1. **State the problem:** We are given the rational function $$f(x) = \frac{2x^2 - 5x + 3}{x^2 + 5x}$$ and need to find its horizontal and vertical asymptotes, and its domain. 2. **Find vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero. - Denominator: $$x^2 + 5x = x(x + 5)$$ - Set denominator equal to zero: $$x(x + 5) = 0$$ - Solutions: $$x = 0$$ 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. **Find horizontal asymptotes:** For rational functions, compare degrees of numerator and denominator. - Degree of numerator: 2 - Degree of denominator: 2 If 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 horizontal asymptote is $$y = \frac{2}{1} = 2$$. 4. **Find the domain:** The domain is all real numbers except where denominator is zero. - Denominator zero at $$x=0$$ and $$x=-5$$ So domain is $$\{x \in \mathbb{R} : x \neq 0, x \neq -5\}$$. **Final answers:** - Vertical asymptotes: $$x=0$$ and $$x=-5$$ - Horizontal asymptote: $$y=2$$ - Domain: $$x \neq 0, x \neq -5$$