leitkurve-30x^2+28x-6

Lösung

leitkurve

- 30 x^{2} + 28 x - 6

y = \frac{13}{24}

Schritte zur Lösung

Rewrite

y = - 30 x^{2} + 28 x - 6

in the standard form

4 (- \frac{1}{120}) (y - \frac{8}{15}) = (x - \frac{7}{15})^{2}

(h, k) = (\frac{7}{15}, \frac{8}{15}), p = - \frac{1}{120}

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{7}{15}, \frac{8}{15})

y-coordinate

y = \frac{8}{15} - p

y = \frac{8}{15} - (- \frac{1}{120})

Fasse zusammen

y = \frac{13}{24}

d i rec t r i x y^{2} - 20 x = 0

d i rec t r i x x^{2} = 100 y

d i rec t r i x x = - \frac{1}{4} (y + 6)^{2} + 7

d i rec t r i x x^{2} + 2 y = 0

d i rec t r i x y = \frac{4 x ^{3} + 11 x ^{2} + 4 x + 21}{x + 3}