1. The problem is to analyze the function $y=2x^3 - x^2$. 2. First, identify the function type: it is a cubic polynomial. 3. To find intercepts: - **y-intercept**: Set $x=0$, then $y=2(0)^3 - (0)^2=0$. - **x-intercepts**: Set $y=0$, solve $2x^3 - x^2=0$. 4. Factor the equation: $$x^2(2x - 1) = 0$$ 5. Set each factor equal to zero: - $x^2=0 \Rightarrow x=0$ - $2x - 1=0 \Rightarrow x=\frac{1}{2}$ 6. So, the x-intercepts are at $x=0$ and $x=\frac{1}{2}$. 7. To find extrema, compute the derivative: $$y' = \frac{d}{dx}(2x^3 - x^2) = 6x^2 - 2x$$ 8. Set derivative equal to zero to find critical points: $$6x^2 - 2x = 0$$ 9. Factor: $$2x(3x - 1) = 0$$ 10. Solve for $x$: - $2x=0 \Rightarrow x=0$ - $3x - 1=0 \Rightarrow x=\frac{1}{3}$ 11. Evaluate $y$ at critical points: - At $x=0$: $y=0$ - At $x=\frac{1}{3}$: $y=2\left(\frac{1}{3}\right)^3 - \left(\frac{1}{3}\right)^2 = 2\cdot\frac{1}{27} - \frac{1}{9} = \frac{2}{27} - \frac{3}{27} = -\frac{1}{27}$ 12. Determine the nature of critical points using the second derivative: $$y'' = \frac{d}{dx}(6x^2 - 2x) = 12x - 2$$ 13. Evaluate $y''$ at critical points: - At $x=0$: $y'' = 12(0) - 2 = -2 < 0$, so local maximum. - At $x=\frac{1}{3}$: $y'' = 12\cdot\frac{1}{3} - 2 = 4 - 2 = 2 > 0$, so local minimum. **Final answer:** - x-intercepts at $x=0$ and $x=\frac{1}{2}$ - y-intercept at $(0,0)$ - Local maximum at $(0,0)$ - Local minimum at $\left(\frac{1}{3}, -\frac{1}{27}\right)$