What is the directrix y=-8(x+1)^2 ?

The directrix y=-8(x+1)^2 is y= 1/32

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directrix y=-8(x+1)^2

directrix

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

y = \frac{1}{32}

Solution steps

Rewrite

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

in the standard form:

4 (- \frac{1}{32}) (y - 0) = (x - (- 1))^{2}

(h, k) = (- 1, 0), p = - \frac{1}{32}

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}{32})

Refine

y = \frac{1}{32}

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

+ 2 x^{2} + 24 x + 76

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

y^{2} + 2 x = 0

y = - (0.009 (x - 220)^{2} - 220)