1. **Problem Statement:** Given the function $$f(x) = \frac{2x^2 - 5x + 3}{x^2 + 5x}$$ (i) Find the horizontal and vertical asymptotes. (ii) Find the domain of the function. --- 2. **Vertical Asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. Denominator: $$x^2 + 5x = x(x + 5)$$ Set denominator equal to zero: $$x(x + 5) = 0$$ So, $$x = 0$$ or $$x = -5$$ Check numerator at these points: - At $$x=0$$: numerator $$2(0)^2 - 5(0) + 3 = 3 \neq 0$$ - At $$x=-5$$: numerator $$2(-5)^2 - 5(-5) + 3 = 2(25) + 25 + 3 = 50 + 25 + 3 = 78 \neq 0$$ Since numerator is not zero at these points, vertical asymptotes are at: $$x = 0$$ and $$x = -5$$ --- 3. **Horizontal Asymptote:** For rational functions, horizontal asymptotes depend on the degrees of numerator and denominator. Degree of numerator: 2 (because of $$2x^2$$) Degree of denominator: 2 (because of $$x^2$$) When degrees are equal, horizontal asymptote is: $$y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$$ Leading coefficient numerator: 2 Leading coefficient denominator: 1 So, $$y = \frac{2}{1} = 2$$ --- 4. **Domain:** The domain of a rational function is all real numbers except where the denominator is zero. From step 2, denominator zero at $$x=0$$ and $$x=-5$$. Therefore, domain is: $$\{x \in \mathbb{R} \mid x \neq 0, x \neq -5\}$$ --- **Final answers:** - Vertical asymptotes: $$x = 0$$ and $$x = -5$$ - Horizontal asymptote: $$y = 2$$ - Domain: $$(-\infty, -5) \cup (-5, 0) \cup (0, \infty)$$