Answer
$$\eqalign{
& {\text{Center }}\left( {0,0} \right) \cr
& {\text{Vertices}}:\left( { - 18, - 4} \right){\text{ and }}\left( {6, - 4} \right) \cr
& {\text{Foci: }}\left( { - 21, - 4} \right){\text{ and }}\left( {9, - 4} \right) \cr
& {\text{Asymptotes: }}y = \pm \frac{3}{4}\left( {x + 6} \right) - 4 \cr
& {\text{domain: }}\left( { - \infty , - 18} \right] \cup \left[ {6,\infty } \right) \cr
& {\text{range: }}\left( { - \infty , + \infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{{\left( {x + 6} \right)}^2}}}{{144}} - \frac{{{{\left( {y + 4} \right)}^2}}}{{81}} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& \frac{{{{\left( {x + 6} \right)}^2}}}{{144}} - \frac{{{{\left( {y + 4} \right)}^2}}}{{81}} = 1,{\text{ then }}a = 12,\,\,b = 9,\,\,\,h = - 6,\,\,\,k = - 4 \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {144 + 81} = 15 \cr
& \cr
& {\text{Characteristics:}} \cr
& {\text{Center: }}\left( {h,k} \right) \cr
& {\text{Center: }}\left( { - 6, - 4} \right) \cr
& {\text{Transverse axis: horizontal}} \cr
& {\text{vertices: }}\left( {h \pm a,k} \right) \cr
& {\text{vertices: }}\left( { - 6 \pm 12, - 4} \right) \cr
& {\text{vertices: }}\left( { - 18, - 4} \right){\text{ and }}\left( {6, - 4} \right) \cr
& {\text{foci:}}\left( {h \pm c,k} \right) \cr
& {\text{foci:}}\left( { - 6 \pm 15, - 4} \right) \cr
& {\text{foci:}}\left( { - 21, - 4} \right){\text{ and }}\left( {9, - 4} \right) \cr
& {\text{asymptotes: }}y = \pm \frac{b}{a}\left( {x - h} \right) + k \cr
& {\text{asymptotes: }}y = \pm \frac{9}{{12}}\left( {x + 6} \right) - 4 \cr
& {\text{asymptotes: }}y = \pm \frac{3}{4}\left( {x + 6} \right) - 4 \cr
& {\text{The domain is: }}\left( { - \infty ,h - a} \right] \cup \left[ {h + a,\infty } \right) \cr
& {\text{domain: }}\left( { - \infty , - 18} \right] \cup \left[ {6,\infty } \right) \cr
& {\text{The range is: }}\left( { - \infty , + \infty } \right) \cr} $$
