What is the foci y=-2(x-3)^2+1 ?

The foci y=-2(x-3)^2+1 is (3, 7/8)

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foci y=-2(x-3)^2+1

foci

y = - 2 (x - 3)^{2} + 1

(3, \frac{7}{8})

Solution steps

Rewrite

y = - 2 (x - 3)^{2} + 1

in the standard form:

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

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

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

p

from the center

(3, 1)

along the y-axis

(3, 1 + p)

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

Refine

(3, \frac{7}{8})

x^{2} + 4 x - 8 y + 12 = 0

y = x^{2} + x + 1

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

y = x^{2} - 2 x^{2}

y = - \frac{1}{8} (x - 2)^{2} + 1