What is y-3x^2+x-25<= 0 ?

The solution to y-3x^2+x-25<= 0 is Parabola with (h,k)=(1/6 , 299/12),p= 1/12

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y-3x^2+x-25<= 0

y - 3 x^{2} + x - 25 \leq 0

(h, k) = (\frac{1}{6}, \frac{299}{12}), p = \frac{1}{12}

Solution steps

y - 3 x^{2} + x - 25 \leq 0

Rewrite

y - 3 x^{2} + x - 25 \leq 0

in the standard form

4 \cdot \frac{1}{12} (y - \frac{299}{12}) = (x - \frac{1}{6})^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{6}, \frac{299}{12}), p = \frac{1}{12}

\frac{x ^{2}}{36} = 4 y

x^{2} - 8 x - 4 y + 24 = 0

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

x^{2} - 4 x - 8 y - 4 = 0

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

What is y-3x^2+x-25<= 0 ?
The solution to y-3x^2+x-25<= 0 is Parabola with (h,k)=(1/6 , 299/12),p= 1/12