## Precalculus (6th Edition)

If there exists an x-intercept, the y-coordinate of that point must be zero. If $y=0,$ $x^{2}-2=0\qquad/$+$2$ $x^{2}=+2$ $x=+\sqrt{2}$ or $x=-\sqrt{2}.$ So, there are two x-intercepts, $(-\sqrt{2},0)$ and $(\sqrt{2},0)$ (and no other)