty } \frac{{2{x^2}}}{{{x^2} - 4}} = 2 \cr
& {\text{Horizontal asymptote }}y = 2 \cr
& \cr
& {\text{*Symmetry}} \cr
& f\left( { - x} \right) = \frac{{2{{\left( { - x} \right)}^2}}}{{{{\left( { - x} \right)}^2} - 4}} = \frac{{2{x^2}}}{{{x^2} - 4}} \cr
& f\left( { - x} \right) = f\left( x \right){\text{ The function is even}} \cr
& {\text{Symmetry about the }}y{\text{ - axis}} \cr
& \cr
& {\text{Graph}} \cr} $$
