1. **State the problem:** Find all values of $x$ such that $g(x) = -1$ and determine the end behavior of $g(x)$ as $x \to -\infty$. 2. **Recall the function $g(x)$:** Given by $$g(x) = \frac{x^2 + 7x - 2}{3x^2 + 1}$$ 3. **Find $x$ such that $g(x) = -1$: Solve** $$\frac{x^2 + 7x - 2}{3x^2 + 1} = -1$$ Multiply both sides by $3x^2 + 1$ (which is always positive since $3x^2 \geq 0$ and $+1$): $$x^2 + 7x - 2 = -1(3x^2 + 1)$$ $$x^2 + 7x - 2 = -3x^2 - 1$$ Bring all terms to one side: $$x^2 + 7x - 2 + 3x^2 + 1 = 0$$ $$4x^2 + 7x - 1 = 0$$ 4. **Solve quadratic equation:** Use quadratic formula: $$x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{-7 \pm \sqrt{49 + 16}}{8} = \frac{-7 \pm \sqrt{65}}{8}$$ 5. **Calculate decimal approximations:** $$\sqrt{65} \approx 8.062$$ So, $$x_1 = \frac{-7 - 8.062}{8} = \frac{-15.062}{8} \approx -1.883$$ $$x_2 = \frac{-7 + 8.062}{8} = \frac{1.062}{8} \approx 0.133$$ 6. **Determine end behavior as $x \to -\infty$:** For rational functions, the end behavior is determined by the leading terms: $$g(x) = \frac{x^2 + 7x - 2}{3x^2 + 1} \approx \frac{x^2}{3x^2} = \frac{1}{3}$$ Therefore, $$\lim_{x \to -\infty} g(x) = \frac{1}{3}$$ **Final answers:** - Values of $x$ where $g(x) = -1$ are approximately $-1.883$ and $0.133$. - End behavior: $\lim_{x \to -\infty} g(x) = \frac{1}{3}$.