przedstaw funkcję f(x)=x (do potęgi 2) -x-2 w postaci kanonicznej i iloczynowej. proszę o pomoc
f(x)=2x²-3x+1
Δ= b² - 4ac = (-3)² - 4*2*1 = 9 -8 = 1
√Δ = √1 = 1
x₁ = (-b-√Δ)/2a = (3-1)/ 4 = 2/4 = 0,5
x₂ = (-b+√Δ)/2a = (3+1)/ 4 = 4/4 = 1
postać iloczynowa
Δ>0
y = a (x -x₁) (x -x ₂)
y = 2(x- 0,5)( x -1)
kanoniczna :
y = a(x-p)² + q
p = -b/2a
q = -Δ/4a
p = 3/4
q = -1/8
y = 2 (x - 3/4 )² -1/8
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f(x)=2x²-3x+1
Δ= b² - 4ac = (-3)² - 4*2*1 = 9 -8 = 1
√Δ = √1 = 1
x₁ = (-b-√Δ)/2a = (3-1)/ 4 = 2/4 = 0,5
x₂ = (-b+√Δ)/2a = (3+1)/ 4 = 4/4 = 1
postać iloczynowa
Δ>0
y = a (x -x₁) (x -x ₂)
y = 2(x- 0,5)( x -1)
kanoniczna :
y = a(x-p)² + q
p = -b/2a
q = -Δ/4a
p = 3/4
q = -1/8
y = 2 (x - 3/4 )² -1/8