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Math Help - Find type, vertex and line of symmetry of the approximating quadratic polynomial.

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    Find type, vertex and line of symmetry of the approximating quadratic polynomial.

    We have the function:
    f(x,y)=32xy^2-20xy^3-3x^3y+4

    Find the approximating quadratic polynomial with the development points: (x_0,y_0)=(0,1)
    I've found it to be:
    z=4+8x+4xy

    Now find the type, vertex and line of symmetry for the graph the approximating quadratic polynomial produces.


    I had to translate it to English, but hopefully it is understandable and if not I'll try to explain the question differently.
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Student57 View Post
    We have the function:
    Now find the type, vertex and line of symmetry for the graph the approximating quadratic polynomial produces.

    You can use the quadric classification table and express the quadric in canonical form.
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    Quote Originally Posted by FernandoRevilla View Post
    You can use the quadric classification table and express the quadric in canonical form.
    possibility for an expanded explanation?
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    MHF Contributor FernandoRevilla's Avatar
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    The quadric is 4xy+8x-z+4=0 . The associated quadratic form is

    q(x,y,z)=4xy=(x,y,z)\begin{pmatrix}{0}&{2}&{0}\\{2  }&{0}&{0}\\{0}&{0}&{0}\end{pmatrix}\begin{pmatrix}  {x}\\{y}\\{z}\end{pmatrix}

    The eigenvalues are \lambda_1=2,\lambda_2=-2,\lambda_3=0. An orthonormal basis of eigenvectors is:

    B=\left\{{(1/\sqrt{2})(1,1,0),(1/\sqrt{2})(1,-1,0),(0,0,1)}\right\}

    The equation of the basis change is:

    G\equiv\begin{pmatrix}{x}\\{y}\\{z}\end{pmatrix}=\  begin{pmatrix}{\;1/\sqrt{2}}&{-1/\sqrt{2}}&{0}\\{1/\sqrt{2}}&{\;\;1/\sqrt{2}}&{0}\\{\;0}&{\;\;0}&{1}\end{pmatrix}\begi  n{pmatrix}{x'}\\{y'}\\{z'}\end{pmatrix}

    and the quadric has the expression:

    2x'^2-2y'^2+(8/\sqrt{2})(x'+y')-z'+4=0

    Using the translation:

    T\equiv\begin{Bmatrix}x'=x''+a\\y'=y''+b\\z'=z''+4  \end{matrix}

    we can cancel the linear terms choosing a=-\sqrt{a},b=-\sqrt{2} . The quadric on the x''y''z'' axis has the equation:

    z''=2x''^2-2y''^2

    that is, a hyperbolic paraboloid. You can easily identify its elemets of simmetry. After, use T^{-1}\circ G^{-1} to find them on the xyz axis.
    Last edited by FernandoRevilla; February 28th 2011 at 04:27 AM.
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