What is the foci y=x^2+3x-16 ?

The foci y=x^2+3x-16 is (-3/2 ,-18)

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foci y=x^2+3x-16

foci

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

(- \frac{3}{2}, - 18)

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 focus lies a distance

p

from the center

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

along the y-axis

(- \frac{3}{2}, - \frac{73}{4} + p)

= (- \frac{3}{2}, - \frac{73}{4} + \frac{1}{4})

Refine

(- \frac{3}{2}, - 18)

y = (5 x - 7)^{2} - 13

4 x^{2} - 12 x - 16 y + 41 = 0

1 - x - x^{2}

4 x^{2} - 20 x - 24 y + 97 = 0

y^{2} = x + 1