1. The problem is to graph the function $y = x^2 + \frac{1}{x}$. 2. This function combines a quadratic term $x^2$ and a rational term $\frac{1}{x}$. 3. Important points to consider: - The function is undefined at $x=0$ because of division by zero. - As $x \to 0^+$, $\frac{1}{x} \to +\infty$ and as $x \to 0^-$, $\frac{1}{x} \to -\infty$. - For large $|x|$, $x^2$ dominates, so $y$ grows large. 4. To find intercepts: - $y$-intercept: none, since $x=0$ is not in the domain. - $x$-intercepts: solve $x^2 + \frac{1}{x} = 0$ or $x^3 + 1 = 0$ which gives $x = -1$. 5. To find extrema, differentiate: $$y' = 2x - \frac{1}{x^2}$$ Set $y' = 0$: $$2x = \frac{1}{x^2} \Rightarrow 2x^3 = 1 \Rightarrow x = \sqrt[3]{\frac{1}{2}}$$ 6. Evaluate $y$ at $x = \sqrt[3]{\frac{1}{2}}$: $$y = \left(\sqrt[3]{\frac{1}{2}}\right)^2 + \frac{1}{\sqrt[3]{\frac{1}{2}}} = \sqrt[3]{\frac{1}{4}} + \sqrt[3]{2}$$ 7. The graph has a vertical asymptote at $x=0$ and grows large positively for large $|x|$. Final answer: The function $y = x^2 + \frac{1}{x}$ has a vertical asymptote at $x=0$, one $x$-intercept at $x=-1$, and one critical point at $x=\sqrt[3]{\frac{1}{2}}$ with value $y=\sqrt[3]{\frac{1}{4}} + \sqrt[3]{2}$.