y=x+1,本身就是直线,斜渐近线还是本身（也可以说没有渐近线）.
我想楼主可能想打y=x+(1/x)或者类似的.如果是这样,当x越来越大,1/x就会趋向于0,相当于消失了,可以看到函数会长得越来越像y=x.因此说他的斜渐进线是y=x
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I know,it's in the Math12 curriculum here in Canada as well.What I'm saying is that your initial function was of the form f(x)=x+(1/p(x)) where p is a polynomial of degree at least 1.The slant asymptote ( also called the oblique asymptote) is y=x.
More generally,if you have a rational function f=a(x)/b(x),first you do long division,and obtain the quotient q(x),and the remainder r(x),then you can write f(x) as q(x)+(r(x)/b(x)) where the degree of b is higher than r.Therefore the remainder term will fade away as x gets larger,then u'll be left with q(x).
I'm guessing that ur initial function was something like y=(x^2+1)/x,you can write it as y=x+(1/x),and therefore the slant asymptote is y=x.
You could totally visualize this without calculus.
