1. **State the problem:** We need to sketch the graph of the function $$y = x^2 + \frac{1}{x}$$. 2. **Formula and rules:** The function is a combination of a quadratic term $$x^2$$ and a rational term $$\frac{1}{x}$$. Important points to consider are domain restrictions (here, $$x \neq 0$$), intercepts, and behavior at extremes. 3. **Find domain:** The function is defined for all $$x$$ except $$x=0$$ because division by zero is undefined. 4. **Find intercepts:** - **y-intercept:** Not defined since $$x=0$$ is not in the domain. - **x-intercepts:** Solve $$x^2 + \frac{1}{x} = 0$$. Multiply both sides by $$x$$ (assuming $$x \neq 0$$): $$x^3 + 1 = 0$$ $$x^3 = -1$$ $$x = -1$$ So, the function crosses the x-axis at $$x = -1$$. 5. **Analyze behavior near $$x=0$$:** - As $$x \to 0^+$$, $$x^2 \to 0$$ and $$\frac{1}{x} \to +\infty$$, so $$y \to +\infty$$. - As $$x \to 0^-$$, $$x^2 \to 0$$ and $$\frac{1}{x} \to -\infty$$, so $$y \to -\infty$$. 6. **End behavior:** - As $$x \to +\infty$$, $$x^2$$ dominates, so $$y \to +\infty$$. - As $$x \to -\infty$$, $$x^2$$ dominates, so $$y \to +\infty$$. 7. **Find critical points:** Calculate derivative: $$y' = 2x - \frac{1}{x^2}$$ Set $$y' = 0$$: $$2x = \frac{1}{x^2}$$ $$2x^3 = 1$$ $$x^3 = \frac{1}{2}$$ $$x = \sqrt[3]{\frac{1}{2}}$$ 8. **Evaluate $$y$$ at critical point:** $$y\left(\sqrt[3]{\frac{1}{2}}\right) = \left(\sqrt[3]{\frac{1}{2}}\right)^2 + \frac{1}{\sqrt[3]{\frac{1}{2}}} = \sqrt[3]{\frac{1}{4}} + \sqrt[3]{2}$$ 9. **Summary:** - Domain: $$x \neq 0$$ - x-intercept at $$x = -1$$ - Vertical asymptote at $$x=0$$ with $$y \to +\infty$$ from right and $$y \to -\infty$$ from left - End behavior $$y \to +\infty$$ as $$x \to \pm \infty$$ - One critical point at $$x = \sqrt[3]{\frac{1}{2}}$$ This information helps sketch the graph accurately.