What is the directrix y=x^2+3x-5 ?

The directrix y=x^2+3x-5 is y=-15/2

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directrix y=x^2+3x-5

directrix

y = x^{2} + 3 x - 5

y = - \frac{15}{2}

Solution steps

Rewrite

y = x^{2} + 3 x - 5

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{29}{4})) = (x - (- \frac{3}{2}))^{2}

(h, k) = (- \frac{3}{2}, - \frac{29}{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{3}{2}, - \frac{29}{4})

y-coordinate

y = - \frac{29}{4} - p

y = - \frac{29}{4} - \frac{1}{4}

Refine

y = - \frac{15}{2}

y = 2 x^{2} + 5 x - 7

x^{2} + 3

y^{2} = - 6 x

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

4 x^{2}