What is the directrix-(x+3)^2-1 ?

The directrix-(x+3)^2-1 is y=-3/4

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directrix-(x+3)^2-1

directrix

- (x + 3)^{2} - 1

y = - \frac{3}{4}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 3, - 1), 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

(- 3, - 1)

y-coordinate

y = - 1 - p

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

Refine

y = - \frac{3}{4}

- x^{2} - 2 x - 3

- x^{2} - 2 x - 1

y + 3 x^{2} = 0

y^{2} - 8 x - 8 y + 24 = 0

y = + x^{2} - 4 x - 32