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

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

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directrix 4-(x-3)^2

directrix

4 - (x - 3)^{2}

y = \frac{17}{4}

Solution steps

Rewrite

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

in the standard form:

4 (- \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}

+ x^{2} - 6 x - 7

lo g_{10} (6) x^{2}

y = \frac{1}{14} x^{2} + 2

x^{2} - 4 x + 4

6 x^{2} + 12 x