What is the foci (x+5)^2+3 ?

The foci (x+5)^2+3 is (-5, 13/4)

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foci (x+5)^2+3

foci

(x + 5)^{2} + 3

(- 5, \frac{13}{4})

Solution steps

Rewrite

y = (x + 5)^{2} + 3

in the standard form:

4 \cdot \frac{1}{4} (y - 3) = (x - (- 5))^{2}

(h, k) = (- 5, 3), p = \frac{1}{4}

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

p

from the center

(- 5, 3)

along the y-axis

(- 5, 3 + p)

= (- 5, 3 + \frac{1}{4})

Refine

(- 5, \frac{13}{4})

- 0.1 x^{2} + 1.2 x + 98.6

y^{2} - 8 x - 6 y - 3 = 0

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

x = 3 - 5 (y + 1)^{2}

y = x^{2} - 6 x + 11