解：∵lim(x->-1/2)y=lim(x->-1/2)[x²/(2x+1)]=∞ ∴x=-1/2是曲線y=x²/(2x+1)的垂直漸近線 設它的斜漸近線為y=ax+b ∵a=lim(x->∞)(y/x) =lim(x->∞)[x/(2x+1)] =lim(x->∞)[1/(2+1/x)] =1/2 b=lim(x->∞)(y-ax) =lim(x->∞)[x²/(2x+1)-x/2] =lim(x->∞)[-x /(4x+2)] =lim(x->∞)[-1/(4+2/x)] =-1/4 ∴它的斜漸近線是y=x/2-1/4。
解：∵lim(x->-1/2)y=lim(x->-1/2)[x²/(2x+1)]=∞ ∴x=-1/2是曲線y=x²/(2x+1)的垂直漸近線 設它的斜漸近線為y=ax+b ∵a=lim(x->∞)(y/x) =lim(x->∞)[x/(2x+1)] =lim(x->∞)[1/(2+1/x)] =1/2 b=lim(x->∞)(y-ax) =lim(x->∞)[x²/(2x+1)-x/2] =lim(x->∞)[-x /(4x+2)] =lim(x->∞)[-1/(4+2/x)] =-1/4 ∴它的斜漸近線是y=x/2-1/4。