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

The foci 3x^2+2x+6 is (-1/3 , 23/4)

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foci 3x^2+2x+6

foci

3 x^{2} + 2 x + 6

(- \frac{1}{3}, \frac{23}{4})

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{12} (y - \frac{17}{3}) = (x - (- \frac{1}{3}))^{2}

(h, k) = (- \frac{1}{3}, \frac{17}{3}), p = \frac{1}{12}

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

p

from the center

(- \frac{1}{3}, \frac{17}{3})

along the y-axis

(- \frac{1}{3}, \frac{17}{3} + p)

= (- \frac{1}{3}, \frac{17}{3} + \frac{1}{12})

Refine

(- \frac{1}{3}, \frac{23}{4})

\frac{1}{8} x^{2} - y = 0

- x^{2} + 4 y - 44 = - 8 x

- x^{2} - 2 x + 3

x - 2 = 3 (y - 5)^{2}

y = (x)^{4} + 6 (x)^{2} - 1