What is y+4x<= x^2+2 ?

The solution to y+4x<= x^2+2 is Parabola with (h,k)=(2,-2),p= 1/4

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y+4x<= x^2+2

y + 4 x \leq x^{2} + 2

(h, k) = (2, - 2), p = \frac{1}{4}

Solution steps

y + 4 x \leq x^{2} + 2

Rewrite

y + 4 x \leq x^{2} + 2

in the standard form

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

Therefore parabola properties are:

(h, k) = (2, - 2), p = \frac{1}{4}

y^{2} - y - x + 1 = 0

9 y^{2} - 16 x^{2} = 81

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

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

x^{2} + y^{2} + 8 x - 2 y - 19 = 0

What is y+4x<= x^2+2 ?
The solution to y+4x<= x^2+2 is Parabola with (h,k)=(2,-2),p= 1/4