Answer
$$\eqalign{
& {\text{Vertices}}:\left( { - 3, - 4} \right){\text{ and }}\left( { - 3,0} \right) \cr
& {\text{Asymptotes: }}y = \pm \frac{2}{3}\left( {x + 3} \right) - 2 \cr
& {\text{domain: }}\left( { - \infty , + \infty } \right) \cr
& {\text{range: }}\left( { - \infty , - 4} \right] \cup \left[ {0,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{{\left( {y + 2} \right)}^2}}}{4} - \frac{{{{\left( {x + 3} \right)}^2}}}{9} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{{\left( {y - k} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{{{\left( {y + 2} \right)}^2}}}{4} - \frac{{{{\left( {x + 3} \right)}^2}}}{9} = 1,{\text{ then }}a = 2,\,\,b = 3,\,\,\,h = - 3,\,\,\,k = - 2 \cr
& \cr
& {\text{Characteristics:}} \cr
& {\text{Transverse axis: horizontal}} \cr
& {\text{vertices: }}\left( {h,k \pm a} \right) \cr
& {\text{vertices: }}\left( { - 3, - 2 \pm 2} \right) \cr
& {\text{vertices: }}\left( { - 3, - 4} \right){\text{ and }}\left( { - 3,0} \right) \cr
& \cr
& {\text{asymptotes: }}y = \pm \frac{a}{b}\left( {x - h} \right) + k \cr
& {\text{asymptotes: }}y = \pm \frac{2}{3}\left( {x + 3} \right) - 2 \cr
& \cr
& {\text{domain: }}\left( { - \infty , + \infty } \right) \cr
& {\text{The range is: }}\left( { - \infty ,k - a} \right] \cup \left[ {k + a,\infty } \right) \cr
& {\text{The range is: }}\left( { - \infty , - 4} \right] \cup \left[ {0,\infty } \right) \cr
& \cr
& {\text{Graph}} \cr} $$