Question

How do you identify all asymptotes or holes for `f(x)=(2x)/(x^2-1)`?

Answer 1

VA: 1,-1
HA: 0
SA: N/A

Explanation:

Vertical asymptote: Set denominator = 0 but factor first
`x^2-1`=`(x-1)(x+1)`
`(0-1)=-1`
`(0+1)=1`

`1` and `-1` are your vertical asymptotes

Horizontal asymptote: When the x with the biggest power in the numerator is less than the x with the biggest power in the numerator, the horizontal asymptote is `0`.

Since a horizontal asymptote exists, then a slanted asymptote does not exist. It would only exist if the x with the biggest power in the numerator is biggest than the x with the biggest power in the denominator.