We note that this is an "east-west opening" hyperbola, with a = 6 and b = 8.

The center of this hyperbola will be at (2, -3), since h = 2 and k = -3 in this example.

The best approach is to ignore the shifting (for now) and figure out the other parameters for the hyperbola.

So I'm assuming (for this part) that the hyperbola is actually `X^2/36+Y^2/64=1`, and I'll use upper case X and Y.

The vertices of the parabola are found when Y = 0. This gives us X = -6 or X = 6.

The asymptotes will have slope `-8/6 = -4/3`, OR `8/6 = 4/3`.

Now we can sketch the asymptotes and the vertices, remembering to shift everything so that the center is (2, -3):

51015-5-10-1551015-5-10-15-20xyOpen image in a new page

Asymptotes, and vertices (-4,-3) and (8,-3).

Now for the hyperbola:

51015-5-10-1551015-5-10-15-20xyOpen image in a new page

"East-West" hyperbola (in green) with its asymptotes (in magenta color).