Answer
Graph
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{ - {x^2} - 4x - 7}}{{x + 3}} \cr
& {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr
& y = \frac{{ - {{\left( 0 \right)}^2} - 4\left( 0 \right) - 7}}{{\left( 0 \right) + 3}} \cr
& y = - \frac{7}{3} \cr
& y{\text{ - intercept }}\left( {0, - \frac{7}{3}} \right) \cr
& {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr
& 0 = \frac{{ - {x^2} - 4x - 7}}{{x + 3}} \cr
& - {x^2} - 4x - 7 = 0 \cr
& {x^2} + 4x + 7 = 0 \cr
& {\text{No real solutions}} \cr
& {\text{No, }}x{\text{ - intercepts}} \cr
& \cr
& *{\text{Find the extrema}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {\frac{{ - {x^2} - 4x - 7}}{{x + 3}}} \right] \cr
& y' = \frac{{\left( {x + 3} \right)\left( { - 2x - 4} \right) - \left( { - {x^2} - 4x - 7} \right)\left( 1 \right)}}{{{{\left( {x + 3} \right)}^2}}} \cr
& y' = \frac{{\left( {x + 3} \right)\left( { - 2x - 4} \right) - \left( { - {x^2} - 4x - 7} \right)\left( 1 \right)}}{{{{\left( {x + 3} \right)}^2}}} \cr
& y' = \frac{{ - 2{x^2} - 4x - 6x - 12 + {x^2} + 4x + 7}}{{{{\left( {x + 3} \right)}^2}}} \cr
& y' = \frac{{ - {x^2} - 6x - 5}}{{{{\left( {x + 3} \right)}^2}}} \cr
& {\text{Let }}y' = 0{\text{ to find critical points}} \cr
& - {x^2} - 6x - 5 = 0 \cr
& {x^2} + 6x + 5 = 0 \cr
& \left( {x + 5} \right)\left( {x + 1} \right) = 0 \cr
& x = - 1,{\text{ }}x = - 5 \cr
& \cr
& *{\text{Find the second derivative}} \cr
& y'' = \frac{d}{{dx}}\left[ {\frac{{ - {x^2} - 6x - 5}}{{{{\left( {x + 3} \right)}^2}}}} \right] \cr
& {\text{Using a CAS }} \cr
& y'' = - \frac{8}{{{{\left( {x + 3} \right)}^3}}} \cr
& \cr
& {\text{Evaluate }}y''\left( x \right){\text{ at }}x = - 1{\text{ and }}x = - 5 \cr
& y''\left( { - 1} \right) = - \frac{8}{{{{\left( { - 1 + 3} \right)}^3}}} < 0,{\text{ relative maximum at }}x = - 1 \cr
& f\left( { - 1} \right) = - 2 \to {\text{Relative maximum at }}\left( { - 1, - 2} \right) \cr
& y''\left( { - 5} \right) = - \frac{8}{{{{\left( { - 5 + 3} \right)}^3}}} > 0,{\text{ relative minimum at }}x = - 5 \cr
& f\left( { - 5} \right) = 6 \to {\text{Relative minimum at }}\left( { - 5,6} \right) \cr
& \cr
& {\text{Let }}y''\left( x \right) = 0 \cr
& - \frac{8}{{{{\left( {x + 3} \right)}^3}}} = 0 \cr
& {\text{No, inflection points}} \cr
& \cr
& {\text{*Calculate the asymptotes}} \cr
& \frac{{ - {x^2} - 4x - 7}}{{x + 3}} \cr
& x + 3 = 0 \cr
& x = - 3 \cr
& {\text{Vertical asymptotes at }}x = - 3 \cr
& \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{ - {x^2} - 4x - 7}}{{x + 3}}} \right) = - \infty \cr
& \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{ - {x^2} - 4x - 7}}{{x + 3}}} \right) = + \infty \cr
& {\text{No horizontal asymptotes}} \cr
& \cr
& {\text{Graph}} \cr} $$
