1. **State the problem:** We need to show algebraically that the curve $C$ with equation $y = x^2 + 3x - 3$ and the line $L$ with equation $y = 5x - 5$ have exactly one point in common. 2. **Set the equations equal:** Since the point(s) of intersection satisfy both equations, set the right-hand sides equal: $$x^2 + 3x - 3 = 5x - 5$$ 3. **Rearrange to form a quadratic equation:** Move all terms to one side: $$x^2 + 3x - 3 - 5x + 5 = 0$$ $$x^2 - 2x + 2 = 0$$ 4. **Use the discriminant to determine the number of solutions:** For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is $$\Delta = b^2 - 4ac$$ Here, $a=1$, $b=-2$, and $c=2$. Calculate: $$\Delta = (-2)^2 - 4(1)(2) = 4 - 8 = -4$$ 5. **Interpret the discriminant:** Since $\Delta < 0$, the quadratic equation has no real solutions, meaning the line and curve do not intersect at any real points if the line equation is $y=5x-5$. 6. **Check the original line equation:** The user gave the line as $y - 5x + 4 = 0$, which rearranges to $$y = 5x - 4$$ Using this correct line equation, set equal to the curve: $$x^2 + 3x - 3 = 5x - 4$$ Rearranged: $$x^2 + 3x - 3 - 5x + 4 = 0$$ $$x^2 - 2x + 1 = 0$$ 7. **Calculate discriminant for corrected line:** $$a=1, b=-2, c=1$$ $$\Delta = (-2)^2 - 4(1)(1) = 4 - 4 = 0$$ 8. **Conclusion:** Since $\Delta = 0$, the quadratic has exactly one real solution, so the curve $C$ and the line $L$ intersect at exactly one point. **Final answer:** The curve and line have exactly one point in common because the quadratic formed by equating their equations has a discriminant of zero.