1. **Graph the inequality $2x + y \geq 8$** - The problem asks to graph the region where $2x + y$ is greater than or equal to 8. - First, rewrite the inequality in terms of $y$: $$y \geq 8 - 2x$$ - The boundary line is $y = 8 - 2x$. - This line has slope $-2$ and y-intercept $8$. - The inequality $y \geq 8 - 2x$ means the region above or on this line. 2. **Divide $\frac{x^2 + x - 2}{3x^2 + 9x + 6}$ by $x - 1$** - First, factor numerator and denominator: $$x^2 + x - 2 = (x + 2)(x - 1)$$ $$3x^2 + 9x + 6 = 3(x^2 + 3x + 2) = 3(x + 1)(x + 2)$$ - The original expression is: $$\frac{(x + 2)(x - 1)}{3(x + 1)(x + 2)}$$ - Divide by $x - 1$ means multiply by its reciprocal: $$\frac{(x + 2)(x - 1)}{3(x + 1)(x + 2)} \times \frac{1}{x - 1} = \frac{(x + 2)(x - 1)}{3(x + 1)(x + 2)(x - 1)}$$ - Cancel common factors $(x - 1)$ and $(x + 2)$: $$= \frac{1}{3(x + 1)}$$ 3. **Sketch the graph of $y = x^2 - 2x - 3$** - This is a quadratic function, a parabola opening upwards because the coefficient of $x^2$ is positive. - Find vertex using formula $x = -\frac{b}{2a}$ where $a=1$, $b=-2$: $$x = -\frac{-2}{2 \times 1} = 1$$ - Find $y$ at vertex: $$y = 1^2 - 2 \times 1 - 3 = 1 - 2 - 3 = -4$$ - Vertex is at $(1, -4)$. - Find roots by solving $x^2 - 2x - 3 = 0$: $$(x - 3)(x + 1) = 0 \Rightarrow x = 3 \text{ or } x = -1$$ - The parabola crosses the x-axis at $x = -1$ and $x = 3$. **Summary:** - Inequality graph: half-plane above line $y = 8 - 2x$. - Division result: $\frac{1}{3(x + 1)}$. - Parabola: vertex at $(1, -4)$, roots at $-1$ and $3$.