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Alex787 [66]
3 years ago
12

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, a

nd 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).

Mathematics
1 answer:
dimaraw [331]3 years ago
5 0

Answer:

The graph of the function is negative on (-infinity, 0)

Step-by-step explanation:

f(x) = (x - 6)(x + 2)³(x + 0)²(x - 4)³     This is the equation you described but put in equation form.

The answer I put is true because if the point (-infinity, 0) is on the line, the line is going to be negative because it reaches to -infinity on the x-axis.

I graphed this equation on the graph below.

If this answer is correct, please make me Brainliest!

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