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Chapter 2: Problem 29

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. $$f(x)=x^{3}-2 x^{2}+x$$

Short Answer

Expert verified
The zeros of the function \(f(x) = x^{3}-2 x^{2}+x\) are \(x=0\), with multiplicity 1, and \(x=1\), with multiplicity 2. The graph crosses the x-axis at \(x=0\) and touches the x-axis at \(x=1\) and turns around.

Step by step solution

01

Find the zeros of the function

To find the zeroes of the function, we set \(f(x) = 0\), thus \(x^{3}-2 x^{2}+x=0\). The equation can be simplified by factoring out \(x\), giving \(x(x^{2}-2x + 1) = 0\). Therefore, \(x=0\) or \(x^{2}-2x + 1 = 0\). Solving for the quadratic \(x^{2}-2x + 1 = 0\) gives \(x = 1\). So the zeros of the function are \(x=0\) and \(x=1\).
02

Determine the multiplicity

The zeros of the function are their corresponding factors. \(x=0\) corresponds to the factor \(x\) and \(x=1\) corresponds to the factor \((x-1)^{2}\). The power of these factors (1 for \(x\) and 2 for \((x-1)\)) is called the multiplicity of the zeros. Therefore, zero \(x=0\) has multiplicity 1, and zero \(x=1\) has multiplicity 2.
03

Determine the behavior at each zero

If the multiplicity is odd, the graph crosses the x-axis at the zero. If it's even, the graph touches the x-axis at the zero and turns around. Therefore, the graph of the function \(f(x) = x(x-1)^{2}\) crosses the x-axis at the zero \(x=0\), and touches the x-axis at the zero \(x=1\) and turns around since the multiplicity is 2.

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