What is the directrix y=4x^2+2x+4 ?

The directrix y=4x^2+2x+4 is y= 59/16

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directrix y=4x^2+2x+4

directrix

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

y = \frac{59}{16}

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{16} (y - \frac{15}{4}) = (x - (- \frac{1}{4}))^{2}

(h, k) = (- \frac{1}{4}, \frac{15}{4}), p = \frac{1}{16}

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{15}{4})

y-coordinate

y = \frac{15}{4} - p

y = \frac{15}{4} - \frac{1}{16}

Refine

y = \frac{59}{16}

y = 2 x^{2} - 3 x

2 x^{2} - 2

2 x^{2} + 3

x^{2} + x + 1

4 + 2 x - x^{2}