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

Lösung

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

(h, k) = (0, - \frac{5}{4}), p = 1

Schritte zur Lösung

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

Rewrite

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

in the standard form:

4 \cdot 1 \cdot (y - (- \frac{5}{4})) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, - \frac{5}{4}), p = 1

2 A^{2} + 3 B = 12

- 2 x^{2} \geq y

f (x) = - (x - 2) (x + 1)

x^{2} + 8 x + k (x) = 0

x^{2} + 6 x + 8 y - 15 = 0