1. **Stating the problem:** We analyze the cubic function $$y = x(x-6)^2$$ which has critical points at $$x=2$$ and $$x=6$$ with given values $$f(2)=8$$ and $$f(6)=0$$. 2. **Find the derivative:** To find where the function increases or decreases, compute $$f'(x)$$. $$f(x) = x(x-6)^2 = x(x^2 - 12x + 36) = x^3 - 12x^2 + 36x$$ Differentiate: $$f'(x) = 3x^2 - 24x + 36$$ 3. **Find critical points:** Solve $$f'(x) = 0$$. $$3x^2 - 24x + 36 = 0$$ Divide both sides by 3: $$x^2 - 8x + 12 = 0$$ Factor: $$(x - 6)(x - 2) = 0$$ So critical points are at $$x=2$$ and $$x=6$$. 4. **Determine increasing/decreasing intervals:** - For $$x < 2$$, pick $$x=0$$: $$f'(0) = 3(0)^2 - 24(0) + 36 = 36 > 0$$, so $$f$$ is increasing. - For $$2 < x < 6$$, pick $$x=4$$: $$f'(4) = 3(16) - 24(4) + 36 = 48 - 96 + 36 = -12 < 0$$, so $$f$$ is decreasing. - For $$x > 6$$, pick $$x=7$$: $$f'(7) = 3(49) - 24(7) + 36 = 147 - 168 + 36 = 15 > 0$$, so $$f$$ is increasing. 5. **Interpretation:** - The function increases on $$(-\infty, 2)$$. - The function decreases on $$(2, 6)$$. - The function increases again on $$(6, \infty)$$. 6. **Horizontal lines $$y = k$$:** - For $$k > 8$$ or $$k < 0$$, the line $$y = k$$ does not intersect the curve because the maximum value is 8 at $$x=2$$ and the minimum on the interval is 0 at $$x=6$$. - For $$0 < k < 8$$, the line $$y = k$$ intersects the curve in three points due to the cubic shape. **Final answer:** - $$f'(x) < 0$$ on $$(2, 6)$$. - The curve has a maximum at $$(2, 8)$$ and a minimum at $$(6, 0)$$. - Horizontal lines $$y = k$$ with $$k > 8$$ or $$k < 0$$ do not intersect the curve.