Question

How do you find vertical, horizontal and oblique asymptotes for `h(x) = (2x^2-5x-12)/(3x^2-11x-4 )`?

Answer 1

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

Explanation:

`"factorise numerator/denominator and simplify"`

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

The denominator of h(x) cannot be zero as this would make h(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),h(x)toc" ( a constant)"`

`"divide terms on numerator/denominator by " x`

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

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

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

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there is no oblique asymptote.
graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}