Question

已知函数y1=mx+n ,y2=ax2+bx+c的图象交于p1(1,-1)和p2(3,1)两点

Answer 1

1.把P1(1,-1)和P2(3,1)两点代入y1=mx+n得m=1,n=-2,y1=x-2 x1+x2=-b/a,x1x2=c/a,（x1-x2)=（x1+x2）-4x1x2=(b-4ac)/a 即(b-4ac)/a=8,又a+b+c=-1,9a+3b+c=1,且抛物线开口向上,即a＞0 解得(a=-1/2舍去）,a=1/2,b=-1,c=-1/2 所以y2=1/2x-x-1/2 2.略 3.A(0,-1/2),△AP1P2的面积=（3+1）*2*1/2-3/2*3*1/2-1/2*1*1/2=3/2