What is the foci y=(x-2)(x+1) ?

The foci y=(x-2)(x+1) is (1/2 ,-2)

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foci y=(x-2)(x+1)

foci

y = (x - 2) (x + 1)

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

Solution steps

Rewrite

y = (x - 2) (x + 1)

in the standard form

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

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

along the y-axis

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

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

Refine

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

x = 5 (y + 3)^{2} + 6

y = \frac{3 \cdot \frac{1}{2} x ^{2}}{2}

5 x^{2} + 3

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

y = 3 x^{2} - 6 x + 5