Answer:
x = - 1, x = 1
Step-by-step explanation:
Given
f(x) = [tex]\frac{2x^2+3x+6}{x^2-1}[/tex]
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
