What is the foci y=3x^2+19x-40 ?

The foci y=3x^2+19x-40 is (-19/6 ,-70)

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foci y=3x^2+19x-40

foci

y = 3 x^{2} + 19 x - 40

(- \frac{19}{6}, - 70)

Solution steps

Rewrite

y = 3 x^{2} + 19 x - 40

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{841}{12})) = (x - (- \frac{19}{6}))^{2}

(h, k) = (- \frac{19}{6}, - \frac{841}{12}), p = \frac{1}{12}

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

p

from the center

(- \frac{19}{6}, - \frac{841}{12})

along the y-axis

(- \frac{19}{6}, - \frac{841}{12} + p)

= (- \frac{19}{6}, - \frac{841}{12} + \frac{1}{12})

Refine

(- \frac{19}{6}, - 70)

(y - 3)^{2} + 24 (x + 2) = 0

2 x^{2}

- 2 x^{2} - 2 x - 2

y = x^{2} + 3 x - 1

3 x^{2} - 12 x + 10