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

The solution to 2x^2y^0+3xy^0-y=0 is Parabola with (h,k)=(-3/4 ,-9/8),p= 1/8

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

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

(h, k) = (- \frac{3}{4}, - \frac{9}{8}), p = \frac{1}{8}

Solution steps

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{9}{8})) = (x - (- \frac{3}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{4}, - \frac{9}{8}), p = \frac{1}{8}

x^{2} - 2 x + 4 y + 9 = 0

y = - (x - 1) \cdot (x - 3)

q^{2} = g (- h + a)

f (x) (a + b) = x^{2}

(x - 5)^{2} = 24 (y - 3)

What is 2x^2y^0+3xy^0-y=0 ?
The solution to 2x^2y^0+3xy^0-y=0 is Parabola with (h,k)=(-3/4 ,-9/8),p= 1/8