What is 0>= 0.1x^2-200q(x)+19000 ?

The solution to 0>= 0.1x^2-200q(x)+19000 is Parabola with (h,k)=(0,95),p=500

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0>= 0.1x^2-200q(x)+19000

0 \geq 0.1 x^{2} - 200 q (x) + 19000

(h, k) = (0, 95), p = 500

Solution steps

0 \geq 0.1 x^{2} - 200 y + 19000

Rewrite

0 \geq 0.1 x^{2} - 200 y + 19000

in the standard form:

4 \cdot 500 (y - 95) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 95), p = 500

(x - 2)^{2} = - 6 (y - 1)

y^{2} - 8 y - 12 x + 76 = 0

x = - \frac{1}{16} (y - 2)^{2} - 3

x^{2} - 6 x + 2 y - 8 = 0

x^{2} - 4 x - 5 y + 7 = 0

What is 0>= 0.1x^2-200q(x)+19000 ?
The solution to 0>= 0.1x^2-200q(x)+19000 is Parabola with (h,k)=(0,95),p=500