What is the foci y=2x^2+5x-3 ?

The foci y=2x^2+5x-3 is (-5/4 ,-6)

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foci y=2x^2+5x-3

foci

y = 2 x^{2} + 5 x - 3

(- \frac{5}{4}, - 6)

Solution steps

Rewrite

y = 2 x^{2} + 5 x - 3

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{49}{8})) = (x - (- \frac{5}{4}))^{2}

(h, k) = (- \frac{5}{4}, - \frac{49}{8}), p = \frac{1}{8}

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

p

from the center

(- \frac{5}{4}, - \frac{49}{8})

along the y-axis

(- \frac{5}{4}, - \frac{49}{8} + p)

= (- \frac{5}{4}, - \frac{49}{8} + \frac{1}{8})

Refine

(- \frac{5}{4}, - 6)

(x + 9)^{2} = - 40 (y - 7)

y = x^{2} - 14 x - 49

4 x - x^{2}

\frac{1}{2} (1 - 2 y - y^{2})

(x + 7)^{2}