Answer :

[tex]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[/tex]

In this case you have find the values of [tex]y(-3)[/tex] and [tex]y(3)[/tex].

[tex]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[/tex]