Odpowiedź:
f ( x) = [tex]log_3 ( x + 9 ) - 1[/tex]
więc
[tex]log_3 ( x + 9) - 1 = 0\\log_3 ( x + 9 ) = 1\\x + 9 = 3\\x = - 6[/tex]A = ( - 6, 0 )
================
x = 0 [tex]log_3 ( 0 + 9 ) - 1 = log_3 9 - 1 = 2 - 1 = 1[/tex]
B = ( 0, 1 )
=============
Asymptota ma równanie x = - 9
C = ( - 9, 0 )
==============
P = 0,5*( - 6 - ( - 9))* ( 1 - 0 ) = 0,5*3*1 = 1,5 j²
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Odpowiedź:
f ( x) = [tex]log_3 ( x + 9 ) - 1[/tex]
więc
[tex]log_3 ( x + 9) - 1 = 0\\log_3 ( x + 9 ) = 1\\x + 9 = 3\\x = - 6[/tex]A = ( - 6, 0 )
================
x = 0 [tex]log_3 ( 0 + 9 ) - 1 = log_3 9 - 1 = 2 - 1 = 1[/tex]
B = ( 0, 1 )
=============
Asymptota ma równanie x = - 9
więc
C = ( - 9, 0 )
==============
P = 0,5*( - 6 - ( - 9))* ( 1 - 0 ) = 0,5*3*1 = 1,5 j²
========================================
Szczegółowe wyjaśnienie: