How do you find the asymptotes for #(2x^2 + x + 2) / (x + 1)#?

1 Answer
Jan 10, 2017

Answer:

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

Explanation:

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

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

As you cannot divide by #0#, #x!=-1#

The vertical asymptote is #x=-1#

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

Let's do a long division

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

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

#color(white)(aaaaa)##0-x+2#

#color(white)(aaaaaaa)##-x-1#

#color(white)(aaaaaaaaa)##0+3#

Therefore,

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

So,

#lim_(x->-oo)(f(x)-(2x-1))=lim_(x->-oo)3/(x+1)=0^-#

#lim_(x->+oo)(f(x)-(2x-1))=lim_(x->+oo)3/(x+1)=0^+#

The slant asymptote is #y=2x-1#

graph{(y-(2x^2+x+2)/(x+1))(y-2x+1)(y-50x-50)=0 [-25.64, 25.67, -12.83, 12.84]}