Question # Find an equation of the tangent plane to the given

Negative numbers and coordinate plane
ANSWERED Find an equation of the tangent plane to the given surface at the specified point.
$$\displaystyle{z}={2}{\left({x}-{1}\right)}^{{2}}+{6}{\left({y}+{3}\right)}^{{2}}+{4},\ {\left({3},-{2},{18}\right)}$$ 2021-08-18
Find an equation of the tangent plane to the given surface at the specified point.
$$\displaystyle{z}={2}{\left({x}-{1}\right)}^{{2}}+{6}{\left({y}+{3}\right)}^{{2}}+{4},{\left({3},-{2},{18}\right)}$$
$$\displaystyle\Rightarrow{z}_{{x}}={4}{\left({x}-{1}\right)}$$
$$\displaystyle={4}{x}-{4}$$,
$$\displaystyle{z}_{{y}}={12}{\left({y}+{3}\right)}$$
$$\displaystyle={12}{y}+{36}$$
At the point (3,-2,18):
$$\displaystyle{z}_{{x}}={4}{x}-{4}$$
$$\displaystyle={4}{\left({3}\right)}-{4}$$
$$\displaystyle={12}-{4}$$
$$\displaystyle={8}$$
$$\displaystyle{z}_{{y}}={12}{y}+{36}$$
$$\displaystyle=-{12}{\left(-{2}\right)}+{36}$$
$$\displaystyle=-{24}+{36}$$
$$\displaystyle={12}$$
Equation of the tangent plane is,
$$\displaystyle{z}-{z}_{{1}}={z}_{{x}}{\left({x}-{x}_{{1}}\right)}+{z}_{{y}}{\left({y}-{y}_{{1}}\right)}$$
$$\displaystyle{z}-{18}={8}{\left({x}-{3}\right)}+{12}{\left({y}-{\left(-{2}\right)}\right)}$$
$$\displaystyle{z}-{18}={8}{x}+{12}{y}$$
$$\displaystyle{z}={8}{x}+{12}{y}+{18}$$