What is the foci (x+1)^2-2 ?

The foci (x+1)^2-2 is (-1,-7/4)

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foci (x+1)^2-2

foci

(x + 1)^{2} - 2

(- 1, - \frac{7}{4})

Solution steps

Rewrite

y = (x + 1)^{2} - 2

in the standard form:

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

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

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

p

from the center

(- 1, - 2)

along the y-axis

(- 1, - 2 + p)

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

Refine

(- 1, - \frac{7}{4})

y = - 4 x^{2}

y = \frac{x ^{2} - 4}{2}

x = (y + 1)^{2} - 4

(y - 2)^{2} = 12 (x - 3)

x^{2} + 14 x + 48