For what values of x, if any, does #f(x) = 1/((x-1)(x-6)) # have vertical asymptotes?

1 Answer
Dec 31, 2015

Answer:

#x=1,6#

Explanation:

Vertical asymptotes occur where the domain is undefined. That is, the spots when an error arises whenever #x# is plugged in. For a rational function such as this one, such an error will occur whenever the denominator of a fraction is equal to #0#.

To find the spots where #f(x)# has vertical asymptotes, set the denominator #(x-1)(x-6)=0#.

#(x-1)(x-6)=0#

#x-1=0#
or
#x-6=0#

#x=1#
or
#x=6#

Check a graph. The vertical asymptotes should occur when #x=1,6#.

graph{1/((x-1)(x-6) [-7.51, 17.8, -6.34, 6.32]}