1. **State the problem:** We need to analyze and simplify the function $$y = \frac{\sqrt{1 + x^2 - 4}}{x}$$ and understand its domain and behavior. 2. **Simplify the expression inside the square root:** $$1 + x^2 - 4 = x^2 - 3$$ So the function becomes: $$y = \frac{\sqrt{x^2 - 3}}{x}$$ 3. **Determine the domain:** - The expression inside the square root must be non-negative: $$x^2 - 3 \geq 0 \implies x^2 \geq 3 \implies |x| \geq \sqrt{3}$$ - Also, the denominator $x$ cannot be zero. Therefore, the domain is: $$(-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$$ 4. **Rewrite the function for $x$ in the domain:** Since $x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$, and $\sqrt{x^2 - 3} = \sqrt{x^2 - 3}$ is always non-negative, For $x > 0$: $$y = \frac{\sqrt{x^2 - 3}}{x} = \frac{\sqrt{x^2 - 3}}{x}$$ For $x < 0$: $$y = \frac{\sqrt{x^2 - 3}}{x} = -\frac{\sqrt{x^2 - 3}}{|x|} = -\frac{\sqrt{x^2 - 3}}{-x} = -\frac{\sqrt{x^2 - 3}}{x}$$ 5. **Analyze behavior:** - As $x \to \pm \infty$, $y \approx \frac{\sqrt{x^2}}{x} = \frac{|x|}{x}$ which approaches $1$ for $x \to +\infty$ and $-1$ for $x \to -\infty$. - The function is undefined for $|x| < \sqrt{3}$ and at $x=0$. 6. **Final simplified form:** $$y = \frac{\sqrt{x^2 - 3}}{x}$$ with domain $$x \in (-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$$. This function has vertical asymptotes at $x = \pm \sqrt{3}$ where the square root expression becomes zero and is undefined inside that interval. **Answer:** $$y = \frac{\sqrt{x^2 - 3}}{x}, \quad x \in (-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$$