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OBJECTIVES:

1. Determine the domain of the variable in a rational expression. 2. Solve rational equations. 3. Recognize the graph of a rational function

1. Determine the domain of the variable in a rational expression. 2. Solve rational equations. 3. Recognize the graph of a rational function

1.
Copyright © 2010 Pearson Education, Inc. All rights reserved

Sec 8.4 - 1

Sec 8.4 - 1

2.
Chapter 8

Rational Expressions and

Functions

Copyright © 2010 Pearson Education, Inc. All rights reserved

Sec 8.4 - 2

Rational Expressions and

Functions

Copyright © 2010 Pearson Education, Inc. All rights reserved

Sec 8.4 - 2

3.
8.4

Equations with Rational

Expressions and Graphs

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Sec 8.4 - 3

Equations with Rational

Expressions and Graphs

Copyright © 2010 Pearson Education, Inc. All rights reserved

Sec 8.4 - 3

4.
8.4 Equations with Rational Expressions and Graphs

Objectives

1. Determine the domain of the variable in a rational

expression.

2. Solve rational equations.

3. Recognize the graph of a rational function.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 4

Objectives

1. Determine the domain of the variable in a rational

expression.

2. Solve rational equations.

3. Recognize the graph of a rational function.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 4

5.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 1 Determining the Domains of Rational

Equations

Find the domain of the equation.

(a) 5 1 11 The domain is { x | x ≠ 0 }.

– =

x 6 2x

The domains of the three rational terms of the equation are, in order,

{ x | x ≠ 0 }, (-∞, ∞), { x | x ≠ 0 }. The intersection of these three domains is all

real numbers except 0, which may be written { x | x ≠ 0 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 5

EXAMPLE 1 Determining the Domains of Rational

Equations

Find the domain of the equation.

(a) 5 1 11 The domain is { x | x ≠ 0 }.

– =

x 6 2x

The domains of the three rational terms of the equation are, in order,

{ x | x ≠ 0 }, (-∞, ∞), { x | x ≠ 0 }. The intersection of these three domains is all

real numbers except 0, which may be written { x | x ≠ 0 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 5

6.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 1 Determining the Domains of Rational

Equations

Find the domain of the equation.

(b) 3 2 6

– = The domain is { x | x ≠ +1 }.

x–1 x+1 x2 – 1

The domains of the three rational terms are, respectively, { x | x ≠ 1 },

{ x | x ≠ –1 }, { x | x ≠ +1 }. The domain of the equation is the intersection

of the three domains, all real numbers except 1 and –1, written { x | x ≠ +1 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 6

EXAMPLE 1 Determining the Domains of Rational

Equations

Find the domain of the equation.

(b) 3 2 6

– = The domain is { x | x ≠ +1 }.

x–1 x+1 x2 – 1

The domains of the three rational terms are, respectively, { x | x ≠ 1 },

{ x | x ≠ –1 }, { x | x ≠ +1 }. The domain of the equation is the intersection

of the three domains, all real numbers except 1 and –1, written { x | x ≠ +1 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 6

7.
8.4 Equations with Rational Expressions and Graphs

Caution on “Solutions”

When each side of an equation is multiplied by a variable expression,

the resulting “solutions” may not satisfy the original equation. You must

either determine and observe the domain or check all potential

solutions in the original equation. It is wise to do both.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 7

Caution on “Solutions”

When each side of an equation is multiplied by a variable expression,

the resulting “solutions” may not satisfy the original equation. You must

either determine and observe the domain or check all potential

solutions in the original equation. It is wise to do both.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 7

8.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 2 Solving an Equation with Rational

Expressions

5 – 1 11

Solve = .

x 6 2x

The domain, which excludes 0, was found in Example 1(a).

5 – 1 11

6x = 6x Multiply by the LCD, 6x.

x 6 2x

5 – 1 11

6x 6x = 6x Distributive property

x 6 2x

30 – x = 33 Multiply.

–x = 3 Subtract 30.

x = –3 Divide by –1.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 8

EXAMPLE 2 Solving an Equation with Rational

Expressions

5 – 1 11

Solve = .

x 6 2x

The domain, which excludes 0, was found in Example 1(a).

