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Substitute $(2,3)$ and $(4,3)$ into the equation to give $$3=4+2b+c\text{and}3=16+4b+c.$$ Subtracting corresponding terms in these equations gives $0=12+2b$. So $$b=-6\text{and}c=3-4-2(-6)=\boxed{11}.$$

OR

The parabola is symmetric about the vertical line through its vertex, and the points $(2,3)$ and $(4,3)$ have the same $y$-coordinate. The vertex has $x$-coordinate $(2+4)/2=3$, so the equation has the form $$y=(x-3{)}^2+k$$ for some constant $k$. Since $y=3$ when $x=4$, we have $3=1^2+k$ and $k=2$. Consequently the constant term $c$ is $$(-3{)}^2+k=9+2=11.$$

OR

The parabola is symmetric about the vertical line through its vertex, so the $x$-coordinate of the vertex is 3. Also, the coefficient of $x^2$ is 1, so the parabola opens upward and the $y$-coordinate of the vertex is 2. We find $c$, the $y$-intercept of the graph by observing that the $y$-intercept occurs 3 units away horizontally from the vertex. On this interval the graph decreased by $3^2=9$ units hence the $y$-intercept is 9 units higher than the vertex, so $c=9+2=\boxed{11}.$