How do you find the vertical, horizontal, and oblique asymptotes of H(x)=(x^4+2x^2+1)/(x^2-x+1)?

1 Answer
May 10, 2018

H(x) has no linear asymptotes. It is asymptotic to the polynomial x^2+x+2.

Explanation:

Given:

H(x) = (x^4+2x^2+1)/(x^2-x+1)

Note that the denominator is always positive (and therefore non-zero) for all real values of x since:

x^2-x+1 = (x-1/2)^2+3/4

Hence H(x) has no vertical asymptote (or hole).

Also note that the degree of the numerator exceeds that of the denominator by 2. Hence this function has no horizontal or oblique asymptote.

We can find a (quadratic) polynomial to which it is asymptotic by dividing and discarding the remainder:

(x^4+2x^2+1)/(x^2-x+1) = ((x^4-x^3+x^2)+(x^3-x^2+x)+(2x^2-2x+2)+x-1)/(x^2-x+1)

color(white)((x^4+2x^2+1)/(x^2-x+1)) = x^2+x+2+(x-1)/(x^2-x+1)

So H(x) is asymptotic to x^2+x+2

graph{(y-(x^4+2x^2+1)/(x^2-x+1))(y-(x^2+x+2)) = 0 [-10.97, 9.03, -7, 43]}