\(y=x^2-4x+4\\
y'=2x-4\\\\
2x-4=0\\
2x=4\\x=2\\
2\in[-3,3]\\\\
y''=2>0 \Rightarrow \text{ There's a minimum at } x=2\)
In this case you have find the values of \(y(-3)\) and \(y(3)\).
\(y(-3)=(-3)^2-4\cdot(-3)+4=9+12+4=25\\
y(3)=3^2-4\cdot3+4=9-12+4=1\\\\
y(-3)>y(3) \Rightarrow y_{max}=25\)
