leitkurve y=3x^2-31x+56-(x-8)

Lösung

leitkurve

y = 3 x^{2} - 31 x + 56 - (x - 8)

y = - \frac{257}{12}

Schritte zur Lösung

Rewrite

y = 3 x^{2} - 31 x + 56 - (x - 8)

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{64}{3})) = (x - \frac{16}{3})^{2}

(h, k) = (\frac{16}{3}, - \frac{64}{3}), 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{16}{3}, - \frac{64}{3})

y-coordinate

y = - \frac{64}{3} - p

y = - \frac{64}{3} - \frac{1}{12}

Fasse zusammen

y = - \frac{257}{12}

d i rec t r i x x^{2} - 5 x + 6

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

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

d i rec t r i x x^{2} = 40 y

d i rec t r i x x^{2} - 6 x - y + 33 = 0