Question

o grafico de f(x)=x²+bx+c, onde b e c sao constantes, passa pelos pontos (0,0) e (1,2). Entao f(-2/3) vale

Answer 1

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Ola Lukito

f(x) = x² + bx + c

f(0) = c = 0
f(1) = 1 + b = 2

b = 1

f(x) = x² + x 

f(-2/3) = 4/9 - 2/3 = 4/9 - 6/9 = - 2/9

pronto

Answer 2

f(x) = x² + bx + c

f(0) = c = 0

f(1) = 1 + b = 2

b = 1

f(x) = x² + x

f(-2/3) = 4/9 - 2/3 = 4/9 - 6/9 = - 2/9