Solve the system of quadratic inequalities: \[ \begin{array}{c}y>x^{2}-1 \\ y \leq-2 x^{2}+5\end{array} \]

To solve the system of quadratic inequalities, we first need to analyze each inequality separately. 1. **First Inequality**: \( y > x^2 - 1 \) This inequality represents a region above the parabola \( y = x^2 - 1 \). The vertex of this parabola is at the point \( (0, -1) \) and opens upwards. Therefore, the area we want is everything above this curve. 2. **Second Inequality**: \( y \leq -2x^2 + 5 \) This represents a region below the parabola \( y = -2x^2 + 5 \). The vertex of this parabola is at \( (0, 5) \) and opens downwards. Thus, we are interested in the area below this curve. Next, we need to find the points of intersection of the two parabolas to understand the overlapping region better. We can set the two expressions for \( y \) equal to each other: \[ x^2 - 1 = -2x^2 + 5 \] Combining like terms: \[ x^2 + 2x^2 = 5 + 1 \quad \Rightarrow \quad 3x^2 = 6 \quad \Rightarrow \quad x^2 = 2 \quad \Rightarrow \quad x = \pm \sqrt{2} \] Now we can find the corresponding \( y \) values by substituting \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) back into either equation. Using the first inequality, \( y = x^2 - 1 \): For \( x = \sqrt{2} \): \[ y = (\sqrt{2})^2 - 1 = 2 - 1 = 1 \] For \( x = -\sqrt{2} \): \[ y = (-\sqrt{2})^2 - 1 = 2 - 1 = 1 \] Thus, the points of intersection are \( (\sqrt{2}, 1) \) and \( (-\sqrt{2}, 1) \). ### Conclusion: The solution to the system of inequalities is the region where: - \( y \) is above the curve \( y = x^2 - 1 \) and, - \( y \) is below the curve \( y = -2x^2 + 5 \). This creates a closed region between the two parabolas from \( x = -\sqrt{2} \) to \( x = \sqrt{2} \). Therefore, the final answer in consideration of the inequalities is: \[ \text{Solution: } \{ (x, y) \mid -\sqrt{2} \leq x \leq \sqrt{2}, \, x^2 - 1 < y \leq -2x^2 + 5 \} \]