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

Question

Madelyn Anderson · Accepted Answer

Horizontal : #y = -1#, in both #larr and rarr# directions.
Vertical :#x = 0 uarr, and uarr x^2-1=0 darr#. See Socratic graphs

Treating y as a function of x^2, the partial fractions are #y=-1+2/x^2+3/(x^2-1)#- The asymptote y = quotient = -1 is horizontal, in both #larr and rarr# directions. The denominators #x = 0 uarr, and uarr x^2-1=0 darr# give vertical asymptotes. graph{(2+x^4)/(x^2-x^4) [-40, 40, -20, 20]} graph{((2+x^4)/(x^2-x^4)-y)(y+1)(x-1-.003y)(x+1+.003y)(x+.001y)=0 [-1.5, 1.5, -15, 15]} Ad hoc Scale y : x is 20 : 1 for showing asymptotes

Chloe Alexander · Answer

undefined Vertical asymptotes occur where the denominator of the rational function equals zero but the numerator does not. In this case, $ x^2 - x^4 = 0 $, so there are vertical asymptotes at $ x = 0 $ and $ x = \pm 1 $. Horizontal asymptotes can be found by looking at the behavior of the function as $ x $ approaches positive or negative infinity. Since the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote can be found by dividing the leading coefficient of the numerator by the leading coefficient of the denominator. Thus, the horizontal asymptote is $ y = 1 $. There are no slant asymptotes in this case.