Step by step
The question
Question: find the asymptotes and holes of f(x) = (x + 3)/(x² − 9).
An asymptote is a line the graph gets closer and closer to, but never reaches.
We draw it dashed.
1 Factor the bottom
The bottom: x² − 9.
x² − 9 is x² minus 3 squared.
A square minus a square is the plus bracket times the minus bracket.
So the bottom is (x + 3)(x − 3).
2 Find the holes
Now look for the same bracket on the top and the bottom.
(x + 3) is on the top and the bottom.
(x + 3)/(x + 3) is 1.
Cross them out.
What is left: 1/(x − 3).
But the x that made the crossed bracket 0 is still not allowed in the question.
The graph just has one missing point there: a hole.
Careful!
A crossed-out bracket gives a hole, not an asymptote.
The graph only misses one point there.
Put x = −3 into what is left: y is −1/6.
So the hole is at (−3, −1/6).
3 Vertical asymptotes
We can never divide by 0.
Where the bottom is 0, the graph shoots up or down.
What is left at the bottom: x − 3 = 0 when x = 3.
Look: at x = 3.1, the bottom is tiny.
So f(x) is about 10: huge.
4 Horizontal asymptote
For a very big x, the biggest power on the top and on the bottom is all that counts.
The rest is tiny next to it.
The top has a smaller power than the bottom.
When x is huge, the bottom is much bigger, so f(x) gets close to 0.
Look: at x = 100, f(x) is about 0.01.
Close to 0.
So the horizontal asymptote is y = 0.
So: hole (−3, −1/6), x = 3 and y = 0.
5 Check
Let's check: put x = 3 in the question's bottom.
It gives 0.
Dividing by 0 is not allowed: an asymptote.
And a very big x, 1000, gives almost exactly 0.
It works!
Let's look back.
First: factor the bottom.
Then: find the holes.
Then: vertical asymptotes.
Then: horizontal asymptote.
Last: check.