How do you find vertical, horizontal and oblique asymptotes for #(x^3 + 1) / (x^2 - 2x + 2)#?

1 Answer
May 28, 2017

Answer:

The oblique asymptote is #y=x+2#
There is no vertical asymptote
No horizontal asymptote

Explanation:

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

The degree of the numerator is greater than the degree of the denominator, so there is an oblique asymptote.

Let's perform a long division

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

#color(white)(aaaaaaaaaaaaaaaaaa)##x^3-2x^2+2x#

#color(white)(aaaaaaaaaaaaaaaaaaa)##0+2x^2-2x+1#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaa)##2x^2+4x+4#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaa)##0+2x-3#

Therefore,

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

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

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

The oblique asymptote is #y=x+2#

The denominator is

#x^2-2x+2=(x-1)^2+1#

#AA x in RR, (x-1)^2+1>0#

There is no vertical asymptote
graph{(y-(x^3-1)/(x^2-2x+2))(y-x-2)=0 [-10.97, 11.53, -2.97, 8.28]}