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

The directrix y= 1/2 (x+3)^2-2 is y=-5/2

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

directrix

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

y = - \frac{5}{2}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 3, - 2), p = \frac{1}{2}

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, - 2)

y-coordinate

y = - 2 - p

y = - 2 - \frac{1}{2}

Refine

y = - \frac{5}{2}

x^{2} + 5 x - 14

(x - 2) = 3 (y - 5)^{2}

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

y = x^{2} - 3 x^{2}

y = 4 x^{2}

What is the directrix y= 1/2 (x+3)^2-2 ?
The directrix y= 1/2 (x+3)^2-2 is y=-5/2