1. **Problem statement:** We are given the function $g(x) = 2x^2 + \frac{1}{x}$ and asked to analyze and plot it. 2. **Function form:** The function is a combination of a quadratic term $2x^2$ and a rational term $\frac{1}{x}$. 3. **Domain:** The function is defined for all $x \neq 0$ because division by zero is undefined. 4. **Key features:** - As $x \to 0^+$, $\frac{1}{x} \to +\infty$ and as $x \to 0^-$, $\frac{1}{x} \to -\infty$. - The quadratic term $2x^2$ grows large positively as $|x|$ increases. 5. **Intercepts:** - To find $y$-intercept, set $x=0$ but $g(x)$ is undefined at $x=0$, so no $y$-intercept. - To find $x$-intercepts, solve $2x^2 + \frac{1}{x} = 0$: $$ 2x^2 + \frac{1}{x} = 0 \\ 2x^3 + 1 = 0 \\ 2x^3 = -1 \\ x^3 = -\frac{1}{2} \\ x = -\sqrt[3]{\frac{1}{2}} = -\frac{\sqrt[3]{1}}{\sqrt[3]{2}} = -\frac{1}{\sqrt[3]{2}} $$ 6. **Summary:** The function has one real root at $x = -\frac{1}{\sqrt[3]{2}}$, no $y$-intercept, and a vertical asymptote at $x=0$. 7. **Plotting:** The function can be plotted to visualize these features. Final answer: The function $g(x) = 2x^2 + \frac{1}{x}$ has domain $x \neq 0$, one root at $x = -\frac{1}{\sqrt[3]{2}}$, and a vertical asymptote at $x=0$.