2. Let the function $f(x)=x^{2}+a x+b$, for any $a, b \in \mathbf{R}$, there always exists $t \in[0,4]$, such that $|f(t)| \geqslant m$ holds, then the maximum value of the real number $m$ is $\qquad$.
Let $M=\max \left\{|f(0)|,|f(4)|,\left|f\left(-\frac{a}{2}\right)\right|\right\}=\max \left\{|b|,|16+4 a+b|,\left|b-\frac{a^{2}}{4}\right|\right\}$,
then $4 M \geqslant|b|+|16+4 a+b|+2\left|b-\frac{a^{2}}{4}\right| \geqslant\left|16+4 a+\frac{a^{2}}{2}\right|$ $=\frac{1}{2}\left|a^{2}+8 a+32\right|=\frac{1}{2}\left|(a+4)^{2}+16\right| \geqslant 8 \Rightarrow M \geqslant 2$, equality holds when $a=-4, b=2$. Therefore, the maximum value of $m$ is 2.