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

1 Answer
Jan 13, 2017

Answer:

Vertcal : #uarr x = 0, uarr x = +-1 darr#

Horizontal : #larr y = -1 rarr#

Explanation:

By actual division,

#y = -1 +(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]}