7.2 abs value function

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  • 7.2 abs value function

    1. 1. Math 20-1 Chapter 7 Absolute Value and Reciprocal Functions Teacher Notes7.2 Absolute Value Function
    2. 2. 7.2 Graph an Absolute Value FunctionAbsolute value is defined as the distance fromzero in the number line. Absolute value of -6 is6, and absolute value of 6 is 6, Both are 6 unitsfrom zero in the number line.Piecewise Definition x if x 0 if 6 then 6 x x if x 0 if –6 then – 6 7.2.1
    3. 3. Express the distance between the two points, -6 and 6, usingabsolute value in two ways. 6 6 6 6
    4. 4. Graph an Absolute Value Function y = | 2x – 4| continuous Method 1: Sketch Using a Table of Values x y = | 2x – 4| y = | 2x – 4| –2 |2(-2) - 4| 8 2x 4 2x 4 0 |2(0) – 4| 4 2 |2(2) - 4| 0 4 |2(4) – 4| 4 6 |2(6) – 4| 8 Domain :{x | x R} The x-intercept occurs at the point (2, 0) Range :{y | y 0, y R} The y-intercept occurs at the point (0, 4)The x-intercept of the linear function is the x-intercept of thecorresponding absolute value function. This point may be called aninvariant point. 7.2.2
    5. 5. Graph an Absolute Value Function y = | 2x – 4| Method 2: Using the Graph of the Linear Function y = 2x - 4 1. Graph y = 2x - 4 Slope = 2 Y-intercept = -4 x-intercept = 2 2. Reflect in the x-axis the part of the graph of y = 2x – 4 that is below the x-axis. 3. Final graph y = |2x – 4| 7.2.3
    6. 6. Express as a piecewise function. y = |2x – 4|Recall the basic definition of absolute value. The expression in the abs is a line x if x 0 with a slope of +1 and x-intercept of 0. x –x if x 0function pieces −(x) x 0 domain x > 0domain x < 0has negative y-values,It must be reflected in the x-axis,multiply by -1 Note that 0 is the invariant point and can be determined by making the expression contained within the absolute value symbols equal to 0. 7.2.4
    7. 7. Express as a piecewise function. y = |2x – 4|Invariant point: =0 expression expressionx- intercept 2x – 4 = 0 Slope positive x=2 function pieces |2x - 4| −(2x - 4) (2x - 4) 2x - 4 – 2 + expression Domain x < 2 Domain x > 2 As a piecewise function y = |2x – 4| would be 2x 4 if x 2 2x 4 (2x 4) if x 2 7.2.5
    8. 8. Be CarefulGraph the absolute value function y = |-2x + 3| and expressit as a piecewise function Slope negative x-intercept 2x 3 0 3 x 2 + (-2x+3) – (-2x+3)Domain :{x | x R} 3 + -Range :{y | y 0, y R} Domain x < 3/2 2 x > 3/2 3 2x 3 if x Piecewise function: 2 2x 3 3 ( 2 x 3) if x 2 7.2.6
    9. 9. True or False:The domain of the function y = x + 2 is always the same as thedomain of the function y = |x + 2|. True The range of the function y = x + 2 is always the same as the range of the function y = |x + 2|. False Suggested Questions: Page 375: 1a, 2, 3, 5a,b, 6a,c,e, 9a,b, 12, 16, 7.2.7
    10. 10. Part B: Abs of Quadratic FunctionsCreate a Table of Values to Compare y = f(x) to y = | f(x) |x y = x2 - 4 y = |x2 - 4|–3 5 5-2 0 00 -4 4 x 2 2 x 2 x 22 0 03 5 5Compare and contrast the domain and range of thefunction and the absolute value of the function.What is the effect of the vertex of the original functionwhen the absolute value is taken on the function? 7.2.8
    11. 11. Graph an Absolute Value Function of the Form f(x) = |ax2 + bx + c| Sketch the graph of the function f ( x ) x 2 3x 4 Express the function as a piecewise function 1. Graph y = x2 – 3x - 4 vertex y x 4 x 1 (1.5, -6.25) x-int ( x 4)( x 1) 0 x 4 or x 1 2. Reflect in the x-axis the part of the graph of Domain :{x | x R} y = x2 – 3x - 4 that is below Range :{y | y 0, y R} the x-axis. How does this compare 3. Final graph y = |x2 – 3x - 4 | to the original function? 7.2.9
    12. 12. Express as a piecewise function. f ( x ) x 2 3x 4 expression = 0 x 2 3x 4 0 Critical points ( x 4)( x 1) 0 x-intercepts x 4 or x 1 a > 0, opens up x 2 3x 4 x 2 3x 4 ( x 2 3x 4) x 2 3x 4 + -1 – 4 + x2 - 3x - 4 expression x < -1 x>4 -1 < x < 4 x 2 3x 4 if -1 x 4 Piecewise function: x 2 3x 4 ( x 2 3x 4) if -1<x 4 7.2.10
    13. 13. Be Careful with DomainGraph the absolute value function y = |-x2 + 2x + 8| and express it as apiecewise function expression = 0 vertex -x 2 2 x 8 0Domain (1, 9)Critical points x 2 2x 8 0 ( x 4)( x 2) 0 x-intercepts x 4 or x 2 a < 0, opens down x2 2x 8 x2 2x 8 x2 2x 8 x < -2 -2 4 -2 < x < 4 x>4 2 x2 2x 8 if -2 x 4 Piecewise function: x 2x 8 ( x2 2 x 8) if -2>x 4 7.2.11
    14. 14. Absolute Value as a Piecewise FunctionMatch the piecewise definition with the graph of an absolute value function. 7.2.12
    15. 15. Suggested Questions:Page 375:4, 7b, 8b,c,d, 10a,c, 11b,d, 13, 15, 20, 7.2.12

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