What is the directrix 3x^2+7x+1 ?

The directrix 3x^2+7x+1 is y=-19/6

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directrix 3x^2+7x+1

directrix

3 x^{2} + 7 x + 1

y = - \frac{19}{6}

Solution steps

Rewrite

y = 3 x^{2} + 7 x + 1

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{37}{12})) = (x - (- \frac{7}{6}))^{2}

(h, k) = (- \frac{7}{6}, - \frac{37}{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{7}{6}, - \frac{37}{12})

y-coordinate

y = - \frac{37}{12} - p

y = - \frac{37}{12} - \frac{1}{12}

Refine

y = - \frac{19}{6}

0.3 + 5.5 t - 4.9 t^{2}

y + 8 x^{2} = 0

- 3 (x - 7)^{2}

(x + 7)^{2} = 5 (y + 3)

y^{2} - 6 y = - 4 x + 3