Find (A) the directrix, (B) the focus and (C) the roots of the parabola y=x^2-5x+4

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`y=x^2-5x+4`

Dirteticx `x=-b/2a=-5/4`

Focus: `(-b/2a;(1-b^2+4ac)/(4a))=(-5/4;-2)`

ERRATA CORRIGE Dirieticx `x=5/4` Focus `(5/4;-2)`

For a parabola `y = ax^2 + bx + c` , the coordinates of the focus are `(-b/(2*a), (1 - (b^2 - 4ac))/(4*a))` and the equation of the directrix is `y = c - (1+b^2)/(4a)` .

For `y=x^2-5x+4` ,

The focus is at `(5/2, (1 - (25 - 16))/4) = (5/2, -2)`

The equation of the directrix is `y = 1 - (1+25)/4 => y = -5`

The roots of the parabola are the solutions of x^2 - 5x + 4 = 0

=> x^2 - 4x - x + 4 = 0

=> x(x - 4) - 1(x - 4) = 0

=> (x - 1)(x - 4) = 0

=> x = 1 and x = 4

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