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Slant asymptotes of rational functions

Algebra 2 ยท 20 problems, easy to harder

โ–ถ Watch the first one solved

How we do it

  1. Compare the powers
  2. Set it up
  3. First part of the answer
  4. Second part of the answer
  5. Read the slant asymptote
  6. Vertical asymptote
  7. Check

The steps for Slant asymptote of f(x) = (x^2 + 1)/(x - 1). Every solution names its steps, so you always know what comes next.

Worked examples

Slant asymptote of f(x) = (x^2 + 1)/(x - 1) Easy

The question

Question: find the slant asymptote of f(x) = (xยฒ + 1)/(x โˆ’ 1).

f(x) = x2 + 1x โˆ’ 1

An asymptote is a line the graph gets closer and closer to.

We draw it dashed.

asymptote: a line the graph gets close to

1 Compare the powers

The top has xยฒ: power 2.

The bottom has x: power 1.

top : power 2, bottom: power 1

The top power is one bigger.

Then there is no flat asymptote.

Instead, the graph gets close to a slanted line: a slant asymptote.

Dividing finds it.

top power one bigger โ†’ a slanted line

2 Set it up

We write it like long division with numbers.

x โˆ’ 1 goes outside, xยฒ + 1 inside.

There is no x part.

We write 0x, so the columns line up.

3 First part of the answer

Divide the first parts: xยฒ divided by x is x.

We write it on top.

Multiply: x times (x โˆ’ 1) is xยฒ โˆ’ x.

We write it under, with a minus: we take it away.

xยฒ take away xยฒ is 0.

That is why we chose x.

0x take away โˆ’x is x.

Bring down the next part: 1.

4 Second part of the answer

Divide the first parts: x divided by x is 1.

We write it on top.

Multiply: 1 times (x โˆ’ 1) is x โˆ’ 1.

We write it under, with a minus: we take it away.

x take away x is 0.

That is why we chose 1.

1 take away โˆ’1 is 2.

2 has no x.

We can't divide it by x.

So 2 is the remainder: what is left over.

So f(x) is x + 1, plus 2/(x โˆ’ 1).

f(x) = x + 1 + 2x โˆ’ 1

5 Read the slant asymptote

When x is big, 2/(x โˆ’ 1) is tiny.

At x = 100 it is 2/99: about 0.02.

Almost 0.

x = 100: 299 โ‰ˆ 0.02

So for a big x, f(x) is close to x + 1.

The slant asymptote is y = x + 1.

slant asymptote : y = x + 1

Careful!

The remainder part is not in the line.

It gets smaller and smaller, so we leave it out.

Careful: not y = x + 1 + 2x โˆ’ 1 โœ—

6 Vertical asymptote

The bottom: x โˆ’ 1 = 0 when x = 1.

x โˆ’ 1 = 0 โ†’ x = 1

We can never divide by 0.

The top is not 0 there, so near it the graph shoots up or down: a vertical asymptote.

รท 0: not allowed

So the asymptotes are y = x + 1 and x = 1.

= y = x + 1, x = 1

7 Check

Let's check with a picture.

The vertical asymptote x = 1, dashed.

The slant asymptote y = x + 1, dashed.

The graph of f: on each side it gets close to both dashed lines.

And with numbers: at x = 100, f is about 101.02, and the line gives 101.

Very close.

It works!

x = 100: f(100) โ‰ˆ 101.02, line: 101

Let's look back.

First: compare the powers.

Then: set it up.

Then: first part of the answer.

Then: second part of the answer.

Then: read the slant asymptote.

Then: vertical asymptote.

Last: check.

Answer
= y = x + 1, x = 1

โ–ถ Watch it solved

Asymptotes of f(x) = (x^2 - x - 6)/(x - 1) Medium

The question

Question: find the asymptotes of f(x) = (xยฒ โˆ’ x โˆ’ 6)/(x โˆ’ 1).

f(x) = x2 โˆ’ x โˆ’ 6x โˆ’ 1

An asymptote is a line the graph gets closer and closer to.

We draw it dashed.

asymptote: a line the graph gets close to

1 Compare the powers

The top has xยฒ: power 2.

The bottom has x: power 1.

top : power 2, bottom: power 1

The top power is one bigger.

Then there is no flat asymptote.

Instead, the graph gets close to a slanted line: a slant asymptote.

Dividing finds it.

top power one bigger โ†’ a slanted line

2 Set it up

We write it like long division with numbers.

x โˆ’ 1 goes outside, xยฒ โˆ’ x โˆ’ 6 inside.

