# What are the asymptote(s) and hole(s), if any, of  f(x) =1/((x-3)(x^3-x^2-x+1))?

Dec 16, 2015

asymptotes:
$x = 3 , - 1 , 1$
$y = 0$

holes:
none

#### Explanation:

$f \left(x\right) = \frac{1}{\left(x - 3\right) \left({x}^{3} - {x}^{2} - x + 1\right)}$

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

$f \left(x\right) = \frac{1}{\left(x - 3\right) \left({x}^{2} - 1\right) \left(x - 1\right)}$

$f \left(x\right) = \frac{1}{\left(x - 3\right) \left(x + 1\right) \left(x - 1\right) \left(x - 1\right)}$; x!=3,-1,1;y!=0

There are no holes for this function since there are no common bracketed polynomials which appear in the numerator and denominator. There are only restrictions which must be stated for each bracketed polynomial in the denominator. These restrictions are the vertical asymptotes. Keep in mind that there is also a horizontal asymptote of $y = 0$.

$\therefore$, the asymptotes are $x = 3$, $x = - 1$, $x = 1$, and $y = 0$.