What is the foci y=2x^2-1+5x ?

The foci y=2x^2-1+5x is (-5/4 ,-4)

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foci y=2x^2-1+5x

foci

y = 2 x^{2} - 1 + 5 x

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

Solution steps

Rewrite

y = 2 x^{2} - 1 + 5 x

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{33}{8})) = (x - (- \frac{5}{4}))^{2}

(h, k) = (- \frac{5}{4}, - \frac{33}{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{33}{8})

along the y-axis

(- \frac{5}{4}, - \frac{33}{8} + p)

= (- \frac{5}{4}, - \frac{33}{8} + \frac{1}{8})

Refine

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

y = 9 x^{2} + 48 x + 64

x^{2} - 12 x + 35

4.905 x^{2} + 8 x - 30

x = - \frac{1}{2} (y - 2)^{2} + 5

(x + 6)^{2} = - 8 (y + 9)