What is the foci y=+0.5(x+2)(x-4) ?

The foci y=+0.5(x+2)(x-4) is (1,-4)

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foci y=+0.5(x+2)(x-4)

foci

y = + 0.5 (x + 2) (x - 4)

(1, - 4)

Solution steps

Rewrite

y = 0.5 (x + 2) (x - 4)

in the standard form

4 \cdot 0.5 (y - (- 4.5)) = (x - 1)^{2}

(h, k) = (1, - 4.5), p = 0.5

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

p

from the center

(1, - 4.5)

along the y-axis

(1, - 4.5 + p)

= (1, - 4.5 + 0.5)

Refine

(1, - 4)

y = 2.1 + 4.4 x - 0.7 x^{2}

y = x^{2} - 10 x - 4

(x - 2)^{2} = 8 (y + 1)

x^{2} + 2

5 x^{2} + 210 x - 34775