Odpowiedź:
a ) f( x) = - x² - 8 x + 1 - postać ogólna
[tex]x_w[/tex] = [tex]\frac{8}{-2} = - 4[/tex]
Δ = (-8)² - 4*(-1)*1 = 64 + 4 = 68
[tex]y_w = \frac{-68}{- 4} = 17[/tex]
W = ( - 4, 17 )
----------------------
a = - 1 < 0
więc
ZW = ( - ∞ , [tex]y_w[/tex] > = ( - ∞ , 17 >
---------------------------------------------
a = - 1 oraz [tex]x_w[/tex] = - 4
f rośnie w ( - ∞ , - 4 ) , a maleje w ( - 4, + ∞ )
========================================
b)
f( x) = 3*(x + 2)*(x - 6 ) - postać iloczynowa
[tex]x_1 = - 2[/tex] [tex]x_2 = 6[/tex]
[tex]x_w = \frac{-2 + 6}{2} = 2[/tex]
[tex]y_w = 3*(2 + 2)*(2 - 6) = 3*4*(-4) = - 48[/tex]
W = ( 2 , - 48 )
---------------------
a = 3 < 0
ZW = < [tex]y_w[/tex] , +∞ ) = < - 48 , + ∞ )
--------------------------------------------
a = 3 i [tex]x_w = 2[/tex]
f maleje w ( - ∞ , 2 ) , a rośnie w ( 2 , + ∞ )
c ) f ( x) = ( x + 2)² - 3 - postać kanoniczna
a = 1 > 0
[tex]x_w = - 2[/tex] [tex]y_w = - 3[/tex]
W = ( -2, - 3)
ZW = < [tex]y_w[/tex] , +∞ ) = < - 3, + ∞ )
a > 0 oraz [tex]x_w = - 2[/tex]
f maleje w ( - ∞ , - 2 ) , a rosnie w ( - 2, + ∞ )
============================================
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Odpowiedź:
a ) f( x) = - x² - 8 x + 1 - postać ogólna
[tex]x_w[/tex] = [tex]\frac{8}{-2} = - 4[/tex]
Δ = (-8)² - 4*(-1)*1 = 64 + 4 = 68
[tex]y_w = \frac{-68}{- 4} = 17[/tex]
W = ( - 4, 17 )
----------------------
a = - 1 < 0
więc
ZW = ( - ∞ , [tex]y_w[/tex] > = ( - ∞ , 17 >
---------------------------------------------
a = - 1 oraz [tex]x_w[/tex] = - 4
więc
f rośnie w ( - ∞ , - 4 ) , a maleje w ( - 4, + ∞ )
========================================
b)
f( x) = 3*(x + 2)*(x - 6 ) - postać iloczynowa
więc
[tex]x_1 = - 2[/tex] [tex]x_2 = 6[/tex]
[tex]x_w = \frac{-2 + 6}{2} = 2[/tex]
[tex]y_w = 3*(2 + 2)*(2 - 6) = 3*4*(-4) = - 48[/tex]
W = ( 2 , - 48 )
---------------------
a = 3 < 0
więc
ZW = < [tex]y_w[/tex] , +∞ ) = < - 48 , + ∞ )
--------------------------------------------
a = 3 i [tex]x_w = 2[/tex]
więc
f maleje w ( - ∞ , 2 ) , a rośnie w ( 2 , + ∞ )
========================================
c ) f ( x) = ( x + 2)² - 3 - postać kanoniczna
więc
a = 1 > 0
[tex]x_w = - 2[/tex] [tex]y_w = - 3[/tex]
W = ( -2, - 3)
----------------------
ZW = < [tex]y_w[/tex] , +∞ ) = < - 3, + ∞ )
---------------------------------------------
a > 0 oraz [tex]x_w = - 2[/tex]
więc
f maleje w ( - ∞ , - 2 ) , a rosnie w ( - 2, + ∞ )
============================================
Szczegółowe wyjaśnienie: