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

Question

1610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-10T21:03:07

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

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]}