1. **State the problem:** We are given the function $$h(x) = \frac{2x^2 - 7x + 3}{x^2 - x - 6}$$ and need to analyze its properties including intercepts, asymptotes, and intervals of increase and decrease. 2. **Factor numerator and denominator:** Numerator: $$2x^2 - 7x + 3 = (2x - 1)(x - 3)$$ Denominator: $$x^2 - x - 6 = (x - 3)(x + 2)$$ 3. **Simplify the function:** $$h(x) = \frac{(2x - 1)(x - 3)}{(x - 3)(x + 2)}$$ Cancel common factor $x - 3$: $$h(x) = \frac{\cancel{(x - 3)}(2x - 1)}{\cancel{(x - 3)}(x + 2)} = \frac{2x - 1}{x + 2}, \quad x \neq 3$$ 4. **Find vertical asymptotes (VA):** Set denominator equal to zero: $$x + 2 = 0 \implies x = -2$$ Note: $x=3$ is a removable discontinuity (hole) since it was canceled. 5. **Find horizontal asymptote (HA):** Since degrees of numerator and denominator are equal (both 1 after simplification), HA is ratio of leading coefficients: $$HA = \frac{2}{1} = 2$$ 6. **Find intercepts:** - **y-intercept:** Evaluate $h(0)$: $$h(0) = \frac{2(0) - 1}{0 + 2} = \frac{-1}{2} = -\frac{1}{2}$$ - **x-intercept:** Set numerator equal to zero: $$2x - 1 = 0 \implies x = \frac{1}{2}$$ 7. **Intervals of increase and decrease:** Since $h(x) = \frac{2x - 1}{x + 2}$, find derivative: $$h'(x) = \frac{(2)(x + 2) - (2x - 1)(1)}{(x + 2)^2} = \frac{2x + 4 - 2x + 1}{(x + 2)^2} = \frac{5}{(x + 2)^2}$$ Because $5 > 0$ and denominator squared is always positive except at $x = -2$ (vertical asymptote), $h'(x) > 0$ for all $x \neq -2$. Therefore, $h(x)$ is increasing on $( -\infty, -2 )$ and $( -2, \infty )$. 8. **Summary:** - Vertical asymptote at $x = -2$ - Removable discontinuity (hole) at $x = 3$ - Horizontal asymptote at $y = 2$ - x-intercept at $x = \frac{1}{2}$ - y-intercept at $y = -\frac{1}{2}$ - Increasing on $( -\infty, -2 ) \cup ( -2, \infty )$