What is the foci y=2x^2-4x+9 ?

The foci y=2x^2-4x+9 is (1, 57/8)

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foci y=2x^2-4x+9

foci

y = 2 x^{2} - 4 x + 9

(1, \frac{57}{8})

Solution steps

Rewrite

y = 2 x^{2} - 4 x + 9

in the standard form

4 \cdot \frac{1}{8} (y - 7) = (x - 1)^{2}

(h, k) = (1, 7), p = \frac{1}{8}

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

p

from the center

(1, 7)

along the y-axis

(1, 7 + p)

= (1, 7 + \frac{1}{8})

Refine

(1, \frac{57}{8})

y = 2 (x - 5)^{2} - 8

y = + 2 x^{2} + 2

\frac{1}{2} (x - 3)^{2} + 4

- x^{2} + 3 x - 5

y = - \frac{1}{3} x^{2}