Question

Determine the range of x+3)}" alt="\sqrt{(x-5)(x+3)}" align="absmiddle" class="latex-formula"> without graphing it.

Answer 1

Consider the function $f(x) = (x-5)(x+3)$. 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 $y \ge -16$.

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 $y \ge 0$. Therefore, the range of $\sqrt{(x-5)(x+3)}$ is $y \ge 0$