x=y^2+4y+4

Lösung

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

(h, k) = (0, - 2), p = \frac{1}{4}

Schritte zur Lösung

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (x - 0) = (y - (- 2))^{2}

Therefore parabola properties are:

(h, k) = (0, - 2), p = \frac{1}{4}

5 y = x^{2}

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

(y + 2)^{2} = 12 x

- f (x) = 3 x^{2} + 2 x - 3

y^{2} - 4 y - 8 x + 20 = 0