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4 minutes. Warm-Up. Evaluate each expression. x – 2, for x = -5 -2x, for x = -1.5 x 2 , for x = -4 -x 2 , for x = -1.2 x 3 , for x = -2 -x 3 , for x = -0.1. 2.3 Intro to Functions. Objectives: State the domain and range of a relation, and tell whether it is a function

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Warm-Up


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    1. 4 minutes Warm-Up Evaluate each expression. • x – 2, for x = -5 • -2x, for x = -1.5 • x2, for x = -4 • -x2, for x = -1.2 • x3, for x = -2 • -x3, for x = -0.1

    2. 2.3 Intro to Functions Objectives: State the domain and range of a relation, and tell whether it is a function Write a function in function notation and evaluate it

    3. Definition of a Function Function: each value of the first variable is paired with exactly one value of the second variable Domain: set of all possible values of the first variable Range: set of all possible values of the second variable

    4. Example 1 State whether the data in each table represents y as a function of x. Explain. function not a function

    5. Vertical-Line Test If every vertical line intersects a given graph at no more than one point, then the graph represents a function. function not a function

    6. Definition of a Relation Relation: each value of the first variable is paired with one or more values of the second variable Domain: set of all possible values of the first variable Range: set of all possible values of the second variable

    7. Example 2 State the domain and range of the relation, and state whether it is a function. { (–7, 5), (4, 12), (8, 23), (16, 8) } domain: { –7, 4, 8, 16} range: { 5, 8, 12, 23 } This is a function because each x-coordinate is paired with only one y-coordinate.

    8. Function Notation If there is a correspondence between values of the domain, x, and values of the range, y, that is a function, then y = f(x), and (x,y) can be written as (x,f(x)). The variable x is called the independent variable. The variable y, or f(x) is called the dependent variable.

    9. Example 3 Evaluate f(x) = –2.5x + 11, where x = –1. f(–1) = –2.5 (–1) + 11 f(–1) = 2.5 + 11 f(–1) = 13.5

    10. Example 4 A gift shop sells a specialty fruit and nut mix at a cost of $2.99 per pound. During the holiday season, you can buy as much of the mix as you like and have it packaged in a decorative tin that costs $4.95. a) Write a linear function to model the total cost in dollars, c, of the tin containing the fruit and nut mix as a function of the number of pounds of the mix, n. c(n) = 4.95 + 2.99n b) Find the total cost of a tin that contains 1.5 pounds of the mix. c(n) = 4.95 + 2.99n c(1.5) = 4.95 + 2.99(1.5) c(1.5) = 9.44 $9.44

    11. Homework p.107 #17-49 odds