1. **State the problem:** We need to determine whether the turning points at $(-5, 275)$ and $(2, -68)$ on the curve $y = 2x^3 + 9x^2 - 60x$ are maxima or minima. 2. **Recall the formula:** Turning points occur where the first derivative $y'$ is zero. To classify them, we use the second derivative test: - If $y''(x) > 0$, the turning point is a local minimum. - If $y''(x) < 0$, the turning point is a local maximum. 3. **Find the first derivative:** $$y = 2x^3 + 9x^2 - 60x$$ $$y' = \frac{d}{dx}(2x^3) + \frac{d}{dx}(9x^2) - \frac{d}{dx}(60x) = 6x^2 + 18x - 60$$ 4. **Find the second derivative:** $$y'' = \frac{d}{dx}(6x^2 + 18x - 60) = 12x + 18$$ 5. **Evaluate the second derivative at each turning point:** - At $x = -5$: $$y''(-5) = 12(-5) + 18 = -60 + 18 = -42$$ Since $y''(-5) < 0$, the turning point at $(-5, 275)$ is a local maximum. - At $x = 2$: $$y''(2) = 12(2) + 18 = 24 + 18 = 42$$ Since $y''(2) > 0$, the turning point at $(2, -68)$ is a local minimum. **Final answer:** - Turning point $(-5, 275)$ is a local maximum. - Turning point $(2, -68)$ is a local minimum.