# Taylor polynomials for multivariable functions

Ok there's something I don't get. I know for instance that the linear polynomial for say f = 91 + 2x + 3y + 8z + Quadratic(x, y, z) + Cubic(x, y, z) ... is 91 + 2x + 3y + 8z if the base point is (0, 0, 0). This is pretty clear. What I don't get is why when you take the base point to be say (1, 2, 3) all of a sudden 91 + 2x + 3y + 8z is no longer the linear approximation. I figure it's because we have to move the graph from (1, 2, 3) to the origin. But I did that, and that didn't seem to work since the constants didnt' work out. Any ideas on the mathematical and intuitive reasoning behind why the linear polynomial for (0, 0, 0) doesn't work? Thanks.

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Dick
Homework Helper
I'm not sure I get your drift here. The tangent plane at (0,0,0) is generally different from the tangent plane at (1,2,3). It's going to have different slopes, right????

HallsofIvy
The linear approximation to f(x,y,z) at $(x_0,y_0,z_0)$ is
$$f(x_0,y_0,z_0)+ \frac{\partial f}{\partial x}(x_0,y_0,z_0)(x-x_0)+ \frac{\partial f}{\partial y}(x_0,y_0,z_0)(y-y_0)+ \frac{\partial f}{\partial z}(x_0,y_0,z_0)(z-z_0)$$