leitkurve y=5x^2+13x-6

Lösung

leitkurve

y = 5 x^{2} + 13 x - 6

y = - \frac{29}{2}

Schritte zur Lösung

Rewrite

y = 5 x^{2} + 13 x - 6

in the standard form

4 \cdot \frac{1}{20} (y - (- \frac{289}{20})) = (x - (- \frac{13}{10}))^{2}

(h, k) = (- \frac{13}{10}, - \frac{289}{20}), p = \frac{1}{20}

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{13}{10}, - \frac{289}{20})

y-coordinate

y = - \frac{289}{20} - p

y = - \frac{289}{20} - \frac{1}{20}

Fasse zusammen

y = - \frac{29}{2}

d i rec t r i x y = 3 x^{2} + 6 x - 12

d i rec t r i x (x - 5)^{2} = 20 (y - 5)

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

d i rec t r i x x = + \frac{17}{4} (y + 2)^{2}

d i rec t r i x (x - 3)^{2} = - 4 (y - 2)