Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `x/(x^2+x-6)`?

Answer 1

`"vertical asymptotes at "x--3" and "x=2`
`"horizontal asymptote at "y=0`

Explanation:

`"let "f(x)=x/(x^2+x-6)`

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 they are vertical asymptotes.

`"solve "x^2+x-6=0rArr(x+3)(x-2)=0`

`x=-3" and "x=2" are the 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)=(x/x^2)/(x^2/x^2+x/x^2-6/x^2)=(1/x)/(1+1/x-6/x^2)`

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

`y=0" 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 are no oblique asymptotes.
graph{x/(x^2+x-6) [-10, 10, -5, 5]}