What is the directrix (x-3)^2-4 ?

The directrix (x-3)^2-4 is y=-17/4

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directrix (x-3)^2-4

directrix

(x - 3)^{2} - 4

y = - \frac{17}{4}

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (3, - 4), 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

(3, - 4)

y-coordinate

y = - 4 - p

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

Refine

y = - \frac{17}{4}

(y + 1.5)^{2} = 4 (x - 2)

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

14 x^{2}

y = \frac{- 4 x ^{2} + 24 x + 161}{25}

x^{2} - 12 x + 35