x+(y+1)^2=2

Lösung

x + (y + 1)^{2} = 2

(h, k) = (2, - 1), p = - \frac{1}{4}

Schritte zur Lösung

x + (y + 1)^{2} = 2

Rewrite

x + (y + 1)^{2} = 2

in the standard form:

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

Therefore parabola properties are:

(h, k) = (2, - 1), p = - \frac{1}{4}

y \leq x^{2} + 4 x + 3

x^{2} - 4 x - 2 y + 11 = 0

y^{2} = 5 x + 1

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

x^{2} + 6 x + 3 y + 12 = 0