How do you find the vertical, horizontal, and oblique asymptotes of #H(x)=(x^4+2x^2+1)/(x^2-x+1)#?
1 Answer
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^2-x+1 = (x-1/2)^2+3/4#
Hence
Also note that the degree of the numerator exceeds that of the denominator by
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
graph{(y-(x^4+2x^2+1)/(x^2-x+1))(y-(x^2+x+2)) = 0 [-10.97, 9.03, -7, 43]}