x=y^2+14y+20

Lösung

x = y^{2} + 14 y + 20

(h, k) = (- 29, - 7), p = \frac{1}{4}

Schritte zur Lösung

x = y^{2} + 14 y + 20

Rewrite

x = y^{2} + 14 y + 20

in the standard form

4 \cdot \frac{1}{4} (x - (- 29)) = (y - (- 7))^{2}

Therefore parabola properties are:

(h, k) = (- 29, - 7), p = \frac{1}{4}