Question

How to find the asymptotes of `f(x)=(x^2+1)/(2x^2-3x-2)`?

Answer 1

vertical asymptotes `x = -1/2 , x = 2`
horizontal asymptotes `y = 1/2`

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.

solve : `2x^2-3x-2 = 0 → (2x+1)(x-2) = 0`

`rArr x = -1/2 , x = 2 " are the asymptotes "`

Horizontal asymptotes occur as `lim_(xtooo) f(x) → 0`

If the degree of the numerator/denominator are equal , as is the case here , both of degree 2. Then the equation is the ratio of the coefficients of the highest powers.

`rArr y = 1/2 " is the asymptote"`

Here is the graph of f(x).
graph{(x^2+1)/(2x^2-3x-2) [-10, 10, -5, 5]}