Question

What are the asymptotes of `f(x)=(x^2+1)/(x^2-9)`?

Answer 1

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

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 they are vertical asymptotes.

`"solve "x^2-9=0rArr(x-3)(x+3)=0`

`rArrx=-3" 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+1/x^2)/(x^2/x^2-9/x^2)=(1+1/x^2)/(1-9/x^2)`

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

`rArry=1" is the asymptote"`
graph{(x^2+1)/(x^2-9) [-10, 10, -5, 5]}