16y^2-x>= 0

Lösung

16 y^{2} - x \geq 0

(h, k) = (0, 0), p = \frac{1}{64}

Schritte zur Lösung

16 y^{2} - x \geq 0

Rewrite

16 y^{2} - x \geq 0

in the standard form:

4 \cdot \frac{1}{64} (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{1}{64}

y^{2} = - 4 (x + 6)

x = 7 y^{2}

y < - x^{2} + 4 x + 5

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

y^{2} + 33 = - 8 x - 23