Question

How do you find the vertical, horizontal or slant asymptotes for `(3x^2+2x-1)/(x^2-4)`?

Answer 1

vertical asymptotes x = ± 2
horizontal asymptote y = 3

Explanation:

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

solve : `x^2-4=0 → (x-2)(x+2)=0 → x=±2`

`rArrx=-2" and " x=2" are the asymptotes "`

Horizontal asymptotes occur as `lim_(x to +- oo) , f(x) to 0`

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

`((3x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)`

as `x to +- oo , y to (3+0-0)/(1-0)`

`rArry=3" is the asymptote "`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both degree 2) hence there are no slant asymptotes.
graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}