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

By actual division,

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]}

To find the vertical asymptotes of the function y = (1 + x^4) / (x^2 - x^4), we need to determine where the denominator becomes zero, except where the numerator is also zero. So, we set the denominator equal to zero:

x^2 - x^4 = 0

Factoring out x^2, we get:

x^2(1 - x^2) = 0

This gives us two critical points: x = 0 and x = ±1. However, we must check whether these critical points make the numerator zero or not. Plugging these values into the numerator, we find that none of them result in a zero numerator.

Therefore, the vertical asymptotes of the function are x = 1 and x = -1.

To find the horizontal asymptote, we observe the degrees of the numerator and the denominator. Both the numerator and the denominator are polynomials. The degree of the numerator is 4, and the degree of the denominator is 4. Since the degree of the numerator is equal to the degree of the denominator, we need to compare the leading coefficients. In this case, both leading coefficients are 1.

Therefore, to find the horizontal asymptote, we divide the leading coefficient of the numerator by the leading coefficient of the denominator:

Horizontal asymptote: y = 1/1 = 1

To find the slant asymptote, we perform polynomial long division or synthetic division. Dividing (1 + x^4) by (x^2 - x^4) results in a quotient of x^2 + 1, with no remainder. Therefore, the slant asymptote is y = x^2 + 1.