Suppose $S = \{x \in \mathbb{R}^n | P(x) = 0\}$, for some polynomial $P$, and I want to find the critical points of $f: S \rightarrow \mathbb{R}$ which takes $(x_1, \ldots, x_n) \mapsto x_1$. It is clear to me that the critical points are contained in the set of solutions of $\{P(x) = 0, \frac{\partial P}{x_2} = \ldots = \frac{\partial P}{x_n} = 0\}$.
I now have a situation where I have a function $g: \mathbb{R^k} \rightarrow \mathbb{R}$, and I embed the graph of $g$ into $\mathbb{R}^n$ taking $(x_1,\ldots, x_{k+1}) \rightarrow (x_1,\ldots, x_{k+1}, 0, \ldots, 0)$. Let us call the graph $G \subseteq \mathbb{R}^n$. Now consider $G \cap S$. I want to obtain the critical points of $f: G \cap S \rightarrow \mathbb{R}$, again taking $(x_1, \ldots, x_n) \mapsto x_1$. I am finding that the critical points in this case are contained in the set of solutions of $\{G, P(x) = 0, \frac{\partial P}{x_2} = \ldots = \frac{\partial P}{x_{k}} = 0\}$. I want to understand why this works. Specifically, why are only $k-1$ partial derivatives enough?
Here are my thoughts. First, $G$ has dimension $k$, so $G \cap S$ has dimension $k-1$, so as a sanity check it makes sense that I would only need $k-1$ partial derivatives. However, what does setting only $k-1$ partial derivatives of $P$ to $0$ even mean geometrically? Any insight would be great.
(assume there is no degeneracy anywhere, and everything is smooth and nice)