`f(x)=ax+b-1/(x+c)=>{("Tiệm cận xiên:"\ y=ax+b),("Tiệm cận đứng:"\ x=-1):}`

Ta có: `{( (0;-1) in y),( (-1;-2) in y):}=>{(b=-1),(-a+b=-2):}=>{(a=1),(b=-1):}`

`=>` `f(x)=x-1-1/(x+1)`

Gọi `M(x; x-1-1/(x+1)) in f(x)`

`=>` `IM=sqrt((x+1)^2+(x-1-1/(x+1)+2)^2)=sqrt((x+1)^2+(x+1-1/(x+1))^2)`

Xét: `h(x)=(x+1)^2+(x+1-1/(x+1))^2=(2x^4+8x^3+10^2+4x+1)/((x+1)^2)`

`=>` `h^'(x)=(4x^4+16x^3+24x^2+16x+2)/((x+1)^3)`

`h^'(x)=0<=> x=-1+- 1/(root[4](2))`

`=>` `IM=sqrt(2sqrt2-2)`