How do you find the asymptotes for #y=(x^3-2x^2-x+2)/(x^2-4)#?

2 Answers
Jan 7, 2017

Answer:

The vertical asymptote is #x=-2#
A hole at #x=2#
The slant asymptote is #y=x-2#
No horizontal asymptote.

Explanation:

The denominator is #x^2-4=(x+2)(x-2)#

Let's do a long division

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

#color(white)(aaaa)##x^3##color(white)(aaaaaa)##-4x##color(white)(aaaaaaa)##∣##x-2#

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

#color(white)(aaaaaa)##-2x^2##color(white)(aaaaaa)##+8#

#color(white)(aaaaaaaaaa)##0##color(white)(aaa)##3x-6#

Therefore,

#(x^3-2x^2-x+2)/(x^2-4)=(x-2)+(3(x-2))/(x^2-4)#

#=(x-2)+(3cancel((x-2)))/((x+2)cancel((x-2))#

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

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

As we cannot divide by #0#, #x!=-2#

The vertical asymptote is #x=-2#

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

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

The slant asymptote is #y=x-2#

graph{(y-(x^3-2x^2-x+2)/(x^2-4))(y-x+2)(y-50x-100)=0 [-23.33, 22.3, -16.34, 6.48]}

Jan 7, 2017

Answer:

Alternative format for long division + graph

Explanation:

Note that I have used place keepers where there is no value solely so that things line up in formatting. Example #0x^2#

#" "x^3-2x^2-x+2#
#color(red)(x)(x^2-4) ->ul( x^3+0x^2-4x) larr" subtract"#
#" "0color(white)(.) -2x^2+3x+2#
#color(red)(-2)(x^2-4)->ul(color(white)(.) -2x^2 +0x+8)larr" subtract"#
#color(red)("Remainder") ->" "0color(white)(.)+3x-6 #

#y" "=" "(x^3-2x-x+2)/(x^2-4)" " =" " (x-2) + (3x-6)/(x^2-4) #

Tony B