What is the directrix y=2x^2-8x+4 ?

The directrix y=2x^2-8x+4 is y=-33/8

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directrix y=2x^2-8x+4

directrix

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

y = - \frac{33}{8}

Solution steps

Rewrite

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

in the standard form

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

(h, k) = (2, - 4), 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

(2, - 4)

y-coordinate

y = - 4 - p

y = - 4 - \frac{1}{8}

Refine

y = - \frac{33}{8}

3 - t^{2}

y = - 8 (x + 1)^{2}

- 8 (y + 1) = (x - 2)^{2}

+ 2 x^{2} + 24 x + 76

y = - 2 x^{2} + 5 x - 6