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

The solution to y^2+3y-4x+2=0 is Parabola with (h,k)=(-1/16 ,-3/2),p=1

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

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

(h, k) = (- \frac{1}{16}, - \frac{3}{2}), p = 1

Solution steps

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

Rewrite

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

in the standard form

4 \cdot 1 \cdot (x - (- \frac{1}{16})) = (y - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{16}, - \frac{3}{2}), p = 1

x^{2} - 9 x^{2} + 20 y - 80 x - 2 = 0

(y + 2)^{2} = 3 (x - \frac{11}{3})

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

15 = \frac{\frac{\frac{x}{5 + R ( x )}}{R ( x )}}{15}

(x - 2)^{2} + (y - 0)^{2} = (x + 3)^{2}

What is y^2+3y-4x+2=0 ?
The solution to y^2+3y-4x+2=0 is Parabola with (h,k)=(-1/16 ,-3/2),p=1