Question: Solve and Graph
1) $y > 3x^2$
2) $y \leq 2x^2 - 5$
3) $y < x^2 + 5x + 6$
1. **State the problem:** We need to analyze and graph the inequalities:
1) $y > 3x^2$
2) $y \leq 2x^2 - 5$
3) $y < x^2 + 5x + 6$
2. **Understand the inequalities:** Each inequality describes a region above or below a parabola.
- $y > 3x^2$ means the region above the parabola $y = 3x^2$.
- $y \leq 2x^2 - 5$ means the region on or below the parabola $y = 2x^2 - 5$.
- $y < x^2 + 5x + 6$ means the region strictly below the parabola $y = x^2 + 5x + 6$.
3. **Analyze each parabola:**
- $y = 3x^2$ is an upward-opening parabola with vertex at $(0,0)$.
- $y = 2x^2 - 5$ is an upward-opening parabola with vertex at $(0,-5)$.
- $y = x^2 + 5x + 6$ can be rewritten by completing the square:
$$y = (x^2 + 5x + \frac{25}{4}) + 6 - \frac{25}{4} = (x + \frac{5}{2})^2 + \frac{-1}{4}$$
So vertex at $(-\frac{5}{2}, -\frac{1}{4})$.
4. **Graphing the inequalities:**
- For $y > 3x^2$, shade the region above the parabola $y=3x^2$.
- For $y \leq 2x^2 - 5$, shade the region on or below the parabola $y=2x^2 - 5$.
- For $y < x^2 + 5x + 6$, shade the region strictly below the parabola $y = x^2 + 5x + 6$.
5. **Summary:**
- Each parabola opens upward.
- The inequalities define regions relative to these parabolas.
- The graph will show three parabolas with shading above or below as per inequalities.
6. **Final answer:** The solution sets are the regions described above for each inequality.