Answer
$z=6x+4y+8$
Work Step by Step
The equation for a tangent plane at $(x_0,y_0,z_0)$ is given by $z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$
Given:
$z=3(x-1)^2+2(y+3)^2+7$
Remember that $z = f(x,y)$
Solve for the partial derivatives:
$f_x=\frac{∂z}{∂x}=6(x-1)$
$f_y=\frac{∂z}{∂y}=4(y+3)$
Evaluate the partials at (2,-2):
$f_x(2,-2) = 6(2-1)=6$
$f_y(2,-2)=4(-2+3)=4$
Substitute all the numbers into the Equation of the Plane at (2,-2,12):
$z-12=6(x-2)+4(y+2)$
$z-12=6x-12+4y+8$
$z=6x+4y+8$
