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Determinati f:R->R daca f(x)=ax^2+bx+c si f(0)=6,f(-1)=15,f(1)=1

Răspuns :

din f(0)=6⇒c=6
din f(-1)=15⇒a-b+c=15
din f(1)=1⇒a+b+c=1
rezolvam sistemul[tex] \left \{ {{a-b+6=15} \atop {a+b+6=1}} \right. \left \{ {{2b=-14} \atop {a=-5-b}} \right. \\ \left \{ {{b=-7} \atop {a=2}} \right. [/tex]
deci f(x)=[tex]2 x^{2} -7x+6[/tex]
f(0)=6⇒c=6
 f(-1)=15⇒a-b+c=15
 f(1)=1⇒a+b+c=1
l
 f(x)=