What is the foci y=2x^2-12x+22 ?

The foci y=2x^2-12x+22 is (3, 33/8)

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foci y=2x^2-12x+22

foci

y = 2 x^{2} - 12 x + 22

(3, \frac{33}{8})

Solution steps

Rewrite

y = 2 x^{2} - 12 x + 22

in the standard form

4 \cdot \frac{1}{8} (y - 4) = (x - 3)^{2}

(h, k) = (3, 4), p = \frac{1}{8}

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

p

from the center

(3, 4)

along the y-axis

(3, 4 + p)

= (3, 4 + \frac{1}{8})

Refine

(3, \frac{33}{8})

y = - x^{2} + 44 x + 1

- 16 t^{2} + 60 t + 3

y = \frac{x ^{3}}{x}

(y - 2)^{2} = 12 (x + 4)

10 x^{2} - 19 x + 5