7. When $x \in[0,2]$, the function $f(x)=a x^{2}+4(a+1) x-3$ reaches its maximum value at $x=2$, the range of values for $a$ is $\qquad$ .
7. $a \geqslant-\frac{1}{2}$. When $a=0$, $f(x)=4x-3$ satisfies the condition; when $a>0$, $-\frac{2(a+1)}{a} \leqslant 0$, which gives $a>0$. Therefore, in summary, $a \geqslant-\frac{1}{2}$.