How do you find the asymptotes for (x^3 + 1) / (x^2 - 2x + 2)x3+1x2−2x+2?

1 Answer
Shwetank Mauria
Feb 19, 2016

There is only one slanting asymptote y=x+2y=x+2

Explanation:

The denominator x^2−2x+2x2−2x+2 cannot be factorized as determinant is (-2)^2-4*1*2=4-8=-4(−2)2−4⋅1⋅2=4−8=−4, a negative number. As such, it is not zero for any real number and there is no vertical asymptote.

As the degree of numerator is just one greater than that of denominator, dividing numerator by denominator we get

(x^3+1)=x(x^2−2x+2)+2(x^2−2x+2)+2x-3(x3+1)=x(x2−2x+2)+2(x2−2x+2)+2x−3 or

(x^3+1)=(x+2)(x^2−2x+2)+2x-3(x3+1)=(x+2)(x2−2x+2)+2x−3

Hence we have a slanting asymptote y=x+2y=x+2

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