The vertex form of [tex]f(x)=ax^2+bx+c[/tex]
[tex]f(x)=a(x-h)^2+k[/tex]
where: [tex]h=\dfrac{-b}{2a};\ k=f(h)[/tex]
We have: [tex]f(x)=x^2-6x+16\to a=1;\ b=-6;\ c=16[/tex]
substitute:
[tex]h=\dfrac{-(-6)}{2\cdot1}=\dfrac{6}{2}=3\\\\k=f(3)=3^2-6\cdot3+16=9-18+16=7[/tex]
The vertex form of f(x):
[tex]f(x)=(x-3)^2+7[/tex]
The value of minimum is equal k.
Therefore: [tex]y_{min}=7[/tex]
