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

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

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directrix 2x^2-5x+5

directrix

2 x^{2} - 5 x + 5

y = \frac{7}{4}

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{8} (y - \frac{15}{8}) = (x - \frac{5}{4})^{2}

(h, k) = (\frac{5}{4}, \frac{15}{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{15}{8})

y-coordinate

y = \frac{15}{8} - p

y = \frac{15}{8} - \frac{1}{8}

Refine

y = \frac{7}{4}

y = \frac{x ^{2} 5 x 8}{x 3}

(x + 5)^{2} = - 20 (y - 2)

x^{2} + 4 x - 3

x^{2} - y = 0

- 4 (y - 10) = (x - 6)^{2}