题目

1.y=f（x）与y=4x+1图像关于y=3对称,求y=f（x）表达式.
2.y=f（x）与y=4x+1图像关于y=x对称,求y=f（x）表达式.3.an等差,a1+a3+a5=2,a3+a5+a7=5,a5+a7+a9=

提问时间：2020-12-10

答案

y = 3与x轴平行,f(x)与y = 4x+1关于y = 3对称,则f(x)与y = 4x+1的倾斜角互补,即斜率互为相反数,f(x)的斜率为-4
y = 3与y = 4x + 1交于(1/2,3),f(x)也过该点,方程为y - 3 = -4(x - 1/2)
y = -4x + 5



y = 4x + 1与y = x交于A(-1/3,-1/3)
在y = 4x+1上取点B(0,1), 关于y = x的对称点为B'(1,0)
AB'即f(x)的方程:(y - 0)/(-1/3 - 0)= (x - 1)/(-1/3 - 1)
y = x/4 - 1/4

(2) - (1):a7 - a1 = 6d = 3
a5 + a7 + a9 = a3 + 2d + a5 + 2d + a7 + 2d = a3 +a5 + a7 + 6d = 5 + 3 = 8