A polynomial function has a root of –6 with multiplicity 1, a root of –2 with multiplicity 3, a root of 0 with multiplicity 2, and a root of 4 with multiplicity 3. If the function has a positive leading coefficient and is of odd degree, which statement about the graph is true?

because the leading coefficient is positive and the highest degree is odd (9) the function is negative for (-inf, -6). Then it crosses the -6 root to become positive until it crosses again the root -2. (it crosses because both roots have an odd multiplicity). then it touches on root 0 (does not cross due to an even multiplicity there), so function remains negative from -2 till root 4 where it crosses into the positive.

Given the above, only the first choice is a correct statement.

MrRoyal MrRoyal · Answer 2 · 2021-11-08T15:41:46+01:00

The true statement about the graph is: (a) the graph of the function is positive on [tex]\mathbf{(-6,-2)}[/tex]

The given parameters are:

[tex]\mathbf{Root = -6, Multiplicity = 1}[/tex]

[tex]\mathbf{Root = -2, Multiplicity = 3}[/tex]

[tex]\mathbf{Root = 0, Multiplicity = 2}[/tex]

[tex]\mathbf{Root = 4, Multiplicity = 3}[/tex]

Add up the multiplicities, to calculate the degree of the polynomial function

[tex]\mathbf{Degree = 1 + 3 + 2 + 3}[/tex]

[tex]\mathbf{Degree = 9}[/tex] --- odd number

The interval of the first root is:

[tex]\mathbf{Interval = (-\infty,-6)}[/tex]

The interval of the second root is:

[tex]\mathbf{Interval = (-6,-2)}[/tex]

A polynomial function with an odd degree, and a positive leading coefficient will start with negative values, until it reaches the smallest root, before it switched to positive values.

This means that, the function increases at: [tex]\mathbf{Interval = (-6,-2)}[/tex]

Hence, option (a) is correct.

Read more about polynomial functions at:

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