leitkurve y=x^2+x-1

Lösung

leitkurve

y = x^{2} + x - 1

y = - \frac{3}{2}

Schritte zur Lösung

Rewrite

y = x^{2} + x - 1

in the standard form

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

(h, k) = (- \frac{1}{2}, - \frac{5}{4}), p = \frac{1}{4}

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}{2}, - \frac{5}{4})

y-coordinate

y = - \frac{5}{4} - p

y = - \frac{5}{4} - \frac{1}{4}

Fasse zusammen

y = - \frac{3}{2}

d i rec t r i x 4 x^{2} - 8 x - 3 y - 2 = 0

d i rec t r i x 0 = (y - 1)^{2} - 16 (x + 4)

d i rec t r i x - \frac{1}{8} (x - 8)^{2} + 2

d i rec t r i x y = 25 x^{2} - 36

d i rec t r i x y = - 3 x^{2} + 12 x - 4