What is the directrix-x^2+5x-7 ?

The directrix-x^2+5x-7 is y=-1/2

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directrix-x^2+5x-7

directrix

- x^{2} + 5 x - 7

y = - \frac{1}{2}

Solution steps

Rewrite

y = - x^{2} + 5 x - 7

in the standard form

4 (- \frac{1}{4}) (y - (- \frac{3}{4})) = (x - \frac{5}{2})^{2}

(h, k) = (\frac{5}{2}, - \frac{3}{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{3}{4})

y-coordinate

y = - \frac{3}{4} - p

y = - \frac{3}{4} - (- \frac{1}{4})

Refine

y = - \frac{1}{2}

(x - 1)^{2} = 8 (y + 3)

+ 3 x^{2}

(x + 1)^{2} = 12 (y - 3)

5 x - x^{2}

y^{2} - 12 x - 12 y + 84 = 0