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Transformation of function

  1. Jan 22, 2005 #1
    Rewrite Function PLZ HELP

    I have a question for translation of functions
    it says the equation of the image is [tex] -2 \sqrt(3x-12) -5 [/tex]
    (note: sorry the square root should be over all of the 3x and -12 only -5 is outside of square root)
    1) the first questions said what is the base function? I wrote [tex] \sqrt(x) [/tex]

    2) the second question said describe the series of transformations, so i wrote
    Vertical stretch by a factor of 2, horizontal compression by a factor of 1/3, reflection in the x-axis, vertical translation 5 units down, horizontal translation 4 units left.

    3) It says write the function in to y=af[k(x-p)]+q form
    I got [tex] -2\sqrt(3(x-(-4)) -5 [/tex] (note: again square root is over everything in brackets except -5 )but i think this is wrong since there is a square root in my answer and no square root in the form that they want. Can someone help me out please :cry: Is only number 3 wrong? How do I rewrite the equation in the correct form?
     
    Last edited: Jan 22, 2005
  2. jcsd
  3. Jan 23, 2005 #2

    Diane_

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    The only problem I can see with your #3 is that it should be (x-4), not (x-(-4)).

    I only see one way to get rid of the square root, and that's more of a trick than anything else, but given the rest of your question it just may be what you need.

    Let f(x) = your base function, [tex]\sqrt{x}[/tex]. Then your function would be

    [tex]y = 2\sqrt{3} f(x - 4) - 5[/tex]

    The square root is still there, it's just hidden in the "f(x)".

    Does that look possible?
     
  4. Jan 23, 2005 #3
    I dont understand
     
  5. Jan 23, 2005 #4

    Diane_

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    All I've done is replace the [tex]\sqrt{x}[/tex] with an f(x). If you replace the functional notation in my expression with [tex]\sqrt{x-4}[/tex] and carry through the algebra, you should end up with your original expression.

    I'm not entirely certain that's what you're looking for, but it's the only way I can see to make the square root go away. It seems like you have access to the answer - can you post that? It may make it easier to see exactly what you need to do.
     
    Last edited: Jan 23, 2005
  6. Jan 23, 2005 #5

    dextercioby

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    How about writing it like that
    [tex] y(x)=-2\sqrt{3x-12}-5=-2\sqrt{3(x-4)}-5=-2\sqrt{3}\sqrt{x-4}-5 [/tex]

    which can be put under the form
    [tex] y(x)=Af(x-4)-5 [/tex]
    where
    [tex] A=-2\sqrt{3} [/tex]
    [tex] f(x)=\sqrt{x} [/tex]

    What do you say now...??

    Daniel.
     
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