What is the foci y=-x^2+2x-5 ?

The foci y=-x^2+2x-5 is (1,-17/4)

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foci y=-x^2+2x-5

foci

y = - x^{2} + 2 x - 5

(1, - \frac{17}{4})

Solution steps

Rewrite

y = - x^{2} + 2 x - 5

in the standard form

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

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

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

p

from the center

(1, - 4)

along the y-axis

(1, - 4 + p)

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

Refine

(1, - \frac{17}{4})

y = 2 x^{2} - 1 + 5 x

y = 9 x^{2} + 48 x + 64

x^{2} - 12 x + 35

4.905 x^{2} + 8 x - 30

x = - \frac{1}{2} (y - 2)^{2} + 5