Consider the function [tex]f(x) = (x-5)(x+3)[/tex]. It is a parabola with a vertex below the x-axis and x-intercepts of 5 and -3. The vertex x-coordinate is in the middle of the two x-intercepts, so the vertex x-coordinate is (5+-3)/2 = 1. The y-coordinate of the vertex is obtained by substituiting the x-coordinate of the vettex inside, which gets you -16. Its vertex is at (1,-16) and it opens upward; therefore, (x-5)(x+3) gets you a range of [tex]y \ge -16[/tex].
But when you take square root, there are no negatives. Anything where the range is negative in (x-5)(x+3) is cut out, and we are left with [tex]y \ge 0[/tex]. Therefore, the range of [tex]\sqrt{(x-5)(x+3)}[/tex] is [tex]y \ge 0[/tex]
