What is the directrix (4-x)^2 ?

The directrix (4-x)^2 is y=-1/4

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directrix (4-x)^2

directrix

(4 - x)^{2}

y = - \frac{1}{4}

Solution steps

Rewrite

y = (4 - x)^{2}

in the standard form:

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

(h, k) = (4, 0), 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

(4, 0)

y-coordinate

y = 0 - p

y = 0 - \frac{1}{4}

Refine

y = - \frac{1}{4}

y = 7 x^{2}

x = (\frac{4}{5} - 3 y) \cdot (\frac{4}{5} + 3 y)

y = (6 x - 1) (3 x - 2)

y = \frac{x ^{2}}{32}

3 y^{2} - 4 x - 6 y + 23 = 0