Find the horizontal and oblique asymptotes, if any, for the given rational function G(x)=(x^4-1)/(3x^2-3x)
simplified ...
f(x) = (x^2 + 1)(x-1)(x+1)/(3x(x-1))
= (x+1)(x^2+1)/(3x)
= (x^3 + x^2 + x + 1)/(3x)
= (1/3) (x^2 + x + 1) + 1/(3x)
clearly there is a vertical asymptote at x = 0
there is also a "hole" at (1, 4/3) , (we divided top and bottom by x-1 )
when x --> ∞
1/3x --> 0
so we left with
(1/3)(x^2 + x + 1) + 0 , which is a paraloba
so we have a "curved" asymptote formed by the parabola
y = (1/3)(x^2 + x + 1)