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

May 10, 2018

$H \left(x\right)$ has no linear asymptotes. It is asymptotic to the polynomial ${x}^{2} + x + 2$.

#### Explanation:

Given:

$H \left(x\right) = \frac{{x}^{4} + 2 {x}^{2} + 1}{{x}^{2} - x + 1}$

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

${x}^{2} - x + 1 = {\left(x - \frac{1}{2}\right)}^{2} + \frac{3}{4}$

Hence $H \left(x\right)$ 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:

$\frac{{x}^{4} + 2 {x}^{2} + 1}{{x}^{2} - x + 1} = \frac{\left({x}^{4} - {x}^{3} + {x}^{2}\right) + \left({x}^{3} - {x}^{2} + x\right) + \left(2 {x}^{2} - 2 x + 2\right) + x - 1}{{x}^{2} - x + 1}$

$\textcolor{w h i t e}{\frac{{x}^{4} + 2 {x}^{2} + 1}{{x}^{2} - x + 1}} = {x}^{2} + x + 2 + \frac{x - 1}{{x}^{2} - x + 1}$

So $H \left(x\right)$ 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]}