What is r(x)-x-2x^2>= 0 ?

The solution to r(x)-x-2x^2>= 0 is Parabola with (h,k)=(-1/4 ,-1/8),p= 1/8

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r(x)-x-2x^2>= 0

r (x) - x - 2 x^{2} \geq 0

(h, k) = (- \frac{1}{4}, - \frac{1}{8}), p = \frac{1}{8}

Solution steps

y - x - 2 x^{2} \geq 0

Rewrite

y - x - 2 x^{2} \geq 0

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{1}{8})) = (x - (- \frac{1}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{4}, - \frac{1}{8}), p = \frac{1}{8}

2 x - x^{2} + r (x) = 0

(x + 3)^{2} + (y - 2) = 13

y = - x^{2} + 22 x - 105

x = \frac{1}{20} y^{2}

x^{2} - 43 y + 348 = 0

What is r(x)-x-2x^2>= 0 ?
The solution to r(x)-x-2x^2>= 0 is Parabola with (h,k)=(-1/4 ,-1/8),p= 1/8