Answer
$$\frac{{{{\left( {x - 3} \right)}^2}}}{9} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1$$
Work Step by Step
$$\eqalign{
& {\text{Vertices }}\left( {0,2} \right){\text{ and }}\left( {6,2} \right) \cr
& {\text{The hyperbola has the standard form,}} \cr
& \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 \cr
& {\text{With}} \cr
& \underbrace {{\text{Vertices }}\left( {0,2} \right){\text{ and }}\left( {6,2} \right)}_{{\text{Vertices }}\left( {h - a,k} \right){\text{ and }}\left( {h + a,k} \right)} \to k = 2 \cr
& h - a = 0 \cr
& h + a = 6 \cr
& 2h = 6 \to h = 3,{\text{ }}a = 3 \cr
& {\text{Asymptotes }}\underbrace {y = \pm \frac{b}{a}\left( {x - h} \right) + k}_{} \cr
& y = \frac{2}{3}x,{\text{ }}y = - \frac{2}{3}x + 4 \cr
& \pm \frac{b}{a} = \pm \frac{2}{3} \cr
& \frac{b}{a} = \frac{2}{3} \cr
& \frac{b}{3} = \frac{2}{3} \to b = 2 \cr
& \underbrace {\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1}_ \downarrow \cr
& \frac{{{{\left( {x - 3} \right)}^2}}}{9} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1 \cr} $$
