What is the directrix 5x^2+4x-9 ?

The directrix 5x^2+4x-9 is y=-197/20

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directrix 5x^2+4x-9

directrix

5 x^{2} + 4 x - 9

y = - \frac{197}{20}

Solution steps

Rewrite

y = 5 x^{2} + 4 x - 9

in the standard form

4 \cdot \frac{1}{20} (y - (- \frac{49}{5})) = (x - (- \frac{2}{5}))^{2}

(h, k) = (- \frac{2}{5}, - \frac{49}{5}), p = \frac{1}{20}

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{2}{5}, - \frac{49}{5})

y-coordinate

y = - \frac{49}{5} - p

y = - \frac{49}{5} - \frac{1}{20}

Refine

y = - \frac{197}{20}

y^{2} + 4 y + 3

(x + 3)^{2} = 4 (y + 4)

y = - 6 (x - 8)^{2} - 4

3 x^{2} + 6 x - 45

- x^{2} - 4 x - 3