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 th function has a negative leading coefficient and is of even degree, which statement about the graph is true?
O The graph of the function is positive on (-0, -5).
O The graph of the function is negative on (-5, 3)!
O The graph of the function is positive on (-0, 1).
O The graph of the function is negative on (3,co).

The graph of the function is negative on (3,0)

Step-by-step explanation:

The limiting behavior of a function describes how a function behaves as x ⇒ ±∞. Its behavior is determined by the degree of a polynomial and the sign of its leading coefficient.

Since the function [f(x)] is an even degree polynomial with negative leading coefficient, then f(x) ⇒ -∞ as x ⇒ ±∞. The graph of the function is negative at the root. Since it has a root of 3, The graph of the function is negative on (3,0)

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

answer: idk

step-by-step explanation: