Question

How do you find the asymptotes for `h(x) = (2x^2-5x-12)/(3x^2-11x-4)`?

Answer 1

vertical asymptote `x = -1/3`
horizontal asymptote y `= 2/3`

Explanation:

First step is to factorise h(x).

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

`rArr h(x) = (2x+3)/(3x+1)`

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

solve : 3x + 1 = 0 → `x = - 1/3" is the asymptote "`

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

divide all terms on numerator/denominator by x

`((2x)/x + 3/x)/((3x)/x + 1/x) = (2 + 3/x )/(3 + 1/x)`

as `xto+-oo , 3/x" and " 1/x to 0`

`rArr y = 2/3" is the asymptote "`

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