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Math Help - Modelling a quadratic - Systems of linear equations

  1. #1
    Ari
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    Modelling a quadratic - Systems of linear equations

    Hi, i cant figure out how to set up this question:
    ax^2 + bx + c passes through (-1,5) and has a horizontal tangent at (1,7) find the coefficents a b and c

    Thanks.
    Last edited by mr fantastic; May 11th 2010 at 07:15 PM. Reason: Edited post title.
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    Quote Originally Posted by Ari View Post
    Hi, i cant figure out how to set up this question:
    ax^2 + bx + c passes through (-1,5) and has a horizontal tangent at (1,7) find the coefficents a b and c

    Thanks.
    a(-1)^2-1b+c=5\rightarrow a-b+c=5

    derivative

    2xa+b=0 when x=-1
    Last edited by dwsmith; May 11th 2010 at 07:57 PM.
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  3. #3
    Ari
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    k im confused, how does that help me find the the values for a,b,c?

    edit: and can you please explain how you got that? i feel very confused atm lol.
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  4. #4
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    Setup the coefficient matrix for the two equations and solve simultaneously.
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  5. #5
    Ari
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    Quote Originally Posted by dwsmith View Post
    Setup the coefficient matrix for the two equations and solve simultaneously.
    i just started this course... which two equations are you talking about?
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    Quote Originally Posted by Ari View Post
    i just started this course... which two equations are you talking about?
    1: a-b+c=5
    2: -2a+b=0
    Last edited by dwsmith; May 11th 2010 at 07:57 PM.
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  7. #7
    Ari
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    Quote Originally Posted by dwsmith View Post
    1: a-b+c=5
    2: 2a+b=0
    oh okay, how did you get that second equations from the

    a-b+c=5\

    edit: nevermind i see it now. its from the originial one.

    now how exactly do i solve them simultaneously? i can get the parametric solutions, but how do i get the number values?
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    a(-1)^2-1b+c=5\rightarrow a-b+c=5

    derivative

    2xa+b=0\rightarrow -2a+b=0 when x=-1
    The 2nd equation came from the original equation.
    Last edited by dwsmith; May 11th 2010 at 07:58 PM.
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  9. #9
    Ari
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    is the answer just going to be a,b,c in parametric form or will there be coefficent values, if so how do i get values for them?
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  10. #10
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    Quote Originally Posted by Ari View Post
    is the answer just going to be a,b,c in parametric form or will there be coefficent values, if so how do i get values for them?
    Reduced row echelon form of the coefficient matrix.

    \begin{bmatrix}<br />
1 & -1 & 1 & 5\\ <br />
-2 & 1 & 0 & 0<br />
\end{bmatrix}
    Last edited by dwsmith; May 11th 2010 at 07:58 PM.
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  11. #11
    Ari
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    Quote Originally Posted by dwsmith View Post
    Reduced row echelon form of the coefficient matrix.

    \begin{bmatrix}<br />
1 & -1 & 1 & 5\\ <br />
2 & 1 & 0 & 0<br />
\end{bmatrix}
    \begin{bmatrix}<br />
1 & -1 & 1 & 5\\ <br />
0 & 3 & -2 & -10<br />
\end{bmatrix}


    is this it or am i wayyyy off?
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    \begin{bmatrix}<br />
1 & -1 & 1 & 5\\ <br />
0 & 1 & 2 & 10<br />
\end{bmatrix}\rightarrow \begin{bmatrix}<br />
1 & 0 & 3 & 15\\ <br />
0 & 1 & 2 & 10<br />
\end{bmatrix}
    Last edited by dwsmith; May 11th 2010 at 07:59 PM.
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  13. #13
    Ari
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    Quote Originally Posted by dwsmith View Post
    \begin{bmatrix}<br />
1 & -1 & 1 & 5\\ <br />
0 & 3 & -2 & -10<br />
\end{bmatrix}


    Close but off a little.
    haha i quickly changed it but i wasnt fast enough. anyway there are 2 equations and 3 unknowns so there are infinietly many solutions?
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  14. #14
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    Yes you can right your solution as you would the nullity with a free variable.

    Note: Not all under determined linear systems are consistent.
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  15. #15
    Ari
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    Quote Originally Posted by dwsmith View Post
    Yes you can right your solution as you would the nullity with a free variable.

    Note: Not all under determined linear systems are consistent.
    okay so

    a = 5 + b - c

    b = -10 + 2

    c = ?
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