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nilai maksimum : 5
f(2)=f(5)=0
f(x)=ax2+bx+5
f(2)=4a+2b+5=0⇒4a+2b=-5⇒8a+4b=-10....(i)
f(5)=16a+4b+5=0⇒16a+4b=-5...(ii)
(i) dan (ii) eliminasi menghasilkan a=5/8 dan b=-5/4
sehingga f(x)=5/8x2 - 5/4 x + 5