Question

For `(x^2+1)/(2x^2-3x-2)`, how do you find horizontal and vertical asymptotes?

Answer 1

Horizontal asymptote is `y=1/2`
Vertical asymptotes are: `x=-1/2` and `x=2`

Explanation:

For the Horizontal asymptote, you look at the degrees of the numerator and denominator. Since the degree of the numerator and denominator are the same, we use a ratio of the leading coefficients.

`y="lead coef."/"lead coef." = 1/2`

So the Horizontal asymptote is `y=1/2`

For the Vertical asymptote, we look at the zeros of the denominator. Set the denominator equal to zero and solve for x.

`2x^2-3x-2 = 0`

`(2x+1)(x-2)=0`

`2x+1=0` and `x-2=0`

`x=-1/2` and `x=2`

So your Vertical asymptotes are:
`x=-1/2` and `x=2`