A polynomial function has a root of –5 with multiplicity 3, a root of 1 with multiplicity 2, and a root of 3 with multiplicity 7. If the function has a negative leading coefficient and is of even degree, which statement about the graph is true?

A. The graph of the function is positive on (-infinity, –5).
B. The graph of the function is negative on (–5, 3).
C. The graph of the function is positive on (-infinity. 1).
D. The graph of the function is negative on (3, infinity).

"The graph of the function is negative on (3, infinity)" is the statement among the choices given in the question that is true about the graph. The correct option among all the options that are given in the question is the fourth option or option "D". I hope that the answer has actually come to your help.

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DelcieRiveria DelcieRiveria · Answer 2 · 2018-10-17T15:03:29+02:00

Answer: The correct option is D, i.e., The graph of function is negative on [tex](3,\infty)[/tex].

Explanation:

If a polynomial have a root p with multiplicity q, then [tex](x-p)^q[/tex] is a factor of the polynomial.

It is given polynomial function has a root of –5 with multiplicity 3, a root of 1 with multiplicity 2, and a root of 3 with multiplicity 7. The leading coefficient is negative.

From the given information the polynomial is,

[tex]P(x)=-(x+5)^3(x-1)^2(x-3)^7[/tex]

The graph of the polynomial shown the figure.

Since the leading coefficient is negative, therefore [tex]p(x)\rightarrow -\infty[/tex] as [tex]x\rightarrow -\infty[/tex] and [tex]p(x)\rightarrow -\infty[/tex] as [tex]x\rightarrow \infty[/tex].

The 3 roots divides the number line in 4 intervals [tex](-\infty,-5),(-5,1),(1,3),(3,\infty)[/tex].

From figure it is noticed that the graph of the function is negative on [tex](3,\infty)[/tex].