What is the directrix x^2+3x-16 ?

The directrix x^2+3x-16 is y=-37/2

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directrix x^2+3x-16

directrix

x^{2} + 3 x - 16

y = - \frac{37}{2}

Solution steps

Rewrite

y = x^{2} + 3 x - 16

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{73}{4})) = (x - (- \frac{3}{2}))^{2}

(h, k) = (- \frac{3}{2}, - \frac{73}{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{73}{4})

y-coordinate

y = - \frac{73}{4} - p

y = - \frac{73}{4} - \frac{1}{4}

Refine

y = - \frac{37}{2}

2 x - x^{2}

y = (x - 12)^{2} - 156

y = x^{2} + 6 x + 11

- x (x - 3)

+ \frac{17}{4} (x + 2)^{2}