Question 760129
{{{(y-1)^2=16(x-4)}}}....it is already written in vertex form; {{{h=4}}} and {{{k=1}}}

so, vertex is at ({{{4}}},{{{1}}})


{{{ drawing( 600, 600, -10, 10, -10, 10,circle(4,1,0.2),locate(4,1,V(4,1)),graph( 600, 600, -10, 10, -10, 10,sqrt(16(x-4))+1,-sqrt(16(x-4))+1)) }}}