y<= 3x^2-6x+1

Lösung

y \leq 3 x^{2} - 6 x + 1

(h, k) = (1, - 2), p = \frac{1}{12}

Schritte zur Lösung

y \leq 3 x^{2} - 6 x + 1

Rewrite

y \leq 3 x^{2} - 6 x + 1

in the standard form

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

Therefore parabola properties are:

(h, k) = (1, - 2), p = \frac{1}{12}

16 y^{2} - x \geq 0

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

x = 7 y^{2}

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

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