Question

How do you graph `f(x)=x/(x^2-9)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

Below

Explanation:

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

Therefore, the vertical asymptotes are `x=3` and `x=-3` since the denominator cannot equal to 0. ie `x^2-9=0` is used to find our asymptote.

For our horizontal asymptote, we look at the degree of our numerator and denominator. Since the degree of the numerator is less than the denominator, then `y=0` is our horizontal asymptote.

Another way you can think of it is it you sub random numbers into the `x` of your numerator and denominator, you will find that the numerator will have a smaller value than your denominator. If you divide a smaller number by a larger number, your resulting value will be quite small, tending to 0 ie the line `y=0`

For our intercepts,
When `x=0`, `y=0`
When `y=0`, `x=0`
Therefore, `(0,0)` is our only intercept

Drawing up our asymptotes (horizontal and vertical) and our intercept, we should be able to draw our graph. Remember that for our asymptotes, it only affects the ends of the graph as it is approaching the asymptotes from above or below. Our graph can actually cross the asymptotes.

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