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

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

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foci y= 1/2 (1-2x-x^2)

foci

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

(- 1, \frac{1}{2})

Solution steps

Rewrite

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

in the standard form

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

(h, k) = (- 1, 1), p = - \frac{1}{2}

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

p

from the center

(- 1, 1)

along the y-axis

(- 1, 1 + p)

= (- 1, 1 + (- \frac{1}{2}))

Refine

(- 1, \frac{1}{2})

y = \frac{1}{4} (x + 3)^{2} - 5

- 2 x^{2} + 3 x

x^{2} - 5 x + 4

- 0.2 t^{2} + 1.6 t + 98.4

y = x^{2} - 6 x - 3