1. **State the problem:** We need to find the roots and the y-intercept of the function $$f(x) = x(x+5)(2x-4)^2(x-4)^2$$ and then fill in the sign table and sketch the graph. 2. **Identify the roots:** The roots are the values of $x$ that make $f(x) = 0$. Set each factor equal to zero: - $x = 0$ - $x + 5 = 0 \Rightarrow x = -5$ - $2x - 4 = 0 \Rightarrow 2x = 4 \Rightarrow x = 2$ - $x - 4 = 0 \Rightarrow x = 4$ So the roots are $x = -5, 0, 2, 4$. 3. **Find the y-intercept:** The y-intercept occurs when $x=0$. Calculate: $$f(0) = 0 \times (0+5) \times (2\times0 - 4)^2 \times (0 - 4)^2 = 0$$ So the y-intercept is at $(0,0)$. 4. **Analyze multiplicity and sign changes:** - Root $x=-5$ has multiplicity 1 (linear factor), so the graph crosses the x-axis here. - Root $x=0$ has multiplicity 1, so the graph crosses the x-axis here. - Root $x=2$ has multiplicity 2 (squared factor), so the graph touches the x-axis and bounces off here. - Root $x=4$ has multiplicity 2, so the graph touches and bounces off here. 5. **Sign table:** | Interval | Test point | Sign of $f(x)$ | |----------------|------------|---------------| | $(-\infty, -5)$| $-6$ | $-$ (since leading term $x^6$ is positive, but check sign of factors: $x$ negative, $x+5$ negative, $(2x-4)^2$ positive, $(x-4)^2$ positive, total negative) | | $(-5, 0)$ | $-1$ | $+$ (one factor changes sign at $-5$) | | $(0, 2)$ | $1$ | $-$ (crosses at 0) | | $(2, 4)$ | $3$ | $+$ (touches at 2, no sign change) | | $(4, \infty)$ | $5$ | $+$ (touches at 4, no sign change) | 6. **Summary:** - Roots: $-5, 0, 2, 4$ - Y-intercept: $(0,0)$ - Sign changes at $-5$ and $0$ - Touches and bounces at $2$ and $4$ 7. **Graph sketch:** - Crosses x-axis at $-5$ and $0$ - Touches x-axis at $2$ and $4$ - Passes through $(0,0)$ - For large $|x|$, since the degree is 6 (even) and leading coefficient positive, $f(x) \to +\infty$ as $x \to \pm \infty$. Final answer: - Roots: $x = -5, 0, 2, 4$ - Y-intercept: $(0,0)$