1. **Problem 1: Find which rational function has a horizontal asymptote at $y = -2$.** 2. The horizontal asymptote of a rational function $h(x) = \frac{P(x)}{Q(x)}$ depends on the degrees of the numerator and denominator polynomials: - If degree of $P(x)$ = degree of $Q(x)$, horizontal asymptote is $y = \frac{\text{leading coefficient of } P(x)}{\text{leading coefficient of } Q(x)}$. - If degree of $P(x) < $ degree of $Q(x)$, horizontal asymptote is $y=0$. - If degree of $P(x) > $ degree of $Q(x)$, no horizontal asymptote (may have oblique asymptote). 3. Check each option: A: $h(x) = \frac{2x^2 - 6}{-4x^2 + 2x - 1}$ - Degrees equal (2 and 2). - Leading coefficients: numerator $2$, denominator $-4$. - Horizontal asymptote: $y = \frac{2}{-4} = -\frac{1}{2} \neq -2$. B: $h(x) = \frac{8x^2 + 3x - 2}{4x^2 + x + 1}$ - Degrees equal (2 and 2). - Leading coefficients: numerator $8$, denominator $4$. - Horizontal asymptote: $y = \frac{8}{4} = 2 \neq -2$. C: $h(x) = \frac{6x^3 + 3x - 2}{4 - 3x^3}$ - Degrees equal (3 and 3). - Leading coefficients: numerator $6$, denominator $-3$ (since $-3x^3$ dominates). - Horizontal asymptote: $y = \frac{6}{-3} = -2$ matches the given asymptote. D: $h(x) = \frac{-3x^3 - 2x + 5}{x^2 + 3x - 2}$ - Degrees: numerator 3, denominator 2. - Since numerator degree $>$ denominator degree, no horizontal asymptote. 4. **Answer for Problem 1:** Option C. 5. **Problem 2: Identify which rational function matches the graph with vertical asymptotes at $x = -2$ and $x = 1$.** 6. Vertical asymptotes occur where the denominator is zero and numerator is nonzero. 7. Check denominators for each option: A: Denominator $(x + 2)(x - 1)$ zeros at $x = -2, 1$ correct. B: Denominator $(x + 2)(x - 1)$ zeros at $x = -2, 1$ correct. C: Denominator $(x + 2)(x - 1)$ zeros at $x = -2, 1$ correct. D: Denominator $(x + 2)(x - 1)$ zeros at $x = -2, 1$ correct. 8. Check numerator zeros to match graph zeros: - The graph shows zeros at $x = 3$ (where function crosses x-axis). A: Numerator $(x - 2)(x - 3)$ zeros at $x=2,3$ (extra zero at 2 not shown in graph). B: Numerator $(x + 2)(x - 3)$ zeros at $x=-2,3$ zero at $-2$ cancels denominator factor, so hole at $x=-2$. C: Numerator $(x - 3)$ zero at $x=3$ matches graph zero. D: Numerator $(x - 1)(x - 3)$ zeros at $x=1,3$ zero at $1$ cancels denominator factor, so hole at $x=1$. 9. The graph shows vertical asymptotes at $x=-2$ and $x=1$, so no holes at these points. - Option B and D have numerator factors that cancel denominator factors, causing holes, so they are unlikely. - Option A has zero at $x=2$ which is not shown in the graph. - Option C has zero at $x=3$ and vertical asymptotes at $x=-2,1$ with no cancellation. 10. **Answer for Problem 2:** Option C. **Final answers:** - Problem 1: Option C - Problem 2: Option C