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Assignment, pencil, red pen, highlighter, textbook, calculator , GP notebook

total:. Assignment, pencil, red pen, highlighter, textbook, calculator , GP notebook. U4D11. Have out:. Bellwork:. Solve for x. c). a). b). +1. 4. 4. +1. x 2 –8x = 4. + 16. + 16. +1. (x – 4) 2 = 20. +1. +1. +1. +2. +1. +2. +1. Absolute Value. Add to your notes:.

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Assignment, pencil, red pen, highlighter, textbook, calculator , GP notebook

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  1. total: Assignment, pencil, red pen, highlighter, textbook, calculator, GP notebook U4D11 Have out: Bellwork: Solve for x. c) a) b) +1 4 4 +1 x2 –8x = 4 + 16 + 16 +1 (x – 4)2 = 20 +1 +1 +1 +2 +1 +2 +1

  2. Absolute Value Add to your notes: absolute value: the distance a number is from zero on a number line. Since distance is nonnegative, the absolute value of a number is always nonnegative. Symbol: | | |4| = 4 |–4| = 4 Example: Given the function f(x) = | x |, evaluate the following values: e) f(0) a) f(–6) b) f(13) c) d) f(–π) f) f(n) = | –6 | = | 0 | = | 13 | = | –π | = | n | = 6 = 0 = 13 = π

  3. PG – 115 Solving Absolute Value Equations For what values of x does: a) | x | = 8 b) | x | = 4 c) | x | = –4 x = –8, 8 x = –4, 4 No Solution These are quick, but let’s solve more complicated problems.

  4. Solving Absolute Value Equations Add to your notes: Example #1: Step: 1) Write two equations and “drop” the absolute values. |x + 1| = 6 x + 1 = 6 x + 1 = – 6 2) Solve each equation. – 1 – 1 – 1 – 1 x = – 7 x = 5

  5. Solving Absolute Value Equations Add to your notes: Example #2: Step: 1) Isolate the absolute value. 4 |x – 3| + 2 = 22 – 2 – 2 4 |x – 3| = 20 4 4 2) Write two equations and “drop” the absolute values. |x – 3| = 5 x – 3 = 5 x – 3 = – 5 + 3 + 3 + 3 + 3 3) Solve each equation. x = 8 x = –2

  6. y 10 8 6 4 2 x 2 4 –2 –4 –2 –4 –6 –8 –10 PG – 114 PARENT GRAPH TOOLKIT Name: Absolute Value Parent Equation: y = | x | Description of Locator: (1, 1) (–1, 1) vertex (h, k) (0, 0) General Equation: y = a|x – h| + k Properties: open up or down horizontal shift Range: Domain: y = +/- a|x – h| + k {y| k ≤ y <} {x| x  R} Wider or narrower (-, ) [k, ) vertical shift

  7. y 10 8 6 4 2 x –2 2 4 6 8 10 –10 –8 –6 –4 –2 –4 –6 –8 –10 Based on the patterns you have observed, graph the next 6 absolute value functions on the worksheet WITHOUT using a graphing calculator.

  8. Finish today's worksheet

  9. old slides

  10. y 8 6 4 2 x –2 –8 –6 –4 2 6 4 8 –2 –4 –6 –8 PG – 113 a) Investigate y = | x |. Be sure to graph the function and include all the elements of a function investigation. y = | x | 4 x–intercept: (0, 0) 3 y–intercept: (0, 0) 2 (-, ) domain: 1 [0, ) range: 0 asymptotes: 1 None 2 3 4

  11. PG – 113 b) Take out your Parent Graph Toolkit. Do you think that any of the equations you already have is the parent of y = | x |? Why or why not? None of them are the parent graphs for y = | x | since their graphs are completely different. It sort of looks like a parabola, but the absolute value function is “V” shaped and does not have a curved vertex like the graph of a parabola. c) How can we change the equation y = | x | to move it up, down, left, or right? How can we stretch it or compress it, or make it open down? Write a general equation for y = | x |. y = a|x – h| + k

  12. y 8 6 4 2 x –2 –8 –6 –4 2 6 4 8 –2 –4 –6 –8 PG – 116 c) Make a sketch of f(x) = |x + 1| and use the graph to show why there are two answers. Identify on your graph what two x–values give a corresponding y–value of 6. f(x) = |x + 1| Shift left 1 unit. y = 6 (5, 6) (–7, 6) |x + 1| = 6

  13. y 8 6 4 2 x –2 –8 –6 –4 2 6 4 8 –2 –4 –6 –8 PG – 117 Graph f(x) = |x – 3| and x = 3 on the same set of coordinate axes. Write a description of the relationship between the two graphs. The graph x = 3 is the line of symmetry for the graph of f(x) = |x – 3|. x = 3 f(x) = |x – 3| Shift right 3 units

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