What is the foci y=x^2+5x-14 ?

The foci y=x^2+5x-14 is (-5/2 ,-20)

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foci y=x^2+5x-14

foci

y = x^{2} + 5 x - 14

(- \frac{5}{2}, - 20)

Solution steps

Rewrite

y = x^{2} + 5 x - 14

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{81}{4})) = (x - (- \frac{5}{2}))^{2}

(h, k) = (- \frac{5}{2}, - \frac{81}{4}), p = \frac{1}{4}

Parabola is symmetric around the y-axis and so the focus lies a distance

p

from the center

(- \frac{5}{2}, - \frac{81}{4})

along the y-axis

(- \frac{5}{2}, - \frac{81}{4} + p)

= (- \frac{5}{2}, - \frac{81}{4} + \frac{1}{4})

Refine

(- \frac{5}{2}, - 20)

y^{2} - 10 y - 12 x + 13 = 0

y^{2} - 4 x = 0

y^{2} = 5 x

2 x^{2} + 3 x - 1

y^{2} = 7 x