Question

How do you graph `y=x/(x^2-9)` using asymptotes, intercepts, end behavior?

Answer 1

The vertical asymptotes are `x=3` and `x=-3`
The horizontal asymptote is `y=0`
No slant asymptotes
Only one intercept at `(0,0)`

Explanation:

Let's factorise the denominator `x^2-9=(x+3)(x-3)`
As we canot divide by `0`, the vertical asymptotes are `x=3` and `x=-3`

As the degree of the numerator is `<` the degree of the denominator, we do not have a slant asymptote.

`lim_(x->-oo)y=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^(-)`

`lim_(x->+oo)y=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^(+)`

So, `y=0` is a horizontal asymptote

The intercept is, when `x=0` `=>` `y=0`
graph{x/(x^2-9) [-11.25, 11.24, -5.63, 5.62]}