Question

How do you find the vertical, horizontal or slant asymptotes for `f(x) = (2x^2-5x-12)/(3x^2-11x-4 )`?

Answer 1

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

Explanation:

The first step is to factorise and simplify f(x).

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

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

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)to c" (a constant)"`

divide numerator/denominator by x

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

as `xto+-oo,f(x)to(2+0)/(3+0)`

`rArry=2/3" is the asymptote"`

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