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Section 1-1: Relations and Functions *Relation: *Domain: *Range: *Function: Example 1: State the domain and range of each relation. Then state whether.

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Presentation on theme: "Section 1-1: Relations and Functions *Relation: *Domain: *Range: *Function: Example 1: State the domain and range of each relation. Then state whether."— Presentation transcript:

1 Section 1-1: Relations and Functions *Relation: *Domain: *Range: *Function: Example 1: State the domain and range of each relation. Then state whether the relation is a function. a.{(-1, 2), (0, 4), (1, 3)} b.{(-3, 1), (-2, 5), (-3, 4), (1, 2)} Set of ordered pairs. X-values that you are allowed to use. Y-values that you get for answers. Each x-value is assigned 1 and only 1 y-value. D: -1, 0, 1R: 2, 4, 3 D: -3, -2, 1R: 1, 5, 4, 2 Yes it is a function. No not a function.

2 To find out if a graph is a function we must use the ___________________. If you can draw a vertical line and it does not pass through more than one point on the graph, then the relation is a function. Example 2: Determine if the graph of each relation represents a function. Explain. a.b. vertical line test If you draw a vertical line, it would intersect the graph in two places. So this graph is not a function. If you draw a vertical line, it would only intersect the graph in 1 place. This graph is a function.

3 Example 3: Evaluate each function for the given value. a.f(-2) if f(x) = 2x 4 - 3x 3 + 4x – 1 b. g(3) if g(x) = |5 - x 3 │ Example 4: Evaluate each function for the given value. a. h(2b) if h(x) = 3x 4 - 2x 3 + x 2 – x b. j(2a - 1) if j(x) = 2x 2 - 5x + 1 * Function Notation: f(x) : read as “f of x” f - tells you what function to go to x - tells you what to plug into the function f(-2) = 2(-2) 4 – 3(-2) 3 + 4(-2) - 1 = 2(16) -3(-8) – 8 – 1 = 32 + 24 – 9 = 47 g(3) = |5 – (3) 3 | = |5 – 27| = |-22| = 22 h(2b) = 3(2b) 4 – 2(2b) 3 + (2b) 2 – 2b = 3(16b 4 ) – 2(8b 3 ) + 4b 2 – 2b = 48b 4 – 16b 3 + 4b 2 – 2b j(2a – 1) = 2(2a – 1) 2 – 5(2a – 1) + 1 = 2(4a 2 – 4a + 1) – 10a + 5 + 1 = 8a 2 – 8a + 2 – 10a + 6 = 8a 2 – 18a + 8

4 *Domain Restrictions: a) b) Example 5: State the domain of each function. a. b. The denominator of any fraction can not be equal to 0. Any expression inside of a radical must be > to 0. x 2 – 6x + 8 0x + 3 > 0 x > -3 D : x > -3

5 Example 6: State the domain and range of each relation. Then, determine if the graph of each relation represents a function. a.b. c. d. R: y < 0 D: x > 0

6 e. f. D: x > 0R: -5 < y < 5


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