Podaj zbiór wartości funkcji f określonej wzorem: a) f(x)=x^2-5x+6 b) f(x)=2x^2+4x-3 c) f(x)=-3x^2+x+1 d) f(x)=-2(x+1/2)^2+3/4 e) f(x)=1/2(x-2)^2 f) f(x)=-(3x-1)(x+2)
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a)
f(x) = x^2 - 5x + 6
a = 1 > 0 - ramiona paraboli są skierowane ku górze
p = -b/(2a) = 5/2 = 2,5
q = f(p) = f(2.5) = (2,5)^2 -5*2,5 + 6 = 6,25 -12,5 + 6 = -0,25 = - 1/4
zatem ZW = < q ; + oo ) = < -1/4; + oo )
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b)
f(x) = 2 x^2 + 4x - 3
a = 2 > 0 - ramiona paraboli są skierowane ku górze
p = -b/(2a) = -4/4 = - 1
q = f(p) = f(-1) = 2*(-1)^2 +4*(-1) - 3 = 2 - 4 - 3 = - 5
zatem
ZW = < -5; + oo )
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c)
f(x) = - 3 x^2 + x + 1
a = - 3 < 0 - ramiona paraboli są skierowane ku dołowi
delta = b^2 - 4ac = 1^2 -4*(-3)*1 = 1 + 12 = 13
q = - delta/(4a) = - 13/( - 12) = 13/12
zatem
ZW = ( - oo; q > = ( - oo; 13/12 >
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d)
f(x) = - 2*(x + 1/2)^2 + 3/4 - postać kanoniczna
a = - 2 < 0 - ramiona skierowane są ku dołowi
q = 3/4
zatem
ZW = ( - oo ; 3/4 >
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e)
f(x) = (1/2)*(x -2)^2 - postać kanoniczna
a = 1/2 > 0 - ramiona paraboli skierowane są ku górze
q = 0
zatem
ZW = < q ; + oo ) = < 0; + oo )
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f)
f(x) = - (3x - 1)*(x +2) = - [ 3 x^2 + 6x - x - 2 ] = - [3 x^2 + 5x - 2 ]
f(x) = - 3 x^2 - 5x + 2
a = -3 < 0 - ramiona paraboli są skierowane ku dołowi
delta = b^2 - 4ac = (-5)^2 - 4*(-3)*2 = 25 + 24 = 49
q = - delta/ (4a) = - 49/(-12) = 49/12 = 4 1/12
zatem
ZW = ( - oo; q > = ( - oo; 4 1/12 >
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Uwaga:
q - obliczano dwoma sposobami:
1) q = f(p) , gdzie p = - b/(2a)
2) q = - delta/(4a), gdzie delta = b^2 - 4ac
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