leitkurve y=x^2+5x-14

Lösung

leitkurve

y = x^{2} + 5 x - 14

y = - \frac{41}{2}

Schritte zur Lösung

Rewrite

y = x^{2} + 5 x - 14

in the standard form

4 \cdot \frac{1}{4} (y - (- \frac{81}{4})) = (x - (- \frac{5}{2}))^{2}

(h, k) = (- \frac{5}{2}, - \frac{81}{4}), p = \frac{1}{4}

Parabola is symmetric around the y-axis and so the directrix is a line parallel to the x-axis, a distance

- p

from the center

(- \frac{5}{2}, - \frac{81}{4})

y-coordinate

y = - \frac{81}{4} - p

y = - \frac{81}{4} - \frac{1}{4}

Fasse zusammen

y = - \frac{41}{2}

d i rec t r i x (x + 1)^{2} = - 3 (y - 6)

d i rec t r i x x = \frac{1}{4} (y + 4)^{2} + 2

d i rec t r i x 2 (x^{2} + 6 x) + 5 y + 8 = 0

d i rec t r i x 10 x^{2}

d i rec t r i x - 4 x^{2} - 2 x + 1