Question

Sa se determine functia f de gradul al doilea daca :
a)f(1)=0, f(0)=1, f(-1)=8.
b) f(3 pe 2)=f(-1 pe 3)=0,f(1)=-4

Answer 1

a) f(x)=a*x²+b*x+c
f(0)=1 => c=1
f(1)=0 => a+b+c=0 => a+b=-1
f(-1)=8 => a-b+c=8 => a-b =7
Adunam cele 2 ecuatii si aflam: 2a=6 => a=3
b=-1-a => b=-1-3 => b=-4
Deci f(x)=3x²-4x+1

b) f(x)=a*x²+b*x+c
f(3/2)=0=> 9a/4+3b/2+c =0 => 9a+6b+4c=0
f(-1/3)=0 => a/9-b/3+c=0 => a-3b+9c=0
f(1)=-4 => a+b+c=-4

a-3b+9c=0 => a=3b-9c si inlocuim in celelalte 2 ecuatii:
9(3b-9c)+6b+4c=0 => 27b-81c+6b+4c=0 => 33b-77c=0 => 3b-7c=0
3b-9c+b+c=-4 => 4b-8c=-4 => b-2c=-1 => b=2c-1
Inlocuim in prima ecuatie si avem: 3(2c-1)-7c=0 => 6c-3-7c=0 => -c-3=0 => c=-3
b=2c-1 => b=-6-1 => b=-7
a=3b-9c => a= 3*(-7) - 9*(-3) => a=27-21 => a=6

Deci f(x)=6x²-7x-3