轴 y=-16x^2+40x+2

解答

轴

y = - 16 x^{2} + 40 x + 2

x = \frac{5}{4}

求解步骤

y = - 16 x^{2} + 40 x + 2

将

y = - 16 x^{2} + 40 x + 2

改写为标准形式

4 (- \frac{1}{64}) (y - 27) = (x - \frac{5}{4})^{2}

因此抛物线性质是:

(h, k) = (\frac{5}{4}, 27), p = - \frac{1}{64}

EN : P a r ab o l e S y mm e t ry A x i s T i tl e 2 Eq

x = \frac{5}{4}

a x i s x = - 2 y + y (y - 2) - 2 y (y - 1) - (y - 3)

a x i s y = - 0.5 x^{2} + 5.5

a x i s y = \frac{1}{2} x^{2} + 9 x + 8

a x i s x = (y - 7)^{2} + 1

a x i s x^{2} + 10 x - 2 y + 21 = 0