1. Chapter 7 ABSOLUTE VALUE AND RECIPROCAL FUNCTIONS

2. EXAMPLE How many triangles are in the diagram below?

3. Chapter 7 7.1 – ABSOLUTE VALUE

4. ABSOLUTE VALUE For a real number a, the absolute value is written as |a| and is a positive number. For example: |5| = 5 |–5| = 5 Absolute value can be used to represent the distance of a number from zero on a real- number line. Evaluate: |3||–7|

5. EXAMPLE Evaluate the following: a) |4| – |–6|b) 5 – 3|2 – 7|c) |–2(5 – 7) 2 + 6| a)|4| – |–6| = 4 – 6 = –2 b) 5 – 3|2 – 7| = 5 – 3|–5| = 5 – 3(5) = 5 – 15 = –10 c) Try it!

6. Independent Practice PG. 363-367 #1, 6, 7(A,C,E), 11, 16

7. HANDOUT Answer the questions on the “Investigating Absolute Value Functions” worksheet to the best of your ability.

8. Chapter 7 7.2 – ABSOLUTE VALUE FUNCTIONS

10. ABSOLUTE VALUE FUNCTIONS For what values of x is the function y = |x| equivalent to y = x?  when x ≥ 0 When x < 0, what is the function represented by y = |x|?  y = –x We can write this as a piecewise function:

11. EXAMPLE Consider the absolute value function y = |2x – 3|. a)Determine the y-intercept and the x-intercept. b)Sketch the graph. c)State the domain and range. d)Express as a piecewise function. a)The y-intercept is at x = 0. y = |2x – 3| y = |2(0) – 3| y = |–3| y = 3 The y-intercept is (0, 3). The x-intercept is at y = 0. 0 = |2x – 3| 0 = 2x – 3 x = 3/2 The x-intercept is at (3/2, 0). b) xy 5 03 3/20 33 45

12. EXAMPLE Consider the absolute value function y = |2x – 3|. a)Determine the y-intercept and the x-intercept. b)Sketch the graph. c)State the domain and range. d)Express as a piecewise function. xy 5 03 3/20 33 45 b) c)D: {x | x E R} R: {y | y ≥ 0, y E R} d) The equation on the right is just y = 2x – 3. What’s the one on the left? It’s just –(2x – 3)! The x-intercept is call an invariant point because it’s a part of both functions.

13. EXAMPLE Consider the absolute value function f(x) = |–x 2 + 2x + 8|. a)Determine the y-intercept and the x-intercepts. b)Sketch the graph. c)State the domain and range. d)Express as a piecewise function a)The y-intercept is at x = 0.  f(0) = |–(0) 2 + 2(0) + 8| = |8| = 8 The x-intercepts are when y = 0.  0 = |–x 2 + 2x + 8|  0 = –x 2 + 2x + 8  0 = –(x – 4)(x + 2)  x = 4  x = –2 b) What’s the vertex of the function f(x) = –x 2 + 2x + 8 Use your calculator, or complete the square: f(x) = –(x 2 – 2x) + 8  f(x) = –(x 2 – 2x + 1 – 1 ) + 8  f(x) = –(x 2 – 2x + 1) + 1 + 8  f(x) = –(x – 1) 2 + 9  The vertex is (1, 9)

14. EXAMPLE Consider the absolute value function f(x) = |–x 2 + 2x + 8|. a)Determine the y-intercept and the x-intercepts. b)Sketch the graph. c)State the domain and range. d)Express as a piecewise function Recall:  y-intercept is (0, 8)  x-intercepts are (4, 0) and (–2, 0)  Vertex is (1, 9) c) D: {x | x E R} R: {y | y ≥ 0, y E R} d)

15. Independent Practice PG. 375-379, #2, 5, 7, 10, 12-14.

16. Chapter 7 7.3 - ABSOLUTE VALUE EQUATIONS

17. ABSOLUTE VALUE EQUATIONS When solving equations that involve absolute value equations you need to consider two cases: Case 1: The expression inside the absolute value symbol is positive or zero. Case 2: The expression inside the absolute value symbol is negative.

18. EXAMPLE Solve: |x – 3| = 7 Consider the equation as a piecewise function: Case 1: x – 3 = 7  x = 10 Case 2: –(x – 3) = 7  x – 3 = –7  x = –4 The solution is x = 10 or x = –4.

19. TRY IT Solve |6 – x| = 2

20. EXAMPLE Solve |2x – 5| = 5 – 3x Consider: What is the x-intercept of y = 2x – 5?  0 = 2x – 5  5 = 2x  x = 5/2 Case 1: (x ≥ 5/2) 2x – 5 = 5 – 3x  5x = 10  x = 2 Case 2: (x < 5/2) –(2x – 5) = 5 – 3x  –2x + 5 = 5 – 3x  x = 0

21. EXAMPLE Solve: |3x – 4| + 12 = 9  |3x – 4| = –3 Is there any possible way that the absolute value of something is equal to –3?  No solution.

22. EXAMPLE Solve: |x – 10| = x 2 – 10x

23. Independent Practice PG. 389-391, #4, 5, 6, 9, 11, 22, 23