What is the directrix y= 1/4 x^2-5/2 x+13/4 ?

The directrix y= 1/4 x^2-5/2 x+13/4 is y=-4

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directrix y= 1/4 x^2-5/2 x+13/4

directrix

y = \frac{1}{4} x^{2} - \frac{5}{2} x + \frac{13}{4}

y = - 4

Solution steps

Rewrite

y = \frac{1}{4} x^{2} - \frac{5}{2} x + \frac{13}{4}

in the standard form

4 \cdot 1 \cdot (y - (- 3)) = (x - 5)^{2}

(h, k) = (5, - 3), p = 1

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

(5, - 3)

y-coordinate

y = - 3 - p

y = - 3 - 1

Refine

y = - 4

4 x^{2} - 25

y = - 6 x^{2} + 37 x + 13

x^{2} = - 24 y

y = - 9 x^{2} + 2

x^{2} - 12 x + 1

What is the directrix y= 1/4 x^2-5/2 x+13/4 ?
The directrix y= 1/4 x^2-5/2 x+13/4 is y=-4