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

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

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

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

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

Solution steps

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

Rewrite

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

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 + 4)

y = x^{2} - 2 x - 8 y

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

(y - 2)^{2} = - 6 (x + 4)

Y^{2} - 8 X + Y + 13 = 0

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