D(h3+n)=tt

Lösung

D (h 3 + n) = tt

(h, k) = (- D n, 0), p = \frac{1}{4}

Schritte zur Lösung

D (h \cdot 3 + n) = tt

Rewrite

D (h \cdot 3 + n) - tt = 0

in the standard form

4 \cdot \frac{1}{4} (h - (- D n)) = (t - 0)^{2}

Therefore parabola properties are:

(h, k) = (- D n, 0), p = \frac{1}{4}

y^{1} = 3 x^{2} + 5 x - 2

3 x^{2} + 5 y = 7 - 2 x

y^{2} - 16 x - 17 = 0

(x - 2) 2 + y^{2} = 16

17 - x + v 169 - x^{2} > 0