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

Question

Christian Antunez · Accepted 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#

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]}

Sadie Alexander · Answer

undefined To find the asymptotes for the function $ f(x) = \frac{{3x^5 + 1}}{{2x^6 + 3x - 1}} $:

1. Horizontal Asymptote:
   - Compare the degrees of the numerator and denominator polynomials. In this case, the degree of the numerator is 5, and the degree of the denominator is 6.
   - If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at y = 0.
   - If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients. Here, the leading coefficient of the numerator is 3 and the leading coefficient of the denominator is 2. So, the horizontal asymptote is at $ y = \frac{3}{2} $.

2. Vertical Asymptotes:
   - Set the denominator equal to zero and solve for x. Any values of x that make the denominator zero will give vertical asymptotes.
   - Solve $ 2x^6 + 3x - 1 = 0 $ for x. This equation may require numerical methods to solve since it is a sixth-degree polynomial equation. Once you find the values of x, those are the vertical asymptotes.