Please solve this---**Question Number:** 11 **Graph Description:** * Type: Graph of a function on a Cartesian coordinate system. * Coordinate Axes: X-axis and Y-axis intersect at the origin O (0,0). The X-axis has ticks labeled at -3, -2, -1, 1, 2, 3. The Y-axis has ticks labeled at -10 and 10. * Curve: A smooth curve that crosses the X-axis at approximately x = -3 and x = 2, and touches the X-axis at x = 0. The curve goes through the origin (0,0). The curve has a local maximum in the second quadrant (between x = -2 and x = -1) and a local minimum in the fourth quadrant (between x = 1 and x = 2). As x approaches negative infinity, the curve goes upwards (towards positive infinity). As x approaches positive infinity, the curve goes upwards (towards positive infinity). **Question Stem:** Which of the following could be the equation of the graph above? **Options:** A) y = x(x - 2)(x + 3) B) y = x^2(x - 2)(x + 3) C) y = x(x + 2)(x - 3) D) y = x^2(x + 2)(x - 3)

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Let's analyze this graph to determine its equation. The graph has x-intercepts at negative 3, 0, and 2. Notice that at x equals 0, the curve touches the x-axis but doesn't cross it, indicating a root with even multiplicity. At x equals negative 3 and x equals 2, the curve crosses the x-axis, indicating roots with odd multiplicity. Also, as x approaches either positive or negative infinity, y approaches positive infinity, suggesting an even-degree polynomial with a positive leading coefficient. Now let's examine the four options. Option A is y equals x times x minus 2 times x plus 3. This has the correct roots, but since the degree is 3, which is odd, the end behavior would be different from our graph. Option B is y equals x squared times x minus 2 times x plus 3. This has roots at negative 3, 0, and 2, with x equals 0 having multiplicity 2. The degree is 4, which is even, and the leading coefficient is positive. This matches our graph perfectly. Options C and D have different roots: 0, negative 2, and 3, which don't match our graph. Therefore, option B is the correct answer. Let's visually compare all four functions. The black curve represents option B, which matches our original graph. The blue curve is option A, which has the same roots but different end behavior because it's an odd-degree polynomial. Notice how it goes up on one side and down on the other. The green and red curves represent options C and D, which have completely different roots at 0, negative 2, and 3. These clearly don't match our original graph. This visual comparison confirms that option B is the only equation that could represent the given graph. Let's analyze the key characteristics of our correct answer, option B. This is a 4th-degree polynomial with a positive leading coefficient of 1. In factored form, it's x squared times x minus 2 times x plus 3, which expands to x to the 4th plus x cubed minus 6x squared. The roots are at x equals negative 3, 0, and 2, with x equals 0 having multiplicity 2. This even multiplicity at x equals 0 explains why the graph touches but doesn't cross the x-axis there. The even degree with positive leading coefficient explains why both ends of the graph point upward as x approaches positive or negative infinity. All these characteristics perfectly match the graph in the question. To summarize, when identifying a polynomial equation from a graph, we need to analyze several key features: the x-intercepts or roots and their multiplicities, the end behavior as x approaches positive and negative infinity, and the local maxima and minima. For this problem, we determined that option B, y equals x squared times x minus 2 times x plus 3, is the correct answer because it has the correct roots at x equals negative 3, 0, and 2. The root at x equals 0 has multiplicity 2, which explains why the graph touches but doesn't cross the x-axis there. The roots at x equals negative 3 and x equals 2 have multiplicity 1, which is why the graph crosses the x-axis at those points. The polynomial has degree 4, which is even, with a positive leading coefficient, explaining why both ends of the graph point upward. All these characteristics perfectly match the given graph, confirming that option B is the correct answer.