What is the foci y=3-2x-x^2 ?

The foci y=3-2x-x^2 is (-1, 15/4)

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foci y=3-2x-x^2

foci

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

(- 1, \frac{15}{4})

Solution steps

Rewrite

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

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{15}{4})

(y - 2)^{2} = 16 (x + 1)

6 x - 8 y + 15 - x^{2} = 0

4 x^{2} - 8 x + 1

x^{2} - 8 x + 2 y + 7 = 0

x^{2} - 4 x - 7