leitkurve 3x^2+4x-3

Lösung

leitkurve

3 x^{2} + 4 x - 3

y = - \frac{53}{12}

Schritte zur Lösung

Rewrite

y = 3 x^{2} + 4 x - 3

in the standard form

4 \cdot \frac{1}{12} (y - (- \frac{13}{3})) = (x - (- \frac{2}{3}))^{2}

(h, k) = (- \frac{2}{3}, - \frac{13}{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{2}{3}, - \frac{13}{3})

y-coordinate

y = - \frac{13}{3} - p

y = - \frac{13}{3} - \frac{1}{12}

Fasse zusammen

y = - \frac{53}{12}

d i rec t r i x y = 600 - 10 x + 0.25 x^{2}

d i rec t r i x 11 x^{2} + 13

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

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

d i rec t r i x 12 (x - 1) = (y - 3)^{2}