Question:

why can't a rational function have both, a horizontal and an oblique asymptote?

could someone give a non-wikipedia answer?

Let f(x) = p(x)/q(x), where p and q are polynomials and reduced to lowest terms. (If p and q have a common factor, then they contribute removable discontinuities ('holes').) 
Write this in cases: 
(i) If deg p(x) ≤ deg q(x), then f(x) is a proper rational function, and lim(x→ ±∞) f(x) = constant. 
If deg p(x) < deg q(x), then these limits equal 0, thus yielding the horizontal asymptote y = 0. 
If deg p(x) = deg q(x), then these limits equal a/b, where a and b are the leading coefficients of p(x) and q(x), respectively. Hence, we have the horizontal asymptote y = a/b. 
Note that there are no obliques asymptotes in this case.  (ii) If deg p(x) > deg q(x), then f(x) is an improper rational function. 
By long division, we can write f(x) = g(x) + r(x)/q(x), where g(x) and r(x) are polynomials and deg r(x) < deg q(x). 
As in (i), note that lim(x→ ±∞) [f(x) - g(x)] = lim(x→ ±∞) r(x)/q(x) = 0. Hence, y = g(x) is an asymptote. (In particular, if deg g(x) = 1, then this is an oblique asymptote.) 
This time, note that there are no horizontal asymptotes.  In summary, the degrees of p(x) and q(x) control which kind of asymptote we have. 
I hope this helps!

Iwould say d but i am not 100% positive