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p = 2/2 = 1
f(1) = 1-2 = -1
W(1,-1)
x²-2x = 0
x = 0 ∨ x = 2
f(0) = 0
kółeczko zamalowane na 0,
2. f(x) = -2x²-8x-6
p = 8/(-4) = -2
f(-2) = -2*4+8*2-6 = 2
f(0) = -6
-2x²-8x - 6 = 0
Δ = 64 - 4*2*6= 16
√Δ = 4
x₁ = (8-4)/(-4) = -1
x₂ = (8+2)/(-4) = -3
kółeczko niezamalowane na punkcie (0,-6)