2x^2-x+4m(x)=0

Lösung

2 x^{2} - x + 4 m (x) = 0

(h, k) = (\frac{1}{4}, \frac{1}{32}), p = - \frac{1}{2}

Schritte zur Lösung

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

Rewrite

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

in the standard form

4 (- \frac{1}{2}) (y - \frac{1}{32}) = (x - \frac{1}{4})^{2}

Therefore parabola properties are:

(h, k) = (\frac{1}{4}, \frac{1}{32}), p = - \frac{1}{2}

y = 6 x - 3 x^{2}

2 x^{2} + 8 y = 15

c \frac{c}{a} = \frac{- 1}{2}

- 5 = - x^{2} + y + 16 x - 16

(x + 2) = (y + 2)^{2}