What is the directrix y=2x^2-19x+9 ?

The directrix y=2x^2-19x+9 is y=-145/4

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directrix y=2x^2-19x+9

directrix

y = 2 x^{2} - 19 x + 9

y = - \frac{145}{4}

Solution steps

Rewrite

y = 2 x^{2} - 19 x + 9

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{289}{8})) = (x - \frac{19}{4})^{2}

(h, k) = (\frac{19}{4}, - \frac{289}{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{19}{4}, - \frac{289}{8})

y-coordinate

y = - \frac{289}{8} - p

y = - \frac{289}{8} - \frac{1}{8}

Refine

y = - \frac{145}{4}

- 2 x^{2}

\frac{x ^{5} + x ^{4} + x ^{3} + x ^{2} + x + 1}{x ^{3} + 1}

y = x^{2} - 15 x

- x^{2} + 8 x - 10

y = - 4 x^{2} - 12 x - 8