x=3(y+4)^2-1

Lösung

x = 3 (y + 4)^{2} - 1

(h, k) = (- 1, - 4), p = \frac{1}{12}

Schritte zur Lösung

x = 3 (y + 4)^{2} - 1

Rewrite

x = 3 (y + 4)^{2} - 1

in the standard form:

4 \cdot \frac{1}{12} (x - (- 1)) = (y - (- 4))^{2}

Therefore parabola properties are:

(h, k) = (- 1, - 4), p = \frac{1}{12}

y = \frac{1}{32} (x + 2)^{2}

x^{2} + y - 6 = 0

f (x) = \frac{2}{11} x^{2} + 6 x

+ 6 x + y^{2} - 2 y - 26 = 0

x = - y^{2} - 5