\(A=-x^2-6x+1\)
\(=-x^2-6x-9+10\)
\(=-\left(x^2+2\cdot x\cdot3+3^2\right)+10\)
\(=-\left(x+3\right)^2+10\)
Ta có: \(\left(x+3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x+3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+3\right)^2+10\le10\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy \(Max_A=10\) khi \(x=-3\)