What is the foci y= 1/4 (x+3)^2-5 ?

The foci y= 1/4 (x+3)^2-5 is (-3,-4)

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foci y= 1/4 (x+3)^2-5

foci

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

(- 3, - 4)

Solution steps

Rewrite

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

in the standard form:

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

(h, k) = (- 3, - 5), p = 1

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

p

from the center

(- 3, - 5)

along the y-axis

(- 3, - 5 + p)

= (- 3, - 5 + 1)

Refine

(- 3, - 4)

- 2 x^{2} + 3 x

x^{2} - 5 x + 4

- 0.2 t^{2} + 1.6 t + 98.4

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

- y^{2} = 6 y + x