Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = (4x^2 - 1 ) / (2x^2 + 5x - 12)`?

Answer 1

vertical asymptotes `x = -4 , x = 3/2`
horizontal asymptote y = 2

Explanation:

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

solve : `2x^2 + 5x - 12 = 0 → (2x-3)(x+4) = 0`

`rArr x = - 4 " and " x = 3/2 " are the asymptotes "`

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

divide all terms on numerator/denominator by `x^2`

`((4x^2)/x^2 - 1/x^2)/((2x^2)/x^2 + (5x)/x^2 - 12/x^2) = (4-1/x^2)/(2+5/x-12/x^2)`

now as x →∞ `1/x^2 , 5/x " and " 12/x^2 → 0`

`rArr y = 4/2 → y = 2 " is the asymptote "`

Slant asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here , hence there are no slant asymptotes.

Here is the graph of the function.
graph{(4x^2-1)/(2x^2+5x-12) [-10, 10, -5, 5]}