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a) f(2) = 5 ---> 2a+b = 5
f(1) = 3 ---> a + b = 3
------------------------------- -
a = 2
subsitusikan a = 2 ke a+b = 3
a+b = 3
2+b = 3
b = 1
rumus fungsinya f(x) = 2x + 1
b) f(3) = -5 ---> 3a+b = -5
f(0) = -6 ---> b = -6
subsitusikan b = -6 ke 3a+b = -5
3a+b = -5
3a -6 = -5
3a = -5+6
3a = 1
a = 1/3
rumus fungsinya f(x) = 1/3 x - 6
c) f(4) = 4 ---> 4a+b = 4
f(2) = 3 ---> 2a + b = 3
------------------------------- -
2a = 1
a = 1/2
subsitusikan a = 1/2 ke 2a+b = 3
2a+b = 3
2(1/2)+b = 3
1 + b = 3
b = 3-1
b = 2
rumus fungsinya f(x) = 1/2 x + 2
2) f(x) = (x+a)+3
f(2) = 2+a+3 = 7
a+5 = 7
a= 2
a) bentuk fungsi f(x) = (x+a)+3
= x+5
b) f(-1) = x + 5
= -1 + 5
= 4
c) f(-2) + f(-1) = (-2+5) + (-1+5)
= 3 + 4
= 7
d) f(2x-5) = (2x-5) + 5
= 2x
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