Answer
$$\eqalign{
& {\text{Vertices }}\left( { \pm 2,0} \right){\text{ and foci}}\left( { \pm \sqrt {13} ,0} \right) \cr
& {\text{Asymptotes: }}y = \frac{3}{2}x{\text{ and }}y = - \frac{3}{2}x \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( { \pm 2,0} \right){\text{ }} \cr
& {\text{The equation of the hyperbola is given by: }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,{\text{ with}} \cr
& {\text{Vertices }}\left( { \pm a,0} \right),\,\,\,a = 2 \cr
& {\text{Then,}} \cr
& {\text{Asymptotes: }}y = \frac{b}{a}x{\text{ and }}y = - \frac{b}{a}x \cr
& \frac{b}{a}x = \frac{3}{2}x \cr
& \frac{b}{2}x = \frac{3}{2}x \cr
& b = 3 \cr
& {\text{The equation of the hyperbola is }} \cr
& \frac{{{x^2}}}{{{{\left( 2 \right)}^2}}} - \frac{{{y^2}}}{{{{\left( 3 \right)}^2}}} = 1 \cr
& \frac{{{x^2}}}{4} - \frac{{{y^2}}}{9} = 1 \cr
& \cr
& {\text{Foci }}\left( { \pm c,0} \right) \cr
& c = \sqrt {{a^2} + {b^2}} = \sqrt {13} \cr
& {\text{Foci }}\left( { \pm \sqrt {13} ,0} \right) \cr
& \cr
& {\text{Graph}} \cr} $$
