x<= y^2+1

Lösung

x \leq y^{2} + 1

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

Schritte zur Lösung

x \leq y^{2} + 1

Rewrite

x \leq y^{2} + 1

in the standard form:

4 \cdot \frac{1}{4} (x - 1) = (y - 0)^{2}

Therefore parabola properties are:

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

2 x = - y^{2}

y = 4 (x + 4)^{2} - 5

y = 4 (x + 4)^{2} - 6

x^{2} - 8 x + 16 = 4 (y - 128)

yy + 3 x = 2