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Quadratic function Help

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Given f(x) = x^2 Sketch the graph of each of the following, state the domain and range.



    - 1/2f(2x+6) -2

    2. Relevant equations


    y=x^2
    3. The attempt at a solution

    I Simply have a question. The question is, that this 2x can be factored out from the brackets correct?

    Factoring that 2 will multiply it with the 1/2 and make the vertical stretch a factor of 1 , esentially, a factor of 0. No?

    Or , would it create -1/2f(2(x+3) -2 , where the horizontal still exists as 2 (1/2) and the vertical still exists as 1/2 ?

    Which one is it?
     
  2. jcsd
  3. Jul 17, 2011 #2

    LCKurtz

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    You need to use parentheses when writing your formulas so we can understand what your formula is. What all is in the denominator??? And after the / your parentheses are unbalanced. Anyway, whatever you actually meant, no, you can not factor the two across the f, if that is what you are asking.
     
  4. Jul 17, 2011 #3

    eumyang

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    I think he/she's asking if you can factor out the 2 in 2x + 6, and the answer is yes, you can, as long as you don't bring it outside the f.

    I'm guessing that he/she meant to write this:
    f(x) = -(1/2)f(2x+6) - 2 = -(1/2)f(2(x+3)) - 2
    or this:
    [itex]f(x) = -\frac{1}{2}f(2x +6) - 2 = -\frac{1}{2}f(2(x +3)) - 2[/itex]

    OP: Maybe you should learn LaTeX.
     
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