Question

How do you find the vertical, horizontal or slant asymptotes for `y=(x^2-1)/(2x^2 + 3x-2)`?

Answer 1

The vertical asymptotes are `x=1/2` and `x=-2`

The horizontal asymptote is `y=1/2`

Explanation:

We facorise `y=((x+1)(x-1))/((2x-1)(x+2))`
As we cannot divide by `0`
So for `x=1/2` and `x=-2` we have vertical asymptotes
To look for horizontal asymptotes, we rewrite
`y` as `y=(1-1/x^2)/(2+3/x-2/x^2)`

the limit `y=(1-0)/(2+0+0)=1/2`
`y->-oo`

the limit `y=(1-0)/(2+0+0)=1/2`
`y->+oo`
So `y=1/2` is a horizontal asymptote