How do you identify all horizontal and slant asymptote for #f(x)=(2x^3-x^2-2x+1)/(x^2+3x+2)#?

1 Answer
Nov 9, 2016

Answer:

The vertical asymptote is #x=-2#
The slant asymptote is #y=2x-7#
There are no horizontal asymptote

Explanation:

The denominator can be factorised as #(x+2)(x+1)#
So the vertcal asymptote is #x=-2#
Let's do a long division to simplify #f(x)#
#color(white)(aaaa)##2x^3-x^2-2x+1##color(white)(aaaa)##∣##x^2+3x+2#
#color(white)(aaaa)##2x^3+6x^2+4x##color(white)(aaaaaaa)##∣##2x-7#
#color(white)(aaaaaa)##0-7x^2-6x+1#
#color(white)(aaaaaaaa)##-7x^2-21x-14#
#color(white)(aaaaaaaaa)##-0+15x+15#

#(2x^3-x^2-2x+1)/(x^2+3x+2)=2x-7+(15cancel(x+1))/((x+2)cancel(x+1))#

#:. y=2x-7# is a slant asymptote
Also, #f(x)=((2x-1)(x+1)(x-1))/((x+2)(x+1))=((2x-1)(x-1))/(x+2)#
#lim_(x->+oo)f(x)=lim_(x->+oo)2x=+oo#
#lim_(x->-oo)f(x)=lim_(x->-oo)2x=-oo#

graph{(y-((2x^3-x^2-2x+1)/(x^2+3x+2)))(y-2x+7)=0 [-83.3, 83.4, -41.7, 41.74]}