1. 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, even, even

2. 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.

3. 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

4. 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

5. 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

6. 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.

7. 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)}
5. y = x – 4 y x 5 -5 y x 5 -5

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

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

10. 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)

11. 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