Example 7. For the 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 $y^{\prime}=-(t-3)^{2}+2$. Then the range of values for $t$ is ).
(A) $t=0$
(B) $0 \leqslant t \leqslant 3$
(C) $t \geqslant 3$
(D) None of the above
$$
\text{Solve} \begin{aligned}
\because y & =-x^{2}+6 x+7 \\
& =-(x-3)^{2}+2, \\
\therefore x=-\frac{b}{2 a} & =3, \text{ and } a=-1<0 .
\end{aligned}
$$

When $3 \in[t, t+2]$, i.e., $1 \leqslant t \leqslant 3$,
$$
y=f(3)=2 \text{ and } y \text{ here }=-(t-3)^{2}+
$$
2 is contradictory.
When $3 \geqslant t+2$, i.e., $t \leqslant 1$,
$$
y \in f(t+2)=-(t-1)^{2}+2 \text{ and }
$$
$y$ group $=-(t-3)^{2}+2$ is contradictory.
When $3 \leqslant t$, i.e., $t \geqslant 3$,
$y_{\text{max}}=f(t)=-(t-3)^{2}+2$ is consistent with the problem. Therefore, the range of $t$ is $t \geqslant 3$. The correct choice is (C).
$$