To determine which equation of \( m(x) \) corresponds to the given description of the curve with two turning points, we need to analyze the behavior of polynomial functions based on their roots (or factors).
The graph begins in quadrant 3, rises to a turning point in quadrant 2, drops to a turning point in quadrant 4, and finally rises through quadrant 1. This suggests that the following must hold:
- The leading term of the polynomial must be positive (as it ultimately rises in quadrant 1).
- There are two turning points, indicating that the polynomial must be of degree 3 or higher. A degree 3 polynomial would typically have either one inflection point or two turning points.
- The x-intercepts of the polynomial (roots) can give us insight into the turning points.
Analyzing the given options:
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m(x)=(x−2)(2x+1)(x+3): This is a degree 3 polynomial with three real roots. The roots are \( x = 2 \), \( x = -\frac{1}{2} \), and \( x = -3 \). This could potentially create a vertical rise and drop pattern, but it has three turning points, which does not fit our description well.
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m(x)=(x+2)(2x−1)(x−3)(x−6): This is a degree 4 polynomial with four real roots, meaning there would be at least three turning points. This does not align with the requirement of only two turning points.
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m(x)=(x+2)(2x−1)(x−3): This is also a degree 3 polynomial. The roots are \( x = -2 \), \( x = \frac{1}{2} \), and \( x = 3 \). This could create a valid pattern. The nature of the roots (one positive, one negative) along with the behavior you described could fit the turning point behavior, depending on their exact positional relationships.
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m(x)=(x−2)(2x+1)(x+3)(x+6): This is a degree 4 polynomial with four roots, and thus it would have at least three turning points. Hence, it does not match our criteria.
From these analyses, the best fitting option that matches the description of the curve with two turning points is:
m(x)=(x+2)(2x−1)(x−3).
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