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Math Help - Transformation - stretches

  1. #1
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    Transformation - stretches

    This stretches is driving me nuts. I don't even know if I am doing the right thing.

    The coordinates for y = f(x) is (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0). Give the new coordinates of y = -f((x+2)).

    Let me try:
    Original coordinates (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0)
    -f(x) (-1, 0), (0, -3), (1, 0), (2, 3), (3, 0)
    f(x) (-2, 0), (0, -3), (2, 0), (4, 3), (6, 0)
    f(x+2) (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0)

    New coordinates for y = -f((x+2)) is (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0).

    Is this right??
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by shenton View Post
    This stretches is driving me nuts. I don't even know if I am doing the right thing.

    The coordinates for y = f(x) is (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0). Give the new coordinates of y = -f((x+2)).

    Let me try:
    Original coordinates (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0)
    -f(x) (-1, 0), (0, -3), (1, 0), (2, 3), (3, 0)
    f(x) (-2, 0), (0, -3), (2, 0), (4, 3), (6, 0)
    f(x+2) (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0)

    New coordinates for y = -f((x+2)) is (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0).

    Is this right??
    y(-4)=-f(1/2(-4+2))=-f(-1)=0

    y(-2)=-f(1/2(-2+2))=-f(0)=-3

    y(0)=-f(1/2(0+2)=-f(1)=0

    y(2)=-f(1/2(2+2)=-f(2)=3

    y(4)=-f(1/2(4+2))=-f(3)=0.

    So yes, you are right.

    RonL
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