Question:
The graph of $y=ax^2 + bx + c$ is a parabola with vertical axis of symmetry.  The vertex of this parabola is $(2,3)$ and the parabola contains the point $(4,4)$.  Find the value of $y$ when $x=6$.

Answer:
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 $a=\frac14$, and our parabola is the graph of \[y=\frac14(x-2)^2 + 3.\] Setting $x=6$ in this gives us  \[y = \frac14(6-2)^2 + 3 = 4+3=\boxed{7}.\]