Odpowiedź:
f(x) =3 x² + 4 x - 4
a) Δ = b²- 4a*c = 16 - 4*3*(-4) = 16 + 48 = 64
√Δ = 8
x1 = [tex]\frac{- 4 - 8}{2*3} = - 2[/tex] x2 = [tex]\frac{- 4 + 8}{6}[/tex] = [tex]\frac{2}{3}[/tex]
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b) p = [tex]\frac{-b}{2a}[/tex] = [tex]\frac{-4}{6} = - \frac{2}{3}[/tex]
q = [tex]\frac{- 8}{4*3}[/tex] = - [tex]\frac{2}{3}[/tex]
W = ( p; q) = ( - [tex]\frac{2}{3}[/tex] ; - [tex]\frac{2}{3}[/tex] )
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c) f(0) = - 4 P =( 0, -4)
d)
e) a = 3 > 0 więc ramiona paraboli są skierowane do góry, więc
f(x) > 0 dla x ∈ ( -∞; - 2) ∪ ( [tex]\frac{2}{3}[/tex] ; +∞ )
f(x) < 0 dla x ∈ ( -2 ; [tex]\frac{2}{3}[/tex] )
f)
f(x) = a*(x -x1)*(x - x2)
f(x) = 3*( x + 2)*( x - [tex]\frac{2}{3}[/tex] ) - p. iloczynowa
f(x) = a*(x - p)²+ q
f(x) = 3*( x + [tex]\frac{2}{3}[/tex] )² - [tex]\frac{2}{3}[/tex] - p. kanoniczna
Szczegółowe wyjaśnienie:
