Question

I need to find the tangent plane at (-1,2,4), and I have a function z = 4x^2-y^2+2y.

Answer 1

Suppose f has a continuous partial derivatives. An equation of the tangent plane to the surface \[z=f(x,y)\] at the point \[P(x_0,y_0,z_0)\] is \[z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\]

Answer 2

So first you should find \[f_x(x_0,y_0) \] and \[f_y(x_0,y_0)\]

Answer 3

How do I find those?

Answer 4

You need to use partial derivatives, which means that you keep y constant when you are trying to find \[f_x(x_0,y_0)\]

Answer 5

Oh okay, is it correct that i've gotten f_x(x_0,y_0) = 8x and f_y(x_0,y_0) = -2y+2 ??

Answer 6

Yes, that is correct. Now you just need to insert it in the function for the tangent plane.

Answer 7

Thank you very much :)