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Finding the inverse of a quadratic function?

  1. Mar 9, 2010 #1
    Hello, I am sorry for removing the template but it is not an actual problem that I need help with. Well it is, but it should be pretty quick and straight forward.

    The problem is located here: http://www.cdli.ca/courses/math3103/unit05_org01_ilo04/images/3-less9.gif [Broken]

    (It's a bit difficult to type out.)

    I'm having difficulty with only 1 step that I need straightened out.

    On the second last line, where did the 1/2 outside of the square root function come from? And Why does it appear there? I am thinking the 1/2 just got moved out of it, because it was under x+3.

    Please help, thank you so much.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 9, 2010 #2

    Dick

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    They wrote (x+3)/-2=(x+3)*2/((-2)*2)=(2x+6)/(-4)=(1/4)*((2x+6)/(-1))=(1/4)*(-2x-6). Then they took sqrt(1/4)=1/2 and moved that out.
     
  4. Mar 10, 2010 #3
    Multiply both numerator and denominator by -2 inside the square root

    [tex] y + 1 = \pm \sqrt{\frac{-2(x + 3)}{- 2 (-2)}[/tex]
     
  5. Mar 10, 2010 #4
    Multiply both numerator and denominator by -2 inside the square root

    [tex] y + 1 = \pm \sqrt{\frac{-2x-6}{4}[/tex]
     
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