Question

抛物线y=ax^2-5ax+4经过三角形ABC三个顶点,点A,C分别在X,Y轴上且BC//X轴,AC=BC,求抛物线的解

Answer 1

y=ax^2-5ax+4,对称轴方程为X=-b/2a=5/2,

点C坐标为(0,4),

BC//x轴,AC=BC,

Y=4,即有4=ax^2-5ax,

ax^2-5ax=0,

a1+x2=5,X1*X2=0,

而|X2-X1|=|BC|,

(X1+X2)^2-4X1*X2=BC^2,

BC=5,AC=5,

则点B坐标为(5,4),点A坐标为(3,0),

0=a*9-5*a*3+4,

a=2/3,

即y=2/3x^2-10/3 x+4