2x^2-x+y<= 1

Lösung

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

(h, k) = (\frac{1}{4}, \frac{9}{8}), p = - \frac{1}{8}

Schritte zur Lösung

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

Rewrite

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

in the standard form

4 (- \frac{1}{8}) (y - \frac{9}{8}) = (x - \frac{1}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{4}, \frac{9}{8}), p = - \frac{1}{8}

x = 6 y^{2}

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

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

15 x^{2} = - 6 y

2 y^{2} + 20 x + 12 y - 1 = 0