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?