What is the directrix 3x^2+x-5 ?

The directrix 3x^2+x-5 is y=-31/6

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directrix 3x^2+x-5

directrix

3 x^{2} + x - 5

y = - \frac{31}{6}

Solution steps

Rewrite

y = 3 x^{2} + x - 5

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{61}{12})) = (x - (- \frac{1}{6}))^{2}

(h, k) = (- \frac{1}{6}, - \frac{61}{12}), p = \frac{1}{12}

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}{6}, - \frac{61}{12})

y-coordinate

y = - \frac{61}{12} - p

y = - \frac{61}{12} - \frac{1}{12}

Refine

y = - \frac{31}{6}

(y - 0)^{2} = - 4 \cdot 0.01 (x - 0)

3 x^{2} - 10 x + 3

2 y^{2} + 8 x = 0

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

y = 3 x^{2} - 12 x + 5