leitkurve y=2x^2+5x+2

Lösung

leitkurve

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

y = - \frac{5}{4}

Schritte zur Lösung

Rewrite

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

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{9}{8})) = (x - (- \frac{5}{4}))^{2}

(h, k) = (- \frac{5}{4}, - \frac{9}{8}), p = \frac{1}{8}

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}{4}, - \frac{9}{8})

y-coordinate

y = - \frac{9}{8} - p

y = - \frac{9}{8} - \frac{1}{8}

Fasse zusammen

y = - \frac{5}{4}

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

d i rec t r i x 4 x^{2} - 8 x + 2 y = 5

d i rec t r i x 88 x - 16 x^{2} - 112

d i rec t r i x - 16 t^{2} + 60 t + 3

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