1. **State the problem:** We are given two equations: $$y = -3x^2 + 2x - 5$$ and $$-x + 2 = -y$$ We want to analyze these equations, find their points of intersection, and understand their graphs. 2. **Rewrite the linear equation:** From $$-x + 2 = -y$$, multiply both sides by $$-1$$ to isolate $$y$$: $$-x + 2 = -y$$ $$\Rightarrow y = x - 2$$ 3. **Set the two expressions for $$y$$ equal to find intersection points:** $$-3x^2 + 2x - 5 = x - 2$$ 4. **Bring all terms to one side:** $$-3x^2 + 2x - 5 - x + 2 = 0$$ $$-3x^2 + (2x - x) + (-5 + 2) = 0$$ $$-3x^2 + x - 3 = 0$$ 5. **Solve the quadratic equation:** $$-3x^2 + x - 3 = 0$$ Multiply both sides by $$-1$$ to simplify: $$\cancel{-3}x^2 + \cancel{-1}x + \cancel{-3} = 0 \Rightarrow 3x^2 - x + 3 = 0$$ 6. **Calculate the discriminant $$\Delta$$:** $$\Delta = b^2 - 4ac = (-1)^2 - 4 \times 3 \times 3 = 1 - 36 = -35$$ 7. **Interpret the discriminant:** Since $$\Delta < 0$$, there are no real solutions, meaning the parabola and the line do not intersect. 8. **Summary:** The quadratic curve $$y = -3x^2 + 2x - 5$$ opens downward, and the line $$y = x - 2$$ does not intersect it because the quadratic equation formed by setting them equal has no real roots. **Final answer:** No real intersection points exist between the two given equations.