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Math Help - How to find f(x) - I suspect an error in the command

  1. #1
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    How to find f(x) - I suspect an error in the command

    Hello
    I want to solve a question, I don't understand why my answer is false so i think there might be an error in the order.

    The quadratic function takes the value of 41 at x=-2 and the value of 20 at x=5. The function is minimalized at x=2.
    I'm asked to give :
    - the A,B,C of y=Ax^2+Bx+C
    - the minimum value of this function, D

    The vertex have coordinates (2;D)
    2=-B/2A
    B=-4A

    I then replaced B in

    41=A(-2)^2 +B(-2)+C
    20=A(5)^2 +B(5)+C

    In the end I've got f(x)= (-3/7)x^2-(12/7)x+(275/7)

    It can't be right because A can't be an negative number if the function has a minimum. Also in this case the vertex is not in x=2 because D=239/7 (it's higher than f(-2)=20)
    But when I calculate
    f(-2)= (-3/7)(-2)^2-(12/7)(-2)+(275/7)
    AND
    f(5)=(-3/7)5^2-(12/7)5+(275/7)
    It takes the value of 41 and 20.

    Is it me confusing everything or there is really an error in the command?
    Last edited by Viou; March 20th 2011 at 02:06 PM.
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  2. #2
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    Quote Originally Posted by Viou View Post
    Hello
    I want to solve a question, I don't understand why my answer is false so i think there might be an error in the order.

    The quadratic function takes the value of 41 at x=-2 and the value of 20 at x=5. The function is minimalized at x=2.
    I'm asked to give :
    - the A,B,C of y=Ax^2+Bx+C
    - the minimum value of this function, D

    The appex have coordinates (2;D)
    2=-B/2A
    B=-4A

    I then replaced B in

    41=A(-2)^2 +B(-2)+C
    20=A(5)^2 +B(5)+C

    In the end I've got f(x)= (-3/7)x^2-(12/7)x+(275/7)

    It can't be right because A can't be an negative number if the function has a minimum. Also in this case the appex is not in x=2 because D=239/7 (it's higher than f(-2)=20)
    But when I calculate
    f(-2)= (-3/7)(-2)^2-(12/7)(-2)+(275/7)
    AND
    f(5)=(-3/7)5^2-(12/7)5+(275/7)
    It takes the value of 41 and 20.

    Is it me confusing everything or there is really an error in the command?
    see this post ...

    http://www.mathhelpforum.com/math-he...le-175161.html
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  3. #3
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    Do you understand that a quadratic function can be written as y= a(x- x_0)^2+ b where x_0 is the x coordinate of the vertex (minimum point)? Knowing that the minimum occurs at x= 2 tells you that you can write the quadratic function as y= a(x- 2)^2+ b. Putting x= -2 and y= 41 gives 41= a(-2-2)^2+ b= 16a+ b. putting x= 5 and y= 20 gives 20= a(5- 2)^2+ b= 9a+ b. You can solve those two equations for a and b.
    Last edited by HallsofIvy; March 21st 2011 at 07:42 AM.
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  4. #4
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    Ok thank you very much! It worked out nice
    And sorry for putting it in the wrong section
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