面积 f(120)=-1/480 x^2+30

解答

面积

f (120) = - \frac{1}{480} x^{2} + 30

无界

求解步骤

y = - \frac{1}{480} x^{2} + 30

将

y = - \frac{1}{480} x^{2} + 30

改写为标准形式:

4 (- 120) (y - 30) = (x - 0)^{2}

因此抛物线性质是:

(h, k) = (0, 30), p = - 120

Parabola's area is always unbounded

无界

a re a (x - 7) (x - 1)

a re a x = y^{2} - 1

a re a (3 x + 2) \cdot (4 x + 1)

a re a y = - 50000000 x^{2} + 300000 x

a re a 5 x^{2} + (- 8) x + 45