What is the foci y=3x^2-7 ?

The foci y=3x^2-7 is (0,-83/12)

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foci y=3x^2-7

foci

y = 3 x^{2} - 7

(0, - \frac{83}{12})

Solution steps

Rewrite

y = 3 x^{2} - 7

in the standard form:

4 \cdot \frac{1}{12} (y - (- 7)) = (x - 0)^{2}

(h, k) = (0, - 7), p = \frac{1}{12}

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

p

from the center

(0, - 7)

along the y-axis

(0, - 7 + p)

= (0, - 7 + \frac{1}{12})

Refine

(0, - \frac{83}{12})

y = p (x - h)^{2} + k

(x - 3)^{2} = - 12 y

y = 3 x^{2} + 6

y = 2 x^{2} - x

x = - \frac{1}{12} (y + 10)^{2} - 6