已知二次函数y=ax2+bx(a≠0),当x取x1,x2(x1≠x2)时,函数值相等,求当x取x1+x2时的函数值.
问题描述:
已知二次函数y=ax2+bx(a≠0),当x取x1,x2(x1≠x2)时,函数值相等,求当x取x1+x2时的函数值.
答
根据题意得ax12+bx1=ax22+bx2,ax12-ax22+bx1-bx2=0,a(x1-x2)(x1+x2)+b(x1-x2)=0,(x1-x2)(ax1+ax2+b)=0,∵x1≠x2,∴ax1+ax2+b=0,即x1+x2=-ba,∴当x=x1+x2=-ba时,y=a×(-ba)2+b×(-ba)=0....