What is the directrix y=(2x+5)^2 ?

The directrix y=(2x+5)^2 is y=-1/16

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directrix y=(2x+5)^2

directrix

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

y = - \frac{1}{16}

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{16} (y - 0) = (x - (- \frac{5}{2}))^{2}

(h, k) = (- \frac{5}{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{5}{2}, 0)

y-coordinate

y = 0 - p

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

Refine

y = - \frac{1}{16}

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

4 t - 9 t^{2}

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

- 2 x^{2} + 12 x - 2

4 x^{2} - 8 x - 3