x=y^2+2y+4

Lösung

x = y^{2} + 2 y + 4

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

Schritte zur Lösung

x = y^{2} + 2 y + 4

Rewrite

x = y^{2} + 2 y + 4

in the standard form

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

Therefore parabola properties are:

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

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

y^{2} = x + 144

2 f (x) = x^{2}

x = y^{2} + 81

y^{2} + 4 x + 2 y - 19 = 0