SOLUTION: 1)degree 3 with a zero 2(of multiplicity 3) such that f(3)=-4
2) degree 3 with a zeros -2(of multiplicity 2) and -4 such that f(-5)=10
3) degree 4 with zeros 1 and 3 (of multipli
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: 1)degree 3 with a zero 2(of multiplicity 3) such that f(3)=-4
2) degree 3 with a zeros -2(of multiplicity 2) and -4 such that f(-5)=10
3) degree 4 with zeros 1 and 3 (of multipli
Log On
Question 916520: 1)degree 3 with a zero 2(of multiplicity 3) such that f(3)=-4
2) degree 3 with a zeros -2(of multiplicity 2) and -4 such that f(-5)=10
3) degree 4 with zeros 1 and 3 (of multiplicity 3) such that f(5)=16 Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! 1. y = a(x-2)^3 P(3,-4)
-4 = a (1)^3
y = -4(x-2)^3
...
2. y = a(x+2)^2(x+4) P(-5,10)
10 = a(-3)^2(-1)
10/-9 = a
y = -10/9(x+2)^2(x+4)
...
3.y = a(x-1)(x-3)^3 P(5,16)
16= a(4)(2)^3
16/32 = a
y = (1/2)(x-1)(x-3)^3
