How do you find vertical, horizontal and oblique asymptotes for #(x^4 -81) /(x^3+3x^2 - x - 3)#?

2 Answers
Feb 20, 2017

Answer:

Invisible hole at (-3, -27/2). Slant asymptote : #x-y+3=0#. Vertical asymptotes : #uarr x = +-1 darr#. See graphs, The third is not to scale but gets graph for great #|y|# near.

Explanation:

#y = (x^4-3^4)/((x+3)(x^2-1))#

#=((x+3)/(x+3))((x-3)(x^2+9))/((x-1)(x+1))#.

Sans the hole at #H(-3, -27/2)#,

#y= ((x-3)(x^2+9))/((x-1)(x+1))#
graph{(x-3)(1+10/(x^2-1)) [-164, 164.1, -82.1, 82]}

graph{((x-3)(1+10/(x^2-1))-y)(y-x-3)(x-1+.01y)(x+1-.01y)=0 [-164, 164.1, -82.1, 82]}

graph{(x-3)(1+10/(x^2-1)) [-16, 16, -160, 160]}

Feb 21, 2017

Answer:

We have vertical asymptote at #x=1# and #x=-1# and oblique asymptote at #y=x-3#

Explanation:

First factorize numerator and denominators

#x^4-81=(x^2)^2-9^2=(x^2-9)(x^2+9)=(x+3)(x-3)(x^2+9)#

and #x^3+3x^2-x-3=x^2(x+3)-1(x+3)=(x^2-1)(x+3)=(x+1)(x-1)(x+3)#

Hence #f(x)=(x^4-81)/(x^3+3x^2-x-3)=((x+3)(x-3)(x^2+9))/((x+1)(x-1)(x+3))#

= #((x-3)(x^2+9))/((x+1)(x-1))=(x^3-3x^2+9x-27)/(x^2-1)#

Observe that although we have cancelled out #x+3#, we have an undefined function #(x^4-81)/(x^3+3x^2-x-3)# at #x=-3#, though close to #x=-3#, on either side, #f(x)# is defined. As such we have a hole at #x=-3#

Further at #x->1# as also for #x_>-1#, #f(x)->+-oo# depending on whether we approach limits from left or right and hence

we have vertical asymptotes at #x=1# and #x=-1#

Here as degree of polynomial in numerator is just one more than that of denominator, we do not have a horizontal asymptote, but we do have a oblique or slant asymptote.

As #f(x)=(x^4-81)/(x^3+3x^2-x-3)=x-3+(10x^2-90)/(x^3+3x^2-x-3)#

= #x-3+(10/x-90/x^3)/(1+3/x-1/x^2-3/x^3)#

and as #x->+-oo#, #y->x-3#

we have an oblique or slant asymptote #y=x-3#
graph{(x^4-81)/(x^3+3x^2-x-3) [-20, 20, -40, 40]}