x=y^2+y+1

Lösung

x = y^{2} + y + 1

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

Schritte zur Lösung

x = y^{2} + y + 1

Rewrite

x = y^{2} + y + 1

in the standard form

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

Therefore parabola properties are:

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

y^{2} = 4 + x

- 2 x^{2} - x + 6 - y = 0

3 x + 5 y = 5 x^{2} + 1

y^{2} = 3 x + 1

(y - 8)^{2} = - 4 (x - 3)