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2.h(-3) = a(-3)+b
10=-3a+b.............pers1
h(-2)=a(-2)+b
8=-2a+b.................pers 2
eliminasikan kedua prs trsbt
-3a+b=10
-2a+b=8
--------------- -
-a=2
a=-2
-2a+b=8
-2(-2)+b=8
4+b=8
b=8-4
b=4
jd h(4)= ax+b
=(-2)(4)+4
=-8+4
=-4
= 6 - 5
= 1
2. Persamaan I:
h(-3) = a(-3) + b = 10
= -3a + b = 10
Persamaan II:
h(-2) = a(-2) + b = 8
= -2a + b = 8
-3a + b = 10
-2a + b = 8
---------- -
-a = 2
a = -2
(Gunakan salah satu persamaan untuk menentukan nilai b)
-3a + b = 10
-3(-2) + b = 10
6 + b = 10
b = 10 - 6
b = 4
h(4) = (-2)(4) + 4
= -8 + 4
= -4
Maka nilai h(4) adalah -4