Question

Se consideră functia f:R->R, f(x)=ax²+bx+c si punctele A(1,16),B(2,9),C(3,4),D(4,1),E(5,0) care apartin graficului functiei. Calculați f(1)•f(2)•f(3)•f(4)•f(5)

Answer 1

A(1,16) => f(1)= 16
Tot f(1)= a+b+c=> a+b+c = 16 => c= 16-a-b
B(2,9) => f(2)= 9
f(2) = 4a+2b+=c => 4a+2b+c = 9 => 4a+2b+16-a-b = 9=> 3a+b = -7=> b= -7-3a
C(3,4) => f(3)=  4
f(3) = 9a+3b+c => 9a+3b+c = 4 => 9a-7-3a+16-a-b = 4=> 5a-b= -5 => b= -5-5a
Dar b= -7-3a => -5-5a=-7-3a => -2a= -2 => a = 1
b= -10
c= 5
f(1)= a+b+c = -10+1+5 =-4
f(2)= 4a+2b+c = 4-20+5 = -11
f(3)= 9a+3b+c = 9-30+5 = 16
f(4)= 16a+4a+c = 16-40+5= 19
f(5)= 25a+5a+c= 25-50+5= -20
f(1)•f(2)•f(3)•f(4)•f(5)= -4*-11*16*19*-20=-267.520