leitkurve y=3x^2+1-(x+4)

Lösung

leitkurve

y = 3 x^{2} + 1 - (x + 4)

y = - \frac{19}{6}

Schritte zur Lösung

Rewrite

y = 3 x^{2} + 1 - (x + 4)

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{37}{12})) = (x - \frac{1}{6})^{2}

(h, k) = (\frac{1}{6}, - \frac{37}{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{1}{6}, - \frac{37}{12})

y-coordinate

y = - \frac{37}{12} - p

y = - \frac{37}{12} - \frac{1}{12}

Fasse zusammen

y = - \frac{19}{6}

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

d i rec t r i x y = 2 x (x - 9) - (x - 9)

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

d i rec t r i x - x^{2} + 14 x - 45

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