Odpowiedź:
f(x) = x² + 6x - 1
a=1, b = 6 ,c = - 1
Postać kanoniczna funkcji kwadratowej
f(x) = a(x - p)² + q ,gdzie p = - b/a i q = - Δ/4a
p = - 6/2 = - 3
Δ = b² - 4ac = 6² - 4 * 1 * (- 1) = 36 + 4 = 40
q = - Δ/4a = - 40/4 = - 10
f(x) = (x + 3)² - 10
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Odpowiedź:
f(x) = x² + 6x - 1
a=1, b = 6 ,c = - 1
Postać kanoniczna funkcji kwadratowej
f(x) = a(x - p)² + q ,gdzie p = - b/a i q = - Δ/4a
p = - 6/2 = - 3
Δ = b² - 4ac = 6² - 4 * 1 * (- 1) = 36 + 4 = 40
q = - Δ/4a = - 40/4 = - 10
f(x) = (x + 3)² - 10