What is 3X^2+12X-3(2Y-4)+12=0 ?

The solution to 3X^2+12X-3(2Y-4)+12=0 is Parabola with (h,k)=(-2,0),p= 1/2

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3X^2+12X-3(2Y-4)+12=0

3 X^{2} + 12 X - 3 (2 Y - 4) + 12 = 0

(h, k) = (- 2, 0), p = \frac{1}{2}

Solution steps

3 X^{2} + 12 X - 3 (2 Y - 4) + 12 = 0

Rewrite

3 X^{2} + 12 X - 3 (2 Y - 4) + 12 = 0

in the standard form

4 \cdot \frac{1}{2} (Y - 0) = (X - (- 2))^{2}

Therefore parabola properties are:

(h, k) = (- 2, 0), p = \frac{1}{2}

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

y^{2} + 16 x + 4 y + 4 = 0

x = - y^{2} + 14

(x + 2)^{2} = - 16 (y - 3)

y > x^{2} + 3 x - 4

What is 3X^2+12X-3(2Y-4)+12=0 ?
The solution to 3X^2+12X-3(2Y-4)+12=0 is Parabola with (h,k)=(-2,0),p= 1/2