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

Feb 10, 2016

$x = - 2$ 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

$\left(x + 2\right) \left(x - 1\right) = 0$

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]}