Question

How do you find the asymptotes for `y=(2x^2+3)/(x^2-6)`?

Answer 1

vertical asymptotes at `x=±sqrt6`
horizontal asymptote at y = 2

Explanation:

The denominator of y cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: `x^2-6=0rArrx^2=6rArrx=±sqrt6`

`rArrx=-sqrt6" and " x=sqrt6" are the asymptotes"`

Horizontal asymptotes occur as

`lim_(xto+-oo),ytoc" (a constant)"`

divide the terms on the numerator/denominator by the highest power of x, that is `x^2`

`((2x^2)/x^2+3/x^2)/(x^2/x^2-6/x^2)=(2+3/x^2)/(1-6/x^2)`

as `xto+-oo,yto(2+0)/(1-0)`

`rArry=2" is the asymptote"`
graph{(2x^2+3)/(x^2-6) [-10, 10, -5, 5]}