What is the foci-16t^2+60t+3 ?

The foci-16t^2+60t+3 is (15/8 , 3791/64)

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foci-16t^2+60t+3

foci

- 16 t^{2} + 60 t + 3

(\frac{15}{8}, \frac{3791}{64})

Solution steps

Rewrite

y = - 16 t^{2} + 60 t + 3

in the standard form

4 (- \frac{1}{64}) (y - \frac{237}{4}) = (t - \frac{15}{8})^{2}

(h, k) = (\frac{15}{8}, \frac{237}{4}), p = - \frac{1}{64}

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

p

from the center

(\frac{15}{8}, \frac{237}{4})

along the y-axis

(\frac{15}{8}, \frac{237}{4} + p)

= (\frac{15}{8}, \frac{237}{4} + (- \frac{1}{64}))

Refine

(\frac{15}{8}, \frac{3791}{64})

y = \frac{x ^{3}}{x}

(y - 2)^{2} = 12 (x + 4)

10 x^{2} - 19 x + 5

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

x^{2} + x - 5