1. **State the problem:** We have the cubic function $$y = x^3 - 8x^2 - 12x + 5$$ and need to find the coordinates of the two stationary points A and B, where the x-coordinate of A is greater than that of B. 2. **Find the stationary points:** Stationary points occur where the derivative $$y'$$ equals zero. 3. **Calculate the derivative:** $$y' = \frac{d}{dx}(x^3 - 8x^2 - 12x + 5) = 3x^2 - 16x - 12$$ 4. **Set the derivative equal to zero to find critical points:** $$3x^2 - 16x - 12 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=3$$, $$b=-16$$, $$c=-12$$. 6. **Calculate the discriminant:** $$\Delta = (-16)^2 - 4 \times 3 \times (-12) = 256 + 144 = 400$$ 7. **Find the roots:** $$x = \frac{16 \pm \sqrt{400}}{2 \times 3} = \frac{16 \pm 20}{6}$$ 8. **Calculate each root:** - $$x_1 = \frac{16 + 20}{6} = \frac{36}{6} = 6$$ - $$x_2 = \frac{16 - 20}{6} = \frac{-4}{6} = -\frac{2}{3}$$ 9. **Assign points:** Since $$x_A > x_B$$, point A has $$x=6$$ and point B has $$x=-\frac{2}{3}$$. 10. **Find corresponding y-values:** - For $$x=6$$: $$y = 6^3 - 8 \times 6^2 - 12 \times 6 + 5 = 216 - 288 - 72 + 5 = -139$$ - For $$x=-\frac{2}{3}$$: $$y = \left(-\frac{2}{3}\right)^3 - 8 \times \left(-\frac{2}{3}\right)^2 - 12 \times \left(-\frac{2}{3}\right) + 5$$ $$= -\frac{8}{27} - 8 \times \frac{4}{9} + 8 + 5 = -\frac{8}{27} - \frac{32}{9} + 13$$ Convert to common denominator 27: $$-\frac{8}{27} - \frac{96}{27} + \frac{351}{27} = \frac{247}{27} \approx 9.15$$ 11. **Final stationary points:** - $$A = (6, -139)$$ - $$B = \left(-\frac{2}{3}, \frac{247}{27}\right)$$ **Answer:** The stationary points are $$A = (6, -139)$$ and $$B = \left(-\frac{2}{3}, \frac{247}{27}\right)$$ with $$x_A > x_B$$.