133k views
0 votes
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?

The graph of the function is positive on (-co, -5).
The graph of the function is negative on (-5,3).
The graph of the function is positive on (-co, 1).
The graph of the function is negative on (3,co).

User Iamsar
by
8.0k points

2 Answers

4 votes
It’s the last one. And the first one... It’s negative on the interval (3,co) and positive on (-co,-5)
A polynomial function has a root of -5 with multiplicity 3, a root of 1 with multiplicity-example-1
User Alficles
by
8.4k points
3 votes

Answer: The graph of the function is positive on (-co, -5).

The graph of the function is negative on (3,co).

Explanation:

We know that the roots are in: -5, 1 and 3.

and after 3, the graph is in the negative side, so between 1 and 3 the graph must be in the positive side, between -5 and 1 the graph must be in the negative side, and between -inifinity and -5 the graph must be in the positive side:

So the statements that are true are:

The graph of the function is positive on (-co, -5).

The graph of the function is negative on (3,co).

User Hiltmon
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories