Function Construction Ad540D

1. **State the problem:** We need to find a function from the given options that has: - A vertical asymptote at $x = 1$ - A horizontal asymptote at $y = -4$ - A zero at $x = -3$ 2. **Recall the properties of rational functions:** - A vertical asymptote occurs where the denominator is zero. - A horizontal asymptote is the value the function approaches as $x \to \pm \infty$. - A zero occurs where the numerator is zero. 3. **Analyze each option:** - Option 1: $y = -4 + \frac{16}{x + 1}$ - Vertical asymptote at $x = -1$ (denominator zero) - Horizontal asymptote at $y = -4$ (since $\frac{16}{x+1} \to 0$ as $x \to \infty$) - Zero at $x$ where $y=0$: $0 = -4 + \frac{16}{x+1} \Rightarrow \frac{16}{x+1} = 4 \Rightarrow x+1=4 \Rightarrow x=3$ - Zero at $x=3$, not $-3$. - Option 2: $y = -4 + \frac{-16}{x + 1}$ - Vertical asymptote at $x = -1$ - Horizontal asymptote at $y = -4$ - Zero at $0 = -4 + \frac{-16}{x+1} \Rightarrow \frac{-16}{x+1} = 4 \Rightarrow x+1 = -4 \Rightarrow x = -5$ - Zero at $x=-5$, not $-3$. - Option 3: $y = -4 + \frac{-16}{x - 1}$ - Vertical asymptote at $x = 1$ (correct) - Horizontal asymptote at $y = -4$ - Zero at $0 = -4 + \frac{-16}{x-1} \Rightarrow \frac{-16}{x-1} = 4 \Rightarrow x-1 = -4 \Rightarrow x = -3$ - Zero at $x = -3$ (correct) - Option 4: $y = -4 + \frac{16}{x - 1}$ - Vertical asymptote at $x = 1$ - Horizontal asymptote at $y = -4$ - Zero at $0 = -4 + \frac{16}{x-1} \Rightarrow \frac{16}{x-1} = 4 \Rightarrow x-1 = 4 \Rightarrow x = 5$ - Zero at $x=5$, not $-3$. 4. **Conclusion:** Only option 3 satisfies all conditions. **Final answer:** $$y = -4 + \frac{-16}{x - 1}$$