Equation of a plane at point (2,-3): if f(x,y)=4x^2+5xy+2y^2, then...

[CarissaF]

if f(x,y)=4x^2+5xy+2y^2
Then an implicit equation for the tangent plane to the graph at point (2,-3) is:

I got an answer of x-2y=0 but it is wrong... I can't figure out why. This is my work.

0=f(2,-3)+f_x(2,-3)(x-2)+f_y(2,-3)(y--3)
f(2,-3)=4
f_x(2,-3)=1
f_y(2,-3)=-2

0=4+1(x-2)-2(y+3)
0=4+x-2-2y-6
4=x-2y

Can anyone spot my mistake?

 

[skaa]

The formula is:
z=f(2,-3)+f_x(2,-3)(x-2)+f_y(2,-3)(y-(-3))

 

[skaa]

The formula for a tangent plane is:
z=f(2,-3)+f_x(2,-3)(x-2)+f_y(2,-3)(y--3)f(2,-3)+f_x(2,-3)(x-2)+f_y(2,-3)(y-(-3))