x=y^2-4y+1

Lösung

x = y^{2} - 4 y + 1

(h, k) = (- 3, 2), p = \frac{1}{4}

Schritte zur Lösung

x = y^{2} - 4 y + 1

Rewrite

x = y^{2} - 4 y + 1

in the standard form

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

Therefore parabola properties are:

(h, k) = (- 3, 2), p = \frac{1}{4}

x - 1 = y^{2}

y^{2} = 20 (x - 1)

y \geq x^{2} + 6

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

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