3 First part of the answer

Divide the first parts: xยฒ divided by x is x.

We write it on top.

Multiply: x times (x โˆ’ 1) is xยฒ โˆ’ x.

We write it under, with a minus: we take it away.

xยฒ take away xยฒ is 0.

That is why we chose x.

โˆ’x take away โˆ’x is 0x.

Bring down the next part: โˆ’6.

4 Second part of the answer

Divide the first parts: 0x divided by x is 0.

We write it on top.

Multiply: 0 times (x โˆ’ 1) is 0.

We write it under, with a minus: we take it away.

0x take away 0x is 0.

That is why we chose 0.

โˆ’6 take away 0 is โˆ’6.

โˆ’6 has no x.

We can't divide it by x.

So โˆ’6 is the remainder: what is left over.

So f(x) is x, minus 6/(x โˆ’ 1).

f(x) = x โˆ’ 6x โˆ’ 1

5 Read the slant asymptote

When x is big, 6/(x โˆ’ 1) is tiny.

At x = 100 it is 6/99: about 0.06.

Almost 0.

x = 100: 699 โ‰ˆ 0.06

So for a big x, f(x) is close to x.

The slant asymptote is y = x.

slant asymptote : y = x

Careful!

The remainder part is not in the line.

It gets smaller and smaller, so we leave it out.

Careful: not y = x โˆ’ 6x โˆ’ 1 โœ—

6 Vertical asymptote

The bottom: x โˆ’ 1 = 0 when x = 1.

x โˆ’ 1 = 0 โ†’ x = 1

We can never divide by 0.

The top is not 0 there, so near it the graph shoots up or down: a vertical asymptote.

รท 0: not allowed

So the asymptotes are y = x and x = 1.

= y = x, x = 1

7 Check

Let's check with a picture.

The vertical asymptote x = 1, dashed.

The slant asymptote y = x, dashed.

The graph of f: on each side it gets close to both dashed lines.

And with numbers: at x = 100, f is about 99.94, and the line gives 100.

Very close.

It works!

x = 100: f(100) โ‰ˆ 99.94, line: 100

Let's look back.

First: compare the powers.

Then: set it up.

Then: first part of the answer.

Then: second part of the answer.

Then: read the slant asymptote.

Then: vertical asymptote.

Last: check.

Answer
= y = x, x = 1

โ–ถ Watch it solved

Asymptotes of f(x) = (3x^2 + 2x - 1)/(x + 2) Harder

The question

Question: find the asymptotes of f(x) = (3xยฒ + 2x โˆ’ 1)/(x + 2).

f(x) = 3x2 + 2x โˆ’ 1x + 2

An asymptote is a line the graph gets closer and closer to.

We draw it dashed.

asymptote: a line the graph gets close to

1 Compare the powers

The top has xยฒ: power 2.

The bottom has x: power 1.

top : power 2, bottom: power 1

The top power is one bigger.

Then there is no flat asymptote.

Instead, the graph gets close to a slanted line: a slant asymptote.

Dividing finds it.

top power one bigger โ†’ a slanted line

2 Set it up

We write it like long division with numbers.

x + 2 goes outside, 3xยฒ + 2x โˆ’ 1 inside.

3 First part of the answer

Divide the first parts: 3xยฒ divided by x is 3x.

We write it on top.

Multiply: 3x times (x + 2) is 3xยฒ + 6x.

We write it under, with a minus: we take it away.

3xยฒ take away 3xยฒ is 0.

That is why we chose 3x.

2x take away 6x is โˆ’4x.

Bring down the next part: โˆ’1.

4 Second part of the answer

Divide the first parts: โˆ’4x divided by x is โˆ’4.

We write it on top.

Multiply: โˆ’4 times (x + 2) is โˆ’4x โˆ’ 8.

We write it under, with a minus: we take it away.

โˆ’4x take away โˆ’4x is 0.

That is why we chose โˆ’4.

โˆ’1 take away โˆ’8 is 7.

7 has no x.

We can't divide it by x.

So 7 is the remainder: what is left over.

So f(x) is 3x โˆ’ 4, plus 7/(x + 2).

f(x) = 3x โˆ’ 4 + 7x + 2

5 Read the slant asymptote

When x is big, 7/(x + 2) is tiny.

At x = 100 it is 7/102: about 0.07.

