Question

Sa se determine functia de gradul 2
f(-2)=11 , f(1)=11, f(3)=6

Answer 1

fie f(x) = ax^2 +bx +c
4a-2b+c=11
a+b+c=11
9a+3b+c=6⇒ c= 6-9a-3b si inlocuim in primele doua relatii:
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4a-2b+6-9a-3b=11
a+b+6-9a-3b=11
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-5a - 5b  = 5 ⇔ a+b = -1 ⇒ b= -a-1
-8a - 2b = 5
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-8a +2a+2 = 5
-6a = 3
a= -1/2
b= 1/2  -  1 = -1/2
c = 6-9a-3b = 6 + 9/2 + 3/2 = 6+6 =12
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f(x) = -x^2 / 2 - x/2 + 12
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Verificare: f(1)= -1/2  - 1/2 +12 = -1+12=11 OK