Question

How do you find the vertical, horizontal or slant asymptotes for `(3x)/(x^2+2)`?

Answer 1

`"horizontal asymptote at "y=0`

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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 thet are vertical asymptotes.

`"solve "x^2+2=0rArrx^2=-2`

`"this has no real solutions hence there are no vertical"`
`"asymptotes"`

`"horizontal asymptotes occur as"`

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

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

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

as `xto+-oo,f(x)to0/(1+0)`

`rArry=0" is the asymptote"`

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