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Chapter 2: Problem 25

Find the zeros for each polynomial function and give the multiplicity for each zero. State whether the graph crosses the \(x\) -axis, or touches the \(x\) -axis and turns around, at each zero. $$f(x)=2(x-5)(x+4)^{2}$$

Short Answer

Expert verified
The polynomial function \(f(x) = 2(x-5)(x+4)^{2}\) has zeros at \(x = 5\) and \(x = -4\). The zero 5 has a multiplicity of 1 and the graph crosses the x-axis at this point. The zero -4 has a multiplicity of 2 and the graph touches the x-axis at this point and turns around.

Step by step solution

01

Identify and state the zeros

To identify the zeros, set the function \(f(x)\) equal to zero and solve for \(x\). For \(f(x) = 2(x-5)(x+4)^{2} = 0\), the zeros can be found by setting each factor equal to zero. Thus, \(x-5 = 0\) yielding \(x = 5\) and \(x+4 = 0\) yielding \(x = -4\). Therefore, the zeros of the function are \(x = 5\) and \(x = -4\).
02

Determine the multiplicity of the zeros

The multiplicity of a zero refers to the power of the factor in the function's factorization. In \(f(x) = 2(x-5)(x+4)^{2}\), the factor \((x-5)\) has a power of 1, so the multiplicity of the zero 5 is 1. The factor \((x+4)\) has a power of 2, so the multiplicity of the zero -4 is 2.
03

Determine whether the graph crosses or touches the x-axis at each zero

The behavior of the graph at a zero is determined by the zero's multiplicity. If the multiplicity is odd, the graph crosses the x-axis at that point. If the multiplicity is even, the graph touches the x-axis at that point and turns around. Because the zero 5 has a multiplicity of 1, which is odd, the graph crosses the x-axis at \(x = 5\). Because the zero -4 has a multiplicity of 2, which is even, the graph touches the x-axis at \(x = -4\) and turns around.

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