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

1 Answer
Apr 26, 2016

Answer:

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]}