Question 1076552
{{{2(x-3)^2+1}}}


Vertex occurs in the exact middle of the roots.  You can solve for {{{2(x-3)^2+1=0}}} and then the average of the roots is the x value for the vertex... but since your function is already given in standard form,  the vertex is  (3,1).


The vertex for {{{y=a(x-h)^2+k}}}  is the point   (h,k).