Question

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

Answer 1

Vertical asymptotes exist at `x=1, 2, or 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=(x-2)(x-1)(x-3)`

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, or 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]}