What is the foci y=x^2+2x+12 ?

The foci y=x^2+2x+12 is (-1, 45/4)

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foci y=x^2+2x+12

foci

y = x^{2} + 2 x + 12

(- 1, \frac{45}{4})

Solution steps

Rewrite

y = x^{2} + 2 x + 12

in the standard form

4 \cdot \frac{1}{4} (y - 11) = (x - (- 1))^{2}

(h, k) = (- 1, 11), p = \frac{1}{4}

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

p

from the center

(- 1, 11)

along the y-axis

(- 1, 11 + p)

= (- 1, 11 + \frac{1}{4})

Refine

(- 1, \frac{45}{4})

x = (y + 4)^{2}

y = 6 x^{2} - 1

1.5 + 17.67 t - 4.905 t^{2}

y = (3 x - 4) (4 x - 3) - 2 x

x^{2} - 4 x - y + 3 = 0