How do you find the asympototes for #y = (-8x^3 - 2x + 12)/( 7x^5 + 9x + 12)#?

1 Answer
Jul 28, 2015

Answer:

The horizontal asymptote is #y = 0#.

There is a vertical asymptote #x=a#, where #a ~= -0.89261052# is the only real root of #7x^5+9x+12 = 0#.

Explanation:

Since the degree of the denominator is greater than the numerator, we find #y->0# as #x->+-oo#. So #y=0# is a horizontal asymptote.

#-8x^3-2x+12 = 0# has a single real root at #x ~= 1.0720202#

#7x^5+9x+12 = 0# has a single real root at #x ~= -0.89261052#

In general, the roots of quintics cannot be expressed in terms of ordinary radicals. In this particular case, the quintic is almost in Bring-Jerrard normal form, so it's relatively easy (believe me) to express the root in terms of the Bring Radical:

Let #x_1 = (7/9)^(1/4)x#

Then:

#0 = 7x^5+9x+12#

#=7((9/7)^(1/4)x_1)^5+9(9/7)^(1/4)x^1+12#

#=9(9/7)^(1/4)x_1^5+9(9/7)^(1/4)x_1+12#

Divide through by #9(9/7)^(1/4)# to get:

#x_1^5+x_1+4/3(7/9)^(1/4) = 0#

So #x_1 = BR(4/3(7/9)^(1/4))#

and #x = (9/7)^(1/4)*x_1 = (9/7)^(1/4)BR(4/3(7/9)^(1/4))#

See: https://en.wikipedia.org/wiki/Bring_radical