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

1. ## 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.

2. Originally Posted by Student57
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.

3. Originally Posted by FernandoRevilla
You can use the quadric classification table and express the quadric in canonical form.
possibility for an expanded explanation?

4. 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.