1. **Problem 7: Find the horizontal asymptote of** $f(x) = \frac{3x^2 + 2x}{x^2 - x - 6}$. 2. **Recall the rule for horizontal asymptotes:** - If degrees of numerator and denominator are equal, horizontal asymptote is $y = \frac{\text{leading coefficient numerator}}{\text{leading coefficient denominator}}$. - Here, degree numerator = degree denominator = 2. 3. **Identify leading coefficients:** - Numerator leading coefficient = 3 - Denominator leading coefficient = 1 4. **Calculate horizontal asymptote:** $$y = \frac{3}{1} = 3$$ 5. **Answer for problem 7:** $y = 3$ (option e). --- 6. **Problem 8: Find the removable point discontinuity (RPD) for** $f(x) = \frac{x+1}{x^2 - 1}$. 7. **Factor denominator:** $$x^2 - 1 = (x-1)(x+1)$$ 8. **Simplify function:** $$f(x) = \frac{x+1}{(x-1)(x+1)}$$ 9. **Cancel common factor $(x+1)$:** $$f(x) = \frac{\cancel{x+1}}{(x-1)\cancel{(x+1)}} = \frac{1}{x-1}, \quad x \neq -1$$ 10. **RPD occurs where factor canceled: $x = -1$.** 11. **Find $y$-value at $x = -1$ by plugging into simplified function:** $$y = \frac{1}{-1 - 1} = \frac{1}{-2} = -\frac{1}{2}$$ 12. **Answer for problem 8:** RPD at $(-1, -\frac{1}{2})$ (option b). --- 13. **Problem 9: Find the x-intercept of** $f(x) = \frac{x+2}{x^2 - 4}$. 14. **Recall x-intercept occurs when numerator = 0:** $$x + 2 = 0 \implies x = -2$$ 15. **Check denominator at $x = -2$ to ensure function is defined:** $$(-2)^2 - 4 = 4 - 4 = 0$$ 16. **Since denominator is zero at $x = -2$, function is undefined there, so no x-intercept at $x = -2$.** 17. **No other zeros in numerator, so no x-intercept.** 18. **Answer for problem 9:** none (option d).