What is the foci y=x^2+6x+5 ?

The foci y=x^2+6x+5 is (-3,-15/4)

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foci y=x^2+6x+5

foci

y = x^{2} + 6 x + 5

(- 3, - \frac{15}{4})

Solution steps

Rewrite

y = x^{2} + 6 x + 5

in the standard form

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

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

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

p

from the center

(- 3, - 4)

along the y-axis

(- 3, - 4 + p)

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

Refine

(- 3, - \frac{15}{4})

x^{2} - 5 x 3

y = (x + 9) (x + 8)

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

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

3 x^{2} + 6 x - 2