a polynomial function has a root of -6 with multiplicity 1, a root of -2 with multiplicity 3, a root of 0…
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?\nthe graph of the function is positive on (-6, -2).\nthe graph of the function is negative on (-∞, 0).\nthe graph of the function is positive on (-2, 4).\nthe graph of the function is negative on (4, ∞).
Answer
Explanation:
Step1: Calculate the degree of the polynomial
The degree is the sum of the multiplicities. So (1 + 3+2 + 3=9) (an odd - degree polynomial). Let the polynomial be (f(x)=a(x + 6)(x + 2)^{3}x^{2}(x - 4)^{3}), where (a>0).
Step2: Analyze the sign of the function in each interval
- For (x<-6), all factors ((x + 6)), ((x + 2)), (x), ((x - 4)) are negative. The number of negative factors is (1+3 + 2+3=9) (odd). Since (a>0), (f(x)<0).
- For (-6<x<-2), ((x + 6)>0), and ((x + 2)), (x), ((x - 4)) are negative. The number of negative factors is (3 + 2+3 = 8) (even). So (f(x)>0).
- For (-2<x<0), ((x + 6)>0), ((x + 2)>0), and (x), ((x - 4)) are negative. The number of negative factors is (2+3 = 5) (odd). So (f(x)<0).
- For (0<x<4), ((x + 6)>0), ((x + 2)>0), (x>0), and ((x - 4)) is negative. The number of negative factors is (3) (odd). So (f(x)<0).
- For (x>4), all factors ((x + 6)), ((x + 2)), (x), ((x - 4)) are positive. So (f(x)>0).
Answer:
The graph of the function is positive on ((-6,-2)).