What is the foci 3x^2-31x+56-(x-8) ?

The foci 3x^2-31x+56-(x-8) is (16/3 ,-85/4)

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foci 3x^2-31x+56-(x-8)

foci

(\frac{16}{3}, - \frac{85}{4})

Solution steps

Rewrite

y = 3 x^{2} - 31 x + 56 - (x - 8)

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{64}{3})) = (x - \frac{16}{3})^{2}

(h, k) = (\frac{16}{3}, - \frac{64}{3}), p = \frac{1}{12}

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

p

from the center

(\frac{16}{3}, - \frac{64}{3})

along the y-axis

(\frac{16}{3}, - \frac{64}{3} + p)

= (\frac{16}{3}, - \frac{64}{3} + \frac{1}{12})

Refine

(\frac{16}{3}, - \frac{85}{4})

What is the foci 3x^2-31x+56-(x-8) ?
The foci 3x^2-31x+56-(x-8) is (16/3 ,-85/4)