I have moved this thread to calculus, since no one is interested in the per-calculus solution.
CB
Without using Calculus- since the position function $x= t^2- 8t+ 15$, is quadratic complete the square:
$x= t^2- 8t+ 16- 16+ 15= (t- 4)^2- 1$. That is a parabola, opening upwards, with vertex at (4, -1). If t< 4, as t increases, x goes downward. The velocity is negative. If t< 4, as t increases, x goes upward. The velocity is positive. At the vertex, t= 4, the velocity is 0.