What is the directrix-8x^2+4x ?

The directrix-8x^2+4x is y= 17/32

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directrix-8x^2+4x

directrix

- 8 x^{2} + 4 x

y = \frac{17}{32}

Solution steps

Rewrite

y = - 8 x^{2} + 4 x

in the standard form

4 (- \frac{1}{32}) (y - \frac{1}{2}) = (x - \frac{1}{4})^{2}

(h, k) = (\frac{1}{4}, \frac{1}{2}), p = - \frac{1}{32}

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

(\frac{1}{4}, \frac{1}{2})

y-coordinate

y = \frac{1}{2} - p

y = \frac{1}{2} - (- \frac{1}{32})

Refine

y = \frac{17}{32}

3 y + 4 x = - 2 x^{2} - 14

- x^{2} - 6 x - 13

3 x^{2} - 6 x + 9

2.6 + 2.2 x - 0.6 x^{2}

3 x^{2} - 6 x + 3