题目

已知抛物线y=ax2+bx+c通过点P(1,1)且在点Q(2,-1)处与直线y=x-3相切,求实数a、b、c的值. 答案：解：∵曲线y=ax2+bx+c过P(1,1)点,∴a+b+c=1.                                ①∵y′=2ax+b,∴y′|x=2=4a+b.又此时切线为y=x-3,∴4a+b=1.                                  ②又曲线过Q(2,-1)点,∴4a+2b+c=-1.                           ③联立①②③得a=3,b=-11,c=9.

查看本题答案和解析