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

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

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directrix y=3x^2+x-1

directrix

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

y = - \frac{7}{6}

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{13}{12})) = (x - (- \frac{1}{6}))^{2}

(h, k) = (- \frac{1}{6}, - \frac{13}{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{13}{12})

y-coordinate

y = - \frac{13}{12} - p

y = - \frac{13}{12} - \frac{1}{12}

Refine

y = - \frac{7}{6}

y = x^{2} - 5 x + 6

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

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

(x - 6)^{2} = 8 (y - 2)

- x^{2} + 6 x + 19