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Math Help - Transforming √(9 - y)

  1. #1
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    Transforming √(9 - y)

    The relation x = √(9 - y) is compressed vertically by a factor of (1/3), then translated 1 unit to the right. Determine the equation of the transformed relation.

    A. x = √(9 - 9y) + 1

    B. x = √(9 - 9y) - 1

    C. x = √[9 (y/9)] + 1

    D. x = √[9 (y/9)] - 1


    This one really beats me. The equation is in the form of an inverse function (unsure if this matters or not) Here is my analysis:

    • Vertical compression by a factor of (1/3)
    • Horizontal translation, 1 unit to the right


    Translation to the right by 1 unit means x=1. Since the function is an inverse function, (A) and (C) is the likely answer since I can see that x = + 1.

    Between (A) and (C), I am lost as I cannot see how a vertical compression of (1/3) factors can change the equation to (9 - 9y) or [9 (y/9)].

    I was thinking more like √(9 (1/3)y) as vertical compression is applied to y-value without taking its reciprocal.

    Please help. Thanks.
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  2. #2
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    Hello, shenton!

    The relation: .x .= .√(9 - y) . Note: x > 0
    is compressed vertically by a factor of (1/3), then translated 1 unit to the right.
    Determine the equation of the transformed relation.
    . . . . . . _____
    A) x = √9 - 9y + 1
    . . . . . . _____
    B) x = √9 - 9y - 1
    . . . . . . ________
    C) x = √9 (y/9) + 1
    . . . . . . ________
    D) x = √9 (y/9) - 1

    I had to baby-talk my way through this one . . .

    Square both sides of the equation:. x .= .9 - y . . x + y .= .9

    We have the right half of a circle centered at the origin with radius 3.


    We have a vertical compression by a factor of 1/3.
    The result is an ellipse:
    . . major axis from (-3,0) to (3,0), minor axis from (-1,0) to (1,0).
    Its equation is: .x/9 + y/1 .= .1


    Translated 1 unit to the right, we'd have: .(x - 1)/9 + y .= .1


    Solve for x: .(x - 1)/9 .= .1 - y

    Multiply by 9: .(x - 1) .= .9 - 9y
    . . . . . . . . . . . . . . . . . . . . .______
    Take square root: .x - 1 .= .√9 - 9y
    . . . . . . . . . . . . . . . .______
    And we have: . x .= .√9 - 9y + 1 . . . Answer-choice: A)


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    Thanks for the detailed workings. This is a little harder than I thought.
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  4. #4
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    Quote Originally Posted by shenton View Post
    The relation x = √(9 - y) is compressed vertically by a factor of (1/3), then translated 1 unit to the right. Determine the equation of the transformed relation.

    A. x = √(9 - 9y) + 1

    B. x = √(9 - 9y) - 1

    C. x = √[9 (y/9)] + 1

    D. x = √[9 (y/9)] - 1


    This one really beats me. The equation is in the form of an inverse function (unsure if this matters or not) Here is my analysis:
    • Vertical compression by a factor of (1/3)
    • Horizontal translation, 1 unit to the right

    Translation to the right by 1 unit means x=1. Since the function is an inverse function, (A) and (C) is the likely answer since I can see that x = + 1.

    Between (A) and (C), I am lost as I cannot see how a vertical compression of (1/3) factors can change the equation to (9 - 9y) or [9 (y/9)].

    I was thinking more like √(9 (1/3)y) as vertical compression is applied to y-value without taking its reciprocal.

    Please help. Thanks.
    There IS a slightly more direct way.

    First when you compress the y axis by a factor of 1/3 you are replacing all y's in the equation with 3y.

    When you translate to the right by 1 unit you are replacing all x's in the equation with x - 1.

    So:
    x = sqrt(9 - y^2)
    becomes:
    x - 1 = sqrt(9 - (3y)^2)


    x = sqrt(9 - 9y^2) + 1

    -Dan
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