What is 3x^2+2x-5y+7=0 ?

The solution to 3x^2+2x-5y+7=0 is Parabola with (h,k)=(-1/3 , 4/3),p= 5/12

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3x^2+2x-5y+7=0

3 x^{2} + 2 x - 5 y + 7 = 0

(h, k) = (- \frac{1}{3}, \frac{4}{3}), p = \frac{5}{12}

Solution steps

3 x^{2} + 2 x - 5 y + 7 = 0

Rewrite

3 x^{2} + 2 x - 5 y + 7 = 0

in the standard form

4 \cdot \frac{5}{12} (y - \frac{4}{3}) = (x - (- \frac{1}{3}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{3}, \frac{4}{3}), p = \frac{5}{12}

y + 2 = x^{2} - x

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

(g (x) 4) = 4 x^{2} - 8 x + 1

0 = 2 x^{2} - y - 12 x + 11

y + 1 = x^{2}

What is 3x^2+2x-5y+7=0 ?
The solution to 3x^2+2x-5y+7=0 is Parabola with (h,k)=(-1/3 , 4/3),p= 5/12