In order to find the equations of the asymptotes you have to put this thing into standard form. Divide the coefficients on the variables by 36 and set it equal then to 1. [tex] \frac{ y^{2} }{4} - \frac{ x^{2} }{9} =1[/tex] where a is 2 (2 squared is 4) and b is 3 (3 squared is 9). The center of your hyperbola is at (0,0). The actual standard form for a hyperbola is this: [tex] \frac{(y-k) ^{2} }{ a^{2} } - \frac{(x-h) ^{2} }{ b^{2} }=1 [/tex] if it has a vertical transverse which yours does (the x and y are switched if you have a horizontal transverse. Only the x and y, though, not the a and the b. They stay where they are no matter the orientation of your transverse). In your particular hyperbola, you have no sets of parenthesis around your x or your y so that means that there is not side to side movement, nor is there up and down movement. Hence, the center hasn't moved from the origin. The equations for the asymptotes are [tex]y=k+/- \frac{a}{b} (x-h)[/tex]. But like I said, since there is no k or h, you just have simple equations of [tex]y= \frac{2}{3}x [/tex] and [tex]y=- \frac{2}{3} x[/tex]
