2x+y^2-y=1

Lösung

2 x + y^{2} - y = 1

(h, k) = (\frac{5}{8}, \frac{1}{2}), p = - \frac{1}{2}

Schritte zur Lösung

2 x + y^{2} - y = 1

Rewrite

2 x + y^{2} - y = 1

in the standard form

4 (- \frac{1}{2}) (x - \frac{5}{8}) = (y - \frac{1}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{5}{8}, \frac{1}{2}), p = - \frac{1}{2}

x + y^{2} - 3 \geq 0

x \frac{x}{5} = \frac{y}{4}

y^{2} = 4 x + 9

4 x + y^{2} = 32

2 x^{2} + y \geq 0