5 – 1 11

6x = 6x Multiply by the LCD, 6x.

x 6 2x

5 – 1 11

6x 6x = 6x Distributive property

x 6 2x

30 – x = 33 Multiply.

–x = 3 Subtract 30.

x = –3 Divide by –1.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 8

9.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 2 Solving an Equation with Rational

Expressions

5 – 1 11

Solve = . Check: Replace x with –3 in the original equation.

x 6 2x

5 – 1 11 Original equation

=

x 6 2x

5 – 1 11 Let x = –3.

= ?

–3 6 2(–3)

– 10 – 1 =

11

?

6 6 –6

– 11 = –

11 True

6 6

The solution is { –3 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 9

EXAMPLE 2 Solving an Equation with Rational

Expressions

5 – 1 11

Solve = . Check: Replace x with –3 in the original equation.

x 6 2x

5 – 1 11 Original equation

=

x 6 2x

5 – 1 11 Let x = –3.

= ?

–3 6 2(–3)

– 10 – 1 =

11

?

6 6 –6

– 11 = –

11 True

6 6

The solution is { –3 }.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 9

10.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 3 Solving an Equation with No Solution

Solve 3 2 6 .

– =

x–1 x+1 x2 – 1

Using the result from Example 1(b), we know that the domain excludes

1 and –1, since these values make one or more of the denominators in the

equation equal 0.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 10

EXAMPLE 3 Solving an Equation with No Solution

Solve 3 2 6 .

– =

x–1 x+1 x2 – 1

Using the result from Example 1(b), we know that the domain excludes

1 and –1, since these values make one or more of the denominators in the

equation equal 0.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 10

11.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 3 Solving an Equation with No Solution

Solve 3 2 6 .

– =

x–1 x+1 x2 – 1

Multiply each side by the LCD, (x –1)(x + 1).

3 – 2 6

(x – 1)(x + 1) = (x – 1)(x + 1)

x–1 x+1 x2 – 1

3 – (x – 1)(x + 1) 2 6

(x – 1)(x + 1) = (x – 1)(x + 1)

x–1 x+1 x2 – 1

Distributive property

3(x + 1) – 2(x – 1) = 6 Multiply.

3x + 3 – 2x + 2 = 6 Distributive property

x+5 = 6 Combine terms.

x = 1 Subtract 5.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 11

EXAMPLE 3 Solving an Equation with No Solution

Solve 3 2 6 .

– =

x–1 x+1 x2 – 1

Multiply each side by the LCD, (x –1)(x + 1).

3 – 2 6

(x – 1)(x + 1) = (x – 1)(x + 1)

x–1 x+1 x2 – 1

3 – (x – 1)(x + 1) 2 6

(x – 1)(x + 1) = (x – 1)(x + 1)

x–1 x+1 x2 – 1

Distributive property

3(x + 1) – 2(x – 1) = 6 Multiply.

3x + 3 – 2x + 2 = 6 Distributive property

x+5 = 6 Combine terms.

x = 1 Subtract 5.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 11

12.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 3 Solving an Equation with No Solution

Solve 3 2 6 .

– =

x–1 x+1 x2 – 1

Since 1 is not in the domain, it cannot be a solution of the equation.

Substituting 1 in the original equation shows why.

Check: 3 – 2 6

=

x–1 x+1 x2 – 1

3 – 2 6

=

1–1 1+1 12 – 1

3 – 2 6

=

0 2 0

Since division by 0 is undefined, the given equation has no solution, and the

solution set is ∅.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 12

EXAMPLE 3 Solving an Equation with No Solution

Solve 3 2 6 .

– =

x–1 x+1 x2 – 1

Since 1 is not in the domain, it cannot be a solution of the equation.

Substituting 1 in the original equation shows why.

Check: 3 – 2 6

=

x–1 x+1 x2 – 1

3 – 2 6

=

1–1 1+1 12 – 1

3 – 2 6

=

0 2 0

Since division by 0 is undefined, the given equation has no solution, and the

solution set is ∅.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 12

13.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 4 Solving an Equation with Rational

Expressions

Solve 4 3 6 .

