wskaż funkcję kwadratową, której miejsca zerowe to -2 i 3. A. y= -x^2-5x+6 B. y= x^2+x-6 C. y=x^2-5x+6 D. y=x^2-x-6   (potrzebuje dokładnego rozw.)

  A. y= -x²-5x+6 -x²-5x+6 = 0 Δ= b² - 4ac = (-5)² - 4*(-1)*6 = 25 + 24 = 49 √Δ = √49 = 7 x₁ = (-b-√Δ)/2a = (5-7)/-2 = -2/-2=1 x₂ = (-b+√Δ)/2a = (5+7)/-2 = 12/-2=-6   B. y= x²+x-6  x²+x-6 = 0 Δ= b² - 4ac = 1² - 4*1*(-6) = 1 + 24 =25 √Δ = √25 = 5 x₁ = (-b-√Δ)/2a = (-1-5)/2 = -6/2 =- 3 x₂ = (-b+√Δ)/2a = (-1+5)/2 = 4/2 = 2   C. y=x²-5x+6 x²-5x+6 = 0 Δ= b² - 4ac =(- 5)² - 4*1*6 = 25 - 24 = 1 √Δ = √1 =1 x₁ = (-b-√Δ)/2a = (5-1)/2 =  4/2 = 2 x₂ = (-b+√Δ)/2a = (5+1)/2 =  6/2 =3   D. y=x²-x-6 x²-x-6=0 Δ= b² - 4ac = (-1)² - 4*1*(-6) = 1 + 24 = 25 √Δ = √25= 5 x₁ = (-b-√Δ)/2a = (1-5)/2 = -4/2 = -2 x₂ = (-b+√Δ)/2a = (1+5)/2 = 6/2 = 3   Odp : D