What is the foci y=+x^2+6x+3 ?

The foci y=+x^2+6x+3 is (-3,-23/4)

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foci y=+x^2+6x+3

foci

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

(- 3, - \frac{23}{4})

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (y - (- 6)) = (x - (- 3))^{2}

(h, k) = (- 3, - 6), p = \frac{1}{4}

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

p

from the center

(- 3, - 6)

along the y-axis

(- 3, - 6 + p)

= (- 3, - 6 + \frac{1}{4})

Refine

(- 3, - \frac{23}{4})

4 (x + 2)^{2} + 5

(y - 8)^{2} = 16 (x - 4)

y = \frac{x ^{3} - 3 x ^{2} - 4 x}{x - 4}

y = s^{2}

3 x^{2} - 31 x + 56 - (x - 8)