Odpowiedź :
Napisz wzór funkcji y=-⅔x²-⅓x+⅓ postaci kanonicznej.
p=-b/2a
q=-Δ/4a
a=-⅔, b=-⅓, c=⅓
p=⅓:(-4/3)
p=1/3*(-3/4)
p=-1/4
Δ=(-⅓)²-4*(-⅔)*⅓
Δ=1/9-4*(-2/9)
Δ=1/9+8/9
Δ=1
q=-1:4*(-2/3)
q=-3/8
postać kanoniczna:
y=a(x-p)²+q
y=-⅔(x+1/4)²-3/8
p=-b/2a
q=-Δ/4a
a=-⅔, b=-⅓, c=⅓
p=⅓:(-4/3)
p=1/3*(-3/4)
p=-1/4
Δ=(-⅓)²-4*(-⅔)*⅓
Δ=1/9-4*(-2/9)
Δ=1/9+8/9
Δ=1
q=-1:4*(-2/3)
q=-3/8
postać kanoniczna:
y=a(x-p)²+q
y=-⅔(x+1/4)²-3/8
y=-⅔x²-⅓x+⅓ postaci kanonicznej
y = ax² +bx +c = a[x+b:2a]² - Δ:4a
a = -⅔
b = -⅓
c =⅓
Δ = b² -4ac
Δ = (-⅓)² -4*(-⅔)(⅓) = 1/9 +8/9 =9/9 =1
Δ=1
y=-⅔x²-⅓x+⅓ = -⅔ [ x + (-⅓):2*(-⅔)]² - 1:4*(-⅔)
y=-⅔x²-⅓x+⅓ = -⅔ [ x + (-⅓):(-4/3)]² - 1:(- 8/3)
y=-⅔x²-⅓x+⅓ = -⅔ [ x + (-⅓)*(-¾)]² - 1*(- 3/8)
y=-⅔x²-⅓x+⅓ = -⅔ [ x + ¼]² + ⅜
y = ax² +bx +c = a[x+b:2a]² - Δ:4a
a = -⅔
b = -⅓
c =⅓
Δ = b² -4ac
Δ = (-⅓)² -4*(-⅔)(⅓) = 1/9 +8/9 =9/9 =1
Δ=1
y=-⅔x²-⅓x+⅓ = -⅔ [ x + (-⅓):2*(-⅔)]² - 1:4*(-⅔)
y=-⅔x²-⅓x+⅓ = -⅔ [ x + (-⅓):(-4/3)]² - 1:(- 8/3)
y=-⅔x²-⅓x+⅓ = -⅔ [ x + (-⅓)*(-¾)]² - 1*(- 3/8)
y=-⅔x²-⅓x+⅓ = -⅔ [ x + ¼]² + ⅜
y=-⅔x²-⅓x+⅓
a=-⅔
b=-⅓
c=⅓
postać kanoniczna: y=a(x-p)²+q
p=-b/2a
p=⅓/2*(-⅔)=⅓*(-¾)=-¼
q=-Δ/4a
Δ=b²-4ac=¹/₉+⁸/₉=1
q=-Δ/4a
q=-1/(-⁸/₃)=³/₈
y=-⅔(x+¼)²+³/₈
a=-⅔
b=-⅓
c=⅓
postać kanoniczna: y=a(x-p)²+q
p=-b/2a
p=⅓/2*(-⅔)=⅓*(-¾)=-¼
q=-Δ/4a
Δ=b²-4ac=¹/₉+⁸/₉=1
q=-Δ/4a
q=-1/(-⁸/₃)=³/₈
y=-⅔(x+¼)²+³/₈
