Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `(x+3)/(x^2-9)`?

Answer 1

`"vertical asymptote at "x=3`
`"horizontal asymptote at "y=0`

Explanation:

`f(x)=(x+3)/(x^2-9)`

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the denominator is non-zero for this value then it is a vertical asymptote.

`"factorise and simplify f(x)"`

`f(x)=(cancel((x+3)))/(cancel((x+3))(x-3))=1/(x-3)`

`"the removal of the factor "(x+3)" from the numerator"`
`"and denominator indicates a hole at x = - 3"`

`"solve " x-3=0rArrx=3" is the asymptote"`

`"horizontal asymptotes occur as"`

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

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

`f(x)=(1/x)/(x/x-3/x)=(1/x)/(1-3/x)`

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

`rArry=0" is the asymptote"`

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