How do you find the horizontal asymptote #y = (1-x^2)/(x-1)#?

Question

Elias Ackerman · Accepted Answer

Because the numerator has degree greater than that of the denominator, there is no horizontal asymptote.

#y = (1-x^2)/(x-1)# # = (-x^2+1)/(x-1)# # = (x(-x+1/x))/(x(1-1/x))# # = (-x+1/x)/(1-1/x)# As #x# increases without bound, #y# decreases without bound and as #x# decreases without bound, #y# increases without bound. The is no number #k#, and no line #y=k# that #y# is getting close to.

Elijah Allen · Answer

#y = (1-x^2)/(x-1) = -x-1# with exclusion #x != 1#

This is a line of slope #-1#. It has no asymptotes.

#y = (1-x^2)/(x-1) = -(x^2-1)/(x-1) = -((x-1)(x+1))/(x-1)# #= -(x+1) = -x-1# with exclusion #x != 1# So this is a line of slope #-1# with an excluded point. It has no asymptotes.

Madelyn Anderson · Answer

undefined To find the horizontal asymptote of $ y = \frac{{1 - x^2}}{{x - 1}} $, we need to examine the behavior of the function as $ x $ approaches positive or negative infinity. By analyzing the degree of the numerator and denominator, we can determine if there is a horizontal asymptote.

Since the degree of the numerator (which is 2) is the same as the degree of the denominator (which is 1), we divide the leading coefficient of the numerator by the leading coefficient of the denominator. The result of this division gives us the equation of the horizontal asymptote, if it exists.

In this case, as $ x $ approaches infinity, the term $ x^2 $ in the numerator dominates, and as $ x $ approaches negative infinity, $ -x^2 $ dominates. Therefore, the horizontal asymptote is $ y = -x $.