1. **State the problem:** Find the roots and y-intercept of the function $$f(x) = x^2 (3x + 6)(x - 3)$$ and sketch its graph. 2. **Find the roots:** Roots occur where $$f(x) = 0$$. Set each factor equal to zero: - $$x^2 = 0 \implies x = 0$$ - $$3x + 6 = 0 \implies 3x = -6 \implies x = -2$$ - $$x - 3 = 0 \implies x = 3$$ So, the roots are $$x = 0, -2, 3$$. 3. **Find the y-intercept:** The y-intercept is the value of $$f(0)$$. Calculate: $$f(0) = 0^2 (3 \cdot 0 + 6)(0 - 3) = 0 \cdot 6 \cdot (-3) = 0$$ So, the y-intercept is at $$y = 0$$. 4. **Sign table:** Analyze the sign of $$f(x)$$ in intervals determined by roots $$-2, 0, 3$$. - For $$x < -2$$: - $$x^2 > 0$$ (always positive except at 0) - $$3x + 6 < 0$$ - $$x - 3 < 0$$ - Product sign: positive $\times$ negative $\times$ negative = positive - For $$-2 < x < 0$$: - $$x^2 > 0$$ - $$3x + 6 > 0$$ - $$x - 3 < 0$$ - Product sign: positive $\times$ positive $\times$ negative = negative - For $$0 < x < 3$$: - $$x^2 > 0$$ - $$3x + 6 > 0$$ - $$x - 3 < 0$$ - Product sign: positive $\times$ positive $\times$ negative = negative - For $$x > 3$$: - $$x^2 > 0$$ - $$3x + 6 > 0$$ - $$x - 3 > 0$$ - Product sign: positive $\times$ positive $\times$ positive = positive 5. **Summary of sign table:** | Interval | Sign of $$f(x)$$ | |----------|-----------------| | $$(-\infty, -2)$$ | Positive | | $$(-2, 0)$$ | Negative | | $$(0, 3)$$ | Negative | | $$(3, \infty)$$ | Positive | 6. **Sketch the graph:** - Crosses x-axis at $$-2, 0, 3$$. - Touches x-axis at $$0$$ with multiplicity 2 (since $$x^2$$ factor), so the graph just touches and bounces off at $$x=0$$. - Y-intercept at $$0$$. - Graph is positive for $$x < -2$$ and $$x > 3$$, negative between $$-2$$ and $$3$$ except at $$0$$ where it touches the axis. **Final answer:** - Roots: $$x = -2, 0, 3$$ - Y-intercept: $$0$$