Question

How do you find vertical, horizontal and oblique asymptotes for `(x^2+x-6)/(x^2-x-6)`?

Answer 1

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

Explanation:

`"let "f(x)=(x^2+x-6)/(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=-2" and "x=3" 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^2/x^2+x/x^2-6/x^2)/(x^2/x^2-x/x^2-6/x^2)=(1+1/x-6/x^2)/(1-1/x-6/x^2)`

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

`y=1" 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^2+x-6)/(x^2-x-6) [-10, 10, -5, 5]}