What is the directrix (x+3)^2=4(y-1) ?

The directrix (x+3)^2=4(y-1) is y=0

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

directrix

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

y = 0

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 3, 1), p = 1

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, 1)

y-coordinate

y = 1 - p

y = 1 - 1

Refine

y = 0

y = x^{2} - 4 x - 3

y = - x^{2} + 6 x

y = x^{2} - 4 x + 5

- 2 x^{2} - 3 x - 1

y = x^{2} - 6 x - 12