What is y-3x^2+2x-36<= 0 ?

The solution to y-3x^2+2x-36<= 0 is Parabola with (h,k)=(1/3 , 107/3),p= 1/12

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y-3x^2+2x-36<= 0

y - 3 x^{2} + 2 x - 36 \leq 0

(h, k) = (\frac{1}{3}, \frac{107}{3}), p = \frac{1}{12}

Solution steps

y - 3 x^{2} + 2 x - 36 \leq 0

Rewrite

y - 3 x^{2} + 2 x - 36 \leq 0

in the standard form

4 \cdot \frac{1}{12} (y - \frac{107}{3}) = (x - \frac{1}{3})^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{3}, \frac{107}{3}), p = \frac{1}{12}

x^{(2)} - 10 x + 8 y + 33 = 0

x^{2} + y = 121

x^{2} + 4 x + y + 3 = 0

y > - x^{2} - 5

3 x^{2} - 6 y = 0

What is y-3x^2+2x-36<= 0 ?
The solution to y-3x^2+2x-36<= 0 is Parabola with (h,k)=(1/3 , 107/3),p= 1/12