1. **Problem statement:** Given the parabola $y = x^2 - 4$ and the line $y = mx + c$ with $m = -4$, find the values of $c$ for which the line intersects the parabola at two distinct points. 2. **Set up the equation for intersection:** To find intersection points, set the parabola equal to the line: $$x^2 - 4 = -4x + c$$ 3. **Rearrange to standard quadratic form:** $$x^2 + 4x - 4 - c = 0$$ 4. **Use the discriminant to determine the number of intersection points:** For a quadratic equation $ax^2 + bx + d = 0$, the discriminant is $$\Delta = b^2 - 4ad$$ Here, $a=1$, $b=4$, and $d = -4 - c$. 5. **Calculate the discriminant:** $$\Delta = 4^2 - 4 \times 1 \times (-4 - c) = 16 + 4(4 + c) = 16 + 16 + 4c = 32 + 4c$$ 6. **Condition for two distinct points:** The line intersects the parabola at two distinct points if and only if $$\Delta > 0$$ 7. **Solve the inequality:** $$32 + 4c > 0$$ Divide both sides by 4: $$\cancel{4}c > \cancel{-32}$$ $$c > -8$$ **Final answer:** The line $y = -4x + c$ intersects the parabola $y = x^2 - 4$ at two distinct points if and only if $$c > -8$$