$13 \cdot 20$ quadratic function $y=-x^{2}+6 x-7$, when $x$ takes values in the range $t \leqslant x \leqslant t+2$, the maximum value of this function is $y$ max $=-(t-3)^{2}+2$, then the range of values for $t$ is
(A) $t \leqslant 0$.
(B) $0 \leqslant t \leqslant 3$.
(C) $t \geqslant 3$.
(D) None of the above.
(China Hubei Province Huanggang Junior High School Mathematics Competition, 1991)
[Solution] Given $y=-x^{2}+6 x-7=-(x-3)^{2}+2$, and when $t \leqslant x \leqslant t+2$, $y_{\text {max }}=-(t-3)^{2}+2$. Then $y$ is a decreasing function when $t \leqslant x \leqslant t+2$.
From the above, we know that when $x \geqslant 3$, $y=-(x-3)^{2}+2$ is a decreasing function. $\therefore \quad t \geqslant 3$.
Therefore, the answer is $(C)$.