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

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

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directrix y=2x^2+3x+1

directrix

y = 2 x^{2} + 3 x + 1

y = - \frac{1}{4}

Solution steps

Rewrite

y = 2 x^{2} + 3 x + 1

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{1}{8})) = (x - (- \frac{3}{4}))^{2}

(h, k) = (- \frac{3}{4}, - \frac{1}{8}), p = \frac{1}{8}

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{3}{4}, - \frac{1}{8})

y-coordinate

y = - \frac{1}{8} - p

y = - \frac{1}{8} - \frac{1}{8}

Refine

y = - \frac{1}{4}

x - 4 y^{2} = 0

x = (- y - 1) (y + 3)

y = 4 x^{2} - 8 x + 1

y = 24 x - 3 x^{2}

x = - 5 y^{2}