f(x)=e^x+2x^2-3x,f(1)=e-1切點(diǎn)為(1,e-1).
f'(x)=e^x+4x-3,f'(1)=e+1.
所求切線方程為：y-e+1=(e+1)(x-1),即y=(e+1)x-2.你上面的幾步我都知道，問(wèn)題是切線方程怎么求的切點(diǎn)是(1,e-1)，切線的斜率是f'(1)=e+1。用點(diǎn)斜式寫(xiě)出切線方程為：y-e+1=(e+1)(x-1)，整理得：y=(e+1)x-2。