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Quadratic Function

  1. Nov 28, 2007 #1
    [SOLVED] Quadratic Function

    1. The problem statement, all variables and given/known data


    solve (f(x)=-3 x^2 + 9 x + 1/4 )
    fx=(-3)(x^2-3 x)+1/4

    fx=(-3)(x^2-3 x+9/4)+1/4+27/4

    fx=(-3)(x^2-3 x+9/4)+28/4
    fx=(-3) (x-3/2)^2+7



    2. Relevant equations



    3. The attempt at a solution


    the solution is already given in the book, but I don't understand why 9/4 was inserted in the third line (bolded). I know at the end of the equation, 27/4 is added to balance out the 9/4.......but why put in 9/4 there?
     
  2. jcsd
  3. Nov 28, 2007 #2

    Kurdt

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    What they've done is simply completed the square. If you have an equation of the form: [itex] x^2+2ax [/itex], you can write it as: [itex] x^2+2ax = (x+a)^2 - a^2 [/itex].
     
    Last edited: Nov 28, 2007
  4. Nov 28, 2007 #3

    rock.freak667

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    f(x)=-3 x^2 + 9 x + 1/4

    they just put 27/4 and -27/4 to make completion easier

    f(x)=-3 x^2 + 9 x -27/4+ 1/4 +27/4
    =-3(x^2-3x+9/4)+7
    =-3(x-3x/2)^2 +7

    expand (x-3x/2)^2 and check it yourself
     
  5. Nov 28, 2007 #4
    yea bro...

    the thing is, how do you come up with 9/4 from to make it easier?

    I'm on purplemath.com right now, trying to figure it out without bothering you guys... hopefully I can figure this one out.
     
  6. Nov 28, 2007 #5

    Kurdt

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    Well heres how I would have done that problem. You're given:

    [tex] f(x) = -3x^2 + 9x + \frac{1}{4} [/tex]
    [tex] f(x) = -3( x^2 -3x) + 1/4 [/tex]

    Now we notice that the term in brackets is of the form [itex] x^2+2ax [/itex], with [itex]a=\frac{-3}{2}[/itex], and so we complete the square [itex] x^2+2ax = (x+a)^2 - a^2 [/itex]:

    [tex] f(x) = -3\left(\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} \right) + \frac{1}{4} [/tex]
    [tex] f(x) = -3\left(x - \frac{3}{2}\right)^2 + \frac{27}{4} + \frac{1}{4} [/tex]
    [tex] f(x) = -3\left(x - \frac{3}{2}\right)^2 + 7 [/tex]
     
    Last edited: Nov 28, 2007
  7. Nov 28, 2007 #6
    okie.........i got it.


    so the point of adding -9/4 inside the brackets is so that you can make it factorable. And since you added a (-), now you have to minus it to even it out.


    thnkx.....can't believe it took me so long.

    anyhow, how do you type the math out like that?
     
    Last edited: Nov 28, 2007
  8. Nov 29, 2007 #7
  9. Nov 29, 2007 #8
    test:

    [tex]x=-3 x^2+9x+1/4[/tex]

    [tex]fx=(-3)(x^2┤-3 x)+1/4\\[/tex]

    [tex]fx=(-3)(x^2┤-3 x+9/4)+1/4+27/4\\[/tex]

    [tex]fx=(-3)(x^2┤-3 x+9/4)+28/4\\[/tex]

    [tex]fx=(-3) (x┤-3/2)^2+7\\[/tex]
     
    Last edited: Nov 29, 2007
  10. Nov 29, 2007 #9

    Kurdt

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    If you want to test then there are plenty of preview websites. Here is one of them.

    http://at.org/~cola/tex2img/index.php
     
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