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I'd also think you'd use the partial derivatives to find out the graph of that plane.

I think you mean \[
\Large L(x,y)= f(x_0,y_0) + f_x(x,y)(x-x_0)+f_y(x,y)(y-y_0)
\]

Since \(z=f(x,y)\)
\[ \large
f_x=\frac{\delta z}{\delta x}
\]

That formula gives you the equation of the plane tangent to \((x_0, y_0)\)

By adjust, I mean get to L(x,y) from f(x,y)...

terms added to f(x,y)?