minimum

The distance from a point on the graph to the origin can be expressed as, in terms of x, as $\sqrt{x^2+(x^2-3)^2}=\sqrt{x^4-6x^2+9+x^2}=\sqrt{x^4-5x^2+9}$

the quartic inside the radical is just a quadratic in terms of x^2, so we can rewrite it in vertex form:

$\sqrt{x^4-5x^2+9}=\sqrt{x^4-5x^2+(\frac{5}{2})^2+9-(\frac{5}{2})^2}=\sqrt{(x^2-\frac{5}{2})^2+\frac{11}{4}}$

since $(x^2-2.5)^2$will always be nonnegative for real numbers, the minimum value for $\sqrt{(x^2-2.5)^2+\frac{11}{4}}$ is just equal to $\boxed{\frac{\sqrt{11}}{2}}$ (specifically when x is equal to $\pm\sqrt{2.5}$)

P.S. my method seems like total overkill for this problem so if anyone has an easier solution, please post it