x=y^2+4y-1

Lösung

x = y^{2} + 4 y - 1

(h, k) = (- 5, - 2), p = \frac{1}{4}

Schritte zur Lösung

x = y^{2} + 4 y - 1

Rewrite

x = y^{2} + 4 y - 1

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 5, - 2), p = \frac{1}{4}

(y + 2)^{2} = 16 (x + 1)

7 x^{2} + y = 5

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

(x + 0)^{2} = 8 (y + 1)

3 f (2 x) = 2 x^{2} - 3 x + 1