1. The problem gives two equations: $y=1-x$ and $y=5-2x-x^2$. We want to analyze or find the points where these two functions intersect. 2. To find the intersection points, set the two expressions for $y$ equal to each other: $$1 - x = 5 - 2x - x^2$$ 3. Rearrange the equation to one side to form a quadratic equation: $$0 = 5 - 2x - x^2 - (1 - x)$$ $$0 = 5 - 2x - x^2 - 1 + x$$ $$0 = 4 - x - x^2$$ 4. Rewrite the quadratic equation in standard form: $$-x^2 - x + 4 = 0$$ Multiply both sides by $-1$ to simplify: $$x^2 + x - 4 = 0$$ 5. Solve the quadratic equation using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=1$, and $c=-4$. 6. Calculate the discriminant: $$\Delta = b^2 - 4ac = 1^2 - 4(1)(-4) = 1 + 16 = 17$$ 7. Find the roots: $$x = \frac{-1 \pm \sqrt{17}}{2}$$ 8. Calculate the corresponding $y$ values using $y=1-x$: For $x_1 = \frac{-1 + \sqrt{17}}{2}$: $$y_1 = 1 - x_1 = 1 - \frac{-1 + \sqrt{17}}{2} = \frac{2 - (-1 + \sqrt{17})}{2} = \frac{3 - \sqrt{17}}{2}$$ For $x_2 = \frac{-1 - \sqrt{17}}{2}$: $$y_2 = 1 - x_2 = 1 - \frac{-1 - \sqrt{17}}{2} = \frac{2 - (-1 - \sqrt{17})}{2} = \frac{3 + \sqrt{17}}{2}$$ 9. Therefore, the intersection points are: $$\left(\frac{-1 + \sqrt{17}}{2}, \frac{3 - \sqrt{17}}{2}\right) \text{ and } \left(\frac{-1 - \sqrt{17}}{2}, \frac{3 + \sqrt{17}}{2}\right)$$