In order to make the function defined, it must have $$x^2 -3x+2\ge 0\implies x \ge 2\quad\text{or}\quad x \le 1$$ When $x \ge 2$, both $x$ and $\sqrt{x^2-3x+2}$ are monotonically increasing. In this case, $y\in[2,+\infty)$. When $x\le 1$, $x$ is monotonically increasing, but $\sqrt{x^2-3x+2}$ is monotonically decreasing. $[1, \frac{3}{2}) \cup [2, +\infty) $