zapisz wzór funkcji w postaci iloczynowej : y = -2(1-x)² + 1
y = -2(1-x)² + 1
-2(1-x)² + 1 = 0
(1-x)² = 0,5
|1-x| = √0,5 = √2/2
1-x = √2/2 ∨ 1-x = -√2/2
x = 1-√2/2 = (2-√2)/2 ∨ x = √2/2 - 1 = (√2-2)/2
y = 0 <=> x = (√2-2)/2 ∨ x = (2-√2)/2
postać iloczynowa:
y = a(x-x₁)(x-x₂)
y = -2[x-(√2-2)/2][x-(2-√2/2)]
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
y=-2(1-x)²+1
y=-2(1-2x+x²)+1=-2+4x-2x²+1=-2x²+4x-1
-2x²+4x-1=0
Δ=4²-4*(-2)*(-1)=16-8=8
Postać iloczynowa funkcji ma postać:
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y = -2(1-x)² + 1
-2(1-x)² + 1 = 0
(1-x)² = 0,5
|1-x| = √0,5 = √2/2
1-x = √2/2 ∨ 1-x = -√2/2
x = 1-√2/2 = (2-√2)/2 ∨ x = √2/2 - 1 = (√2-2)/2
y = 0 <=> x = (√2-2)/2 ∨ x = (2-√2)/2
postać iloczynowa:
y = a(x-x₁)(x-x₂)
y = -2[x-(√2-2)/2][x-(2-√2/2)]
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
y=-2(1-x)²+1
y=-2(1-2x+x²)+1=-2+4x-2x²+1=-2x²+4x-1
-2x²+4x-1=0
Δ=4²-4*(-2)*(-1)=16-8=8
Postać iloczynowa funkcji ma postać: