1. **Problem:** Graph and identify the pieces of the function $$f(x) = \frac{x^2 + 2x - 8}{x + 1}$$ 2. **Find x-intercepts:** Set numerator equal to zero: $$x^2 + 2x - 8 = 0$$ Factor: $$ (x + 4)(x - 2) = 0 $$ So, $$x = -4$$ or $$x = 2$$ are zeros of numerator. 3. **Find vertical asymptotes (VA) and holes:** Denominator is $$x + 1$$. Check if $$x + 1$$ is a factor of numerator: Factor numerator: $$x^2 + 2x - 8 = (x + 4)(x - 2)$$ No $$x + 1$$ factor, so no hole. Vertical asymptote at $$x = -1$$. 4. **Find horizontal asymptote (HA):** Degree numerator = 2, degree denominator = 1. Since numerator degree > denominator degree, no horizontal asymptote. Instead, there is an oblique asymptote. 5. **Find oblique asymptote:** Divide numerator by denominator: $$\frac{x^2 + 2x - 8}{x + 1}$$ Use polynomial division: $$x^2 + 2x - 8 \div (x + 1) = x + 1$$ remainder $$-9$$ So, $$f(x) = x + 1 - \frac{9}{x + 1}$$ Oblique asymptote is $$y = x + 1$$. 6. **Domain:** All real numbers except where denominator is zero: $$x \neq -1$$. 7. **Range:** Since there is an oblique asymptote and a vertical asymptote, range is all real numbers except the value of $$y$$ at the hole or asymptote. No hole, so range is all real numbers. **Summary:** - x-intercepts: $$-4, 2$$ - Vertical asymptote: $$x = -1$$ - Horizontal asymptote: none - Oblique asymptote: $$y = x + 1$$ - Hole: none - Domain: $$(-\infty, -1) \cup (-1, \infty)$$ - Range: $$(-\infty, \infty)$$ --- 2. **Problem:** Solve $$\frac{x^2 - 9x + 9}{x + \frac{1}{2}} + 2 = 0$$ Rewrite: $$\frac{x^2 - 9x + 9}{x + \frac{1}{2}} = -2$$ Multiply both sides by $$x + \frac{1}{2}$$ (note $$x \neq -\frac{1}{2}$$): $$x^2 - 9x + 9 = -2\left(x + \frac{1}{2}\right)$$ Expand right side: $$x^2 - 9x + 9 = -2x - 1$$ Bring all terms to left: $$x^2 - 9x + 9 + 2x + 1 = 0$$ Simplify: $$x^2 - 7x + 10 = 0$$ Factor: $$(x - 5)(x - 2) = 0$$ Solutions: $$x = 5$$ or $$x = 2$$ Check domain restriction $$x \neq -\frac{1}{2}$$, both valid. --- 3. **Problem:** Factor $$\frac{x^2 - 2x - 15}{3x^2 - 14x - 5}$$ Factor numerator: $$x^2 - 2x - 15 = (x - 5)(x + 3)$$ Factor denominator: Find factors of $$3 \times (-5) = -15$$ that sum to $$-14$$: $$-15$$ and $$1$$ Rewrite denominator: $$3x^2 - 15x + x - 5$$ Group: $$(3x^2 - 15x) + (x - 5) = 3x(x - 5) + 1(x - 5)$$ Factor: $$(3x + 1)(x - 5)$$ Simplify fraction: $$\frac{(x - 5)(x + 3)}{(3x + 1)(x - 5)}$$ Cancel $$x - 5$$: $$\frac{\cancel{(x - 5)}(x + 3)}{(3x + 1)\cancel{(x - 5)}}$$ Final: $$\frac{x + 3}{3x + 1}$$ --- 4. **Problem:** Factor $$\frac{12x^3}{x^2 + 14x + 45} \div \frac{3x^3 - 6x^2}{x^2 + 7x - 18}$$ Rewrite division as multiplication by reciprocal: $$\frac{12x^3}{x^2 + 14x + 45} \times \frac{x^2 + 7x - 18}{3x^3 - 6x^2}$$ Factor denominators and numerators: $$x^2 + 14x + 45 = (x + 5)(x + 9)$$ $$x^2 + 7x - 18 = (x + 9)(x - 2)$$ $$3x^3 - 6x^2 = 3x^2(x - 2)$$ Rewrite expression: $$\frac{12x^3}{(x + 5)(x + 9)} \times \frac{(x + 9)(x - 2)}{3x^2(x - 2)}$$ Cancel common factors: - Cancel $$x + 9$$ - Cancel $$x - 2$$ - Cancel $$x^2$$ from numerator and denominator: $$12x^3 = 12x^2 \times x$$ $$3x^2$$ in denominator Cancel $$x^2$$: $$\frac{12x}{(x + 5)} \times \frac{1}{3} = \frac{12x}{3(x + 5)}$$ Simplify coefficient: $$\frac{12x}{3(x + 5)} = \frac{4x}{x + 5}$$ --- 5. **Problem:** Solve $$\frac{8}{x + 2} - \frac{5}{3} = \frac{x - 1}{3}$$ Multiply both sides by $$3(x + 2)$$ to clear denominators: $$3(x + 2) \times \left(\frac{8}{x + 2} - \frac{5}{3}\right) = 3(x + 2) \times \frac{x - 1}{3}$$ Distribute: $$3 \times 8 - (x + 2) \times 5 = (x + 2)(x - 1)$$ Simplify left: $$24 - 5x - 10 = (x + 2)(x - 1)$$ $$14 - 5x = x^2 + 2x - x - 2 = x^2 + x - 2$$ Bring all terms to one side: $$0 = x^2 + x - 2 - 14 + 5x$$ $$0 = x^2 + 6x - 16$$ Solve quadratic: $$x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times (-16)}}{2} = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm \sqrt{100}}{2}$$ $$x = \frac{-6 \pm 10}{2}$$ Solutions: $$x = \frac{-6 + 10}{2} = 2$$ $$x = \frac{-6 - 10}{2} = -8$$ Check domain restrictions: $$x \neq -2$$, both valid. --- **Final answers:** 1. x-int: $$-4, 2$$ VA: $$x = -1$$ HA: none Hole: none Domain: $$(-\infty, -1) \cup (-1, \infty)$$ Range: $$(-\infty, \infty)$$ 2. $$x = 5, 2$$ 3. $$\frac{x + 3}{3x + 1}$$ 4. $$\frac{4x}{x + 5}$$ 5. $$x = 2, -8$$