Steply

Asymptotes of f(x) = (x^2 - 4)/(x^2 + x - 2)

Step by step, with the reason for every step · Rational functions · Algebra 2

Step by step

The question

Question: find the asymptotes and holes of f(x) = (x² − 4)/(x² + x − 2).

f(x) = x2 − 4x2 + x − 2

An asymptote is a line the graph gets closer and closer to, but never reaches.

We draw it dashed.

asymptote: a line the graph gets close to, but never reaches

1 Factor the top and bottom

The top: x² − 4.

top : x2 − 4

x² − 4 is x² minus 2 squared.

A square minus a square is the plus bracket times the minus bracket.

a² − b² = (a + b)(a − b)

So the top is (x + 2)(x − 2).

= (x + 2)(x − 2)

The bottom: x² + x − 2.

bottom : x2 + x − 2

We need two numbers.

They multiply to −2, and they add to 1.

? × ? = −2, ? + ? = 1

They multiply to a minus number, so one number is plus and one is minus.

They add to a plus number, so the bigger one is plus.

× is − → one +, one −; bigger one +

We list the pairs that make 2, smallest first.

We take the smaller from the bigger.

We need 1.

1 and 2 give 1.

Yes!

With the signs: −1 and 2.

1 × 2: 2 − 1 = 1 ✓

So the bottom is (x − 1)(x + 2).

= (x − 1)(x + 2)

2 Find the holes

Now look for the same bracket on the top and the bottom.

f(x) = (x + 2)(x − 2)(x − 1)(x + 2)

(x + 2) is on the top and the bottom.

(x + 2)/(x + 2) is 1.

Cross them out.

What is left: (x − 2)/(x − 1).

f(x) = x − 2x − 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.

crossed out → a hole (one missing point)

Careful!

A crossed-out bracket gives a hole, not an asymptote.

The graph only misses one point there.

Careful: not x = −2 is an asymptote ✗

Put x = −2 into what is left: y is 4/3.

So the hole is at (−2, 4/3).

= hole: x = −2, y = 43

3 Vertical asymptotes

We can never divide by 0.

Where the bottom is 0, the graph shoots up or down.

÷ 0: not allowed

What is left at the bottom: x − 1 = 0 when x = 1.

x − 1 = 0 → x = 1

Look: at x = 1.1, the bottom is tiny.

So f(x) is about −9: huge.

x = 1.1: bottom 0.1, f(x) ≈ −9

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.

a big x: only the biggest powers count

The same power on top and bottom.

The x's cancel, and the numbers in front give 1.

f(x) ≈ x2x2 = 1

Look: at x = 100, f(x) is about 0.99.

Close to 1.

x = 100: f(100) ≈ 0.99

So the horizontal asymptote is y = 1.

y = 1

So: hole (−2, 4/3), x = 1 and y = 1.

= hole (−2, 43), x = 1, y = 1

5 Check

Let's check: put x = 1 in the question's bottom.

It gives 0.

Dividing by 0 is not allowed: an asymptote.

x = 1: bottom = 0 ÷ 0

And a very big x, 1000, gives almost exactly 1.

It works!

x = 1000: f(1000) ≈ 0.999 close to 1

Let's look back.

First: factor the top and bottom.

Then: find the holes.

Then: vertical asymptotes.

Then: horizontal asymptote.

Last: check.

Answer
= hole (−2, 43), x = 1, y = 1

Your turn

Try this one yourself, then check:

asymptotes of f(x) = x2 + x − 6x2 + x − 6
Show the answer
hole (2, 1), hole (−3, 1), y = 1

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