What is the foci y=3x^2-12x+8 ?

The foci y=3x^2-12x+8 is (2,-47/12)

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foci y=3x^2-12x+8

foci

y = 3 x^{2} - 12 x + 8

(2, - \frac{47}{12})

Solution steps

Rewrite

y = 3 x^{2} - 12 x + 8

in the standard form

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

(h, k) = (2, - 4), p = \frac{1}{12}

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

p

from the center

(2, - 4)

along the y-axis

(2, - 4 + p)

= (2, - 4 + \frac{1}{12})

Refine

(2, - \frac{47}{12})

x = (y + 3)^{2}

x^{2} - 6 x + 5 y = 34

y = 2 x (x - 9) - (x - 9)

- 0.0000848 x^{2} - 0.006005 x + 1.004

x^{2} + 16 y = 0