What is 3x^2+4x+y+2=0 ?

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

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

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

(h, k) = (- \frac{2}{3}, - \frac{2}{3}), p = - \frac{1}{12}

Solution steps

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

(h, k) = (- \frac{2}{3}, - \frac{2}{3}), p = - \frac{1}{12}

(X + 10)^{2} = 3 (y - 12)

x - y^{2} = - 3

x^{2} = \frac{6 r ( x )}{3} - \frac{1}{2}

x^{2} + 3 x - \frac{y}{2} = 7

x^{2} + 6 x - 3 y + 4 = 0

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