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Math Help - Quadratic function (HELP)

  1. #1
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    Question Quadratic function (HELP)

    Could you please help me to answer this question:

    The Quadratic function which takes the value 41 at x=-2 and the value 20 at x = 5 and is minimized at x = 2 is

    y = __x2 - __x +__

    the minimum value of this function is ____.



    ---Thank you in advance---
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  2. #2
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    Re: Quadratic function (HELP)

    Write it as \displaystyle y = a\,x^2 + b\,x + c, then \displaystyle \frac{dy}{dx} = 2a\,x + b.

    You know that when x = 2, [tex]\displaystyle \frac{dy}{dx} = 0[/math] due to the function being minimised there. You also know that when x = -2, y = 41, and when x = 5, y = 20. Substitute all this information into the equations so that you get equations in terms of a,b,c to solve simultaneously. See how you go.
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  3. #3
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    Re: Quadratic function (HELP)

    I try the first part is this one is correct?
    y=0, x=2

    0=2ax +b
    0 = 2(2)x + b
    0 =4x + b
    b =4x

    and when I try the other this is my outcome and I don't know what to do next
    y= 41, x=-2

    41=2(-2)x+b
    41= -4x + b
    41= -4x + 4x (I'd replaced it to 4x since I got b=4x)
    41 = 0 (and then it turn out to 0)

    next is

    y=20, x=5

    20 = 2(5)x +b
    20 = 10x +b
    20 = 10x + 4x
    20 = 14x

    could you pls help me more. I'm so sorry If I don't get it right
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  4. #4
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    Re: Quadratic function (HELP)

    Why do you still have x's in your equations when you have let them equal the numerical values? Remember, you are trying to find a,b,c NOT x.
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  5. #5
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    Re: Quadratic function (HELP)

    oh yes, you're right!

    I done the first part and I ended up here:

    0=2ax +b
    0 = 2(2)a + b
    0 =4a + b
    b = -4a

    y=41 x=-2
    41=2(-2)a+b
    41= -4a + b
    41= -4a - 4a (I'd replaced it to 4a since I got b= -4a)
    41 = -8a
    -41/8 = a

    is this correct?
    Last edited by Renrie; March 27th 2013 at 07:29 AM.
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    Re: Quadratic function (HELP)

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