1. **State the problem:** Graph the quadratic inequalities: a) $y > -x^2 - 3x - 4$ b) $-2y \geq x^2 - 1$ 2. **Rewrite each inequality in graphing form (vertex form) by completing the square if needed.** **a)** Start with $y > -x^2 - 3x - 4$ Rewrite the right side: $$-x^2 - 3x - 4 = -(x^2 + 3x) - 4$$ Complete the square inside the parentheses: Take half of 3, which is $\frac{3}{2}$, square it: $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ Add and subtract $\frac{9}{4}$ inside the parentheses: $$-(x^2 + 3x + \frac{9}{4} - \frac{9}{4}) - 4 = -\left( (x + \frac{3}{2})^2 - \frac{9}{4} \right) - 4$$ Distribute the minus sign: $$- (x + \frac{3}{2})^2 + \frac{9}{4} - 4 = - (x + \frac{3}{2})^2 - \frac{7}{4}$$ So the inequality becomes: $$y > - (x + \frac{3}{2})^2 - \frac{7}{4}$$ This is the graphing form: a parabola opening downward with vertex at $\left(-\frac{3}{2}, -\frac{7}{4}\right)$. 3. **b)** Start with $-2y \geq x^2 - 1$ Divide both sides by $-2$, remembering to reverse the inequality sign because dividing by a negative number: $$y \leq -\frac{1}{2}x^2 + \frac{1}{2}$$ This is already in graphing form with vertex at $(0, \frac{1}{2})$ and parabola opening downward. 4. **Graph shape description:** - For (a), the parabola opens downward with vertex at $\left(-\frac{3}{2}, -\frac{7}{4}\right)$ and the region above the parabola is shaded (since $y$ is greater than the parabola). - For (b), the parabola opens downward with vertex at $(0, \frac{1}{2})$ and the region below the parabola is shaded (since $y$ is less than or equal to the parabola). 5. **Summary:** - (a) $y > - (x + \frac{3}{2})^2 - \frac{7}{4}$ - (b) $y \leq -\frac{1}{2}x^2 + \frac{1}{2}$ These forms are ready for graphing on a calculator or by hand. Final answers: $$\boxed{y > - (x + \frac{3}{2})^2 - \frac{7}{4}}$$ $$\boxed{y \leq -\frac{1}{2}x^2 + \frac{1}{2}}$$