How do you find the vertical, horizontal or slant asymptotes for #f(x) = (x^2 + x + 3) /( x-1)#?

1 Answer
Jan 8, 2017

Answer:

The vertical asymptote is #x=1#
The slant asymptote is #y=x+2#
No horizontal asymptote

Explanation:

The domain of #f(x)# is #D_f(x)=RR-{1}#

As we cannot divide by #0#, #x!=1#

The vertical asymptote is #x=1#

As the degree of the numerator is #># the degree of the denominator, we have a slant asymptote.

Let's do a long division

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

#color(white)(aaaa)##x^2-x##color(white)(aaaaaaaa)##∣##x+2#

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

#color(white)(aaaaaa)##+2x-2#

#color(white)(aaaaaaa)##+0+5#

Therefore,

#f(x)=(x^2+x+3)/(x-1)=(x+2)+5/(x-1)#

#lim_(x->-oo)(f(x)-(x+2))=lim_(x->-oo)5/x=0^-#

#lim_(x->+oo)(f(x)-(x+2))=lim_(x->+oo)5/x=0^+#

The slant asymptote is #y=x+2#

graph{(y-(x^2+x+3)/(x-1))(y-x-2)=0 [-32.48, 32.44, -16.2, 16.32]}