2. If real numbers $x, y$ satisfy $y^{2}=4 x$, then the range of $\frac{y}{x+1}$ is $\qquad$ .
2. $[-1,1]$.

It is easy to know that the focus of the parabola is $(0,1)$, and the directrix is $x=-1$.

By the definition of a parabola, we have
$$
y^{2}=4 x \Leftrightarrow \sqrt{y^{2}+(x-1)^{2}}=|x+1| \text {, }
$$

Thus, $\left|\frac{y}{x+1}\right| \leqslant \frac{\sqrt{y^{2}+(x-1)^{2}}}{|x+1|}=1$.
When $x=1, y= \pm 2$, $\frac{y}{x+1}$ can respectively reach the maximum value 1 and the minimum value -1.
Therefore, the range of $\frac{y}{x+1}$ is $[-1,1]$.