What is the directrix y=2x^2+7x+3 ?

The directrix y=2x^2+7x+3 is y=-13/4

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directrix y=2x^2+7x+3

directrix

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

y = - \frac{13}{4}

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{25}{8})) = (x - (- \frac{7}{4}))^{2}

(h, k) = (- \frac{7}{4}, - \frac{25}{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{7}{4}, - \frac{25}{8})

y-coordinate

y = - \frac{25}{8} - p

y = - \frac{25}{8} - \frac{1}{8}

Refine

y = - \frac{13}{4}

y = \frac{6 x}{7} \frac{4 x}{3}

\frac{4 x ^{3} + 11 x ^{2} + 4 x + 21}{x + 3}

- \frac{1}{3} (x - 3) = (y + 5)^{2}

+ 2.5 t^{2} + 6 t

y = x^{2} + 6 x + 8