Question

抛物线y==ax^2+bx+c 当x=3时的最大值为10 他的图像在x轴上截得线段为4求解析式

Answer 1

y = ax^2+bx+c

y' = 2ax + b =0

x = -b/2a = 3

=> b = -6a

y(3) = 10

10 = 9a + 3b + c

= -9a + c

c = 10 + 9a

y(4) = 0

0= 16a + 4b + c

= 16a -24a + 10+9a

a = -10

b = 60

c = -80

ie

y = -10x^2+60x -80