Vertical Asymptote 0Edc77

1. **State the problem:** We need to find which rational function has a vertical asymptote at $x = -2$. 2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is not zero at that point. 3. **Check each function's denominator for $x = -2$: ** - For $f(x) = \frac{x + 2}{3x^2 + 1}$, denominator is $3x^2 + 1$. Substitute $x = -2$: $$3(-2)^2 + 1 = 3(4) + 1 = 12 + 1 = 13 \neq 0$$ No vertical asymptote at $x = -2$. - For $f(x) = \frac{3x}{x - 2}$, denominator is $x - 2$. Substitute $x = -2$: $$-2 - 2 = -4 \neq 0$$ No vertical asymptote at $x = -2$. - For $f(x) = \frac{3x}{x + 2}$, denominator is $x + 2$. Substitute $x = -2$: $$-2 + 2 = 0$$ Denominator is zero, check numerator: $$3(-2) = -6 \neq 0$$ So, vertical asymptote at $x = -2$. - For $f(x) = \frac{1 - 2x}{4 + x}$, denominator is $4 + x$. Substitute $x = -2$: $$4 + (-2) = 2 \neq 0$$ No vertical asymptote at $x = -2$. 4. **Conclusion:** The function with vertical asymptote at $x = -2$ is $$f(x) = \frac{3x}{x + 2}$$