x^2+y+12x-10y=12

Lösung

x^{2} + y + 12 x - 10 y = 12

(h, k) = (- 6, - \frac{16}{3}), p = \frac{9}{4}

Schritte zur Lösung

x^{2} + y + 12 x - 10 y = 12

Rewrite

x^{2} + y + 12 x - 10 y = 12

in the standard form

4 \cdot \frac{9}{4} (y - (- \frac{16}{3})) = (x - (- 6))^{2}

Therefore parabola properties are:

(h, k) = (- 6, - \frac{16}{3}), p = \frac{9}{4}

\frac{y}{x} < x - 1

(x - 3)^{2} = 5 (y + 2)

x^{2} - 2 - 8 \leq o

2 y^{2} - 6 x = 0

0.59 x^{2} - x + z = 0