What is the foci 1/2 x^2-7x+7 ?

The foci 1/2 x^2-7x+7 is (7,-17)

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foci 1/2 x^2-7x+7

foci

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

(7, - 17)

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{2} (y - (- \frac{35}{2})) = (x - 7)^{2}

(h, k) = (7, - \frac{35}{2}), p = \frac{1}{2}

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

p

from the center

(7, - \frac{35}{2})

along the y-axis

(7, - \frac{35}{2} + p)

= (7, - \frac{35}{2} + \frac{1}{2})

Refine

(7, - 17)

(2 x - 5)^{2}

y = + x^{2} + 6 x + 3

4 (x + 2)^{2} + 5

(y - 8)^{2} = 16 (x - 4)

y = \frac{x ^{3} - 3 x ^{2} - 4 x}{x - 4}