1. **State the problem:** Find the points of intersection between the quadratic function and the linear function for the first pair: $$y = 2x^2 - 3x + 1$$ and $$y = x - 1$$. 2. **Set the equations equal to find intersection points:** $$2x^2 - 3x + 1 = x - 1$$ 3. **Bring all terms to one side:** $$2x^2 - 3x + 1 - x + 1 = 0$$ $$2x^2 - 4x + 2 = 0$$ 4. **Simplify the quadratic equation:** $$\cancel{2}x^2 - \cancel{4}x + \cancel{2} = 0 \implies x^2 - 2x + 1 = 0$$ 5. **Factor the quadratic:** $$x^2 - 2x + 1 = (x - 1)^2 = 0$$ 6. **Solve for x:** $$x - 1 = 0 \implies x = 1$$ 7. **Find y by substituting x into the linear equation:** $$y = 1 - 1 = 0$$ 8. **Conclusion:** The curves intersect at the point $$(1, 0)$$. This process finds the intersection points by equating the quadratic and linear functions, simplifying, factoring, and solving for $x$, then substituting back to find $y$.