How do you find the horizontal asymptote y = (1-x^2)/(x-1)y=1−x2x−1?

2 Answers
Jim H · George C.
Sep 18, 2015

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

Explanation:

y = (1-x^2)/(x-1)y=1−x2x−1

= (-x^2+1)/(x-1)=−x2+1x−1

= (x(-x+1/x))/(x(1-1/x))=x(−x+1x)x(1−1x)

= (-x+1/x)/(1-1/x)=−x+1x1−1x

As xx increases without bound, yy decreases without bound
and
as xx decreases without bound, yy increases without bound.

The is no number kk, and no line y=ky=k that yy is getting close to.

George C.
Sep 18, 2015

y = (1-x^2)/(x-1) = -x-1y=1−x2x−1=−x−1 with exclusion x != 1x≠1

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

Explanation:

y = (1-x^2)/(x-1) = -(x^2-1)/(x-1) = -((x-1)(x+1))/(x-1)y=1−x2x−1=−x2−1x−1=−(x−1)(x+1)x−1

= -(x+1) = -x-1=−(x+1)=−x−1

with exclusion x != 1x≠1

So this is a line of slope -1−1 with an excluded point.

It has no asymptotes.

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