Question

znajdz postać kanoniczną podanych funkcji:

y= 2x2 -5x+7

y= x2 -4x+8

y= - 2x2+4x+9

y= 3x2-4x

Answer 1

y = 2 x^2 - 5 x + 7

a = 2 , b = - 5, c = 7

p = - b/(2a) = 5/4

delta = b^2 - 4ac = ( -5)^2 - 4*2*7 = 25 - 56 = - 31

q = - delta/ (4a) = 31/8

y = a*(x - p)^2 + q

czyli odp>

y = 2*( x - 5/4)^2 + 31/8

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y = x^2 - 4 x + 8

 

p = 4/2 = 2

delta = ( -4)^2 - 4*1*8 = 16 - 32 = - 16

q = 16/4 = 4

zatem

Odp.  y = ( x - 2)^2 + 4

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y = - 2 x^2 + 4 x + 9

 

p = - 4 /( -4) = 1

delta = 4^2 - 4*(-2)*9 = 16 + 72 = 88

q = - 88/( -8) = 11

Odp.  y = -2*( x - 1)^2 + 11

==============================

y = 3 x^2 - 4 x

 

p = 4/6 = 2/3

delta = ( -4)^2 - 4*3*0 = 16

q = - 16/12 = 4/3

Odp. y = 3*( x - 2/3)^2  + 4/3

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Answer 2

y= 2x2 -5x+7

delta=25-56=-31

p=-b/2a=5/4

q=-delta/4a=31/8

y=a(x-p)^2+q

y=2(x-5/4)^2+31/8

 

y= x2 -4x+8

delta=16-32=-16

p=4/2=2

q=16/4=4

y=(x-2)^2+4

 

y= - 2x2+4x+9

delta=16+72=88

p=-4/-4=1

q=-86/-8=11

y=-2(x-1)^2+11

 

y= 3x2-4x

delta=16-0=16

p=4/6=2/3

q=-16/12=-4/3

y=3(x-2/3)^2-4/3