leitkurve y=x^2+2x-1-(2x^2-3x+6)

Lösung

leitkurve

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

y = - \frac{1}{2}

Schritte zur Lösung

Rewrite

y = x^{2} + 2 x - 1 - (2 x^{2} - 3 x + 6)

in the standard form

4 (- \frac{1}{4}) (y - (- \frac{3}{4})) = (x - \frac{5}{2})^{2}

(h, k) = (\frac{5}{2}, - \frac{3}{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{5}{2}, - \frac{3}{4})

y-coordinate

y = - \frac{3}{4} - p

y = - \frac{3}{4} - (- \frac{1}{4})

Fasse zusammen

y = - \frac{1}{2}

d i rec t r i x y = 2250 (0.015 - 0.001 x)^{2} + 6000 x

d i rec t r i x 2 x^{2}

d i rec t r i x y = 12 x^{2} - 72 x + 77

d i rec t r i x y = - 7 (x - 8)^{2} + 6

d i rec t r i x y = x^{2} - 9 x + 20