pthdgd
2x^2-x-1=0
(x-1)(2x+1)=0
x=1=>y=2: x=-1/2=> y=1/2
td gd
A(1;2);B(-1/2;1/2)
b.
C(m,2m^2);∆: x-y+1=0
S∆sbc max =>sAd(c,∆) max
|m-(2m^2)+1|/√(1+1) max
dk m€(-1/2;1)
F(m)=-2m^2+m+1=
(2m+1)(1-m)>0
|f(x)|=-2m^2+m+1=-2(m-1/4)^2+9/8
khi m=1/4
C(1/4;1/8)