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Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
Rational
Expressions and
Applications
14
Slide - 2Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
1. Solve direct variation problems.
2. Solve inverse variation problems.
Objectives
7.8 Variation
Slide - 3Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
Direct Variation
y varies directly as x if there exists a constant k
such that
y = kx.
Solve Direct Variation Problems
Slide - 4Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
Example
Suppose y varies directly as x, and y = 600 when x = 12.
Find y when x = 25.
Solve Direct Variation Problems
The first part of the problem gives us the information we need to
find the constant k. Then, we can find y when x = 25.
Since y varies directly as x,
y = kx
600 = k · 12
50 = k
Now we can solve for y.
y = 50x
y = 50 · 25
y = 1250
Slide - 5Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
.
k
y
x
Inverse Variation
y varies inversely as x if there exists a constant k
such that
Solve Inverse Variation Problems
Slide - 6Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
Example
At one Toys R Us store, the supply and demand for the latest
singing stuffed animal have an inverse relationship. When the
manufacturer supplies 500 stuffed animals, the demand is for
4000. Find the demand when the manufacturer supplies the
store with 2000 stuffed animals.
Solve Inverse Variation Problems
Let D = # of stuffed animals demanded, and
S = # of stuffed animals supplied.
Since supply and demand have an inverse relationship,
.
k
D
S

Slide - 7Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
500
 
  
 
Use D = 4000 and S = 500. Solve for D when S = 2000.
The first part of the problem gives us the information we need to
find the constant k. Then we can find D when S = 2000.
Example (concluded)
Solve Inverse Variation Problems
4000
500
k
 500
2,000,000 = k
2,000,000
D
S

2,000,000
2000
D 
1000D 
When the manufacturer supplies the store with 2000
stuffed animals, the demand is for 1000.

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Section 14.8 variation

  • 1. Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G 2 Rational Expressions and Applications 14
  • 2. Slide - 2Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G 1. Solve direct variation problems. 2. Solve inverse variation problems. Objectives 7.8 Variation
  • 3. Slide - 3Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Direct Variation y varies directly as x if there exists a constant k such that y = kx. Solve Direct Variation Problems
  • 4. Slide - 4Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Example Suppose y varies directly as x, and y = 600 when x = 12. Find y when x = 25. Solve Direct Variation Problems The first part of the problem gives us the information we need to find the constant k. Then, we can find y when x = 25. Since y varies directly as x, y = kx 600 = k · 12 50 = k Now we can solve for y. y = 50x y = 50 · 25 y = 1250
  • 5. Slide - 5Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G . k y x Inverse Variation y varies inversely as x if there exists a constant k such that Solve Inverse Variation Problems
  • 6. Slide - 6Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G Example At one Toys R Us store, the supply and demand for the latest singing stuffed animal have an inverse relationship. When the manufacturer supplies 500 stuffed animals, the demand is for 4000. Find the demand when the manufacturer supplies the store with 2000 stuffed animals. Solve Inverse Variation Problems Let D = # of stuffed animals demanded, and S = # of stuffed animals supplied. Since supply and demand have an inverse relationship, . k D S 
  • 7. Slide - 7Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G 500        Use D = 4000 and S = 500. Solve for D when S = 2000. The first part of the problem gives us the information we need to find the constant k. Then we can find D when S = 2000. Example (concluded) Solve Inverse Variation Problems 4000 500 k  500 2,000,000 = k 2,000,000 D S  2,000,000 2000 D  1000D  When the manufacturer supplies the store with 2000 stuffed animals, the demand is for 1000.