Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& x{\text{ - Intercepts }}\left( {0,0} \right){\text{ and }}\left( { - 1,0} \right) \cr
& {\text{ relative minimum at }}\left( { - 0.5358, - 0.0669} \right) \cr
& {\text{Inflection point }}\left( { - 11.044, - 0.9402} \right) \cr
& {\text{Vertical asymptotes}}{\text{, }}x = 2,{\text{ }}x = - 2 \cr
& {\text{Horizontal asymptote }}y = - 1 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^2} + x}}{{4 - {x^2}}} \cr
& {\text{Find the }}y{\text{ intercept}}{\text{, }}x = 0 \cr
& f\left( 0 \right) = \frac{{{{\left( 0 \right)}^2} + \left( 0 \right)}}{{4 - {{\left( 0 \right)}^2}}} = 0 \cr
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr
& 0 = \frac{{{x^2} + x}}{{4 - {x^2}}} \cr
& {x^2} + x = 0 \cr
& x\left( {x + 1} \right) = 0 \cr
& x = 0,{\text{ }}x = - 1 \cr
& x{\text{ - Intercepts }}\left( {0,0} \right){\text{ and }}\left( { - 1,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2} + x}}{{4 - {x^2}}}} \right] \cr
& f'\left( x \right) = \frac{{\left( {4 - {x^2}} \right)\left( {2x + 1} \right) - \left( {{x^2} + x} \right)\left( { - 2x} \right)}}{{{{\left( {4 - {x^2}} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{8x + 4 - 2{x^3} - {x^2} + 2{x^3} + 2{x^2}}}{{{{\left( {4 - {x^2}} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{{x^2} + 8x + 4}}{{{{\left( {4 - {x^2}} \right)}^2}}} \cr
& {\text{Let }}f'\left( x \right) = 0 \cr
& \frac{{{x^2} + 8x + 4}}{{{{\left( {4 - {x^2}} \right)}^2}}} = 0 \cr
& {x^2} + 8x + 4 = 0 \cr
& {\text{By the quadratic formula we obtain}} \cr
& {x_1} = - 4 - 2\sqrt 3 \approx - 7.4641,{\text{ }}{x_2} = - 4 + 2\sqrt 3 \approx - 0.5358 \cr
& \cr
& {\text{Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2} + 8x + 4}}{{{{\left( {4 - {x^2}} \right)}^2}}}} \right] \cr
& f''\left( x \right) = \frac{{{{\left( {4 - {x^2}} \right)}^2}\left( {2x + 8} \right) + 4x\left( {{x^2} + 8x + 4} \right)\left( {4 - {x^2}} \right)}}{{{{\left( {4 - {x^2}} \right)}^4}}} \cr
& f''\left( x \right) = \frac{{\left( {4 - {x^2}} \right)\left( {2x + 8} \right) + 4x\left( {{x^2} + 8x + 4} \right)}}{{{{\left( {4 - {x^2}} \right)}^3}}} \cr
& f''\left( x \right) = \frac{{8x + 32 - 2{x^3} - 8{x^2} + 4{x^3} + 32{x^2} + 16x}}{{{{\left( {4 - {x^2}} \right)}^3}}} \cr
& f''\left( x \right) = \frac{{2{x^3} + 24{x^2} + 24x + 32}}{{{{\left( {4 - {x^2}} \right)}^3}}} \cr
& {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }} \cr
& f''\left( { - 7.4641} \right) < 0,{\text{ then there is a relative maximum at }} \cr
& \left( { - 7.4641,f\left( { - 7.4641} \right)} \right) \to \left( { - 7.4641, - 0.9330} \right) \cr
& and \cr
& f''\left( { - 0.5358} \right) > 0,{\text{ then there is a relative minimum at }} \cr
& \left( { - 0.5358,f\left( { - 0.5358} \right)} \right) \to \left( { - 0.5358, - 0.0669} \right) \cr
& \cr
& *{\text{Find the Inflection points let }}f''\left( x \right) = 0 \cr
& \frac{{2{x^3} + 24{x^2} + 24x + 32}}{{{{\left( {4 - {x^2}} \right)}^3}}} = 0 \cr
& 2{x^3} + 24{x^2} + 24x + 32 = 0 \cr
& x = - 11.044 \cr
& f\left( { - 11.044} \right) = \frac{{{{\left( { - 11.044} \right)}^2} + \left( { - 11.044} \right)}}{{4 - {{\left( { - 11.044} \right)}^2}}} \approx - 0.9402 \cr
& {\text{Inflection point }}\left( { - 11.044, - 0.9402} \right) \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& f\left( x \right) = \frac{{{x^2} + x}}{{4 - {x^2}}} \cr
& 4 - {x^2} = 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} + x}}{{4 - {x^2}}}} \right) = - 1 \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{{x^2} + x}}{{4 - {x^2}}}} \right) = - 1 \cr
& {\text{Horizontal asymptote }}y = - 1 \cr
& \cr
& {\text{Graph}} \cr} $$
