What is the foci (x+3)^2-8 ?

The foci (x+3)^2-8 is (-3,-31/4)

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foci (x+3)^2-8

foci

(x + 3)^{2} - 8

(- 3, - \frac{31}{4})

Solution steps

Rewrite

y = (x + 3)^{2} - 8

in the standard form:

4 \cdot \frac{1}{4} (y - (- 8)) = (x - (- 3))^{2}

(h, k) = (- 3, - 8), p = \frac{1}{4}

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

p

from the center

(- 3, - 8)

along the y-axis

(- 3, - 8 + p)

= (- 3, - 8 + \frac{1}{4})

Refine

(- 3, - \frac{31}{4})

- x^{2} + 3 x

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

y = - x^{2} + 64 x - 292

y = x^{2} - 12 x + 5

y = - 2 x + x (x - 2) - 2 x (x - 1) - (x - 3)