Question

# Find an equation of the tangent plane to the given surface at the specified point. z=2(x-1)^2+6(y+3)^2+4,\ (3,-2,18)

Negative numbers and coordinate plane
Find an equation of the tangent plane to the given surface at the specified point.
$$z=2(x-1)^2+6(y+3)^2+4,\ (3,-2,18)$$

2021-05-19
Find an equation of the tangent plane to the given surface at the specified point.
$$z=2(x-1)^2+6(y+3)^2+4,(3,-2,18)$$
$$\Rightarrow z_x=4(x-1)$$
$$=4x-4$$,
$$z_y=12(y+3)$$
$$=12y+36$$
At the point (3,-2,18):
$$z_x=4x-4$$
$$=4(3)-4$$
$$=12-4$$
$$=8$$
$$z_y=12y+36$$
$$=-12(-2)+36$$
$$=-24+36$$
$$=12$$
Equation of the tangent plane is,
$$z-z_1=z_x(x-x_1)+z_y(y-y_1)$$
$$z-18=8(x-3)+12(y-(-2))$$
$$z-18=8x+12y$$
$$z=8x+12y+18$$