Question

what would be the extrema point for y=x^2-6x+3? also max or min?

Answer 1

you have a parabola that faces up because the leading coefficient is positive (it is 1) so your extreme value will be a minimum

Answer 2

find it my making \(x=-\frac{b}{2a}\) the first coordinate of the vertex
in your case you get
\[x=-\frac{-6}{2}=3\] replace \(x\) by 3 to find your minimum