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a. f(1) = a(1)² +b(1) = 1
a + b = 1
f(3) = a(3)²+b(3)=0
= 9a +3b = 0
eleminasi
a + b = 1 | x3 | 3a +3b =3
9a +3b = 0|x1 | 9a +3b = 0 -
-6a = 3
a = -2
subtitusi
a+b = 1
-2 +b=1
b= 1+2
b=3