What is the directrix y=(x^2-4x-1)/7 ?

The directrix y=(x^2-4x-1)/7 is y=-69/28

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directrix y=(x^2-4x-1)/7

directrix

y = \frac{x ^{2} - 4 x - 1}{7}

y = - \frac{69}{28}

Solution steps

Rewrite

y = \frac{x ^{2} - 4 x - 1}{7}

in the standard form

4 \cdot \frac{7}{4} (y - (- \frac{5}{7})) = (x - 2)^{2}

(h, k) = (2, - \frac{5}{7}), p = \frac{7}{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

(2, - \frac{5}{7})

y-coordinate

y = - \frac{5}{7} - p

y = - \frac{5}{7} - \frac{7}{4}

Refine

y = - \frac{69}{28}

y^{2} - 28 x = 0

3 y^{2} = - 4 x

(x + 3)^{2} = 4 (y - 1)

y = x^{2} - 4 x - 3

y = - x^{2} + 6 x