# 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

#### Answer:

#### Explanation:

Note that:

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

The quadratic factor

#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

The only factor in the denominator that can be zero is

The factors that we cancelled were also at

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

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