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Algebra 1 Relations and Functions A Relation is a set of ordered pairs. The Domain of a relation is the set of first coordinates of the ordered pairs.

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Presentation on theme: "Algebra 1 Relations and Functions A Relation is a set of ordered pairs. The Domain of a relation is the set of first coordinates of the ordered pairs."— Presentation transcript:

1 Algebra 1 Relations and Functions A Relation is a set of ordered pairs. The Domain of a relation is the set of first coordinates of the ordered pairs. The Range is the second coordinates. A Function is a relation that assigns exactly one value in the range to each value in the domain. (Each X has only one Y) Domain = X’s = Input Range = Y’s = Output

2 Algebra 1 Determine whether each relation is a function. a. {(4, 3), (2, –1), (–3, –3), (2, 4)} The domain value 2 corresponds to two range values, –1 and 4. b. {(–4, 0), (2, 12), (–1, –3), (1, 5)} There is no value in the domain that corresponds to more than one value of the range. Lesson 5-2 The relation is not a function. The relation is a function. Relations and Functions Additional Examples

3 Algebra 1 Lesson 5-2 Use the vertical-line test to determine whether the relation {(3, 2), (5, –1), (–5, 3), (–2, 2)} is a function. A vertical line would not pass through more than one point, so the relation is a function. Step 1: Graph the ordered pairs on a coordinate plane. Step 2: Pass a pencil across the graph. Keep your pencil straight to represent a vertical line. Relations and Functions Additional Examples

4 Algebra 1 Vertical Line Test

5 Algebra 1 Make a table for ƒ(x) = 0.5x + 1. Use 1, 2, 3, and 4 as domain values. Lesson 5-2 x0.5x + 1f(x)f(x) 10.5(1) + 11.5 20.5(2) + 12 30.5(3) + 12.5 40.5(4) + 13 Relations and Functions Additional Examples

6 Algebra 1 Evaluate the function rule ƒ(g) = –2g + 4 to find the range for the domain {–1, 3, 5}. The range is {–6, –2, 6}. ƒ(g) = –2g + 4 ƒ(5) = –2(5) + 4 ƒ(5) = –10 + 4 ƒ(5) = –6 ƒ(g) = –2g + 4 ƒ(–1) = –2(–1) + 4 ƒ(–1) = 2 + 4 ƒ(–1) = 6 ƒ(g) = –2g + 4 ƒ(3) = –2(3) + 4 ƒ(3) = –6 + 4 ƒ(3) = –2 Lesson 5-2 Relations and Functions Additional Examples


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