Odpowiedź:
x²+4x+2 =========f(x) = ax²+bx+c
p = -b/2a = -4/2 = -2
q = -Δ/4a = -8/4 = -2
Δ = b²-4ac = 16-8=8
postać kanoniczna f(x) = a(x-p)²+q
f(x)= 1 (x+2)² -2 = (x+2)² - 2
b
-x²+10x-9
Δ = 100 -4 *(-1*-9) = 100-36 = 64
√Δ = 8
x₁ = (-b-√Δ)/2a = (-10-8)/-2 = 9
x₂ = (-b+√Δ)/2a = (-10+8)-2 = 1
postać iloczynowa
f(x) =a (x - x₁) * (x-x₂)
f(x) = -1 (x-9)*(x-1) = -(x-9)*(x-1)
Szczegółowe wyjaśnienie:
