Răspunsuri

2014-05-19T18:47:24+03:00
R,f(x)=2x²-5x+8
Deci:
f(1)=2-5+8
f(1)=5=f(3)
f(2)=2·2²-2·5+8
f(2)=8-10+8
f(2)=-2+8
f(2)=6 =>f(2) ≠2
2014-05-19T23:43:02+03:00
Fie f=a x^{2} +bx+c
f(1)=a+b+c=1,
   f(3)=9a+3b+c=1,f(2)=4a+2b+c

De aici rezolvi sistemul cu Cramer