What is the directrix y=x^2-13x+40 ?

The directrix y=x^2-13x+40 is y=-5/2

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directrix y=x^2-13x+40

directrix

y = x^{2} - 13 x + 40

y = - \frac{5}{2}

Solution steps

Rewrite

y = x^{2} - 13 x + 40

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{9}{4})) = (x - \frac{13}{2})^{2}

(h, k) = (\frac{13}{2}, - \frac{9}{4}), p = \frac{1}{4}

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{13}{2}, - \frac{9}{4})

y-coordinate

y = - \frac{9}{4} - p

y = - \frac{9}{4} - \frac{1}{4}

Refine

y = - \frac{5}{2}

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

\frac{x ( x + 5 )}{7}

y = + x^{2} - 18 x + 73

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

3 x^{2} + 5 x - 4