Odpowiedź:
f(x) = x² - 4x + 5
a = 1 , b = - 4 , c = 5
p = - b/2a = 4/2 = 2
q = - Δ/4a = - (b² - 4ac)/4 = - [(- 4)² - 4 * 1 * 5]/4 = - (16 - 20)/4 =
= - (- 4)/4 = 4/4 = 1
postać kanoniczna
f(x) = a(x - p)² + q = (x - 2)² + 1
W - współrzędne iwerzchołka = (p , q) = (2 , 1 )
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Odpowiedź:
f(x) = x² - 4x + 5
a = 1 , b = - 4 , c = 5
p = - b/2a = 4/2 = 2
q = - Δ/4a = - (b² - 4ac)/4 = - [(- 4)² - 4 * 1 * 5]/4 = - (16 - 20)/4 =
= - (- 4)/4 = 4/4 = 1
postać kanoniczna
f(x) = a(x - p)² + q = (x - 2)² + 1
W - współrzędne iwerzchołka = (p , q) = (2 , 1 )