Zapisz wzór funkcji w postaci kanonicznej; y= x2 + 6x+ 9
y= x² + 6x+ 9
p=-b/2a
p= -6/2=-3
q=-Δ/4a
Δ=6²-4*1*9=0
q=0
y=a(x-p²)+q
y=(x+3)²
y = x^2 + 6x + 9
y = a(x-p^2 + q
a = 1
p = -b/2a = -6/2
p = -3
q = D(delta)/4a
D = b^2 - 4ac = 36-36 = 0
q = 0
Postać kanoniczna:
y = a(x-p)^2 + q
y = (x+3)^2
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
y= x² + 6x+ 9
p=-b/2a
p= -6/2=-3
q=-Δ/4a
Δ=6²-4*1*9=0
q=0
y=a(x-p²)+q
y=(x+3)²
y = x^2 + 6x + 9
y = a(x-p^2 + q
a = 1
p = -b/2a = -6/2
p = -3
q = D(delta)/4a
D = b^2 - 4ac = 36-36 = 0
q = 0
Postać kanoniczna:
y = a(x-p)^2 + q
y = (x+3)^2