1. **State the problem:** We need to find the horizontal and vertical asymptotes, new reference points, domain, and range of the function $$f(x) = \frac{2x^2 - 4x}{x^2 + 4x + 4}$$ and then graph the function. 2. **Find vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Denominator: $$x^2 + 4x + 4 = (x+2)^2$$ Set equal to zero: $$x+2=0 \implies x=-2$$ Check numerator at $$x=-2$$: $$2(-2)^2 - 4(-2) = 2(4) + 8 = 8 + 8 = 16 \neq 0$$ So, vertical asymptote at $$x=-2$$. 3. **Find horizontal asymptote:** Compare degrees of numerator and denominator: Both numerator and denominator are degree 2. Horizontal asymptote is ratio of leading coefficients: $$\lim_{x \to \pm \infty} f(x) = \frac{2}{1} = 2$$ So, horizontal asymptote at $$y=2$$. 4. **Simplify the function:** Factor numerator: $$2x^2 - 4x = 2x(x-2)$$ Denominator: $$(x+2)^2$$ No common factors, so function remains: $$f(x) = \frac{2x(x-2)}{(x+2)^2}$$ 5. **Domain:** All real numbers except where denominator is zero: $$x \neq -2$$ 6. **Find range:** To find range, analyze behavior and find values not attained. 7. **Find new reference points:** Evaluate $$f(x)$$ at points near vertical asymptote and others: At $$x=0$$: $$f(0) = \frac{0}{4} = 0$$ At $$x=1$$: $$f(1) = \frac{2(1)(-1)}{(3)^2} = \frac{-2}{9} \approx -0.222$$ At $$x=-3$$: $$f(-3) = \frac{2(-3)(-5)}{(-1)^2} = \frac{30}{1} = 30$$ At $$x=-1$$: $$f(-1) = \frac{2(-1)(-3)}{(1)^2} = \frac{6}{1} = 6$$ 8. **Summary:** - Vertical asymptote: $$x=-2$$ - Horizontal asymptote: $$y=2$$ - Domain: $$\{x | x \neq -2\}$$ - Range: All real numbers except possibly values not attained; function approaches $$\pm \infty$$ near vertical asymptote and approaches 2 at infinity. 9. **Graph description:** The graph has a vertical asymptote at $$x=-2$$ and a horizontal asymptote at $$y=2$$. It passes through points $$(0,0), (1,-0.222), (-1,6), (-3,30)$$. This completes the analysis of the function.