I am confused about a solution of the following problem:
Let $f$ be a function from $\mathbb{R^2}\rightarrow \mathbb{R}$ such that $f$ $\in C^1$. Furthermore, $9x-14y-2z=20$ is the plane tangent to the graphic of the function at the point $(2,-1,f(2,-1))$ find the value of the function and its gradient at (2,-1)
In the solution I was consulting, they start off from the statement that the vector $(f_x,f_y,-1)$ is normal to the graphic at the appropriate point, from thereafter they compare it with the normal vector given to us by the equation of the plane. I am confused as to how they infered that; why the -1?
Let $g(x,y,z)=f(x,y)-z$. Then$$\text{graphic of }f=\{(x,y,z)\in\Bbb R^3\mid g(x,y,z)=0\}.$$But the tangent plane to a level surface $S=\{(x,y,z)\in\Bbb R^3\mid g(x,y,z)=c\}$ at a point $(a,b,c)\in S$ (assuming that it is a regular point of that surface) is the plane passing through $(a,b,c)$ which is orthogonal to the gradient of $\varphi$ at $(a,b,c)$, which is $\bigl(\varphi_x(a,b,c),\varphi_y(a,b,c),\varphi_z(a,b,c)\bigr)$. But$$\nabla g(a,b,c)=\bigl(f_x(a,b),f_y(a,b),-1\bigr).$$