Răspuns :
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
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
[tex]Fie f=a x^{2} +bx+c[/tex]
[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
[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