Use the image to answer the question.

Question

A plotted curve with 3 inflection points is labeled m left parenthesis x right parenthesis. The plot starts in quadrant 2, descends to the x-axis, rises to a turning point in quadrant 1, descends to a turning point in quadrant 4 and rises through quadrant 1.

Consider the graph of m(x). Given that (x+1)
is a factor of m(x), which of the following could be the multiplicity of (x+1)
?

Option #1: −1


Option #2: 0

Option #3: 1

Option #4: 2

(1 point)
Option #
gives a possible multiplicity of (x+1)
.

Answer 1

To determine the possible multiplicity of the factor \((x+1)\) in the polynomial \(m(x)\), we can analyze the behavior of the graph at the point \(x = -1\).

Here are the key points to consider regarding the multiplicity of a factor:

• If \((x + 1)\) has an odd multiplicity, the graph will cross the x-axis at \(x = -1\).
• If \((x + 1)\) has an even multiplicity, the graph will touch the x-axis at \(x = -1\) but will not cross it.

From your description of the graph:

1. The graph starts in quadrant 2, which means it is above the x-axis initially.
2. It then descends to the x-axis, indicating it crosses the x-axis (x = -1) at this point.
3. Given that the graph ascends after reaching the x-axis at some point, it suggests that this location is indeed a crossing point rather than just touching.

Since the factor \((x + 1)\) results in crossing the x-axis at \(x = -1\), this corresponds to an odd multiplicity. Possible odd multiplicities include 1, 3, 5, etc.

Among the options given:

• Option #1: \(-1\) (not valid)
• Option #2: \(0\) (not valid)
• Option #3: \(1\) (valid; odd multiplicity)
• Option #4: \(2\) (not valid; even multiplicity)

Thus, the only option that gives a possible multiplicity of \((x + 1)\) is:

Option #3: 1.