What is the directrix y=x^2-5x+4 ?

The directrix y=x^2-5x+4 is y=-5/2

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directrix y=x^2-5x+4

directrix

y = x^{2} - 5 x + 4

y = - \frac{5}{2}

Solution steps

Rewrite

y = x^{2} - 5 x + 4

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{9}{4})) = (x - \frac{5}{2})^{2}

(h, k) = (\frac{5}{2}, - \frac{9}{4}), p = \frac{1}{4}

Parabola is symmetric around the y-axis and so the directrix is a line parallel to the x-axis, a distance

- p

from the center

(\frac{5}{2}, - \frac{9}{4})

y-coordinate

y = - \frac{9}{4} - p

y = - \frac{9}{4} - \frac{1}{4}

Refine

y = - \frac{5}{2}

y = x^{2} - x - 6

+ x^{2} + 12

y = 5 x^{2} + 210 x - 34775

y^{2} - 6 y - 4 x + 5 = 0

(x + 2)^{2} = - 6 (y - 1)