What is the foci 4x^2+2x+4 ?

The foci 4x^2+2x+4 is (-1/4 , 61/16)

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foci 4x^2+2x+4

foci

4 x^{2} + 2 x + 4

(- \frac{1}{4}, \frac{61}{16})

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{16} (y - \frac{15}{4}) = (x - (- \frac{1}{4}))^{2}

(h, k) = (- \frac{1}{4}, \frac{15}{4}), p = \frac{1}{16}

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

p

from the center

(- \frac{1}{4}, \frac{15}{4})

along the y-axis

(- \frac{1}{4}, \frac{15}{4} + p)

= (- \frac{1}{4}, \frac{15}{4} + \frac{1}{16})

Refine

(- \frac{1}{4}, \frac{61}{16})

y^{2} + 16 x = 0

y = 16 - x^{2}

y = 196 x^{2} - 112 x - 240

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

y = x^{2} - 5 x 3