Consider the graph of p(x). Given that (x−3) is a factor of p(x), is the multiplicity of (x−3) even or odd? Why?
(1 point)
Responses
The multiplicity of (x−3) is odd because the associated zero, x=3, is odd.
The multiplicity of left parenthesis x minus 3 right parenthesis is odd because the associated zero, x equals 3 , is odd.
The multiplicity of (x−3) is even because the graph of p(x) touches the x-axis at (3,0) and then turns around.
The multiplicity of left parenthesis x minus 3 right parenthesis is even because the graph of p ( x ) touches the x -axis at left parenthesis 3 comma 0 right parenthesis and then turns around.
The multiplicity of (x−3) is even because the associated zero, x=3, is odd.
The multiplicity of left parenthesis x minus 3 right parenthesis is even because the associated zero, x equals 3 , is odd.
The multiplicity of (x−3) is odd because the graph of p(x) touches the x-axis at (3,0) and then turns around.
The answer is: The multiplicity of (x−3) is odd because the associated zero, x=3, is odd.
The correct response is:
The multiplicity of (x−3) is odd because the associated zero, x=3, is odd.
To determine whether the multiplicity of (x−3) is even or odd, we need to consider the behavior of the graph of p(x) at the zero x=3.
If the graph touches the x-axis at x=3 and then turns around, without crossing the x-axis again, then the multiplicity of (x−3) is odd. This means the function p(x) has a single factor of (x−3) in its factored form.
If the graph crosses the x-axis at x=3, indicating that the zero x=3 occurs twice or an even number of times, then the multiplicity of (x−3) is even. This means the function p(x) has at least two factors of (x−3) in its factored form.
From the given information, it is stated that (x−3) is a factor of p(x). However, there is no mention of whether the graph touches or crosses the x-axis at x=3. Therefore, we cannot definitively determine whether the multiplicity of (x−3) is even or odd based on the given information.