anonymous
  • anonymous
I need to find the tangent plane at (-1,2,4), and I have a function z = 4x^2-y^2+2y.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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 \[z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\]
anonymous
  • anonymous
So first you should find \[f_x(x_0,y_0) \] and \[f_y(x_0,y_0)\]
anonymous
  • anonymous
How do I find those?

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anonymous
  • anonymous
You need to use partial derivatives, which means that you keep y constant when you are trying to find \[f_x(x_0,y_0)\]
anonymous
  • anonymous
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 ??
anonymous
  • anonymous
Yes, that is correct. Now you just need to insert it in the function for the tangent plane.
anonymous
  • anonymous
Thank you very much :)

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