Having a critical point "at infinity" makes sense when you're looking at the Riemann sphere, which you can think of as the union of $\mathbb C$ with a single point that we call $\infty$, and which is useful in complex dynamics (which includes the study of the Mandelbrot set and Julia sets). If you want to evaluate a rational function $\phi(z) = f(z)/g(z)$ whose numerator and denominator share no common factors, at a point $c\in\mathbb C$ where $g(c) = 0$, we define $\phi(c) = \infty$.
We say that a rational function $\phi(z)$ has a critical point at $\alpha\in\mathbb C\cup\{\infty\}$ if for all Möbius transformations $f(z)$ such that $\beta = f^{-1}(\alpha)\neq \infty$ and $\phi^f(\beta) = (f^{-1}\circ\phi\circ f)(\beta)\neq \infty$, $(\phi^f)'(\beta) = (f^{-1}\circ\phi\circ f)'(\beta) = 0$. One can show that this is a extension of the familiar definition of a critical point: if $\alpha\neq\infty$ and $\phi(\alpha)\neq\infty$, then $\alpha$ is a critical point of $\phi$ (according to the definition in the previous sentence) iff $\phi'(\alpha) = 0$. Note that we when refer to a derivative in this context, it is a formal derivative; for example, we are formally defining $ \phi'(z) = \frac{g(z)f'(z) - f(z)g'(z)}{(g(z))^2}. $ Now, the following lemma is quite handy:
Lemma. Let $\phi$ be a rational function and let $f$ be a Möbius transformation. Then $\forall \alpha\in\mathbb C\cup\{\infty\}$, $\alpha$ is a critical point of $\phi$ iff $\beta = f^{-1}(\alpha)$ is a critical point of $\phi^f$.
Using this lemma, we can verify Isaac's suggestion that every polynomial of degree at least two has infinity as a critical point. Consider a polynomial function $\phi(z) = a_n z^n + \cdots + a_1z + a_0$, where $n\geq 2$. Define the Möbius transformation $f(z) = 1/z$. Then $f^{-1}(\infty) = 0$, and \begin{align*} \phi^f(z) & = (f^{-1}\circ\phi\circ f)(z) \\ & = \frac{1}{\tfrac{a_n}{z^n} + \cdots + \tfrac{a_1}{z} + a_0} \\[3pt] & = \frac{z^n}{a_n + \cdots + a_1z^{n-1} + a_0z^n}. \end{align*} By writing out the formal derivative $(\phi^f)'(z)$, we can check that $(\phi^f)'(0) = 0$, so clearly $0$ is a critical point of $\phi^f$. It then follows from the lemma that $\infty$ is a critical point of $\phi$.