1. **Problem statement:** Find the intersection points of the parabolas in part (a): $$y = x^2 + 4x - 1$$ $$y = x^2 - 2x + 5$$ 2. **Set the two equations equal to find intersection points:** $$x^2 + 4x - 1 = x^2 - 2x + 5$$ 3. **Simplify by subtracting $x^2$ from both sides:** $$\cancel{x^2} + 4x - 1 = \cancel{x^2} - 2x + 5$$ which simplifies to $$4x - 1 = -2x + 5$$ 4. **Bring all terms involving $x$ to one side and constants to the other:** $$4x + 2x = 5 + 1$$ $$6x = 6$$ 5. **Solve for $x$:** $$x = \frac{6}{6} = 1$$ 6. **Find corresponding $y$ by substituting $x=1$ into one of the original equations, for example $y = x^2 + 4x - 1$:** $$y = 1^2 + 4(1) - 1 = 1 + 4 - 1 = 4$$ 7. **Conclusion:** The parabolas intersect at the point $$(1, 4)$$ This is the only intersection point because the quadratic terms canceled out, leaving a linear equation with a single solution.