What is the foci y=(x^3-3x^2-4x)/(x-4) ?

The foci y=(x^3-3x^2-4x)/(x-4) is (-1/2 ,0)

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foci y=(x^3-3x^2-4x)/(x-4)

foci

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

(- \frac{1}{2}, 0)

Solution steps

Rewrite

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

in the standard form

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

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

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

p

from the center

(- \frac{1}{2}, - \frac{1}{4})

along the y-axis

(- \frac{1}{2}, - \frac{1}{4} + p)

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

Refine

(- \frac{1}{2}, 0)

y = s^{2}

3 x^{2} - 31 x + 56 - (x - 8)

y^{2} = - 25 x

(x + 2)^{2} = 20 (y - 5)

y^{2} = - x

What is the foci y=(x^3-3x^2-4x)/(x-4) ?
The foci y=(x^3-3x^2-4x)/(x-4) is (-1/2 ,0)