What is the foci 16t^2+48t+4 ?

The foci 16t^2+48t+4 is (-3/2 ,-2047/64)

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foci 16t^2+48t+4

foci

16 t^{2} + 48 t + 4

(- \frac{3}{2}, - \frac{2047}{64})

Solution steps

Rewrite

y = 16 t^{2} + 48 t + 4

in the standard form

4 \cdot \frac{1}{64} (y - (- 32)) = (t - (- \frac{3}{2}))^{2}

(h, k) = (- \frac{3}{2}, - 32), p = \frac{1}{64}

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

p

from the center

(- \frac{3}{2}, - 32)

along the y-axis

(- \frac{3}{2}, - 32 + p)

= (- \frac{3}{2}, - 32 + \frac{1}{64})

Refine

(- \frac{3}{2}, - \frac{2047}{64})

- 0.4 x^{2} - 5

x^{2} + 6 x = 4 y - 5

18 x^{2} + y = 0

(x + 1)^{2} = - 3 (y - 6)

y = 3 x^{2} - 11 x + 10