Final answer:
The correct statement about the graph is that the graph of the polynomial function is negative on the interval (3, ∞) because the function has an even degree, a negative leading coefficient, and the final root at 3 has an odd multiplicity causing a sign change.
Step-by-step explanation:
The polynomial function is described as having a root at – 5 with multiplicity 3, a root at 1 with multiplicity 2, and a root at 3 with multiplicity 7. The function has a negative leading coefficient and is of even degree. Because the function has an even degree and a negative leading coefficient, it will open downwards, and the ends of the graph will point downwards. Since the leading coefficient is negative, as we move from left to right on the graph, every time we pass through a root with an odd multiplicity, the sign of the function value will change. When the multiplicity is even, the sign of the function value will not change.
Here is how the function behaves around each root:
- At the root – 5 (odd multiplicity of 3), the graph will cross the x-axis, causing a sign change in the function value.
- At the root 1 (even multiplicity of 2), the function's graph will just touch the x-axis and turn back without changing the sign of the function value.
- Finally, at the root 3 (odd multiplicity of 7), the graph will again cross the x-axis, resulting in another sign change in the function value.
Now, analyzing the intervals:
- Between – 5 and 1, since the sign changed at – 5 and there is no sign change at 1, the graph is positive on this interval.
- Between 1 and 3, since there is no sign change at 1, the function maintains its positivity until it reaches the next root at 3.
- Since the function is positive at 1 and has no sign change until 3, and then the sign changes at 3 (due to odd multiplicity), the graph of the function must be negative on (3, ∞).
This means the correct statement about the graph is that the graph of the function is negative on (3, ∞), which corresponds to option 2.