Question

3 de la testul 3, va rog

Answer 1

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☺Testul 3, exercitiul 3.

f : R → R

f(x) = -3x² + 8x - 4

f(x) ∩ Ox ⇒ y = 0 ⇒ f(x) = 0

-3x² + 8x - 4 = 0

a = -3  ;  b = 8  ;  c = -4

Δ = b² - 4ac =  8² - 4 · (-3) · (-4) = 64 - 48 = 16 = 4²

x₁,₂ = ( -b ± √Δ ) / 2a

x₁ = ( -8 + 4 ) / -6 = -4 / -6 = 2 / 3  ⇒ A ( 2/3 , 0 )

x₂ = ( -8 - 4 ) / -6 = -12 / -6 = 2  ⇒ B ( 2 , 0 )

f(x) ∩ Oy ⇒ x = 0 ⇒ f(0) = -3 · 0² + 8 · 0 - 4 = -4  ⇒ C ( 0 , -4 )

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