What is the directrix y=x^2+x-1 ?

The directrix y=x^2+x-1 is y=-3/2

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directrix y=x^2+x-1

directrix

y = x^{2} + x - 1

y = - \frac{3}{2}

Solution steps

Rewrite

y = x^{2} + x - 1

in the standard form

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

(h, k) = (- \frac{1}{2}, - \frac{5}{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{1}{2}, - \frac{5}{4})

y-coordinate

y = - \frac{5}{4} - p

y = - \frac{5}{4} - \frac{1}{4}

Refine

y = - \frac{3}{2}

4 x^{2} - 8 x - 3 y - 2 = 0

0 = (y - 1)^{2} - 16 (x + 4)

(x + 6)^{2} = 12 (y + 1)

- \frac{1}{8} (x - 8)^{2} + 2

y = 25 x^{2} - 36