What is the foci y=x^2+12x+26 ?

The foci y=x^2+12x+26 is (-6,-39/4)

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foci y=x^2+12x+26

foci

y = x^{2} + 12 x + 26

(- 6, - \frac{39}{4})

Solution steps

Rewrite

y = x^{2} + 12 x + 26

in the standard form

4 \cdot \frac{1}{4} (y - (- 10)) = (x - (- 6))^{2}

(h, k) = (- 6, - 10), p = \frac{1}{4}

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

p

from the center

(- 6, - 10)

along the y-axis

(- 6, - 10 + p)

= (- 6, - 10 + \frac{1}{4})

Refine

(- 6, - \frac{39}{4})

(y - 3)^{2} = 4 (x + 2)

y = 5 x^{2} + 40 x + 79

x^{2} + 3 x - 2

x^{2} + 3 \cdot x + 2

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