How to find the asymptotes of #y =(1-x^2)/( x-1)# ?

1 Answer
Truong-Son N.
Feb 2, 2016

Notice how you can actually simplify this.

#y = (-(x^2 - 1))/(x-1)#

The numerator is a difference of two squares. #x^2 - 1 = (x+1)(x-1)#.

#= (-(x + 1)cancel((x-1)))/cancel(x-1)#

#= color(blue)(-x - 1)#

Asymptotes are generally found when the denominator of the fractional equation would be #0#. Since there are no longer points where you have to divide by #0#, there are no asymptotes.

However, since we did just cancel out #x - 1#, we do have one removable discontinuity at #(1,-2)#, since the original denominator would be undefined when #x - 1 = 0# (can't divide by #0#), and at #color(green)(x = 1)#, #color(green)(y) = -(1) - 1 = color(green)(-2)#.

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