Question

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?
The graph of the function is positive on (–6, –2).
The graph of the function is negative on (infinity, 0).
The graph of the function is positive on (–2, 4).
The graph of the function is negative on (4, infinity)

Answer 1

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.

Answer 2

\(\color{#0cbb34}{\text{Originally Posted by}}\) @daddsu
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.
\(\color{#0cbb34}{\text{End of Quote}}\)
e