[tex]f(x)=ax^2+bx+c\\\\\text{the vertex}\\\\V(h,\ k)\\\\h=\dfrac{-b}{2a},\ k=f(h)[/tex]
We have:
[tex]f(x)=x^2-6x+3\\\\a=1,\ b=-6,\ c=3\\\\h=\dfrac{-(-6)}{2(1)}=\dfrac{6}{2}=3\\\\k=f(3)=3^2-6(3)+3=9-18+3=-6[/tex]
Rearrange into vertex form
x^2 - 6x + 3
= (x - 3)^2 - 9 + 3
= (x - 3)^2 - 6
This is vertex form ( y = a(x - b)^2 + c is general vertex form with vertex
= (b, c) ).
So here the vertex is (3, -6) answer
