What is the foci (x+5)^2+4 ?

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

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foci (x+5)^2+4

foci

(x + 5)^{2} + 4

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

Solution steps

Rewrite

y = (x + 5)^{2} + 4

in the standard form:

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

(h, k) = (- 5, 4), p = \frac{1}{4}

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

p

from the center

(- 5, 4)

along the y-axis

(- 5, 4 + p)

= (- 5, 4 + \frac{1}{4})

Refine

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

y = - 8 x^{2}

y^{2} = \frac{16}{3} x

\frac{( x - 3 ) ^{2}}{8} + 5

(x + 5)^{2} + 3

- 0.1 x^{2} + 1.2 x + 98.6