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

1 Answer
Feb 10, 2016

Answer:

#x=-2# and #x=1#

Explanation:

Vertical asymptotes will occur whenever the denominator of the function is equal to #0#.

So, the vertical asymptotes are at the values of #x# that satisfy

#(x+2)(x-1)=0#

Just like when solving a quadratic equation, we set both of these terms equal to #0# individually. When terms are being multiplied by one another to equal #0#, at least one of the terms must also equal #0#.

So, in setting both term equal to #0#, we find the #2# values where there are vertical asymptotes:

#x+2=0" "=>" "color(red)(x=-2#

#x-1=0" "=>" "color(red)(x=1#

We can check a graph of the function:

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