Question: Consider the following polynomial function. f(x) = (x + 3)^2(x - 2)^3(x - 1) Step 3 of 3: Find…

Consider the following polynomial function.

f(x) = (x + 3)^2(x - 2)^3(x - 1)

Step 3 of 3: Find the zero(s) at which f “flattens out”. Express the zero(s) as ordered pair(s).

Select the number of zero(s) at which f “flattens out”.

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none 1 2 3 4

Solution

To determine the zero(s) at which \( f \) “flattens out,” we need to identify the roots of the polynomial function that have even multiplicities. A zero with an even multiplicity indicates that the graph of the function touches the x-axis without crossing it, which is what is meant by “flattening out.” Given the polynomial function: \[ f(x) = (x + 3)^2 (x - 2)^3 (x - 1) \] Step 1: Identify the Zeros and Their Multiplicities The zeros of the function are the values of \( x \) that make \( f(x) = 0 \). These are found by setting each factor equal to zero: \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] Step 2: Determine the Multiplicity of Each Zero The multiplicity of a zero is the exponent of the corresponding factor in the polynomial: \[ \begin{cases} (x + 3)^2 \quad \Rightarrow \quad \text{Multiplicity of } -3 \text{ is } 2 \\ (x - 2)^3 \quad \Rightarrow \quad \text{Multiplicity of } 2 \text{ is } 3 \\ (x - 1) \quad \Rightarrow \quad \text{Multiplicity of } 1 \text{ is } 1 \\ \end{cases} \] Step 3: Identify Zeros with Even Multiplicity A zero with an even multiplicity will cause the graph to touch the x-axis without crossing it: \[ -3 \quad (\text{Multiplicity } 2) \quad \text{is even} \] \[ 2 \quad (\text{Multiplicity } 3) \quad \text{is odd} \] \[ 1 \quad (\text{Multiplicity } 1) \quad \text{is odd} \] Step 4: Express the Flattening Zeros as Ordered Pairs Only \( x = -3 \) has an even multiplicity. Therefore, the zero where \( f \) “flattens out” is: \[ (-3, 0) \] Step 5: Answer to the Multiple-Choice Question There is 1 zero at which \( f \) “flattens out.”