What is the directrix y= 1/3 x^2+6x+10 ?

The directrix y= 1/3 x^2+6x+10 is y=-71/4

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directrix y= 1/3 x^2+6x+10

directrix

y = \frac{1}{3} x^{2} + 6 x + 10

y = - \frac{71}{4}

Solution steps

Rewrite

y = \frac{1}{3} x^{2} + 6 x + 10

in the standard form

4 \cdot \frac{3}{4} (y - (- 17)) = (x - (- 9))^{2}

(h, k) = (- 9, - 17), p = \frac{3}{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

(- 9, - 17)

y-coordinate

y = - 17 - p

y = - 17 - \frac{3}{4}

Refine

y = - \frac{71}{4}

F = \frac{m v ^{2}}{r}

- x^{2} + 6 x + 1

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

x^{2} = 8 (y + 4)

y = 2 x (x - 9) - (x - 9)

What is the directrix y= 1/3 x^2+6x+10 ?
The directrix y= 1/3 x^2+6x+10 is y=-71/4