– =

a2 – 9 2(a2 – 2a – 3) 2

a + 4a + 3

Factor each denominator to find the LCD, 2(a + 3)(a – 3)(a + 1).

The domain excludes –3, 3, and –1.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 13

EXAMPLE 4 Solving an Equation with Rational

Expressions

Solve 4 3 6 .

– =

a2 – 9 2(a2 – 2a – 3) 2

a + 4a + 3

Factor each denominator to find the LCD, 2(a + 3)(a – 3)(a + 1).

The domain excludes –3, 3, and –1.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 13

14.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 4 Solving an Equation with Rational

Expressions

Solve 4 3 6 .

– =

2 2 2

a –9 2(a – 2a – 3) a + 4a + 3

Multiply each side by the LCD, 2(a + 3)(a – 3)(a + 1).

4 – 3

2(a + 3)(a – 3)(a + 1)

(a + 3)(a – 3) 2(a – 3)(a + 1)

6

= 2(a + 3)(a – 3)(a + 1)

(a + 3)(a + 1)

4 · 2(a + 1) – 3(a + 3) = 6 · 2 (a – 3) Distributive property

8a + 8 – 3a – 9 = 12a – 36 Distributive property

5a – 1 = 12a – 36 Combine terms

35 = 7a Subtract 5a; Add 36.

5 = a Divide by 7.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 14

EXAMPLE 4 Solving an Equation with Rational

Expressions

Solve 4 3 6 .

– =

2 2 2

a –9 2(a – 2a – 3) a + 4a + 3

Multiply each side by the LCD, 2(a + 3)(a – 3)(a + 1).

4 – 3

2(a + 3)(a – 3)(a + 1)

(a + 3)(a – 3) 2(a – 3)(a + 1)

6

= 2(a + 3)(a – 3)(a + 1)

(a + 3)(a + 1)

4 · 2(a + 1) – 3(a + 3) = 6 · 2 (a – 3) Distributive property

8a + 8 – 3a – 9 = 12a – 36 Distributive property

5a – 1 = 12a – 36 Combine terms

35 = 7a Subtract 5a; Add 36.

5 = a Divide by 7.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 14

15.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 4 Solving an Equation with Rational

Expressions

Solve 4 3 6 .

– =

a2 – 9 2(a2 – 2a – 3) 2

a + 4a + 3

Note that 5 is in the domain; substitute 5 in for a in the original equation to

check that the solution set is { 5 }.

5 = a

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 15

EXAMPLE 4 Solving an Equation with Rational

Expressions

Solve 4 3 6 .

– =

a2 – 9 2(a2 – 2a – 3) 2

a + 4a + 3

Note that 5 is in the domain; substitute 5 in for a in the original equation to

check that the solution set is { 5 }.

5 = a

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 15

16.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 5 Solving an Equation That Leads to a

Quadratic Equation

Solve 4 = 2 – 8x .

2x + 1 x 2x + 1

1

Since the denominator cannot equal 0, – 2 is excluded from

the domain, as is 0.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 16

EXAMPLE 5 Solving an Equation That Leads to a

Quadratic Equation

Solve 4 = 2 – 8x .

2x + 1 x 2x + 1

1

Since the denominator cannot equal 0, – 2 is excluded from

the domain, as is 0.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 16

17.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 5 Solving an Equation That Leads to a

Quadratic Equation

Solve 4 = 2 – 8x .

2x + 1 x 2x + 1

Multiply each side by the LCD, x(2x + 1).

x(2x + 1) 4 = x(2x + 1) 2 – 8x

2x + 1 x 2x + 1

4x = 2(2x + 1) – 8x2 Distributive property

4x = 4x + 2 – 8x2 Distributive property

0 = –8x2 + 2 Subtract 4x; standard form.

0 = –2(4x2 – 1) Factor.

0 = –2(2x + 1)(2x – 1) Factor.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 17

EXAMPLE 5 Solving an Equation That Leads to a

Quadratic Equation

Solve 4 = 2 – 8x .

