1. The problem asks us to determine for which function the value of $y$ approaches $-\infty$ as $x$ approaches $\infty$. 2. Let's analyze each function one by one: 3. For $y = -\frac{x^2}{x}$: Simplify the expression: $$y = -\frac{x^2}{x} = -\cancel{\frac{x^2}{x}} = -x$$ As $x \to \infty$, $-x \to -\infty$. 4. For $y = \sqrt{x - 6}$: As $x \to \infty$, $x - 6 \to \infty$, so $\sqrt{x - 6} \to \infty$, not $-\infty$. 5. For $y = \frac{x}{6 - x}$: Rewrite denominator: $$6 - x = -(x - 6)$$ As $x \to \infty$, denominator $6 - x \to -\infty$, numerator $x \to \infty$. So, $$y = \frac{x}{6 - x} = \frac{x}{-(x - 6)} = -\frac{x}{x - 6}$$ Divide numerator and denominator by $x$: $$y = -\frac{\cancel{x}}{\cancel{x} - \frac{6}{x}} = -\frac{1}{1 - 0} = -1$$ So as $x \to \infty$, $y \to -1$, not $-\infty$. 6. For $y = -\sqrt{x - 6}$: As $x \to \infty$, $\sqrt{x - 6} \to \infty$, so $-\sqrt{x - 6} \to -\infty$. 7. Conclusion: Functions $y = -\frac{x^2}{x}$ (which simplifies to $y = -x$) and $y = -\sqrt{x - 6}$ both approach $-\infty$ as $x \to \infty$. Since the question is multiple choice and these are separate options, the correct answers are: - $y = -\frac{x^2}{x}$ - $y = -\sqrt{x - 6}$