1. **State the problem:** Find the x-intercept, vertical asymptote (VA), horizontal asymptote (HA), hole, y-intercept, domain, and range of the function $$f(x) = \frac{2x^2 + 8x + 6}{x^2 + 6x + 5}$$ 2. **Factor numerator and denominator:** $$2x^2 + 8x + 6 = 2(x^2 + 4x + 3) = 2(x+3)(x+1)$$ $$x^2 + 6x + 5 = (x+5)(x+1)$$ 3. **Simplify the function:** $$f(x) = \frac{2(x+3)(x+1)}{(x+5)(x+1)}$$ Cancel the common factor $(x+1)$: $$f(x) = \frac{2\cancel{(x+3)}\cancel{(x+1)}}{(x+5)\cancel{(x+1)}} = \frac{2(x+3)}{x+5}$$ 4. **Identify holes:** Since $(x+1)$ was canceled, there is a hole at $x = -1$. 5. **Find vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and not canceled: $$x+5=0 \Rightarrow x = -5$$ 6. **Find horizontal asymptotes:** Degree numerator = degree denominator = 1 Horizontal asymptote is ratio of leading coefficients: $$y = \frac{2}{1} = 2$$ 7. **Find x-intercepts:** Set numerator equal to zero: $$2(x+3) = 0 \Rightarrow x = -3$$ 8. **Find y-intercept:** Evaluate $f(0)$: $$f(0) = \frac{2(0+3)}{0+5} = \frac{6}{5} = 1.2$$ 9. **Domain:** All real numbers except where denominator zero: $$x \neq -5, -1$$ 10. **Range:** Since there is a hole at $x=-1$, $f(-1)$ is undefined. Horizontal asymptote $y=2$ is approached but not crossed. Range is all real numbers except possibly $y$ value at hole. Calculate hole $y$ value: $$\lim_{x \to -1} f(x) = \frac{2(-1+3)}{-1+5} = \frac{2(2)}{4} = 1$$ So $y=1$ is missing from range. **Final answers:** - x-int: $-3$ - VA: $x = -5$ - HA: $y = 2$ - Hole: $x = -1$ - y-int: $\frac{6}{5}$ - Domain: $\{x \in \mathbb{R} : x \neq -5, -1\}$ - Range: $\{y \in \mathbb{R} : y \neq 1\}$