What is the foci-(x-3)^2-4 ?

The foci-(x-3)^2-4 is (3,-17/4)

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foci-(x-3)^2-4

foci

- (x - 3)^{2} - 4

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

Solution steps

Rewrite

y = - (x - 3)^{2} - 4

in the standard form:

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

(h, k) = (3, - 4), p = - \frac{1}{4}

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

p

from the center

(3, - 4)

along the y-axis

(3, - 4 + p)

= (3, - 4 + (- \frac{1}{4}))

Refine

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

y = + (x - 7)^{2} + 8

(y - 5)^{2} = - 8 x

(2 x - 3)^{2}

\frac{x ^{2}}{6}

y = x^{2} + 12 x + 26