leitkurve y=6x^2-29x-5

Lösung

leitkurve

y = 6 x^{2} - 29 x - 5

y = - \frac{481}{12}

Schritte zur Lösung

Rewrite

y = 6 x^{2} - 29 x - 5

in the standard form

4 \cdot \frac{1}{24} (y - (- \frac{961}{24})) = (x - \frac{29}{12})^{2}

(h, k) = (\frac{29}{12}, - \frac{961}{24}), p = \frac{1}{24}

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{29}{12}, - \frac{961}{24})

y-coordinate

y = - \frac{961}{24} - p

y = - \frac{961}{24} - \frac{1}{24}

Fasse zusammen

y = - \frac{481}{12}

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

d i rec t r i x y = - 0.23 x^{2} + 1.87 x + 1.5

d i rec t r i x 3 y^{2} = 24 x

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

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