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f(1)=1
Maka 1=a(1)^2+b(1)
1=a+b
f(3)=0
Maka 0=a(3)^2+b(3)
0=9a+3b
Kedua pers. dieleminasi
1=a+b ......... x3
0=9a+3b
3=-6a
a=-1/2
b=3/2
24b.
Bentuk fungsi sbb:
f:x->-1/2x^2 + 3/2x
24c.
f(-5)=-1/2(-5)^2 + 3/2(-5)
=-25/2+(-15/2)
=-50/2
=-25