Question

1)如图所示,二次函数y=ax^2+bx+c图象交x轴于点A、B,交y轴于点C,若OA:OC:OB=4:1:1,求ac+

Answer 1

OA:OC:OB=4:1:1,

假设比例系数为k k>0

OA=4k

OC=k

OB=k

c=-k

x1=-4k

x2=-k

x1+x2=-b/a

-5k=-b/a

5k=b/a

x1x2=c/a

4k^2=-k/a

a=-k/4

b=-5k^2/4

y=-k/4*x^2-5k^2/4x-k

=-k/4(x^2+5kx+4k)

=-k/4*(x+k)(x+4k)

把(0,c)

代入

-k^2=-k

k=1

ac+b

=k^2/4-5k^2/4

=-k^2=-1