1. **State the problem:** We need to sketch the function $$y = (x-1)(x^2 + x + 1)$$. 2. **Recall the formula and rules:** This is a product of a linear term and a quadratic term. To understand the graph, we find intercepts, critical points, and behavior. 3. **Find the x-intercepts:** Set $$y=0$$: $$0 = (x-1)(x^2 + x + 1)$$ This implies either $$x-1=0$$ or $$x^2 + x + 1=0$$. 4. Solve $$x-1=0$$: $$x=1$$. 5. Solve $$x^2 + x + 1=0$$: Calculate discriminant $$\Delta = b^2 - 4ac = 1^2 - 4\cdot1\cdot1 = 1 - 4 = -3 < 0$$. Since $$\Delta < 0$$, no real roots here. 6. **Therefore, the only x-intercept is at $$x=1$$.** 7. **Find the y-intercept:** Set $$x=0$$: $$y = (0-1)(0^2 + 0 + 1) = (-1)(1) = -1$$. 8. **Find critical points:** Differentiate using product rule: $$y = (x-1)(x^2 + x + 1)$$ $$y' = (x-1)'(x^2 + x + 1) + (x-1)(x^2 + x + 1)'$$ $$= 1 \cdot (x^2 + x + 1) + (x-1)(2x + 1)$$ 9. Simplify: $$y' = x^2 + x + 1 + (x-1)(2x + 1)$$ $$= x^2 + x + 1 + (2x^2 + x - 2x -1)$$ $$= x^2 + x + 1 + 2x^2 - x - 1$$ $$= (x^2 + 2x^2) + (x - x) + (1 - 1) = 3x^2$$ 10. Set $$y' = 0$$: $$3x^2 = 0 \implies x=0$$. 11. **Evaluate $$y$$ at $$x=0$$:** $$y(0) = (0-1)(0^2 + 0 + 1) = -1$$. 12. **Summary:** - x-intercept at $$x=1$$ - y-intercept at $$y=-1$$ - Critical point at $$(0,-1)$$ - Since $$y' = 3x^2 \geq 0$$ for all $$x$$, the function is increasing except at $$x=0$$ where slope is zero. 13. **Sketch notes:** The graph touches the x-axis at $$x=1$$ and has a minimum or flat point at $$(0,-1)$$. Final answer: The function $$y = (x-1)(x^2 + x + 1)$$ has one real root at $$x=1$$, a y-intercept at $$-1$$, and a critical point at $$(0,-1)$$ where the slope is zero. The derivative shows the function is increasing everywhere else.