Answer
$${\text{Foci}}\left( { - \sqrt {41} ,0} \right){\text{ and }}\left( {\sqrt {41} ,0} \right)$$
$${\text{Vertices}}\left( { - 5,0} \right){\text{ and }}\left( {5,0} \right)$$
$${\text{Center}}\left( {0,0} \right)$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1 \cr
& {\text{This equation is written in standard form}} \cr
& \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1 \Rightarrow \underbrace {\frac{{{x^2}}}{{{5^2}}} - \frac{{{y^2}}}{{{4^2}}} = 1}_{\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1} \to a = 5,{\text{ }}b = 4 \cr
& c = \sqrt {{a^2} + {b^2}} \cr
& c = \sqrt {{5^2} + {4^2}} \Rightarrow \boxed{c = \sqrt {41} } \cr
& {\text{Characteristics of the hyperbola }} \cr
& {\text{Center}}\left( {0,0} \right) \cr
& \underbrace {{\text{Vertices}}\left( { - a,0} \right){\text{ and }}\left( {a,0} \right)}_ \downarrow \cr
& {\text{Vertices}}\left( { - 5,0} \right){\text{ and }}\left( {5,0} \right) \cr
& \underbrace {{\text{Foci}}\left( { - c,0} \right){\text{ and }}\left( {c,0} \right)}_ \Downarrow \cr
& {\text{Foci}}\left( { - \sqrt {41} ,0} \right){\text{ and }}\left( {\sqrt {41} ,0} \right) \cr
& {\text{Graph:}} \cr} $$
