Question

How do you find the asymptote(s) or hole(s) of `f(x) = (x+3)/(x^2-9)`?

Answer 1

asymptote at `x=3`

hole at `x=-3`

Explanation:

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

Factor the denominator, it is the difference of squares:

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

The factor(s) in the denominator that will cancel out, in this case `x+3` are holes.

The factor(s) in the denominator that will not cancel out, in this case `x-3` are asymptotes.

Here is the graph, it shows the asymptote, but not the hole at `x=-3`

graph{(x+3)/(x^2-9) [-6.67, 13.33, -4.44, 5.56]}

Answer 2

VA: `x=3`

HA: `y=0`

Explanation:

We have the following expression

`(x+3)/color(blue)((x^2-9))`

What I have in blue is a Difference of Squares of the form

`a^2-b^2`, where `a=x` and `b=3`, which is factored as

`(a+b)(a-b)`

Now, our new expression is

`(x+3)/color(blue)((x+3)(x-3))`

which simplifies to

`1/(x-3)`

Vertical asymptotes are found by setting the denominator equal to zero, because this is the value for which the function is undefined.

`x-3=0`

`=>color(red)(x=3)` (Vertical Asymptote)

Horizontal asymptotes are found by comparing the degrees. Since there is a higher degree in the denominator, we have a horizontal asymptote at

`=>color(darkviolet)(y=0)` (Horizontal Asymptote)

We have a vertical asymptote at `x=3`, and a horizontal asymptote at `y=0`.

Hope this helps!