f(x)=(sin²x+cos²x)+√3sinxcosx+cos²x
=1+√3/2*sin2x+(1+cos2x)/2
=√3/2sin2x+1/2*cos2x+3/2
=sin2xcosπ/6+cos2xsinπ/6+3/2
=sin(2x+π/6)+3/2
=sin[2(x+π/12)]+3/2
所以向左移π/12单位
然后把横坐标缩小成原来的1/2
最后再向上移3/2个单位
f(x)=(sin²x+cos²x)+√3sinxcosx+cos²x
=1+√3/2*sin2x+(1+cos2x)/2
=√3/2sin2x+1/2*cos2x+3/2
=sin2xcosπ/6+cos2xsinπ/6+3/2
=sin(2x+π/6)+3/2
=sin[2(x+π/12)]+3/2
所以向左移π/12单位
然后把横坐标缩小成原来的1/2
最后再向上移3/2个单位