What is -x^2+11x+y-10<0 ?

The solution to -x^2+11x+y-10<0 is Parabola with (h,k)=(11/2 ,-81/4),p= 1/4

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-x^2+11x+y-10<0

- x^{2} + 11 x + y - 10 < 0

(h, k) = (\frac{11}{2}, - \frac{81}{4}), p = \frac{1}{4}

Solution steps

- x^{2} + 11 x + y - 10 < 0

Rewrite

- x^{2} + 11 x + y - 10 < 0

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{81}{4})) = (x - \frac{11}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{11}{2}, - \frac{81}{4}), p = \frac{1}{4}

6 (y - 15) = - 2 (x^{2})

3 f (x) - 1 = x^{2} - 1

3 y^{2} + 3 x = 0

1 6^{2} + 25 y^{2} - 64 x + 150 y - 11 = 0

(y - 0)^{2} = - 4 (x - 0)

What is -x^2+11x+y-10<0 ?
The solution to -x^2+11x+y-10<0 is Parabola with (h,k)=(11/2 ,-81/4),p= 1/4