1. **State the problem:** We need to find the features of the graph of the function $$f(x) = \frac{5x^2}{3x^2 + 7x}$$ including horizontal asymptotes, vertical asymptotes, x-intercepts, y-intercepts, and holes. 2. **Horizontal asymptote:** For rational functions, compare degrees of numerator and denominator. - Degree numerator = 2 - Degree denominator = 2 Since degrees are equal, horizontal asymptote is $$y = \frac{\text{leading coefficient numerator}}{\text{leading coefficient denominator}} = \frac{5}{3}$$. 3. **Vertical asymptote:** Set denominator equal to zero and solve: $$3x^2 + 7x = 0$$ Factor: $$x(3x + 7) = 0$$ So, $$x=0$$ or $$x=-\frac{7}{3}$$. Check if these cause holes or vertical asymptotes by seeing if numerator is zero at these points. - Numerator at $$x=0$$ is $$5(0)^2=0$$, so both numerator and denominator are zero at $$x=0$$, indicating a hole. - Numerator at $$x=-\frac{7}{3}$$ is $$5\left(-\frac{7}{3}\right)^2 > 0$$, so vertical asymptote at $$x=-\frac{7}{3}$$. 4. **Hole:** Since numerator and denominator both zero at $$x=0$$, factor and simplify: $$f(x) = \frac{5x^2}{x(3x+7)} = \frac{5x}{3x+7}, x \neq 0$$ Evaluate limit at $$x=0$$: $$f(0) = \frac{5(0)}{3(0)+7} = 0$$ So hole at $$(0,0)$$. 5. **x-intercept:** Set numerator equal to zero: $$5x^2=0 \Rightarrow x=0$$ But $$x=0$$ is a hole, so no x-intercept. 6. **y-intercept:** Evaluate $$f(0)$$: Undefined (hole), so no y-intercept. **Final answers:** - Horizontal asymptote: $$y=\frac{5}{3}$$ - Vertical asymptote: $$x=-\frac{7}{3}$$ - x-intercept: No x-intercept - y-intercept: No y-intercept - Hole: $$(0,0)$$