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Quick quadratics question

  1. Aug 2, 2011 #1
    1. The problem statement, all variables and given/known data

    y=-3f(2-x)-2 for the root function...

    2. Relevant equations

    3. The attempt at a solution

    If im given a function like, y=-3f(2-x)-2 for the root function...

    Do i factor out the negetive on the x?

    Making it.. y=-3f(-(x-2)-2 . making a reflection on both the x/y axis's ? Or is that wrong?
  2. jcsd
  3. Aug 2, 2011 #2
    Simple yes/no answer? :)
  4. Aug 2, 2011 #3


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    Homework Helper

    I don't want to be rude, but I have no idea what you're trying to do. Is y = f(x)? Is your equation:

    f(x) = 3f(2-x) - 2?

    What are you trying to do with this?
  5. Aug 3, 2011 #4


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    Staff Emeritus
    Science Advisor

    "Yes, No" to what question? You titled this "Quick Quadratics Question", but there is no quadratic in the problem. You say you are given the "root" function -3(2- x)- 2. What do you mean by "root function"? and , as gb7nash aked what are you trying to do with it?
  6. Aug 3, 2011 #5
    Perhaps the thing to do is to solve for x in terms of y? I don't fully understand this either. You could isolate x on the left and that might be what is being asked? Not sure? will look forward to the correct solution.
  7. Aug 3, 2011 #6
    ::: (sqrt)2(x-4) +1

    Is this a horizontal or vertical stretch/compression?

    I said it was a vertical, since its outside the bracket.
  8. Aug 3, 2011 #7


    Staff: Mentor

    Please stop using your (sqrt) notation, with parenthes around "sqrt". Put the parentheses around the expression whose square root you're taking. With your notation it's impossible to tell whether the +1 is inside the radical or outside.

    Is this your function?
    y = sqrt(2(x - 4) + 1)
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