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

The directrix 2x^2+5x is y=-13/4

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directrix 2x^2+5x

directrix

2 x^{2} + 5 x

y = - \frac{13}{4}

Solution steps

Rewrite

y = 2 x^{2} + 5 x

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{25}{8})) = (x - (- \frac{5}{4}))^{2}

(h, k) = (- \frac{5}{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{5}{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{1}{14} x^{2} + 2

2 y^{2} = 16 x

y^{2} + 8 x + 2 y + 33 = 0

2 x^{2} - x + 1

y^{2} = x