What is the foci y=x^2+7x+10 ?

The foci y=x^2+7x+10 is (-7/2 ,-2)

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foci y=x^2+7x+10

foci

y = x^{2} + 7 x + 10

(- \frac{7}{2}, - 2)

Solution steps

Rewrite

y = x^{2} + 7 x + 10

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{9}{4})) = (x - (- \frac{7}{2}))^{2}

(h, k) = (- \frac{7}{2}, - \frac{9}{4}), p = \frac{1}{4}

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

p

from the center

(- \frac{7}{2}, - \frac{9}{4})

along the y-axis

(- \frac{7}{2}, - \frac{9}{4} + p)

= (- \frac{7}{2}, - \frac{9}{4} + \frac{1}{4})

Refine

(- \frac{7}{2}, - 2)

y = (2 x - 1) (4 x - 1)

(y + 6)^{2} = 12 (x + 1)

(x + 7)^{2} = 5 (y + 3)

x^{2} - 8 y = 0

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