Describing a Transformation of Trigonometric Functions

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Nov 9, 2021.

  1. nycmathguy

    nycmathguy

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    Section 4.5

    I need help with 58 parts (a - c). Seeking a detailed reply with steps.

    20211107_201433.jpg
     
    nycmathguy, Nov 9, 2021
    #1
  2. nycmathguy

    MathLover1

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    58.

    g(x)=4-sin(2x+pi/2)

    a.
    descriebe transformations

    Compare it to g(x)=a*sin(bx+c)+d

    .a=-1, b=2, c=pi/2, d=4

    Amplitude =| a|=1

    Period=2pi/b=2pi/2=pi

    vertical shift: 4 units up

    b.
    sketch the graph
    [​IMG]
    c.
    use function notation to write g(x) in terms of f(x)

    g(x) =4-f(2x+pi/2)
     
    MathLover1, Nov 9, 2021
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Thanks. Nice-done.
     
    nycmathguy, Nov 10, 2021
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  4. nycmathguy

    nycmathguy

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    Question 54

    g(x) =sin(2x + pi)

    Part (a)

    Compare given function to g(x) = a•sin(bx + c) + d.

    Note: a = 1, b = 2, c = pi, d = 0

    Amplitude = A = | 1 | = 1

    Period = P = 2pi/b = 2pi/2 = pi

    No vertical shift as d = 0 indicates.

    Part (b)

    20211112_200540.jpg

    Part (c)

    g(x) = 4 - f(2x + pi)
     
    nycmathguy, Nov 13, 2021
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  5. nycmathguy

    nycmathguy

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    Question 56

    We can write the given function like this:

    g(x) = 1 + cos(x + pi)

    We compare the given function with
    g(x) = d + a•cos(bx + pi).

    Note: a = 1, b = 1, c = pi, d = 1

    There's a vertical shift upward as indicated by d = 1.

    Part (b)

    20211112_201350.jpg

    Part (c)

    g(x) = 1 + f(x + pi)
     
    nycmathguy, Nov 13, 2021
    #5
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