Minimum of Quadratic

First, we must decide whether the minimum is obtained at the vertex of the parabola, or at the edge of the domain. Since the coefficient of $x^2$ is positive, the vertex would be a minimum, so check that first.

Completing the square gives $f(x) = (x-8)^2 + \cdots$ , so that the vertex would be located at $x = 8$ . Clearly this is outside the domain. Therefore the minimum is reached at an edge of the domain, i.e. $x = -1$ or $x = 1$ .

Since the function decreases toward the vertex at $x = 8$ , the minimum will be at $x = 1$ . So $-31 = 1^2 - 16\cdot 1 - 2a, \\ a = \frac{1 - 16 - (-31)}{2} = \boxed{8}.$