1. **State the problem:** We want to analyze and understand the function $$y = \frac{1 + x^2}{x}$$. 2. **Rewrite the function:** We can split the fraction to simplify understanding: $$y = \frac{1}{x} + \frac{x^2}{x} = \frac{1}{x} + x$$ 3. **Identify domain restrictions:** The function is undefined at $$x = 0$$ because division by zero is not allowed. This creates a vertical asymptote at $$x = 0$$. 4. **Behavior near the vertical asymptote:** - As $$x \to 0^+$$, $$\frac{1}{x} \to +\infty$$ and $$x \to 0^+$$, so $$y \to +\infty$$. - As $$x \to 0^-$$, $$\frac{1}{x} \to -\infty$$ and $$x \to 0^-$$, so $$y \to -\infty$$. 5. **End behavior (asymptotes at infinity):** - As $$x \to +\infty$$, $$\frac{1}{x} \to 0$$ and $$x \to +\infty$$, so $$y \approx x$$ which grows without bound. - As $$x \to -\infty$$, $$\frac{1}{x} \to 0$$ and $$x \to -\infty$$, so $$y \approx x$$ which decreases without bound. 6. **Intercepts:** - **y-intercept:** None, since $$x=0$$ is not in the domain. - **x-intercept:** Set $$y=0$$: $$0 = \frac{1}{x} + x \implies x = -\frac{1}{x} \implies x^2 = -1$$ which has no real solutions. So, no x-intercepts. 7. **Summary:** The function has a vertical asymptote at $$x=0$$, no x- or y-intercepts, and behaves like $$y \approx x$$ for large $$|x|$$. **Final answer:** The function $$y = \frac{1 + x^2}{x}$$ has a vertical asymptote at $$x=0$$ and no real zeros. It behaves like $$y = x$$ for large $$|x|$$.