1. **State the problem:** We are given the cubic function $$y = -\frac{3x^3}{3} - 2x^2 + 5x - 7$$ which simplifies to $$y = -x^3 - 2x^2 + 5x - 7$$. We need to: - Find the coordinates where the function is optimized (turning points). - Determine the nature of these turning points (maxima or minima). - Find the function values at these points. - Sketch the function. 2. **Find critical points:** To find turning points, compute the first derivative and set it to zero. $$y' = \frac{d}{dx}(-x^3 - 2x^2 + 5x - 7) = -3x^2 - 4x + 5$$ Set $$y' = 0$$: $$-3x^2 - 4x + 5 = 0$$ Multiply both sides by -1 for simplicity: $$3x^2 + 4x - 5 = 0$$ 3. **Solve quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=3$$, $$b=4$$, $$c=-5$$. Calculate discriminant: $$\Delta = 4^2 - 4 \times 3 \times (-5) = 16 + 60 = 76$$ So, $$x = \frac{-4 \pm \sqrt{76}}{6} = \frac{-4 \pm 2\sqrt{19}}{6} = \frac{-2 \pm \sqrt{19}}{3}$$ 4. **Calculate approximate values:** $$\sqrt{19} \approx 4.3589$$ So, $$x_1 = \frac{-2 + 4.3589}{3} \approx 0.7863$$ $$x_2 = \frac{-2 - 4.3589}{3} \approx -2.1196$$ 5. **Determine nature of turning points:** Use the second derivative: $$y'' = \frac{d}{dx}(-3x^2 - 4x + 5) = -6x - 4$$ Evaluate at $$x_1$$: $$y''(0.7863) = -6(0.7863) - 4 = -4.7178 - 4 = -8.7178 < 0$$ Since $$y'' < 0$$, this point is a local maximum. Evaluate at $$x_2$$: $$y''(-2.1196) = -6(-2.1196) - 4 = 12.7176 - 4 = 8.7176 > 0$$ Since $$y'' > 0$$, this point is a local minimum. 6. **Find function values at turning points:** Calculate $$y(x_1)$$: $$y(0.7863) = - (0.7863)^3 - 2(0.7863)^2 + 5(0.7863) - 7$$ $$= -0.4859 - 1.2366 + 3.9315 - 7 = -4.7910$$ Calculate $$y(x_2)$$: $$y(-2.1196) = -(-2.1196)^3 - 2(-2.1196)^2 + 5(-2.1196) - 7$$ $$= -(-9.529) - 2(4.494) - 10.598 - 7 = 9.529 - 8.988 - 10.598 - 7 = -17.057$$ 7. **Summary:** - Local maximum at approximately $$(0.79, -4.79)$$ - Local minimum at approximately $$(-2.12, -17.06)$$ 8. **Sketching the function:** - The cubic has one local max and one local min. - It passes through these turning points. - The function decreases to the left of the minimum and increases to the right of the maximum. This matches the typical cubic shape with one peak and one valley.