Respuesta :

Cs010416
ax^2+bx+c
x^2-4x+3
(x-3)(x-1)=0
x=3, x=1 when y=0
The graph of the parabola has a minimum because a is positive.
(3+1)/2=2 so y will be at a minimum when x=2
f(x)=y=2^2-(4*2)+3
=4-8+3

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