What is the foci y=4x^2+8x-11 ?

The foci y=4x^2+8x-11 is (-1,-239/16)

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foci y=4x^2+8x-11

foci

(- 1, - \frac{239}{16})

Solution steps

Rewrite

y = 4 x^{2} + 8 x - 11

in the standard form

4 \cdot \frac{1}{16} (y - (- 15)) = (x - (- 1))^{2}

(h, k) = (- 1, - 15), p = \frac{1}{16}

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

p

from the center

(- 1, - 15)

along the y-axis

(- 1, - 15 + p)

= (- 1, - 15 + \frac{1}{16})

Refine

(- 1, - \frac{239}{16})