2018-2019学年人教B版必修1 2.2.3 待定系数法 作业
2018-2019学年人教B版必修1 2.2.3 待定系数法 作业第2页

答案C

5设函数f(x)={■(x^2+bx+c"," x≤0"," @2"," x>0"," )┤若f(-1)=f(0),f(-2)=-2,则关于x的方程f(x)=x的解的个数为0(  )

A.1 B.2 C.3 D.4

解析由f(-1)=f(0),f(-2)=-2,

  可得{■(1"-" b+c=c"," @4"-" 2b+c="-" 2"," )┤

  解得{■(b=1"," @c="-" 4"," )┤

  故f(x)={■(x^2+x"-" 4"," x≤0"," @2"," x>0"." )┤

  令f(x)=x,解得x=2或x=-2.

答案B

6抛物线y=ax2+bx+c与x轴相交于点A与点B,与y轴相交于点C,如果OB=OC=1/2OA,那么b的值为0(  )

A.-2 B.-1 C.-1/2 D.1/2

解析由图象可知c>0,且B(c,0),A(-2c,0).

  设f(x)=a(x-c)(x+2c),

  则a(x-c)(x+2c)=ax2+bx+c,

  即ax2+acx-2ac2=ax2+bx+c.

  故{■(ac=b"," @"-" 2ac^2=c"," )┤即ac=-1/2,b=-1/2.

答案C

7已知一次函数的图象经过(5,-2)和(3,4),则这个函数的解析式为    .

解析设一次函数为y=kx+b(k≠0),

  则有{■(5k+b="-" 2"," @3k+b=4"," )┤解得{■(k="-" 3"," @b=13"." )┤

答案y=-3x+13

8如图所示为二次函数y=ax2+bx+c的图象,则该函数的解析式为    .