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

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

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directrix-(y+3)^2-1

directrix

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

x = - \frac{3}{4}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 1, - 3), p = - \frac{1}{4}

Parabola is symmetric around the x-axis and so the directrix is a line parallel to the y-axis, a distance

- p

from the center

(- 1, - 3)

x-coordinate

x = - 1 - p

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

Refine

x = - \frac{3}{4}

7 x^{2} - 3 x + 10 - (- 4 x^{2} + 6 x - 4)

y = - 2 x^{2} + 64 x - 90

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

4 x^{2} - 6 x - 2

4 x^{2} + 16 x - 3