smc

Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of $y=x^2-8x+c$ to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So, $$\begin{array}{rl}0& =b^2-4ac\\ 0& =(-8{)}^2-4c\\ c& =64\\ c& =16\end{array}$$ Therefore $c=16$, and our answer is $\boxed{16}$.