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

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

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foci y=-2x^2+x-1

foci

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

(\frac{1}{4}, - 1)

Solution steps

Rewrite

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

in the standard form

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

(h, k) = (\frac{1}{4}, - \frac{7}{8}), p = - \frac{1}{8}

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

p

from the center

(\frac{1}{4}, - \frac{7}{8})

along the y-axis

(\frac{1}{4}, - \frac{7}{8} + p)

= (\frac{1}{4}, - \frac{7}{8} + (- \frac{1}{8}))

Refine

(\frac{1}{4}, - 1)

4 x^{2} + 2 x + 4

y^{2} + 16 x = 0

y = 16 - x^{2}

y = 196 x^{2} - 112 x - 240

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