What is the directrix y=(x+1)^2+5 ?

The directrix y=(x+1)^2+5 is y= 19/4

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directrix y=(x+1)^2+5

directrix

y = (x + 1)^{2} + 5

y = \frac{19}{4}

Solution steps

Rewrite

y = (x + 1)^{2} + 5

in the standard form:

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

(h, k) = (- 1, 5), 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

(- 1, 5)

y-coordinate

y = 5 - p

y = 5 - \frac{1}{4}

Refine

y = \frac{19}{4}

t^{2} + 2

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

- 4 x^{2} + 8 x + 1

y = 2 x^{2} + 3 x + 1

x - 4 y^{2} = 0