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Warm-Up: 1. 2. Solve 8a – (15 – 3.2) = a + (5 2 – 13).

Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

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Page 1: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Warm-Up:

1.

2. Solve 8a – (15 – 3.2) = a + (52 – 13).

Page 2: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

1.6 RelationsAlgebra AB

Page 3: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

You solved equations with one or two variables.

• Represent relations.

• Interpret graphs of relations.

1.6 Relations

Page 4: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Example 1A. Express the relation {(4, 3), (–2, –1), (2, –4), (0, –4)} as

a table, a graph, and a mapping. Table List the x-coordinates in the first column and the corresponding y-coordinates in the second column.

Page 5: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Example 1

Graph Graph each ordered pair on a coordinate plane.

A. Express the relation {(4, 3), (–2, –1), (2, –4), (0, –4)} as

a table, a graph, and a mapping.

Page 6: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Example 1

Mapping List the x-values in the domain and the y-values in the range. Draw an arrow from the x-value to the corresponding y-value.

Domain Range

A. Express the relation {(4, 3), (–2, –1), (2, –4), (0, –4)} as

a table, a graph, and a mapping.

B. Determine the domain and range for the relation

{(4, 3), (–2, –1), (2, –4), (0, –4)}.

Page 7: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Check yourself.Express the relation {(3, –2), (4, 6), (5, 2), (–1, 3)} as a mapping.A. C.

B. D.

Page 8: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Interpreting a MappingB. Determine the domain and range of the

relation {(3, –2), (4, 6), (5, 2), (–1, 3)}.

A. D = {–1, 3, 4, 5}; R = {–2, 2, 3, 6}

B. D = {–2, 2, 3, 6}; R = {–1, 3, 4, 5}

C. D = {–1, 3}; R = {–2, 2}

D. D = {4}; R = {4}

Page 9: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Word Problems.CLIMATE In warm climates, the average amount of electricity used rises as the daily average temperature increases, and falls as the daily average temperature decreases. Identify the independent and the dependent variables for this function.

Page 10: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Word ProblemIndependent and Dependent Variables

The number of calories you burn increases as the number of minutes that you walk increases. Identify the independent and the dependent variables for this function.

Page 11: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Check your self.

A. The number of new members is the independent variable. The dues is the dependent variable.

B. Membership dues is the independent variable. The number of new members is the dependent variable.

C. x is the independent variable. y is the dependent variable.

D. Both variables are independent.

In a particular club, as membership dues increase, the number of new members decreases. Identify the independent and dependent variable in this function.

Page 12: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Double check.

A. The length of the side is independent, and the the area of the square is dependent.

B. The area is independent, and the side length is dependent.

C. Both variables are independent.

D. Both variables are dependent.

The area of a square increases as the length of a side increases. Identify the independent and dependent variable in this function.

Page 13: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Lets tell a story.The graph represents the temperature in Ms. Ling’s classroom on a winter school day. Describe what is happening in the graph.

Page 14: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

Which story is true?

A. Macy is doing bobs.

B. Macy’s speed increases as she crosses the length of the pool, but then decreases to zero when she turns around at the end of each lap.

C. Macy is swimming at a constant speed.

D. Macy’s speed continues to decrease.

The graph represents Macy’s speed as she swims laps in a pool. Describe what is happening in the graph.

Page 15: Warm-Up: Solve 8a – (15 – 3.2) = a + (5 2 – 13)

HomeworkO 1.6 Practice worksheet (ALL)