September 30, 2011 At the end of today, you will be able to: Analyze and graph relations. Find functional values. Warm-up: Evaluate the expression if: a = -1, b = 3, c = -2, d = 1. c + d a2 – 5a + 3 2b2 + b + 7 HW 2.1: Pg. 60 #1, 3, 18-34 even, 42-54 even

Lesson 2.1 Relations and Functions A relation is a set of ordered pairs: {(0, 1), (-2, 2), (1, 4), (2, 1)} The domain is all the x values in a relation. The range is all the y values in a relation. {-2, 0, 1, 2} {1, 2, 4} Write the sets in order and do not repeat any values in the set.

A function is a relation in which each element of the domain has only one range. There are several ways to determine whether a relation is a function. Mapping and the Vertical Line Test

Mapping Example 1: Determine whether the relation {(4, 2), (2, 3), (6, 1), (-2, 1)} is a function. Domain Range This is a function because each domain has only ONE range -2 2 4 6 1 2 3

Mapping Example 2: (0, 7), (-1, 5), (1, 6), (-1, 4), (2, 7) 1 2 4 5 6 7 You could also see that it is not a function because it has an x value repeating in the domain. This is NOT a function, because -1 has two elements in the range: 4, 5

The Vertical Line Test - If a vertical line intersects the graph only once, then it is a function . Ex. 2 Ex. 1: y x 5 -5 y x 5 -5 The graph is not a function because it hits it 3 times. The graph is a function because it only hits the vertical line once.

Individual Practice – Determine whether the relation is a function Individual Practice – Determine whether the relation is a function. Explain why. {(-3, 2), (0, -2), (-2, 1), (-3, -1) } {(1, 2), (2, 2), (3, 2), (4, 2)} 3. 4. 5. y = x – 4 y x 5 -5 y x 5 -5

Types of Functions There are two types of special functions: Onto Each element in range goes to an element in domain. Domain Range 1 A 2 B C One-to-one Each element in domain pairs to 1 unique element in range. Domain Range 1 A 2 B C

Example: What type of function is {(-6,-1), (-5,-9), (-3,-7), (-1,7), (6,-9)}?

Example 3: Function Notation f(x) f(x) is the same as “y” Instead of having a column with “y” we put “f(x)” f(x) (x, f(x)) x f(x) = 2x + 3 y (x, y) y = 2x + 3 -2 -1 1 2 f(-2) = 2(-2) + 3 -1 (-2, -1) f(-1) = 2(-1) + 3 1 (-1, 1) f(0) = 2(0) + 3 3 (0, 3) f(1) = 2(1) + 3 5 (1, 5) f(-1) = 2(2) + 3 7 (2, 7)

Example 3) Evaluating functions If f(x) = x2 – x and g(x) = -2x + 10, Find: a) f(4) b) g(3.5) c) f(2z) d) f(-5) Substitute x = 4 in f(x), f(4) = (4)2 – 4 f(4) = 16 – 4 f(4) = 8