Question

How do you find vertical, horizontal and oblique asymptotes for `(3x^2) /( x^2 - 9)`?

Answer 1

vertical asymptotes x = ± 3
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 - 9 =0 → (x-3)(x+3) = 0 → x = ± 3`

`rArr x = -3 , x = 3 " are the asymptotes "`

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

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

`(3x^2)/x^2/(x^2/x^2 - 9/x^2 )= 3/(1-9/x^2)`

as `x to+- oo , 9/x^2 to 0" and " y to 3/1`

`rArr y = 3 " is the asymptote "`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(3x^2)/(x^2-9) [-10, 10, -5, 5]}