1. **Stating the problem:** We need to solve the system of inequalities: $$y > x^2$$ and $$y < -x^2 + 1$$ using graphing and one indicator point. 2. **Understanding the inequalities:** - The first inequality $$y > x^2$$ means the solution region is above the parabola $$y = x^2$$. - The second inequality $$y < -x^2 + 1$$ means the solution region is below the parabola $$y = -x^2 + 1$$. 3. **Graphing the boundary curves:** - Plot $$y = x^2$$, a parabola opening upwards. - Plot $$y = -x^2 + 1$$, a parabola opening downwards with vertex at $$(0,1)$$. 4. **Finding the intersection points:** Set $$x^2 = -x^2 + 1$$ to find where the parabolas meet: $$x^2 + x^2 = 1$$ $$2x^2 = 1$$ $$x^2 = \frac{1}{2}$$ $$x = \pm \frac{1}{\sqrt{2}}$$ 5. **Choosing an indicator point:** Pick a point between the intersection points, for example, the origin $$(0,0)$$. 6. **Testing the indicator point:** - Check if $$0 > 0^2$$ i.e. $$0 > 0$$ (False) - Check if $$0 < -0^2 + 1$$ i.e. $$0 < 1$$ (True) Since the first inequality is false at $$(0,0)$$, this point is not in the solution region. 7. **Conclusion:** The solution region is the set of points above $$y = x^2$$ and below $$y = -x^2 + 1$$, between $$x = -\frac{1}{\sqrt{2}}$$ and $$x = \frac{1}{\sqrt{2}}$$. **Final answer:** $$\{(x,y) \mid x \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), x^2 < y < -x^2 + 1\}$$