-x^2+6x+y-12=0

Lösung

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

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

Schritte zur Lösung

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

Rewrite

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

in the standard form

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

Therefore parabola properties are:

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

y > x^{2} - 4

y^{2} + 8 x - 16 = 0

y^{2} - 5 x + 6 y + 13 = 0

\frac{x}{2} x - 3 y = 6

x^{2} + y \geq 0