What is the directrix y=2x^2+5x-7 ?

The directrix y=2x^2+5x-7 is y=-41/4

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directrix y=2x^2+5x-7

directrix

y = 2 x^{2} + 5 x - 7

y = - \frac{41}{4}

Solution steps

Rewrite

y = 2 x^{2} + 5 x - 7

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{81}{8})) = (x - (- \frac{5}{4}))^{2}

(h, k) = (- \frac{5}{4}, - \frac{81}{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{5}{4}, - \frac{81}{8})

y-coordinate

y = - \frac{81}{8} - p

y = - \frac{81}{8} - \frac{1}{8}

Refine

y = - \frac{41}{4}

x^{2} + 3

y^{2} = - 6 x

(x - 1)^{2} = 16 (y - 2)

4 x^{2}

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