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bonjour

f(x)= ( x - 3 )² + 2

f (x) = x² - 6 x + 9 + 2

f(x) = x² - 6 x + 11

x² - 6 x + 11 = 0

Δ = ( - 6 )² - 4 ( 1 * 11 ) = 36 - 44 = - 12

Δ < 0 donc pas de solution

x² - 6 x + 11 = 6

x² - 6 x + 11 - 6 = 0

x² - 6 x + 5 = 0

Δ = 36 - 20 = 16 = 4 ²

x 1 = ( 6 - 4 ) /2 = 2/2 = 1

x 2 = ( 6 + 4 ) /2 = 10/2 = 5

antécédents = 1 et 5

tu fais pareil pour 8

g (x) = - 2 ( x - 1 )² + 8

g (x) = - 2 ( x² - 2 x + 1 ) + 8

g ( x) = - 2 x² + 4 x - 2 + 8

g (x) = - 2 x² + 4 x + 6

- 2 x² + 4 x + 6 = 0

Δ = 16 - 4 ( - 2 *6 ) = 16 + 48 = 64 = 8 ²

x 1 = ( - 4 - 8 ) / - 4 = - 12 / - 4 = 3

x 2 = ( - 4 + 8 ) / - 4 = 4 / - 4 = - 1

tu fais pareil pour les autres

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