jmc

Because the parabola has vertex $(2,3)$, it is the graph of $$y=a(x-2{)}^2+3$$ for some number $a$. In order for the graph to contain the point $(4,4)$, we must also have $$4=a(4-2{)}^2+3=4a+3,$$ so $a=\frac{1}{4}$, and our parabola is the graph of $$y=\frac{1}{4}(x-2{)}^2+3.$$ Setting $x=6$ in this gives us $$y=\frac{1}{4}(6-2{)}^2+3=4+3=\boxed{7}.$$