3. Let the quadratic function $y=x^{2}+m x+n$ have its vertex on $y=-4$, and the graph intersects with the lines $y=x+2$, $y=1-\frac{1}{2} x$ each at one point within $0 \leqslant x \leqslant 2$. Then the range of values for $m$ is $\qquad$.
3. $-8 \leqslant m \leqslant-2 \sqrt{6}$ or $4(\sqrt{2}-1) \leqslant$ $m \leqslant 2 \sqrt{5}$.

Since the vertex is on $y = -4$, we have $n = \frac{m^{2}}{4} - 4$. The quadratic function can be expressed as $y = (x + \frac{m}{2})^{2} - 4$. According to the problem, the vertex should be between points $P_{1}, P_{2}$ or $P_{3}, P_{4}$. It is easy to find that when the curve passes through $P_{1}$ and $(0,2)$, $m = -2 \sqrt{6}$ (note that the x-coordinate of the vertex is $-\frac{m}{2} > 0$); when the curve passes through $P_{2}$ and $(2,0)$, $m = -8$, so, $-8 \leqslant m \leqslant -2 \sqrt{6}$. Similarly, the other set of solutions can be found.