What is the foci y=3x^2-15x+6 ?

The foci y=3x^2-15x+6 is (5/2 ,-38/3)

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foci y=3x^2-15x+6

foci

y = 3 x^{2} - 15 x + 6

(\frac{5}{2}, - \frac{38}{3})

Solution steps

Rewrite

y = 3 x^{2} - 15 x + 6

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{51}{4})) = (x - \frac{5}{2})^{2}

(h, k) = (\frac{5}{2}, - \frac{51}{4}), p = \frac{1}{12}

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

p

from the center

(\frac{5}{2}, - \frac{51}{4})

along the y-axis

(\frac{5}{2}, - \frac{51}{4} + p)

= (\frac{5}{2}, - \frac{51}{4} + \frac{1}{12})

Refine

(\frac{5}{2}, - \frac{38}{3})

- 12 y + 10 = x^{2} - 4 x + 14

(x + 4)^{2} = - 6 (y - 3)

(y + 1)^{2} = 2 (x - 5)

y = 5 x^{2} - 12 x + 7

4 x = y^{2} - 4 y