Question

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Answer 1

Răspuns:

Explicație pas cu pas:

f:R --> R ; f(x) = ax²+bx+c ; a≠0

f(2) = 2  => f(2) = 4a+2b+c

f(1) = -1  => f(1) = a -b +c

{4a+2b+c = 2  

{a -b +c = -1  I·2  =>

b = a+c+1

{4a+2b+c = 2

{2a-2b+2c = -2

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6a+3c = 0   I:3  => 2a + c = 0

c = -2a

a = 1 ; c = -2  ; b = 0  => f(x) = x²-2

a = 2 ; c = -4 ; b = -1  => f(x) = 2x²-x -4

a = 3 ; c = -6 ; b = -2 => f(x) = 3x²-2x-6

a = 4 ; c = -8 ; b = -3 => f(x) = 4x²-3x-8

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a = k ; c = -2k ; b = 1-k => f(x) = kx²+(1-k)x -2k