x=y^2+3y+4

Lösung

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

(h, k) = (\frac{7}{4}, - \frac{3}{2}), p = \frac{1}{4}

Schritte zur Lösung

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

Rewrite

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

in the standard form

4 \cdot \frac{1}{4} (x - \frac{7}{4}) = (y - (- \frac{3}{2}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{7}{4}, - \frac{3}{2}), p = \frac{1}{4}

y^{2} - 12 x + 8 y = - 40

y^{2} = 6 x - 4 y - 34

6 x + y^{2} - 2 y - 26 = 0

x^{2} - 10 x - 4 y + 21 = 0

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