Answer
$$\eqalign{
& {\text{Center }}\left( {0,0} \right) \cr
& {\text{Vertices}}:\left( { \pm 4,0} \right) \cr
& {\text{Foci: }}\left( { \pm 5,0} \right) \cr
& {\text{Asymptotes: }}y = \pm \frac{3}{4}x \cr
& {\text{domain: }}\left( { - \infty , - 4} \right] \cup \left[ {4,\infty } \right) \cr
& {\text{range: }}\left( { - \infty , + \infty } \right){\text{ }} \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1,{\text{ then }}a = 4,\,\,b = 3 \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {25} = 5 \cr
& \cr
& {\text{Therefore,}} \cr
& {\text{Center }}\left( {0,0} \right) \cr
& {\text{Vertices: }}\left( { \pm a,0} \right):\left( { \pm 4,0} \right) \cr
& {\text{Foci: }}\left( { \pm c,0} \right):\left( { \pm 5,0} \right) \cr
& {\text{Asymptotes: }}y = \pm \frac{b}{a}x \cr
& {\text{Asymptotes: }}y = \pm \frac{3}{4}x \cr
& {\text{The domain of the hyperbola is }}\left( { - \infty ,a} \right] \cup \left[ {a,\infty } \right) \cr
& {\text{domain: }}\left( { - \infty , - 4} \right] \cup \left[ {4,\infty } \right) \cr
& {\text{The range of the hyperbola is }}\left( { - \infty , + \infty } \right) \cr
& \cr
& {\text{Graph}} \cr} $$
