What is 9x^2-4(y-2)-36>= 0 ?

The solution to 9x^2-4(y-2)-36>= 0 is Parabola with (h,k)=(0,-7),p= 1/9

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9x^2-4(y-2)-36>= 0

9 x^{2} - 4 (y - 2) - 36 \geq 0

(h, k) = (0, - 7), p = \frac{1}{9}

Solution steps

9 x^{2} - 4 (y - 2) - 36 \geq 0

Rewrite

9 x^{2} - 4 (y - 2) - 36 \geq 0

in the standard form:

4 \cdot \frac{1}{9} (y - (- 7)) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - 7), p = \frac{1}{9}

x^{2} - 13 x + 42 = o

y \frac{y + 1}{x + 0} = 3

f (x) = x^{2} + 12 x + 27

5 y^{2} - 35 x - 40 x - 25 = 0

y^{2} + 12 y - 5 x + 16 = 0

What is 9x^2-4(y-2)-36>= 0 ?
The solution to 9x^2-4(y-2)-36>= 0 is Parabola with (h,k)=(0,-7),p= 1/9