Answer
$$\eqalign{
& {\text{Vertices }}\left( { \pm 4,0} \right){\text{ and foci}}\left( { \pm 6,0} \right) \cr
& {\text{Asymptotes: }}y = \frac{{\sqrt 5 }}{2}x{\text{ and }}y = - \frac{{\sqrt 5 }}{2}x \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( { \pm 4,0} \right){\text{ and foci}}\left( { \pm 6,0} \right) \cr
& {\text{The equation of the hyperbola is given by: }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,{\text{ with}} \cr
& {\text{Vertices }}\left( { \pm a,0} \right){\text{ and Foci }}\left( { \pm c,0} \right) \cr
& {\text{Then,}} \cr
& a = 4{\text{ and }}c = 6 \cr
& b = \sqrt {{c^2} - {a^2}} \cr
& b = \sqrt {{6^2} - {4^2}} \cr
& b = 2\sqrt 5 \cr
& \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{{x^2}}}{{{{\left( 4 \right)}^2}}} - \frac{{{y^2}}}{{{{\left( {2\sqrt 5 } \right)}^2}}} = 1 \cr
& \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{20}} = 1 \cr
& \cr
& {\text{Asymptotes: }}y = \frac{b}{a}x{\text{ and }}y = - \frac{b}{a}x \cr
& {\text{Asymptotes: }}y = \frac{{\sqrt 5 }}{2}x{\text{ and }}y = - \frac{{\sqrt 5 }}{2}x \cr
& \cr
& {\text{Graph}} \cr} $$
