Oznacza to,że W( 1,0) z postaci kanonicznej: y= −3(x−1)2 +0 = −3( x2 −2x +1) = −3x2 +6x −3 b= 6 c= −3
x₀=1 P(1,0) Δ=0
y=-3x²+bx+c
x₀=-b/2a
1=-b/-6
-6=-b
b=6
a=-3 b=6 c=?
P(1,0)
0=-3+6+c
c=-3
y=-3x²+6x-3
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Oznacza to,że W( 1,0) z postaci kanonicznej: y= −3(x−1)2 +0 = −3( x2 −2x +1) = −3x2 +6x −3 b= 6 c= −3
x₀=1 P(1,0) Δ=0
y=-3x²+bx+c
x₀=-b/2a
1=-b/-6
-6=-b
b=6
a=-3 b=6 c=?
P(1,0)
0=-3+6+c
c=-3
y=-3x²+6x-3