Question

Zapisz wzór funkcji kwadratowej w postaci kanonicznej

a)y= x^2-x-9

b)y=x^2-x+1/4

c)y=x^2-5x+5

d)y=x^2-5x

e)y=x^2+7x

f)y=3x^2-3x-1/4

g)y=-2x^2+4x+5

h)y=4x^2-12x+5

i)y=2x^2+x+1

Answer 1

Aby otrzymać postać kanoniczną, tj. y=a(x-p)² +q:

⇒ liczymy deltę

⇒ otrzymujemy p= -b/2a;  q= -Δ/4a

 

a) y= x²-x-9

a=1, bo (1)*x²

b=-1, bo (-1)x

c=-9

Δ=1-4*1*(-9)=1+36=37

p= 1/2

q= -37/4= -9 i 1/4

 

y=(x-1/2)² - 9 i 1/4

 

b) y=x²-x+1/4

Δ= 1-4*1*1/4=0

p=1/2

q=0

 

y=(x-1/2)²

 

c) y=x²-5x+5

Δ=25-4*1*5=25-20=5

p=5/2

q= -5/4

 

y=(x-5/2)² - 5/4

 

d) y=x²-5x

Δ=25-4*1*0=25

p=5/2

q=-25/4

 

y=(x-5/2)²-25/4

 

e) y=x²+7x

Δ=49-4*1*0=49

p=-7/2

q=-49/4

 

y=(x+7/2)²-49/4

 

f)y=3x²-3x-1/4

Δ=9-4*3*(-1/4)=12

p=3/6=1/2

q=-12/12=-1

 

y=3(x-1/2)²-1

 

g)y=-2x²+4x+5

Δ=16-4*(-2)*5=56

p=-4/-4=1

q=-56/-8=7

 

y=-2(x-1)+7

 

h)y=4x²-12x+5

Δ=(-12)² - 4*4*5=64

p=12/8

q=-64/16

 

y=4(x-12/8)²-64/16

 

i)y=2x²+x+1

Δ=1-4*2*1=1-8=-7

p=-1/4

q=7/8

 

y=2(x+1/4)²+7/8

Answer 2

Δ = b²-4ac

p = -b/2a

q = -Δ/4a

y = a(x-p)²+q

 

a)y= x^2-x-9

a=1 b=-1 c=-9

Δ = (-1)²-4*1*(-9) = 1+36 = 37

p = -½

q = -37/4 = -9¼

y = 1(x+½)²-9¼

 

b)y=x^2-x+¼

a=1 b=-1 c=¼

Δ = (-1)²-4*1*¼ = 1-1 = 0

p = ½

q = 0/4 = 0

y = 1(x-½)²

 

c)y=x^2-5x+5

a=1 b=-5 c=5

Δ = (-5)²-4*1*5 = 25-20 = 5

p = 5/2 = 2½

q = -5/4 = -1¼

y = 1(x-2½)-1¼

 

d)y=x^2-5x

a=1 b=-5 c=0

Δ = (-5)²-4*1*0 = 25

p = 5/2 = 2½

q = -25/4 = -6¼

y = 1(x-2½)-6¼

 

e)y=x^2+7x

a=1 b=7 c=0

Δ = 7²-4*1*0

Δ = 49

p = -7/2 = -3½

q = -49/4 = -12¼

y =1(x+3½)-12¼

 

f)y=3x^2-3x-1/4

a=3 b=-3 c=-¼

Δ=(-3)²-4*3*(-¼) = 9+3 = 12

p = 3/6 = ½

q = -12/12 = -1

y = 3(x-½)-1

 

g)y=-2x^2+4x+5

a=-2 b=4 c=5

Δ = 4²-4*(-2)*5 = 16+40 = 56

p = -4/-4 = 1

q = -56/-8 = 7

y = -2(x-1)+7

 

h)y=4x^2-12x+5

a=4 b=-12 c=5

Δ = (-12)²-4*4*5 = 144-80 = 64

p = 12/8 = 1½

q = -64/16 = -4

y = 4(x-1½)-4

 

i)y=2x^2+x+1

a=2 b=1 c=1

Δ = 1²-4*2*1

Δ = 1-8 = -7

p = -¼

q = 7/8

y = 2(x+¼)+7/8