Question 973317
None.  The asymptote could be quadratic or to a parabola, but not a line in your function example.


You can find the parabolic asymptote if you want, just doing the indicated polynomial division.  The quotient, without the remainder, is the asymptote.





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Assumption:  you have {{{(x^3-2x^2+1)/(x-2)}}}.



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Note carefully:  No horizontal asymptote either.  Degrees of numerator and denominator are not equal.