22. Let $f(x)=x^{2}-2 a x+2$, when $x \in[-1,+\infty]$, $f(x) \geqslant a$, find the range of values for $a$.


22. $f(x)=(x-a)^{2}+2-a^{2}$, vertex is $\left(a, 2-a^{2}\right)$, then for $x \in[-1,+\infty)$, $f(x) \geqslant a$ has the following two cases:
(1) $\left\{\begin{array}{l}a<-1 \\ f(-1) \geqslant a\end{array}\right.$
(2) $\left\{\begin{array}{l}-1 \leqslant a \\ 2-a^{2} \geqslant a\end{array}\right.$
(1) gives $-3 \leqslant a<-1$, (2) gives $-1 \leqslant a \leqslant 1$, hence the solution to the original problem is $a \in[-3,1]$