Question

Zapisz w postaci kanonicznej :

a) y=4(x-2)(x+1)

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

c) y=6x(x+1)

na dziś.

Answer 1

a]

y=4(x-2)(x+1)

y=4(x²+x-2x-2)

y=4x²-4x-8

Δ=b²-4ac=16+128=144

p=-b/2a=4/8=½

q=-Δ/4a=-144/16=-9

y=4(x-½)²-9

b]

y=-x²+2x-1

Δ=4-4=0

p=-2/-2=1

q=0/-4=0

y=-(x-1)²

c]

y=6x(x+1)

y=6x²+6x

Δ=36

p=-6/12=-½

q=-36/24=-1,5

y=6(x+½)²-1,5

Answer 2

a) y= 4(x-2)(x+1)

  y= (4x-8) (x+1)

 y= 4x^2 + 4x - 8x - 8

y= 4x^2 - 4x-8

 

delta= 16 - 4 * 4 * -8

delta= 144

 

p= 4/8= 1/2

q= -144/16 = -9

 

y= 4( x-1/2)^2 - 9

 

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

delta= 4- 4 * -1 * -1

delta= 0

 

p= -2/-2 = 1

q= 0/4= 0

 

y= -(x-1)^2

 

 

c) y= 6x(x+1)

y= 6x^2 + 6

 

delta= 36- 4*6*0

delta= 36

 

p= -6/12 = -1/2

q= -36/24 = -6/4 = -1 i 1/2

 

y=  6(x+ 1/2)^2 - 6/4