How do you find the asymptotes for #f(x)=(3x^5 + 1) / (2x^6 + 3x -1)#?

1 Answer
Jan 12, 2016

Answer:

#f(x)# has horizontal asymptote #y=0# and vertical asymptotes
#x=x_1# and #x=x_2#, where:

#x_1 ~~ 0.3324335502431692846#

#x_2 ~~ -1.1414892917449508403#

Explanation:

Let #g(x) = 3x^5+1# and #h(x) = 2x^6+3x-1#

#f(x)=(3x^5+1)/(2x^6+3x-1) = g(x)/(h(x))#

#g(x)# is of degree #5# and #h(x)# is of degree #6#, so:

#f(x) = g(x)/(h(x))->0# as #x->+-oo#

So #f(x)# has horizontal asymptote #y = 0#

#f(x)# will have vertical asymptotes wherever #h(x) = 0# and #g(x) != 0#

#h(x) = 2x^6+3x-1# has #2# changes of sign, so #1# or #2# positive zeros.

#h(-x) = 2x^6-3x-1# has #1# change of sign, so #h(x)# has one negative zero.

Since the coefficients of #h(x)# are Real, any non-Real zeros will occur in Complex conjugate pairs, so there are an even number of non-Real zeros.

So we can deduce that #h(x)# has exactly one positive, one negative and four non-Real zeros.

#h(x) = 0# has no solution expressible in terms of #n#th roots, but we can find good approximations using Newton's method.

#h'(x) = 12x^5+3#

Starting with an initial approximation #a_0#, iterate using the formula:

#a_(i+1) = a_i - (h(a_i))/(h'(a_i)) = a_i - (2a_i^6+3a_i-1)/(12a_i^5+3)#

Putting this into a spreadsheet and using initial approximations #a_0=1# and #a_0=-1#, I got the following approximations after a few iterations:

#x_1 ~~ 0.3324335502431692846#

#x_2 ~~ -1.1414892917449508403#

Note also that the only Real zero of #g(x)# is when:

#x=root(5)(-1/3)~~-0.80274#

So neither of the zeros of #h(x)# are zeros of #g(x)# and both of the zeros of #h(x)# are vertical asymptotes of #f(x)#.

graph{(3x^5+1)/(2x^6+3x-1) [-10, 10, -5, 5]}