1. **State the problem:** We need to analyze the rational function $$f(x) = \frac{2x^3 + 3x^2 - 3x - 2}{2x^3 - 7x^2 + 2x + 3}$$ including simplification, domain, asymptotes, and behavior. 2. **Formula and rules:** For rational functions $$\frac{P(x)}{Q(x)}$$ where $P$ and $Q$ are polynomials, vertical asymptotes occur where $Q(x)=0$ (denominator zero) and horizontal asymptotes depend on degrees and leading coefficients. 3. **Factor numerator and denominator:** - Numerator: $2x^3 + 3x^2 - 3x - 2$ Try factoring by grouping: $$2x^3 + 3x^2 - 3x - 2 = (2x^3 + 3x^2) + (-3x - 2) = x^2(2x + 3) -1(3x + 2)$$ Not matching, try rearranging: $$2x^3 - 3x + 3x^2 - 2 = (2x^3 - 3x) + (3x^2 - 2)$$ Try rational root theorem for numerator: Test $x=1$: $$2(1)^3 + 3(1)^2 - 3(1) - 2 = 2 + 3 - 3 - 2 = 0$$ So $x=1$ is a root. Divide numerator by $(x-1)$: Using synthetic division: Coefficients: 2, 3, -3, -2 Bring down 2 Multiply 2*1=2, add to 3=5 Multiply 5*1=5, add to -3=2 Multiply 2*1=2, add to -2=0 remainder Quotient: $2x^2 + 5x + 2$ Factor $2x^2 + 5x + 2$: Find factors of $2*2=4$ that sum to 5: 4 and 1 $$2x^2 + 4x + x + 2 = 2x(x+2) + 1(x+2) = (2x+1)(x+2)$$ So numerator factors as: $$ (x-1)(2x+1)(x+2) $$ - Denominator: $2x^3 - 7x^2 + 2x + 3$ Try rational root theorem: Test $x=1$: $$2 - 7 + 2 + 3 = 0$$ So $x=1$ is a root. Divide denominator by $(x-1)$: Coefficients: 2, -7, 2, 3 Bring down 2 Multiply 2*1=2, add to -7 = -5 Multiply -5*1 = -5, add to 2 = -3 Multiply -3*1 = -3, add to 3 = 0 remainder Quotient: $2x^2 - 5x - 3$ Factor $2x^2 - 5x - 3$: Find factors of $2*(-3) = -6$ that sum to -5: -6 and 1 $$2x^2 - 6x + x - 3 = 2x(x-3) + 1(x-3) = (2x+1)(x-3)$$ So denominator factors as: $$ (x-1)(2x+1)(x-3) $$ 4. **Simplify the function:** $$f(x) = \frac{(x-1)(2x+1)(x+2)}{(x-1)(2x+1)(x-3)}$$ Cancel common factors: $$f(x) = \frac{\cancel{(x-1)}\cancel{(2x+1)}(x+2)}{\cancel{(x-1)}\cancel{(2x+1)}(x-3)}$$ 5. **Domain and discontinuities:** - Original denominator zero at $x=1$, $x=-\frac{1}{2}$ (from $2x+1=0$), and $x=3$. - After simplification, $x=1$ and $x=-\frac{1}{2}$ are removable discontinuities (holes). - Vertical asymptote at $x=3$. 6. **Horizontal asymptote:** Degrees numerator and denominator after simplification are both 1 (linear). Leading coefficients numerator: 1 (from $x+2$) Leading coefficient denominator: 1 (from $x-3$) So horizontal asymptote is: $$y = \frac{1}{1} = 1$$ 7. **Final simplified function:** $$f(x) = \frac{x+2}{x-3}$$ with holes at $x=1$ and $x=-\frac{1}{2}$. **Answer:** $$f(x) = \frac{x+2}{x-3}, \quad x \neq 1, -\frac{1}{2}$$ Vertical asymptote at $x=3$, horizontal asymptote at $y=1$, holes at $x=1$ and $x=-\frac{1}{2}$.