1. **State the problem:** Find the coordinates of the turning point of the curve given by the function $$y = \frac{x^2}{2} + 8x$$. 2. **Recall the formula and rules:** The turning point of a curve given by a function $y=f(x)$ occurs where the derivative $\frac{dy}{dx}$ equals zero. This is because the slope of the tangent line at the turning point is zero. 3. **Find the derivative:** $$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^2}{2} + 8x\right) = \frac{d}{dx}\left(\frac{1}{2}x^2\right) + \frac{d}{dx}(8x) = x + 8$$ 4. **Set the derivative equal to zero to find critical points:** $$x + 8 = 0$$ $$x = -8$$ 5. **Find the y-coordinate of the turning point by substituting $x = -8$ back into the original function:** $$y = \frac{(-8)^2}{2} + 8(-8) = \frac{64}{2} - 64 = 32 - 64 = -32$$ 6. **Conclusion:** The turning point of the curve is at the coordinates $$(-8, -32)$$.