1. **State the problem:** We are given the cubic function $$f(x) = -0.3x^3 - 0.3x^2 + 1.8x$$ and want to analyze its behavior including intercepts and extrema. 2. **Formula and rules:** To find intercepts, set $$f(x) = 0$$ for x-intercepts and evaluate $$f(0)$$ for the y-intercept. To find extrema (local maxima and minima), find the critical points by setting the first derivative $$f'(x)$$ equal to zero. 3. **Find the first derivative:** $$f'(x) = \frac{d}{dx}(-0.3x^3 - 0.3x^2 + 1.8x) = -0.9x^2 - 0.6x + 1.8$$ 4. **Find critical points by solving $$f'(x) = 0$$:** $$-0.9x^2 - 0.6x + 1.8 = 0$$ Divide both sides by -0.3 to simplify: $$\cancel{-0.3} \times (3x^2 + 2x - 6) = 0 \Rightarrow 3x^2 + 2x - 6 = 0$$ 5. **Solve quadratic equation:** Use quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=3$$, $$b=2$$, $$c=-6$$: $$x = \frac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-6)}}{2 \times 3} = \frac{-2 \pm \sqrt{4 + 72}}{6} = \frac{-2 \pm \sqrt{76}}{6}$$ 6. **Simplify $$\sqrt{76}$$:** $$\sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}$$ 7. **Final critical points:** $$x = \frac{-2 \pm 2\sqrt{19}}{6} = \frac{-1 \pm \sqrt{19}}{3}$$ 8. **Approximate values:** $$\sqrt{19} \approx 4.36$$ So, $$x_1 = \frac{-1 + 4.36}{3} \approx 1.12$$ $$x_2 = \frac{-1 - 4.36}{3} \approx -1.79$$ 9. **Find y-values at critical points:** Calculate $$f(1.12)$$ and $$f(-1.79)$$: $$f(1.12) = -0.3(1.12)^3 - 0.3(1.12)^2 + 1.8(1.12) \approx -0.3(1.41) - 0.3(1.25) + 2.02 = -0.42 - 0.38 + 2.02 = 1.22$$ $$f(-1.79) = -0.3(-1.79)^3 - 0.3(-1.79)^2 + 1.8(-1.79) \approx -0.3(-5.73) - 0.3(3.20) - 3.22 = 1.72 - 0.96 - 3.22 = -2.46$$ 10. **Find x-intercepts by solving $$f(x) = 0$$:** $$-0.3x^3 - 0.3x^2 + 1.8x = 0$$ Factor out $$x$$: $$x(-0.3x^2 - 0.3x + 1.8) = 0$$ So, $$x=0$$ or solve quadratic: $$-0.3x^2 - 0.3x + 1.8 = 0$$ Divide by -0.3: $$x^2 + x - 6 = 0$$ Solve: $$x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ 11. **Solutions:** $$x = 2$$ or $$x = -3$$ 12. **Y-intercept:** $$f(0) = 0$$ **Summary:** - X-intercepts at $$x = -3, 0, 2$$ - Y-intercept at $$y = 0$$ - Local minimum near $$x \approx -1.79$$ with $$f(x) \approx -2.46$$ - Local maximum near $$x \approx 1.12$$ with $$f(x) \approx 1.22$$