1. The problem is to analyze the curve given by the function $$y = x^4 - 4x^3$$. 2. We want to find key features such as intercepts and extrema. 3. First, find the intercepts: - The y-intercept is found by evaluating $$y$$ at $$x=0$$: $$y(0) = 0^4 - 4 \cdot 0^3 = 0$$. - The x-intercepts are found by solving $$x^4 - 4x^3 = 0$$. 4. Factor the equation for x-intercepts: $$x^3(x - 4) = 0$$ This gives solutions $$x=0$$ and $$x=4$$. 5. Next, find the extrema by computing the first derivative: $$y' = \frac{d}{dx}(x^4 - 4x^3) = 4x^3 - 12x^2$$. 6. Set the derivative equal to zero to find critical points: $$4x^3 - 12x^2 = 0$$ Factor out $$4x^2$$: $$4x^2(x - 3) = 0$$ So critical points are at $$x=0$$ and $$x=3$$. 7. Determine the nature of these critical points using the second derivative: $$y'' = \frac{d}{dx}(4x^3 - 12x^2) = 12x^2 - 24x$$. 8. Evaluate $$y''$$ at critical points: - At $$x=0$$: $$y''(0) = 0$$ (inconclusive, test with first derivative sign change or higher derivatives). - At $$x=3$$: $$y''(3) = 12 \cdot 9 - 24 \cdot 3 = 108 - 72 = 36 > 0$$, so $$x=3$$ is a local minimum. 9. Check the behavior around $$x=0$$: - For $$x < 0$$, say $$x=-1$$, $$y'(-1) = 4(-1)^3 - 12(-1)^2 = -4 - 12 = -16 < 0$$. - For $$0 < x < 3$$, say $$x=1$$, $$y'(1) = 4 - 12 = -8 < 0$$. - For $$x > 3$$, say $$x=4$$, $$y'(4) = 4 \cdot 64 - 12 \cdot 16 = 256 - 192 = 64 > 0$$. Since $$y'$$ does not change sign at $$x=0$$ (negative on both sides), $$x=0$$ is a point of inflection. 10. Summary: - x-intercepts at $$x=0$$ and $$x=4$$. - y-intercept at $$y=0$$. - Local minimum at $$x=3$$ with $$y(3) = 3^4 - 4 \cdot 3^3 = 81 - 108 = -27$$. - Inflection point at $$x=0$$. Final answer: The curve $$y = x^4 - 4x^3$$ has intercepts at (0,0) and (4,0), a local minimum at (3,-27), and an inflection point at (0,0).