5.2s+0.1r0.7(r+0.2)+3.2=-3s

解答

5.2 s + 0.1 r 0.7 (r + 0.2) + 3.2 = - 3 s

(h, k) = (- 0.39015 \dots, - 0.1), p = - 29.28571 \dots

求解步骤

5.2 s + 0.1 r \cdot 0.7 (r + 0.2) + 3.2 = - 3 s

将

5.2 s + 0.1 r \cdot 0.7 (r + 0.2) + 3.2 = - 3 s

改写为标准形式

4 (- 29.28571 \dots) (s - (- 0.39015 \dots)) = (r - (- 0.1))^{2}

因此抛物线性质是:

(h, k) = (- 0.39015 \dots, - 0.1), p = - 29.28571 \dots

3 y = - x^{2} + 4

4 (- \frac{1}{8}) (x - \frac{57}{8}) = (y - \frac{5}{4})^{2}

2 x^{2} - 20 x + 2 m (x) > 0

x^{2} + 2 x - y = 2

4 x^{2} + 4 y = 25