Equations with Rational Expressions and Graphs

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Sharp Tutor Mon, Jan 17, 2022 07:06 PM UTC

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. Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 8.4 - 1

2. Chapter 8
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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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