Question

Sa se deterimine functia de gradul al 2-lea , stiind ca verifica conditiile : 
f(1)=f(3)=1
f(2)=2

Answer 1

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

Answer 2

$Fie f=a x^{2} +bx+c$
[tex]f(1)=a+b+c=1, f(3)=9a+3b+c=1,f(2)=4a+2b+c [/tex]
De aici rezolvi sistemul cu Cramer