How do you find the asymptotes for #(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20)#?

1 Answer
Jul 29, 2016

Answer:

Horizontal asymptote (left and right): #y = 1#

Vertical asymptotes at #x=1#, #x=2#, #x=-2#

Hole at #(5, 9/7)#

Explanation:

Let's attempt to factorise the numerator and denominator first.

#color(white)()#
Numerator

Substituting #x=-1# in the numerator we get:

#1+3-21-43+60 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#x^4-3x^3-21x^2+43x+60 = (x+1)(x^3-4x^2-17x+60)#

Substituting #x=3# in the remaining cubic, we get:

#27-36-51+60 = 0#

So #x=3# is a zero and #(x-3)# a factor:

#x^3-4x^2-17x+60=(x-3)(x^2-x-20) = (x-3)(x-5)(x+4)#

In summary:

#x^4-3x^3-21x^2+43x+60 = (x+1)(x-3)(x-5)(x+4)#

#color(white)()#
Denominator

Substituting #x=1# in the denominator we get:

#1-6+1+24-20 = 0#

So #x=1# is a zero and #(x-1)# a factor:

#x^4-6x^3+x^2+24x-20 = (x-1)(x^3-5x^2-4x+20)#

The remaining cubic factors by grouping:

#x^3-5x^2-4x+20#

#=(x^3-5x^2)-(4x-20)#

#=x^2(x-5)-4(x-5)#

#=(x^2-4)(x-5)#

#=(x-2)(x+2)(x-5)#

In summary:

#x^4-6x^3+x^2+24x-20 = (x-1)(x-2)(x+2)(x-5)#

#color(white)()#
Together

#(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20)#

#=((x+1)(x-3)color(red)(cancel(color(black)((x-5))))(x+4))/((x-1)(x-2)(x+2)color(red)(cancel(color(black)((x-5)))))#

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

with exclusion #x != 5#

Looking at the leading terms of the numerator and denominator, the degrees and coefficients are identical, so there is a horizontal asymptote (left and right): #y = 1#

When #x=1# or #+-2# the denominator is zero and the numerator non-zero, so these values of #x# correspond to vertical asymptotes.

Substituting the value #x=5# into the simplified rational function, we find the coordinate of the hole as: #(5, 9/7)#

graph{(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20) [-41.83, 38.17, -10.24, 29.76]}