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

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

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foci 3x^2+19x-40

foci

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^{2} + 4 x = 7

x^{2} - 4 x

(x - 11)^{2}

0 = (y - 1)^{2} - 16 (x + 4)

y^{2} - 6 y - 8 x + 17 = 0