Step by step
The question
Question: find the asymptotes and holes of f(x) = (x² + 2x)/(x² − 4).
An asymptote is a line the graph gets closer and closer to, but never reaches.
We draw it dashed.
1 Factor the top and bottom
The top: x² + 2x.
Every part can be divided by x.
We take it out in front.
So the top is x(x + 2).
The bottom: x² − 4.
x² − 4 is x² minus 2 squared.
A square minus a square is the plus bracket times the minus bracket.
So the bottom is (x + 2)(x − 2).
2 Find the holes
Now look for the same bracket on the top and the bottom.
(x + 2) is on the top and the bottom.
(x + 2)/(x + 2) is 1.
Cross them out.
What is left: x/(x − 2).
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 = −2 into what is left: y is 1/2.
So the hole is at (−2, 1/2).
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 − 2 = 0 when x = 2.
Look: at x = 2.1, the bottom is tiny.
So f(x) is about 21: 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 same power on top and bottom.
The x's cancel, and the numbers in front give 1.
Look: at x = 100, f(x) is about 1.02.
Close to 1.
So the horizontal asymptote is y = 1.
So: hole (−2, 1/2), x = 2 and y = 1.
5 Check
Let's check: put x = 2 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 1.
It works!
Let's look back.
First: factor the top and bottom.
Then: find the holes.
Then: vertical asymptotes.
Then: horizontal asymptote.
Last: check.