Question 338921
did you mean  {{{y=4* sqrt (2*x-6)+1}}}??
--
{{{y-1=4*sqrt(2*x-6)}}}
{{{(y-1)^2=16*(2*x-6)=16*2*(x-3)=32*(x-3)}}}
--
in this form {{{(y-1)^2=32*(x-3)}}} its easy to see that the vertex is at (3,1)