3x^2+4y=12

Lösung

3 x^{2} + 4 y = 12

(h, k) = (0, 3), p = - \frac{1}{3}

Schritte zur Lösung

3 x^{2} + 4 y = 12

Rewrite

3 x^{2} + 4 y = 12

in the standard form:

4 (- \frac{1}{3}) (y - 3) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 3), p = - \frac{1}{3}

y^{2} = 3 x - 3

y < x^{2} + 1

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

2 f (x) = x^{2} - 5 x + 4

3 (4 + y^{2}) = 5 z