leitkurve 3x^2+19x-40

Lösung

leitkurve

3 x^{2} + 19 x - 40

y = - \frac{421}{6}

Schritte zur Lösung

Rewrite

y = 3 x^{2} + 19 x - 40

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{841}{12})) = (x - (- \frac{19}{6}))^{2}

(h, k) = (- \frac{19}{6}, - \frac{841}{12}), p = \frac{1}{12}

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{19}{6}, - \frac{841}{12})

y-coordinate

y = - \frac{841}{12} - p

y = - \frac{841}{12} - \frac{1}{12}

Fasse zusammen

y = - \frac{421}{6}

d i rec t r i x (y - 10)^{2} = - 12 (x + 2)

d i rec t r i x x^{2} + 4 x - 8 y + 12 = 0

d i rec t r i x x^{(2)} - 6 x + 5 y = - 34

d i rec t r i x x^{2} - 3 x - 4

d i rec t r i x (x - 4)^{2} = 8 (y + 2)