Answer
$$\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can see that the orientation of the transverse}} \cr
& {\text{axis is horizontal }}\left( {{\text{along the x - axis}}} \right){\text{ and the hyperbola is centered }} \cr
& {\text{at the Origin}}.{\text{ Then, the equation is of the form}} \cr
& \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& and \cr
& {\text{Vertices }}\left( { \pm a,0} \right),\,\,{\text{foci }}\left( { \pm c,0} \right) \cr
& {\text{Vertices }}\left( { \pm 4,0} \right),\,\,{\text{foci }}\left( { \pm 5,0} \right) \cr
& a = 4,\,\,\,c = 5 \cr
& b = \sqrt {{c^2} - {a^2}} = \sqrt {25 - 16} = 3 \cr
& \cr
& {\text{The equation of the hyperbola is}} \cr
& \frac{{{x^2}}}{{{{\left( 4 \right)}^2}}} - \frac{{{y^2}}}{{{{\left( 3 \right)}^2}}} = 1 \cr
& \frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1 \cr} $$
