# How do you use the taylor series to find a quadratic approximation of 2x^2 - 3xy + x at (1,1)?

Feb 15, 2017

$= x \left(2 x - 3 y + 1\right)$

#### Explanation:

The 2-D Taylor Expansion about $\left(\alpha , \beta\right)$ is:

 T(x,y)=f(alpha,beta)+(x-alpha )f_(x)(alpha,beta)+(y-beta)f_{y}(alpha,beta)+{1}/{2!} ( (x-alpha)^{2}f_(x x)(alpha,beta)+2(x-alpha)(y-beta)f_{xy}(alpha,beta)+(y-beta)^{2}f_{yy}(alpha,beta) )+\cdots

We therefore need the following partial derivatives:

${f}_{x} = 4 x - 3 y + 1 , q \quad {f}_{x} \left(1 , 1\right) = 2$

${f}_{y} = - 3 x , q \quad {f}_{y} \left(1 , 1\right) = - 3$

${f}_{x y} = {f}_{y x} = - 3 , q \quad {f}_{x y} \left(1 , 1\right) = - 3$

${f}_{x x} = 4 , q \quad {f}_{x x} \left(1 , 1\right) = 4$

${f}_{y y} = 0 , q \quad {f}_{y y} \left(1 , 1\right) = 0$

This means that, working to only the quadratic terms:

 T(x,y) =0+2(x-1 )-3(y-1)+{1}/{2!} ( 4(x-1)^{2}+2(-3)(x-1)(y-1)+0 (y-1)^{2} )

$= 2 x - 2 - 3 y + 3 + 2 {x}^{2} - 4 x + 2 - 3 x y + 3 x + 3 y - 3$

$= x \left(2 x - 3 y + 1\right)$