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

The directrix y=(2x+3)^2 is y=-1/16

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directrix y=(2x+3)^2

directrix

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

y = - \frac{1}{16}

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{16} (y - 0) = (x - (- \frac{3}{2}))^{2}

(h, k) = (- \frac{3}{2}, 0), p = \frac{1}{16}

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

(- \frac{3}{2}, 0)

y-coordinate

y = 0 - p

y = 0 - \frac{1}{16}

Refine

y = - \frac{1}{16}

3 x^{2} + 7 x + 1

0.3 + 5.5 t - 4.9 t^{2}

y + 8 x^{2} = 0

- 3 (x - 7)^{2}

(x + 7)^{2} = 5 (y + 3)