prosze o pomoc ! dział - postac iloczynowa i miejsca zerowe funkcjii kwadratowej
zad:
dana funkcja f (x) = 1/2 ( x + 4 )do kwadratu - 2
zapisz w postaci iloczynowej
( / - ułamek)
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f(x) = (1/2)*( x + 4)^2 - 2
f(x) = a *( x - p)^2 + q
czyli
a = 1/2
p = - 4
q = - 2
ale
p = -b/(2a)
czyli
p = - b / 1 = - b
b = - p = 4
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q = - delta/ (4a) = - delta/2
- delta = 2*q = 2*(-2) = - 4
delta = 4
x1 = [ - b - p(delty)]/(2a) = [ - 4 - 2] /1 = - 6
x2 = [ - b + p(delty)]/(2a) = [ - 4 + 2]/1 = - 2
f(x) = a*(x -x1)*(x - x2) = (1/2)*( x - (-6))*(x - (- 2))
Odp.
f(x) = (1/2)*(x + 6)*( x + 2)
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