A polynomial with no linear terms has (0, 0) as a critical point? My professor said in lecture that if you see a polynomial function with no linear (1-degree) terms, such as $f(x, y) = x^3 - y^2 - xy + 1$, then you can immediately tell that one of its critical points is $(0, 0)$. She said that it's because of something to do with all the terms going to zero, leaving $f(x, y) = 1$ as the closest linear approximation at $(0, 0)$. I still don't understand how this means $(0, 0)$ is a critical point. Surely if there was a linear term, it would also go to zero as $(x, y)$ went to $(0, 0)$? Could someone please explain this to me?
 A: If
$$f(x,y)=f_0+ax+by+\dots$$
(dots meaning terms of degree at least two), then
$$
f'_x(x,y)=a+\dots
,\qquad
f'_y(x,y)=b+\dots
$$
(dots meaning terms of degree at least one).
So $f'_x(0,0)=a$ and $f'_y(0,0)=b$.
Thus $f'_x$ and $f'_y$ are both zero at the origin if (and only if) there are no linear terms to begin with (i.e., $a=b=0$).
A: If you had terms like $ax$ and $by$ in your function like $f(x, y) = x^3 - y^2 - xy + 1$, then it should be
$$f(x, y) = x^3 - y^2 - xy + ax+by+1$$
you see that $\dfrac{\partial f}{\partial x}=3x^2-y+a=0$ and $\dfrac{\partial f}{\partial y}=-2y-x+b=0$. This system doesn't get zero as a critical point, since it has constant term. Now let $a=b=0$, so in the system
\begin{cases}
3x^2-y=0,\\
-2y-x=0.
\end{cases}
clearly $(0,0)$ is critical point of it. That's all.
A: This whole thing works in $\Bbb R^n$:
Suppose
$P(x_1, x_2, \ldots, x_n) \in \Bbb R[x_1, x_2, \ldots, x_n] \tag 1$
is a polynomial in the $n$ variables $x_1$, $x_2$, $\ldots$, $x_n$, and that $P(x_1, x_2, \ldots, x_n)$ has no linear terms.  Then we may we may write $P$ as
$P(x_1, x_2, \ldots, x_n) = \sum_i x_iQ_i(x_1, x_2, \ldots, x_n) + P_0, \tag 2$
where the constant term $Q_{i0} \in \Bbb R$ of each $Q_i(x_1, x_2, \ldots, x_n) \in \Bbb R[x_1, x_2, \ldots, x_n]$ vanishes and $P_0 \in \Bbb R$.  A critical point of $P(x_1, x_2, \ldots, x_n)$ is defined as a point $(c_1, c_2, \ldots, c_n) \in \Bbb R^n$ where 
$\dfrac{\partial P(c_1, c_2, \ldots, c_n)}{\partial x_i} = 0, \tag 3$
for every $x_i$, $1 \le i \le n$.  Now from (2),
$\dfrac{\partial P(c_1, c_2, \ldots, c_n)}{\partial x_i} = \sum_i (Q_i(x_1, x_2, \ldots, x_n) + x_i \dfrac{\partial Q_i(x_1, x_2, \ldots, x_n)}{\partial x_i}); \tag 4$
since $Q_{i0} = 0$ for all $i$, we have
$Q_i(0, 0, \ldots, 0) = 0, \tag 5$
and so
$\dfrac{\partial P(0, 0, \ldots, 0)}{\partial x_i} = \sum_i (Q_i(0, 0, \ldots, 0) + 0 \cdot \dfrac{\partial Q_i(0, 0, \ldots, 0)}{\partial x_i}) = 0; \tag 6$
this shows $(0, 0, \ldots, 0)$ is a critical point of $P(x_1, x_2, \ldots, x_n)$; the corresponding critical value is evidently $P_0$.
In other words, $P(x_1, x_2, \ldots, x_n)$ is, to second order in the $x_i$, the constant $P_0$ near $(0, 0, \ldots, 0)$.
If there were a linear term in $P(x_1, x_2, \ldots, x_n)$, it would be of the form
$\sum_1^n P_{1i}x_i \to 0 \; \text{as} \; (x_1, x_2, \ldots, x_n) \to (0, 0, \ldots, 0), \tag 7$
but if any $P_{1i} \ne 0$, it would leave behind a non-vanishing derivative 
$\dfrac{\partial P(0, 0, \ldots, 0)}{\partial x_i} = P_{0i}, \tag 8$
and $(0, 0, \ldots, 0)$ wouldn't be critical for $P(x_1, x_2, \ldots, x_n)$.
