How do you find the asymptote(s) or holes of # f(x) = (x^2-1) / ((x-1)(x^3-2x^2+2x-1)#?

1 Answer
Feb 14, 2018

Answer:

#f(x)# has a horizontal asymptote #y=0# and a vertical asymptote #x=1#. It has no slant asymptote or any holes.

Explanation:

Note that:

#x^3-2x^2+2x-1 = (x-1)(x^2-x+1)#

The quadratic factor #x^2-x+1# is always positive, as you can see by checking its discriminant, or by completing the square:

#x^2-x+1 = (x-1/2)^2+3/4#

So we find:

#f(x) = (x^2-1)/((x-1)(x^3-2x^2+2x-1))#

#color(white)(f(x)) = (color(red)(cancel(color(black)((x-1))))(x+1))/(color(red)(cancel(color(black)((x-1))))(x-1)(x^2-x+1))#

#color(white)(f(x)) = (x+1)/((x-1)(x^2-x+1))#

Note that the denominator is of higher degree than the numerator. Hence we can deduce that #f(x)# has a horizontal asymptote #y=0#.

The only factor in the denominator that can be zero is #(x-1)#, occurring when #x=1#, when the numerator (in the simplified expression) is non-zero. We can deduce that #f(x)# has a vertical asymptote #x=1#.

The factors that we cancelled were also at #x=1#, so would have resulted in a hole instead of an asymptote if the extra factor in the denominator were not present. As it is, we have a vertical asymptote there.

There are no slant (a.k.a. oblique) asymptotes. Those only happen when the degree of the numerator exceeds the denominator by exactly #1#.

graph{(x^2-1)/((x-1)(x^3-2x^2+2x-1)) [-12.66, 12.65, -6.33, 6.33]}