What is the foci y=x^2+2x+4 ?

The foci y=x^2+2x+4 is (-1, 13/4)

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foci y=x^2+2x+4

foci

y = x^{2} + 2 x + 4

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

Solution steps

Rewrite

y = x^{2} + 2 x + 4

in the standard form

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

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

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

p

from the center

(- 1, 3)

along the y-axis

(- 1, 3 + p)

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

Refine

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

y^{2} + 2 y + 1 = - 2 x + 4

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

y = + x^{2} + 6 x

y = - x^{2} + 14 x + 49

2 x^{2} - 4 x + 15