Step by step
The question
Question: graph f(x) = (x + 2)/(x + 2)(x − 1).
An asymptote is a line the graph gets closer and closer to, but never reaches.
We draw it dashed.
1 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: 1/(x − 1).
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 −1/3.
So the hole is at (−2, −1/3).
2 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 − 1 = 0 when x = 1.
Look: at x = 1.1, the bottom is tiny.
So f(x) is about 10: huge.
3 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.
4 Find the intercepts
The top that is left has no x.
It is never 0, so the graph never crosses the x-axis.
Put x = 0 in: f(0) is −1.
The graph crosses the y-axis at (0, −1).
5 Draw the graph
A few points on both sides of x = 1.
We put each x into f.
Draw the grid.
Draw the asymptotes dashed: x = 1 is vertical.
y = 0 is flat.
Plot the intercepts and the points.
Join the points with a smooth curve.
It gets close to the dashed lines, but never touches them.
The hole: an open circle at (−2, −1/3).
6 Check
Let's check one more point: put x = −4 in the question.
We get −1/5.
The point at x = −4 is on our curve.
It works!
Let's look back.
First: find the holes.
Then: vertical asymptotes.
Then: horizontal asymptote.
Then: find the intercepts.
Then: draw the graph.
Last: check.