3x^2+2x-3y+5=0

Lösung

3 x^{2} + 2 x - 3 y + 5 = 0

(h, k) = (- \frac{1}{3}, \frac{14}{9}), p = \frac{1}{4}

Schritte zur Lösung

3 x^{2} + 2 x - 3 y + 5 = 0

Rewrite

3 x^{2} + 2 x - 3 y + 5 = 0

in the standard form

4 \cdot \frac{1}{4} (y - \frac{14}{9}) = (x - (- \frac{1}{3}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{3}, \frac{14}{9}), p = \frac{1}{4}

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

x^{2} - 2 y = 16

x = - y^{2} + 40 y - 371

x^{2} - 7 x + k (x) = 0

x - 5 = y^{2}