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

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

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

foci

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

(- \frac{3}{2}, 0)

Solution steps

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{1}{4})) = (x - (- \frac{3}{2}))^{2}

(h, k) = (- \frac{3}{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{3}{2}, - \frac{1}{4})

along the y-axis

(- \frac{3}{2}, - \frac{1}{4} + p)

= (- \frac{3}{2}, - \frac{1}{4} + \frac{1}{4})

Refine

(- \frac{3}{2}, 0)

3 x^{2} - 16 x + 5

y^{2} - 12 y + 6 x + 36 = 0

x^{2} - 6 x - 20 y - 11 = 0

y = 3 x^{2} - 36 x + 81

(y - 6)^{2}