Almost 0.

x = 100: 7102 โ‰ˆ 0.07

So for a big x, f(x) is close to 3x โˆ’ 4.

The slant asymptote is y = 3x โˆ’ 4.

slant asymptote : y = 3x โˆ’ 4

Careful!

The remainder part is not in the line.

It gets smaller and smaller, so we leave it out.

Careful: not y = 3x โˆ’ 4 + 7x + 2 โœ—

6 Vertical asymptote

The bottom: x + 2 = 0 when x = โˆ’2.

x + 2 = 0 โ†’ x = โˆ’2

We can never divide by 0.

The top is not 0 there, so near it the graph shoots up or down: a vertical asymptote.

รท 0: not allowed

So the asymptotes are y = 3x โˆ’ 4 and x = โˆ’2.

= y = 3x โˆ’ 4, x = โˆ’2

7 Check

Let's check with a picture.

The vertical asymptote x = โˆ’2, dashed.

The slant asymptote y = 3x โˆ’ 4, dashed.

The graph of f: on each side it gets close to both dashed lines.

And with numbers: at x = 100, f is about 296.07, and the line gives 296.

Very close.

It works!

x = 100: f(100) โ‰ˆ 296.07, line: 296

Let's look back.

First: compare the powers.

Then: set it up.

Then: first part of the answer.

Then: second part of the answer.

Then: read the slant asymptote.

Then: vertical asymptote.

Last: check.

Answer
= y = 3x โˆ’ 4, x = โˆ’2

โ–ถ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. slant asymptote of f(x) = (x^2 + 1)/(x - 1)
    Answer
    = y = x + 1, x = 1
  2. slant asymptote of f(x) = (x^2 + 3)/(x + 1)
    Answer
    = y = x โˆ’ 1, x = โˆ’1
  3. slant asymptote of f(x) = (x^2 - 5)/(x - 2)
    Answer
    = y = x + 2, x = 2
  4. slant asymptote of f(x) = (x^2 + 2x + 4)/(x + 1)
    Answer
    = y = x + 1, x = โˆ’1
  5. slant asymptote of f(x) = (x^2 - 3x + 1)/(x - 1)
    Answer
    = y = x โˆ’ 2, x = 1
  6. slant asymptote of f(x) = (x^2 + x - 1)/(x + 2)
    Answer
    = y = x โˆ’ 1, x = โˆ’2
  7. slant asymptote of f(x) = (x^2 + 4)/(x - 3)
    Answer
    = y = x + 3, x = 3
  8. slant asymptote of f(x) = (x^2 - x)/(x + 1)
    Answer
    = y = x โˆ’ 2, x = โˆ’1

Medium

  1. asymptotes of f(x) = (x^2 - x - 6)/(x - 1)
    Answer
    = y = x, x = 1
  2. asymptotes of f(x) = (2x^2 + 3x)/(x + 2)
    Answer
    = y = 2x โˆ’ 1, x = โˆ’2
  3. asymptotes of f(x) = (x^2 + 5x + 2)/(x + 3)
    Answer
    = y = x + 2, x = โˆ’3
  4. asymptotes of f(x) = (2x^2 - x + 1)/(x - 1)
    Answer
    = y = 2x + 1, x = 1
  5. asymptotes of f(x) = (x^2 - 2x - 2)/(x - 3)
    Answer
    = y = x + 1, x = 3
  6. asymptotes of f(x) = (-x^2 + 2x + 1)/(x - 1)
    Answer
    = y = โˆ’x + 1, x = 1

Harder

  1. asymptotes of f(x) = (3x^2 + 2x - 1)/(x + 2)
    Answer
    = y = 3x โˆ’ 4, x = โˆ’2
  2. asymptotes of f(x) = (2x^2 + 5x - 4)/(x + 3)
    Answer
    = y = 2x โˆ’ 1, x = โˆ’3
  3. asymptotes of f(x) = (x^2 - 4x + 7)/(x - 2)
    Answer
    = y = x โˆ’ 2, x = 2
  4. asymptotes of f(x) = (-2x^2 + 3x + 4)/(x - 2)
    Answer
    = y = โˆ’2x โˆ’ 1, x = 2
  5. slant asymptote of f(x) = (2x^2 + 1)/(x + 1)
    Answer
    = y = 2x โˆ’ 2, x = โˆ’1
  6. asymptotes of f(x) = (x^2 + 6)/(x + 2)
    Answer
    = y = x โˆ’ 2, x = โˆ’2