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

Jan 13, 2017

Vertcal : $\uparrow x = 0 , \uparrow x = \pm 1 \downarrow$

Horizontal : $\leftarrow y = - 1 \rightarrow$

#### Explanation:

By actual division,

$y = - 1 + \frac{1 + {x}^{2}}{{x}^{2} - {x}^{4}}$

y = quotient and x = zeros of the denominator give the asymptotes.

They are

y = -1, x = 0 and x = +-1.

graph{(y(x^2-x^4)-1-x^2)(y+1)x=0 [-10, 10, -5, 5]}

graph{y(x^2-x^4)-1-x^2=0 [-39.7, 39.7, -19.85, 19.84]}