y^2-4y-x+3=0

Lösung

y^{2} - 4 y - x + 3 = 0

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

Schritte zur Lösung

y^{2} - 4 y - x + 3 = 0

Rewrite

y^{2} - 4 y - x + 3 = 0

in the standard form

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

Therefore parabola properties are:

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

(x + 3)^{2} = 24 (y + 3)

y - 2 x^{2} \geq 2

y - 3 x^{2} = 6

2 x^{2} - 8 y - 12 = 0

x^{2} \leq y