What is the foci y=5x^2+40x+79 ?

The foci y=5x^2+40x+79 is (-4,-19/20)

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foci y=5x^2+40x+79

foci

y = 5 x^{2} + 40 x + 79

(- 4, - \frac{19}{20})

Solution steps

Rewrite

y = 5 x^{2} + 40 x + 79

in the standard form

4 \cdot \frac{1}{20} (y - (- 1)) = (x - (- 4))^{2}

(h, k) = (- 4, - 1), p = \frac{1}{20}

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

p

from the center

(- 4, - 1)

along the y-axis

(- 4, - 1 + p)

= (- 4, - 1 + \frac{1}{20})

Refine

(- 4, - \frac{19}{20})

x^{2} + 3 x - 2

x^{2} + 3 \cdot x + 2

y = x^{2} - 4 x + 2

y = x^{2} - 4 x + 1

y = 2 x^{2} - 12 x + 22