What is the foci 2x^2+4x-16 ?

The foci 2x^2+4x-16 is (-1,-143/8)

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foci 2x^2+4x-16

foci

2 x^{2} + 4 x - 16

(- 1, - \frac{143}{8})

Solution steps

Rewrite

y = 2 x^{2} + 4 x - 16

in the standard form

4 \cdot \frac{1}{8} (y - (- 18)) = (x - (- 1))^{2}

(h, k) = (- 1, - 18), p = \frac{1}{8}

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

p

from the center

(- 1, - 18)

along the y-axis

(- 1, - 18 + p)

= (- 1, - 18 + \frac{1}{8})

Refine

(- 1, - \frac{143}{8})

- x^{2} + 4 x

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

- \frac{1}{2} (x - 4)^{2} + 3

(x - 6)^{2} = + 8 (y - 2)

y = \frac{1}{8} (x^{2} - 4 x - 12)