[tex]\begin{gathered} \text{parabola} \\ y=2x^2-8x+8 \end{gathered}[/tex][tex]\begin{gathered} x\text{ coordinate of a vetex is given by} \\ h=-\frac{b}{2a} \\ and\text{ the y coordiante is given by} \\ k=f(h) \\ \text{where} \\ f(x)=ax^2+bx+c \end{gathered}[/tex][tex]\begin{gathered} In\text{ this case, we can s}ee\text{ that} \\ a=\text{ 2, b=-8, c=8} \\ \text{hence,} \\ h=-\frac{-8}{2(2)}\Rightarrow h=\frac{8}{4}\Rightarrow h=2 \\ \text{then} \\ k=f(2)\Rightarrow k=2(2)^2-8(2)+8 \\ k=2(4)-16+8 \\ k=8-8 \\ k=0 \\ \text{Therefore, the vertex (h,k) is} \\ (h,k)=(2,0) \end{gathered}[/tex][tex]\begin{gathered} \text{you can s}ee\text{ that the minimum is the same as the vertex. Then, the minimun is in} \\ (2,0) \end{gathered}[/tex]