2x + 1 x 2x + 1

Multiply each side by the LCD, x(2x + 1).

x(2x + 1) 4 = x(2x + 1) 2 – 8x

2x + 1 x 2x + 1

4x = 2(2x + 1) – 8x2 Distributive property

4x = 4x + 2 – 8x2 Distributive property

0 = –8x2 + 2 Subtract 4x; standard form.

0 = –2(4x2 – 1) Factor.

0 = –2(2x + 1)(2x – 1) Factor.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 17

18.
8.4 Equations with Rational Expressions and Graphs

EXAMPLE 5 Solving an Equation That Leads to a

Quadratic Equation

Solve 4 = 2 – 8x .

2x + 1 x 2x + 1

2x + 1 = 0 or 2x – 1 = 0 Zero-factor property

x=– 1 or x= 1

2 2

Because – 1 is not in the domain of the equation, it is not a solution. Check

2

that the solution set is 1 .

2

0 = –2(2x + 1)(2x – 1)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 18

EXAMPLE 5 Solving an Equation That Leads to a

Quadratic Equation

Solve 4 = 2 – 8x .

2x + 1 x 2x + 1

2x + 1 = 0 or 2x – 1 = 0 Zero-factor property

x=– 1 or x= 1

2 2

Because – 1 is not in the domain of the equation, it is not a solution. Check

2

that the solution set is 1 .

2

0 = –2(2x + 1)(2x – 1)

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 18

19.
8.4 Equations with Rational Expressions and Graphs

Graph of f (x) = 1x

The domain of this function

includes all real numbers except

x = 0. Thus, there will be no

point on the graph with x = 0.

The vertical line with equation

x = 0 is called a vertical

asymptote of the graph.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 19

Graph of f (x) = 1x

The domain of this function

includes all real numbers except

x = 0. Thus, there will be no

point on the graph with x = 0.

The vertical line with equation

x = 0 is called a vertical

asymptote of the graph.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 19

20.
8.4 Equations with Rational Expressions and Graphs

Graph of f (x) = 1x

The horizontal line with equation

y = 0 is called a horizontal

Notice the closer positive values of

x are to 0, the larger y is. Similarly,

the closer negative values of x are

to 0, the smaller (more negative) y

is. Plot several points to verify this

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 20

Graph of f (x) = 1x

The horizontal line with equation

y = 0 is called a horizontal

Notice the closer positive values of

x are to 0, the larger y is. Similarly,

the closer negative values of x are

to 0, the smaller (more negative) y

is. Plot several points to verify this

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 20

21.
8.4 Equations with Rational Expressions and Graphs

Graph of g(x) = x –2

–3

There is no point on the graph

for x = 3 because 3 is excluded

from the domain. The dashed line

x = 3 represents the vertical

asymptote and is not part of the

Notice the graph gets closer to the

vertical asymptote as the x-values

get closer to 3.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 21

Graph of g(x) = x –2

–3

There is no point on the graph

for x = 3 because 3 is excluded

from the domain. The dashed line

x = 3 represents the vertical

asymptote and is not part of the

Notice the graph gets closer to the

vertical asymptote as the x-values

get closer to 3.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 21

22.
8.4 Equations with Rational Expressions and Graphs

Graph of g(x) = x –2

–3

Observe the y-values as the

x-values get closer to the vertical

asymptote (from both sides).

As the x-values get closer to the

vertical asymptote from the left,

the y-values get larger and as the

x-values get closer to the vertical

asymptote from the right, the

y-values get smaller (more negative).

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 22

Graph of g(x) = x –2

–3

Observe the y-values as the

x-values get closer to the vertical

asymptote (from both sides).

As the x-values get closer to the

vertical asymptote from the left,

the y-values get larger and as the

x-values get closer to the vertical

asymptote from the right, the

y-values get smaller (more negative).

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 22

23.
8.4 Equations with Rational Expressions and Graphs

Graph of g(x) = x –2

–3

Again, y = 0 is a horizontal

Plot several points to verify

this graph.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 23

Graph of g(x) = x –2

–3

Again, y = 0 is a horizontal

Plot several points to verify

this graph.

Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 8.4 - 23