Question: A cartographer is analyzing a terrain modeled by the function $ h(x) = x^3 - 3x^2 + 2x $, and wants to find the sum of all $ x $-coordinates where the terrain has horizontal slope.

["Question: A cartographer is analyzing a terrain modeled by the function $ h(x) = x^3 - 3x^2 + 2x $, and wants to find the sum of all $ x $-coordinates where the terrain has a horizontal slope. Find this sum.", "When a terrain modeled by a function has a horizontal slope, it means the derivative of the function is zero at that point. In this case, the terrain is described by the cubic function:", "$$\nh(x) = x^3 - 3x^2 + 2x\n$$", "To find where the slope is horizontal, we compute the first derivative of $ h(x) $:", "$$\nh'(x) = \frac{d}{dx}(x^3 - 3x^2 + 2x) = 3x^2 - 6x + 2\n$$", "The horizontal slopes occur at the values of $ x $ where $ h'(x) = 0 $. Thus, we solve the quadratic equation:", "$$\n3x^2 - 6x + 2 = 0\n$$", "We apply the quadratic formula:", "$$\nx = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6}\n$$", "Simplify $ \sqrt{12} = 2\sqrt{3} $, so:", "$$\nx = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}\n$$", "Thus, the two $ x $-coordinates where the terrain has a horizontal slope are:", "$$\nx_1 = \frac{3 - \sqrt{3}}{3}, \quad x_2 = \frac{3 + \sqrt{3}}{3}\n$$", "We are asked to find the sum of these $ x $-coordinates:", "$$\nx_1 + x_2 = \frac{3 - \sqrt{3}}{3} + \frac{3 + \sqrt{3}}{3} = \frac{(3 - \sqrt{3}) + (3 + \sqrt{3})}{3} = \frac{6}{3} = 2\n$$", "Therefore, the sum of all $ x $-coordinates where the terrain has a horizontal slope is:", "$$\n\boxed{2}\n$$", "This result highlights a useful property of polynomials: the sum of roots of a quadratic equation $ ax^2 + bx + c = 0 $ is $ -\frac{b}{a} $. Here, $ -\frac{-6}{3} = 2 $, confirming our answer efficiently."]