Question

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

Answer 1

`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]}