Question: 1. y > x^2 - x - 6
2. y \leq x^2 + 7x + 10
Graph the quadratic inequality and shade the region that contains the solution (the graph need to be a picture)
1. **State the problem:**
We need to graph the system of quadratic inequalities:
$$y > x^2 - x - 6$$
and
$$y \leq x^2 + 7x + 10$$
2. **Understand the inequalities:**
- The first inequality is strict ($>$), so the boundary curve $y = x^2 - x - 6$ is not included.
- The second inequality is inclusive ($\leq$), so the boundary curve $y = x^2 + 7x + 10$ is included.
3. **Graph the boundary curves:**
- For $y = x^2 - x - 6$, this is a parabola opening upwards.
- For $y = x^2 + 7x + 10$, this is also a parabola opening upwards.
4. **Find the intersection points of the two parabolas:**
Set
$$x^2 - x - 6 = x^2 + 7x + 10$$
Simplify:
$$-x - 6 = 7x + 10$$
$$-x - 7x = 10 + 6$$
$$-8x = 16$$
$$x = \frac{16}{-8} = -2$$
Find $y$ at $x = -2$:
$$y = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0$$
So the parabolas intersect at $(-2, 0)$.
5. **Determine the solution region:**
- For $y > x^2 - x - 6$, shade above the parabola $y = x^2 - x - 6$.
- For $y \leq x^2 + 7x + 10$, shade below or on the parabola $y = x^2 + 7x + 10$.
The solution region is where these shaded areas overlap.
6. **Summary:**
The solution is the region above the parabola $y = x^2 - x - 6$ and at or below the parabola $y = x^2 + 7x + 10$.
**Final answer:** The graph shows two parabolas intersecting at $(-2,0)$ with the solution region shaded between them, above the first parabola and below the second.