For what values of x, if any, does f(x) = 1/((x-2)(x-1)(x-3))  have vertical asymptotes?

Apr 26, 2016

Vertical asymptotes exist at $x = 1 , 2 , \mathmr{and} 3$

Explanation:

A vertical asymptote is caused in a rational function when there is a zero in the denominator while the numerator is non-zero. For our function that corresponds to :

$0 = \left(x - 2\right) \left(x - 1\right) \left(x - 3\right)$

which means that any of the factors on the right hand side could be zero. By inspection we see that this happens when

$x = 1 , 2 , \mathmr{and} 3$

Therefore, these are the values where the vertical asymptotes exist. Graphing the function confirms our results.

graph{1/((x-3)(x-2)(x-1)) [-1, 5, -55, 55]}