What is the foci y=x^2-12x+36 ?

The foci y=x^2-12x+36 is (6, 1/4)

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foci y=x^2-12x+36

foci

y = x^{2} - 12 x + 36

(6, \frac{1}{4})

Solution steps

Rewrite

y = x^{2} - 12 x + 36

in the standard form

4 \cdot \frac{1}{4} (y - 0) = (x - 6)^{2}

(h, k) = (6, 0), p = \frac{1}{4}

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

p

from the center

(6, 0)

along the y-axis

(6, 0 + p)

= (6, 0 + \frac{1}{4})

Refine

(6, \frac{1}{4})

- x^{2} + 16 x - 94

2 x^{2} + 24 x - 16

(y + 1.5)^{2} = 4 (x - 2)

\frac{1}{2} x^{2} - 7 x + 7

(2 x - 5)^{2}