Since your hyperbola has the "y" term listed first, it opens "up and down.'
It's also centered at (0,0)
So the form is: y^2/b^2 - x^2/a^2 = 1
In your case, b^2 = 16, so b = SQRT(16) = 4 and a^2 = 9, so, a = SQRT(9) = 3
So, the equations for the asymptotes are y = +(b/a)x and y = -(b/a)x
So we have, in your case .....y = +(4/3)x and y = -(4/3)x
And there you are!!
+128460 
