Question

How do you find the asymptotes for `f(x) =(x^2 - 4)/(9 - x^2)`?

Answer 1

Vertical asymptotes at `x=3` and `x=-3` and horizontle asymptote at `y=-1`

Explanation:

To find all the asymptotes for function `y=(x^2-4)/(9-x^2)` let us first start with vertical asymptotes, which are given by putting denominator equal to zero i.e. here `9-x^2=0` i.e. `(3-x)(3+x)=0`. Hence vertical asymptotes are `x=3` and `x=-3`.

Further as in `y=(x^2-4)/(9-x^2)`, highest degree in numerator is for `x^2` and for denominator `-x^2`, hence we have a horizontal asymptote at `y=x^2/(-x^2)=-1`

graph{(x^2-4)/(9-x^2) [-10, 10, -5, 5]}