11. Given the parabola $y=x^{2}+m x+n$ passes through the point $(2,-1)$, and intersects the $x$-axis at points $A(a, 0)$ and $B(b, 0)$. If $P$ is the vertex of the parabola, find the equation of the parabola that minimizes the area of $\triangle P A B$.
Three, 11. Given that the parabola $y=x^{2}+m x+n$ passes through the point $(2,-1)$, we have
$$
4+2 m+n=-1 \Rightarrow 2 m+n=-5 \text {. }
$$

Since the parabola $y=x^{2}+m x+n$ intersects the $x$-axis at points $A(a, 0), B(b, 0)$, we have,
$$
a+b=-m, a b=n \text {. }
$$

Thus, $|A B|=|a-b|=\sqrt{(a+b)^{2}-4 a b}$ $=\sqrt{m^{2}-4 n}$.
Given that the vertex $P\left(-\frac{m}{2}, \frac{4 n-m^{2}}{4}\right)$.
Therefore, $S_{\triangle P A B}=\frac{1}{2}|a-b|\left|\frac{4 n-m^{2}}{4}\right|$
$$
=\frac{1}{8}\left(m^{2}-4 n\right)^{\frac{3}{2}} \text {. }
$$

Notice that,
$$
m^{2}-4 n=m^{2}+4(5+2 m)=(m+4)^{2}+4 \text {. }
$$

Thus, when $m=-4$, $S_{\triangle P A B}$ reaches its minimum value.
At this time, $n=-5-2 m=3$.
Therefore, the parabola $y=x^{2}-4 x+3$ is the one we are looking for.