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