What is the directrix (x+1)^2 ?

The directrix (x+1)^2 is y=-1/4

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directrix (x+1)^2

directrix

(x + 1)^{2}

y = - \frac{1}{4}

Solution steps

Rewrite

y = (x + 1)^{2}

in the standard form:

4 \cdot \frac{1}{4} (y - 0) = (x - (- 1))^{2}

(h, k) = (- 1, 0), 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

(- 1, 0)

y-coordinate

y = 0 - p

y = 0 - \frac{1}{4}

Refine

y = - \frac{1}{4}

2 y^{2} + x + 20 y + 51 = 0

- 8 (x - 5) = (y + 1)^{2}

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

y^{2} - 8 y + 4 x = 0

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