1. **State the problem:** Identify the roots and the y-intercept of the function $$f(x) = (x+4)^2 (x+1) (x-2)^2$$. 2. **Find the roots:** The roots are the values of $x$ that make $f(x) = 0$. Set each factor equal to zero: - $x+4=0 \Rightarrow x=-4$ - $x+1=0 \Rightarrow x=-1$ - $x-2=0 \Rightarrow x=2$ So the roots are $x = -4, -1, 2$. 3. **Find the y-intercept:** The y-intercept occurs when $x=0$. Calculate: $$f(0) = (0+4)^2 (0+1) (0-2)^2 = 4^2 \times 1 \times (-2)^2 = 16 \times 1 \times 4 = 64$$ 4. **Summary:** - Roots: $-4, -1, 2$ - Y-intercept: $(0,64)$ 5. **Sign table:** | Interval | $(-\infty, -4)$ | $-4$ | $(-4, -1)$ | $-1$ | $(-1, 2)$ | $2$ | $(2, \infty)$ | |----------------|------------------|------|------------|------|-----------|-----|---------------| | $(x+4)^2$ | $+$ | $0$ | $+$ | $+$ | $+$ | $+$ | $+$ | | $(x+1)$ | $-$ | $-$ | $-$ | $0$ | $+$ | $+$ | $+$ | | $(x-2)^2$ | $+$ | $+$ | $+$ | $+$ | $+$ | $0$ | $+$ | | $f(x)$ | $-$ | $0$ | $-$ | $0$ | $+$ | $0$ | $+$ | Explanation: - Squares $(x+4)^2$ and $(x-2)^2$ are always non-negative and zero only at their roots. - The sign of $f(x)$ depends on $(x+1)$ since the other factors are squared. 6. **Graph sketch notes:** - The graph touches the x-axis at $x=-4$ and $x=2$ (because of squared factors, the graph bounces off the axis). - The graph crosses the x-axis at $x=-1$ (linear factor). - The graph passes through $(0,64)$ on the y-axis. Final answer: **Roots:** $-4, -1, 2$ **Y-intercept:** $(0,64)$