A plotted curve with 4 inflection points is labeled p left parenthesis x right parenthesis. The plot starts in quadrant 3, rises steeply to a turning point in quadrant 2, before dropping and rising twice in quadrant 1.
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 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 odd 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 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 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.
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1 answer
The correct response is:
The multiplicity of (x−3) is odd because the graph of p(x) touches the x-axis at (3,0) and then turns around.
This is the correct choice because when a polynomial has a root with an odd multiplicity, the graph will cross the x-axis at that point (making it an inflection point). If it had even multiplicity, the graph would touch the x-axis and turn around without crossing it, which is inconsistent with the description provided.
