Answer
$${\text{Vertices}}\left( {1, - 3} \right){\text{ and }}\left( {3, - 3} \right)$$
$${\text{Foci}}\left( {2 - \sqrt {10} , - 3} \right){\text{ and }}\left( {2 + \sqrt {10} , - 3} \right)$$
$${\text{Center}}\left( {2, - 3} \right)$$
Work Step by Step
$$\eqalign{
& 9{x^2} - {y^2} - 36x - 6y + 18 = 0 \cr
& {\text{Group terms}} \cr
& \left( {9{x^2} - 36x} \right) - \left( {{y^2} + 6y} \right) = - 18 \cr
& 9\left( {{x^2} - 4x} \right) - \left( {{y^2} + 6y} \right) = - 18 \cr
& {\text{Completing the square}} \cr
& 9\left( {{x^2} - 4x + 4} \right) - \left( {{y^2} + 6y + 9} \right) = - 18 + 9\left( 4 \right) - 9 \cr
& 9{\left( {x - 2} \right)^2} - {\left( {y + 3} \right)^2} = 9 \cr
& {\text{Divide the equation by 9}} \cr
& {\left( {x - 2} \right)^2} - \frac{{{{\left( {y + 3} \right)}^2}}}{9} = 1 \cr
& {\text{This equation has the standard form}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{Therefore,}} \cr
& \underbrace {{{\left( {x - 2} \right)}^2} - \frac{{{{\left( {y + 3} \right)}^2}}}{9} = 1}_{\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1} \cr
& h = 2,{\text{ }}k = - 3,{\text{ }}a = 1,{\text{ }}b = 3,{\text{ }}c = \sqrt {{b^2} + {a^2}} = \sqrt {10} \cr
& {\text{Characteristics of the hyperbola }} \cr
& {\text{Orientation}}:{\text{ Horizontal transverse axis}} \cr
& {\text{Center}}\left( {h,k} \right) \to {\text{Center}}\left( {2, - 3} \right) \cr
& \underbrace {{\text{Vertices}}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right)}_ \Downarrow \cr
& {\text{Vertices}}\left( {1, - 3} \right){\text{ and }}\left( {3, - 3} \right) \cr
& \underbrace {{\text{Foci}}\left( {h - c,k} \right){\text{ and }}\left( {h + c,k} \right)}_ \Downarrow \cr
& {\text{Foci}}\left( {2 - \sqrt {10} , - 3} \right){\text{ and }}\left( {2 + \sqrt {10} , - 3} \right) \cr
& \underbrace {{\text{Asymptotes}}:{\text{ }}y = \pm \frac{b}{a}\left( {x - h} \right) + k}_ \Downarrow \cr
& {\text{Asymptotes}}:{\text{ }}y = \pm 3\left( {x - 2} \right) - 3 \cr
& \cr
& {\text{Graph:}} \cr} $$