Answer
$$\eqalign{
& y{\text{ - intercept }}\left( {0,0} \right) \cr
& x{\text{ - intercept }}\left( { - 1,0} \right) \cr
& {\text{Relative minimum at }}\left( { - 3, - \frac{2}{3}} \right) \cr
& {\text{Relative maximum at }}\left( {1,2} \right) \cr
& {\text{inflection points }} \left( { - 4,75, - 0.586} \right),\left( { - 0.3054,0.8982} \right),\left( {2.0641,1.688} \right) \cr
& {\text{No vertical asymptotes}} \cr
& {\text{Horizontal asymptote }}y = 0 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{4x + 4}}{{{x^2} + 3}} \cr
& {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr
& f\left( 0 \right) = \frac{{4\left( 0 \right) + 4}}{{{{\left( 0 \right)}^2} + 3}} \cr
& f\left( 0 \right) = 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{{4x + 4}}{{{x^2} + 3}} \cr
& 4x + 4 = 0 \cr
& x = - 1 \cr
& x{\text{ - intercept }}\left( { - 1,0} \right) \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{4x + 4}}{{{x^2} + 3}}} \right] \cr
& f'\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( 4 \right) - \left( {4x + 4} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{4{x^2} + 12 - 8{x^2} - 8x}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{12 - 4{x^2} - 8x}}{{{{\left( {{x^2} + 3} \right)}^2}}} \cr
& {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr
& \frac{{12 - 4{x^2} - 8x}}{{{{\left( {{x^2} + 3} \right)}^2}}} = 0 \cr
& {x^2} + 2x - 3 = 0 \cr
& \left( {x + 3} \right)\left( {x - 1} \right) = 0 \cr
& x = - 3,{\text{ }}x = 1 \cr
& \cr
& {\text{Find the second derivative}} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{12 - 4{x^2} - 8x}}{{{{\left( {{x^2} + 3} \right)}^2}}}} \right] \cr
& f''\left( x \right) = \frac{{{{\left( {{x^2} + 3} \right)}^2}\left( { - 8x - 8} \right) - 2\left( {12 - 4{x^2} - 8x} \right)\left( {{x^2} + 3} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^4}}} \cr
& {\text{Simplifying}} \cr
& f''\left( x \right) = \frac{{\left( {{x^2} + 3} \right)\left( { - 8x - 8} \right) - 2\left( {12 - 4{x^2} - 8x} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr
& f''\left( x \right) = \frac{{ - 8{x^3} - 8{x^2} - 24x - 24 + 16{x^3} + 32{x^2} - 48x}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr
& f''\left( x \right) = \frac{{8{x^3} + 24{x^2} - 72x - 24}}{{{{\left( {{x^2} + 3} \right)}^3}}} \cr
& \cr
& {\text{Evaluate the second derivative at the critical points}} \cr
& *f''\left( { - 3} \right) = \frac{{8{{\left( { - 3} \right)}^3} + 24{{\left( { - 3} \right)}^2} - 72\left( { - 3} \right) - 24}}{{{{\left( {{{\left( { - 3} \right)}^2} + 3} \right)}^3}}} = \frac{1}{9} > 0 \cr
& {\text{Relative minimum at }}\left( { - 3,f\left( { - 3} \right)} \right) \to \left( { - 3, - \frac{2}{3}} \right) \cr
& *f''\left( 1 \right) = \frac{{8{{\left( 1 \right)}^3} + 24{{\left( 1 \right)}^2} - 72\left( 1 \right) - 24}}{{{{\left( {{{\left( 1 \right)}^2} + 3} \right)}^3}}} = - 1 < 0 \cr
& {\text{Relative maximum at }}\left( {1,f\left( 1 \right)} \right) \to \left( {1,2} \right) \cr
& \cr
& *{\text{Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr
& \frac{{8{x^3} + 24{x^2} - 72x - 24}}{{{{\left( {{x^2} + 3} \right)}^3}}} = 0 \cr
& 8{x^3} + 24{x^2} - 72x - 24 = 0 \cr
& {\text{Solving by a graphing utility we obtain}} \cr
& {x_1} \approx - 4.75,{\text{ }}{x_2} \approx 2.0641,{\text{ }}{x_3} = - 0.3054 \cr
& {\text{The inflection points are:}} \cr
& \left( { - 4,75, - 0.586} \right),\left( { - 0.3054,0.8982} \right),\left( {2.0641,1.688} \right) \cr
& {\text{*Calculate the asymptotes}} \cr
& f\left( x \right) = \frac{{4x + 4}}{{{x^2} + 3}} \cr
& {\text{The denominator is never zero}}{\text{, then there are no}} \cr
& {\text{vertical asymptotes}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{4x + 4}}{{{x^2} + 3}}} \right) = 0 \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{4x + 4}}{{{x^2} + 3}}} \right) = 0 \cr
& {\text{Horizontal asymptote }}y = 0 \cr
& \cr
& {\text{Graph}} \cr} $$
