Apolynomial 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).

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

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

Step-by-step 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).

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.