Find the vertical asymptote of f(x)=2x^2+3x+6/x^2-1 I'm having trouble with this one, seems simple tho I just don't want to make a stupid mistake,,, And here are the choices:

x = - 1, x = 1

Step-by-step explanation:

Given

f(x) =

The denominator cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non zero for these values then they are vertical asymptotes.

x² - 1 = 0 ← difference of squares

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

x - 1 = 0 ⇒ x = 1

x + 1 = 0 ⇒ x = - 1

x = - 1 and x = 1 are vertical asymptotes