What is x-4y^2>= 0.2-x-7 ?

The solution to x-4y^2>= 0.2-x-7 is Parabola with (h,k)=(-3.4,0),p= 1/8

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x-4y^2>= 0.2-x-7

x - 4 y^{2} \geq 0.2 - x - 7

(h, k) = (- 3.4, 0), p = \frac{1}{8}

Solution steps

x - 4 y^{2} \geq 0.2 - x - 7

Rewrite

x - 4 y^{2} \geq 0.2 - x - 7

in the standard form:

4 \cdot \frac{1}{8} (x - (- 3.4)) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (- 3.4, 0), p = \frac{1}{8}

x^{2} = y - 7

x^{2} - 10 x + 6 y = 1

x^{2} + (y - 4) 2 = 25

\frac{x ^{2}}{9} + \frac{y}{16} = 1

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

What is x-4y^2>= 0.2-x-7 ?
The solution to x-4y^2>= 0.2-x-7 is Parabola with (h,k)=(-3.4,0),p= 1/8