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For the polynomial function \( f(x)=x^{4}+4 x^{3}+4 x^{2} \), answer the parts a through e a. Use the Leading Coefficient Tesi to delermine the graph's end behavior A. The graph of \( f(x) \) falls to the left and lails to the right B. The graph of \( f(x) \) rises to the leff and falls to the right C. The graph of \( \mathrm{f}(\mathrm{x}) \) rises to the lell and rises to the right D. The graph of \( f(x) \) falls to the left and rises to the right b. Find the \( x \)-intercept(s). State whether the graph crosses the \( x \)-axis, or louches the \( x \)-axis and turns arcund, at each intercept The \( x \)-infercopt(s) islaro \( 0,-2 \). (Iype an integer or a decimal, Use a comma to sepatate answers as needed. Type each answer only once.) At which \( x \)-intercept(5) does the graph cross the \( x \)-axis? Select the correct choice below and, if necessary. fill in the answer box to complele your choice. A. The \( x \)-inlorcept \( [s \) ] at which the graph ciosses the \( x \)-axis is/are \( \square \) . (Type an inleger or a decimal. Use a comma lo separate answers as needed. Type each answer only once.) B. There are no x-intercepts at which the graph crosses the \( x \)-axis.

Ask by Bird Maxwell. in the United States
Mar 10,2025

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Answer

**a. End Behavior** The graph of \( f(x) \) rises on both ends. **b. \(x\)-Intercepts and Behavior** The \(x\)-intercepts are at \( -2 \) and \( 0 \). At both intercepts, the graph touches the \(x\)-axis without crossing it.

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The Deep Dive

The polynomial function \( f(x) = x^4 + 4x^3 + 4x^2 \) has a leading coefficient of 1 (the coefficient of \( x^4 \)), which is positive. Since it's an even-degree polynomial, we can conclude that the graph will rise to the right (positive infinity) and rise to the left (positive infinity). Therefore, the correct answer for the end behavior is C: The graph of \( f(x) \) rises to the left and rises to the right. The \( x \)-intercepts can be found by setting \( f(x) = 0 \). Factoring gives \( f(x) = x^2(x^2 + 4x + 4) = x^2(x + 2)^2 \). This results in \( x \)-intercepts at \( 0 \) and \( -2 \). Since the factors of \( (x + 2)^2 \) have an even exponent, the graph touches the \( x \)-axis at \( -2 \) and turns around, while it crosses the axis at \( 0 \). So, we can summarize: - The \( x \)-intercepts are \( 0, -2 \). - At the \( x \)-intercept \( 0 \), the graph crosses the \( x \)-axis. - At the \( x \)-intercept \( -2 \), the graph touches the \( x \)-axis and turns around. Therefore, the answer to the intercept question is: A. The \( x \)-intercept \( [0] \) at which the graph crosses the \( x \)-axis is/are \( 0 \).

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