Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0, - \frac{1}{4}} \right) \cr
& {\text{No }}x{\text{ - intercept}}{\text{.}} \cr
& {\text{Relative maximum at}}\left( {0, - \frac{1}{4}} \right) \cr
& {\text{No inflection points}} \cr
& {\text{Horizontal asymptote }}y = 1 \cr
& {\text{Vertical asymptotes}}{\text{, }}x = 2,{\text{ }}x = - 2 \cr} $$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2} + 1}}{{{x^2} - 4}} \cr
& {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr
& y = \frac{{{0^2} + 1}}{{{0^2} - 4}} \cr
& y = - \frac{1}{4} \cr
& y{\text{ - intercept }}\left( {0, - \frac{1}{4}} \right) \cr
& {\text{Find the }}x{\text{ intercept}}{\text{, let }}y = 0 \cr
& 0 = \frac{{{x^2} + 1}}{{{x^2} - 4}} \cr
& {x^2} + 1 = 0,{\text{ No real solutions }} \cr
& {\text{No }}x{\text{ - intercept}}{\text{.}} \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {\frac{{{x^2} + 1}}{{{x^2} - 4}}} \right] \cr
& y' = \frac{{\left( {{x^2} - 4} \right)\left( {2x} \right) - \left( {{x^2} + 1} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr
& y' = \frac{{2{x^3} - 8x - 2{x^3} - 2x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr
& y' = \frac{{ - 10x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr
& {\text{Let }}y' = 0 \cr
& \frac{{ - 10x}}{{{{\left( {{x^2} - 4} \right)}^2}}} = 0 \cr
& x = 0 \cr
& \cr
& {\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ {\frac{{ - 10x}}{{{{\left( {{x^2} - 4} \right)}^2}}}} \right] \cr
& y'' = \frac{{{{\left( {{x^2} - 4} \right)}^2}\left( { - 10} \right) + 10x\left( 2 \right)\left( {{x^2} - 4} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^4}}} \cr
& y'' = \frac{{\left( {{x^2} - 4} \right)\left( { - 10} \right) + 10x\left( 2 \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 4} \right)}^3}}} \cr
& y'' = \frac{{ - 10{x^2} + 40 + 40{x^2}}}{{{{\left( {{x^2} - 4} \right)}^3}}} \cr
& y'' = \frac{{30{x^2} + 40}}{{{{\left( {{x^2} - 4} \right)}^3}}} \cr
& {\text{Evaluate }}y''\left( 0 \right) \cr
& y''\left( 0 \right) = \frac{{30{{\left( 0 \right)}^2} + 40}}{{{{\left( {{{\left( 0 \right)}^2} - 4} \right)}^3}}} = - \frac{5}{8} < 0 \cr
& {\text{Relative maximum at }}\left( {0,f\left( 0 \right)} \right) \to \left( {0, - \frac{1}{4}} \right) \cr
& \cr
& *{\text{Find the Inflection points}} \cr
& y'' = \frac{{30{x^2} + 40}}{{{{\left( {{x^2} - 4} \right)}^3}}} \cr
& \frac{{30{x^2} + 40}}{{{{\left( {{x^2} - 4} \right)}^3}}} = 0 \cr
& 30{x^2} + 40 = 0,{\text{ No real solutions}} \cr
& {\text{No inflection points}} \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& y = \frac{{{x^2} + 1}}{{{x^2} - 4}} \cr
& {x^2} - 4 = 0,{\text{ }}x = \pm 2 \cr
& {\text{Vertical asymptotes}}{\text{, }}x = 2,{\text{ }}x = - 2 \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + 1}}{{{x^2} - 4}}} \right) = 1 \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^2} + 1}}{{{x^2} - 4}}} \right) = 1 \cr
& {\text{Horizontal asymptote }}y = 1 \cr
& \cr
& {\text{Graph}} \cr} $$
