Find the minimum value of the parabola y=x^2-2 Simplify your answer and write it as a proper fraction improper fraction, or square

The minimum value of a parabola $y=ax^{2}+bx+c$ is given by the formula $-\frac{b^{2}-4ac}{4a}$. In this case, $a=1$, $b=0$, and $c=-2$. Substituting these values into the formula, we get $-\frac{0^{2}-4(-2)}{4(1)}=2$. Therefore, the minimum value of the parabola $y=x^{2}-2$ is 2.