What is 2>= ((4y)/3)^2+x ?

The solution to 2>= ((4y)/3)^2+x is Parabola with (h,k)=(2,0),p=-9/64

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2>= ((4y)/3)^2+x

2 \geq (\frac{4 y}{3})^{2} + x

(h, k) = (2, 0), p = - \frac{9}{64}

Solution steps

2 \geq (\frac{4 y}{3})^{2} + x

Rewrite

2 \geq (\frac{4 y}{3})^{2} + x

in the standard form:

4 (- \frac{9}{64}) (x - 2) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (2, 0), p = - \frac{9}{64}

x^{2} + 3 x + y = 0

f (x) = 5 (x - 9)^{2}

7 x^{2} + 3 x + h (x) = 0

x^{2} - 2 m (x) + 5 m (x) = 0

x, 2 m (x)^{2} + 6 x = 0

What is 2>= ((4y)/3)^2+x ?
The solution to 2>= ((4y)/3)^2+x is Parabola with (h,k)=(2,0),p=-9/64