How Do You Find the Z-Component of the Normal Vector in a Tangent Plane Problem?

[Loppyfoot]

Homework Statement

Consider the function f(x,y) = 4-x^2+3y^2 + y.

Let S be the surface described by the equation z= f(x,y) where f(x,y) is given above. Find an equation for the plane tangent to S at the point (-1,0,3)

The Attempt at a Solution

Ok, SO i solved for the gradient of F; <-2x,6y+1>. I understand that to find the normal vector to the tangent plane, I need to plug in the points (-1,0,3) into the gradient, BUT what I get are only the x and y values for the normal vector. Where does the z value for normal vector come from, in order to solve the implicit equation, ax+by+cz=d?

Thanks!

 

[vela]

When you have a surface defined implicitly by F(x,y,z)=0, the normal is given by ∇F. You can convert your explicit surface z=f(x,y) simply by writing F(x,y,z)=z-f(x,y)=0.

 

[Loppyfoot]

Duh. Thank you!