What is the foci x^2-13x+40 ?

The foci x^2-13x+40 is (13/2 ,-2)

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foci x^2-13x+40

foci

x^{2} - 13 x + 40

(\frac{13}{2}, - 2)

Solution steps

Rewrite

y = x^{2} - 13 x + 40

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{9}{4})) = (x - \frac{13}{2})^{2}

(h, k) = (\frac{13}{2}, - \frac{9}{4}), p = \frac{1}{4}

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

p

from the center

(\frac{13}{2}, - \frac{9}{4})

along the y-axis

(\frac{13}{2}, - \frac{9}{4} + p)

= (\frac{13}{2}, - \frac{9}{4} + \frac{1}{4})

Refine

(\frac{13}{2}, - 2)

- 2 x^{2} - 8 x - 12

\frac{4 x ^{3} - 7 x ^{2} - 11 x + 5}{4 x + 5}

y = 3 x^{2} - 5 x

2 x^{2} - 12 x + 14 - 14 y = 0

x = - 3 y^{2} + 5