How do you find the slant asymptote of # (2x^3+2x)/(x^2-1) #?

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Answer 1 · 2018-07-16T19:01:39

The slant asymptote is #y=2x#.

Let #f(x)=(2x^2+2x)/(x^2-1)#

First, perform a long division

#color(white)(aaaa)##2x^3+0x^2+2x##color(white)(aaaa)##|##x^2-1#

#color(white)(aaaa)##2x^3+0x^2-2x##color(white)(aaaa)##|##2x#

#color(white)(aaaa)##0x^3+0x^2+4x#

Therefore,

#f(x)=2x+(4x)/(x^2-1)#

The slant asymptote is #y=2x#

#lim_(x->+oo)f(x)-2x=lim_(x->+oo)(4x)/(x^2-1)=lim_(x->+oo)4/x=0^+#

#lim_(x->-oo)f(x)-2x=lim_(x->-oo)(4x)/(x^2-1)=lim_(x->-oo)4/x=0^-#

graph{(y-(2x^3+2x)/(x^2-1))(y-2x)=0 [-35.66, 37.4, -15.9, 20.63]}