Question

1) Calcule o zero da função f(x)= x^2 -2x -8 (Bhaskara) 

2)Dada f(x)= 6x -4, calcule: f(-2) -f(0) + f(3)

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Answer 1

1)
Fórmula de Bhaskara: [-b ± raiz( b^2-4•a•c)]/2•a

f(x) = x^2 - 2x - 8
a = 1 ; b = -2 e c = -8

[ -(-2) ± raiz( (-2)^2 - 4•1•(-8)]/ 2•1
[ 2 ± raiz ( 4 + 32) ]/ 2
[ 2 ± raiz (36)]/ 2 —> [2 ± 6]/2
x1 = (2 + 6)/2 = 8/2 = 4
x2 = ( 2 - 6)/2 = -4/2 = -2

Raizes => x1 = 4 e x2 = -2

2)
f(x) = 6x - 4
f(-2) = 6(-2) - 4 = -12 - 4 = -16
f(0) = 6(0) - 4 = -4
f(3) = 6(3) - 4 = 18 - 4 = 14

f(-2) - f(0) + f(3) = -16 - 4 + 14 = -20 + 14 = -6

Espero ter ajudado!