This is equation of hyperbola in general form which you need to change it to standard form to see the information asked. You need to complete the square for both x and y.
3(x^2+2x) - 12(y^2-4y) = 93
3(x^2+2x+1) - 12(y^2-4y+4) = 93 + 3 -48
3(x+1)^2 - 12(y-2)^2 = 48
divide both side by 48
(x+1)^2/16 - (y-2)^2/4 = 1
now center is at (-1, 2)
vertices (-5,2) and (3,2)
Slopes of the asymptotes - 1/2 and 1/2
