To determine which graph represents the polynomial \( f(x) = -x(x+2)^2 \), we first need to analyze the function.
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Identify the Zeros: The zeros of the polynomial are found by setting \( f(x) = 0 \): \[ -x(x+2)^2 = 0 \] This gives us: \[ x = 0 \quad \text{and} \quad (x+2)^2 = 0 \implies x = -2 \] Thus, the zeros of the function are \( x = 0 \) and \( x = -2 \). The zero at \( x = -2 \) has multiplicity 2 because of the squared term.
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Behavior at the Zeros:
- At \( x = 0 \), the graph crosses the x-axis because this zero has odd multiplicity (1).
- At \( x = -2 \), the graph touches the x-axis but does not cross it because this zero has even multiplicity (2).
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End Behavior: Since the leading coefficient of the polynomial is negative (as indicated by the minus sign in front), the ends of the graph will both point downwards:
- As \( x \to -\infty \), \( f(x) \to +\infty \)
- As \( x \to +\infty \), \( f(x) \to -\infty \)
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Sketching the Graph:
- The graph will start high on the left side (as \( x \to -\infty \)) and will approach the x-axis at \( x = -2 \), touching it there.
- Then, it will go up to cross the x-axis at \( x = 0 \) before heading downwards as \( x \) increases beyond 0.
Now, based on this analysis, the correct graph should:
- Touch at \( x = -2 \)
- Cross at \( x = 0 \)
- Start high on the left and end low on the right.
Among the options given, select the graph that matches these characteristics to conclude.