## Precalculus (6th Edition) Blitzer

Step 1. Given the function $y=\sqrt {27-3x^2}$, the domain requirement is that $27-3x^2\geq0$ Step 2. Rewrite the above inequality; we have $3x^2-27\leq0$ and $3(x+3)(x-3)\leq0$ Step 3. The boundary points are $x=-3,3$; use test points to examine the signs of the left side across boundary points. We have $...(+)...(-3)...(-)...(3)...(+)...$ Thus the solutions are that $-3\leq x\leq 3$, which explains why the graph is restricted to a domain of $[-3,3]$