Problem 29

Question

In Exercises \(25-32,\) 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. $$f(x)=x^{3}-2 x^{2}+x$$

Step-by-Step Solution

Verified
Answer
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. At \(x=0\), the graph crosses the x-axis and at \(x=1\), the graph touches the x-axis and turns around.
1Step 1: Factoring the Polynomial
Start by factoring the given polynomial function: \(f(x)=x^{3}-2 x^{2}+x\). Factor out the greatest common factor which is \(x\) : \[x(x^{2}-2x+1)\]
2Step 2: Solve for Zeroes
Set each factor equal to zero and solve for \(x\): \(x=0\) and \(x^{2}-2x+1=0\). The quadratic factor can be factored further to \( (x-1)^2 = 0\), where the roots are \(x=0\) and \(x=1\)
3Step 3: Determine Multiplicity and Graph Behavior
The root \(x=0\) occurs only once, so its multiplicity is 1, which means the graph will cross the x-axis at \(x=0\). The root \(x=1\) occurs twice, so its multiplicity is 2. When the multiplicity is even, the graph touches the x-axis and turns around at \(x = 1\).