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

##### 1 Answer

Feb 10, 2016

#### Answer:

#### Explanation:

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

So, the vertical asymptotes are at the values of

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

Just like when solving a quadratic equation, we set *both* of these terms equal to

So, in setting both term equal to

#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]}