Encontramos os zeros da função - raízes - fazendo [tex]y=0[/tex]
Com isso,
[tex]x^2-4x+4=0\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot1\cdot4\\\\\Delta=16-16\\\\\Delta=0\\\\x=\frac{-b\pm\sqrt{\Delta}}{2a}\Rightarrow x=\frac{-(-4)\pm\sqrt{0}}{2\cdot1}\Rightarrow x=\frac{4\pm0}{2}\\\\[/tex]
[tex]\begin{cases}x'=\frac{4+0}{2}\Rightarrow\boxed{x'=2}\\x''=\frac{4-0}{2}\Rightarrow\boxed{x''=2} \end{cases}\\\\\boxed{\boxed{S=\left\{2\right\}}}[/tex]
